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8202019 03[Anal Add Math CD]
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1
Additional Mathematics SPM Chapter 3
copy Penerbitan Pelangi Sdn Bhd
1 (b) (c) (d) (e) and (h)
2 f ( x) = 4 x2 ndash 8 x + 6
When x = ndash2
f (ndash2) = 4(ndash2)2 ndash 8(ndash2) + 6
= 16 + 16 + 6
= 38
3 f ( x) = x2 ndash 3 x + 2
When f ( x) = 0
x2 ndash 3 x + 2 = 0
( x ndash 1)( x ndash 2) = 0
x ndash 1 = 0 or x ndash 2 = 0
x = 1 or x = 2
4 f ( x) = ndash3 x2 + 5 x ndash 1
When f ( x) = 1
ndash3 x2 + 5 x ndash 1 = 1
ndash3 x2 + 5 x ndash 1 ndash 1 = 0
3 x2 ndash 5 x + 2 = 0 (3 x ndash 2)( x ndash 1) = 0
3 x ndash 2 = 0 or x ndash 1 = 0
x =2mdash3
or x = 1
5 (a) (b)
(c) (d)
(e) (f)
6 (a) Two different real roots
(b) One real root or two similar real roots
(c) No real roots
7 (a) The minimum value is 3
(b) The maximum value is 4
(c) The minimum value is ndash10
(d) f ( x) = 2983091 1mdash2
( x ndash 3)2 +1mdash4 983092
= ( x ndash 3)2 +1mdash2
Therefore the minimum value is1mdash2
(e) f ( x) =1mdash3
[6 ndash ( x + 1)2] + 5
= 2 ndash 1mdash3
( x + 1)2 + 5
= ndash1mdash3
( x + 1)2 + 7
Therefore the maximum value is 7
(f) The minimum value is 3
8 (a) f ( x) = x2 ndash 4 x + 2
= x2 ndash 4 x + 983089 4mdash2 983090
2
ndash 983089 4mdash2 983090
2
+ 2
= ( x ndash 2)2 ndash 4 + 2
= ( x ndash 2)2 ndash 2
Hence the minimum value is ndash2
(b) f ( x) = 2 x2 + 6 x ndash 5
= 2( x2 + 3 x) ndash 5
= 2983091 x2 + 3 x + 983089 3mdash2 983090
2
ndash 983089 3mdash2 983090
2
983092 ndash 5
= 2983091983089 x + 3mdash2 983090
2
ndash9mdash4 983092 ndash 5
= 2983089 x + 3mdash2 983090
2
ndash9mdash2
ndash 5
= 2983089 x + 3mdash2 983090
2
ndash19
ndashndashndash2
Hence the minimum value is ndash19
ndashndashndash2
(c) f ( x) = x2 + 5 x
= x2 + 5 x + 983089 5mdash2 983090
2
ndash 983089 5mdash2 983090
2
= 983089 x +5mdash2 983090
2
ndash25
ndashndashndash4
Hence the minimum value is ndash25
ndashndashndash4
CHAPTER
3 Quadratic Functions
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(d) f ( x) = 6 x ndash x2
= ndash( x2 ndash 6 x)
= ndash983091 x2 ndash 6 x + 983089 6mdash2 983090
2
ndash 983089 6mdash2 983090
2
983092 = ndash[( x ndash 3)2 ndash 9]
= ndash ( x ndash 3)2 + 9
Hence the maximum value is 9
(e) f ( x) = 3 ndash 4 x ndash x2
= ndash x2 ndash 4 x + 3
= ndash( x2 + 4 x) + 3
= ndash983091 x2 + 4 x + 983089 4mdash2 983090
2
ndash 983089 4mdash2 983090
2
983092 + 3
= ndash[( x + 2)2 ndash 4] + 3
= ndash( x + 2)2 + 4 + 3
= ndash( x + 2)2 + 7
Hence the maximum value is 7
(f) f ( x) = 4 x ndash 2 x2
= ndash2 x2 + 4 x
= ndash2( x2
ndash 2 x) = ndash2983091 x2 ndash 2 x + 983089 2
mdash2 983090
2
ndash 983089 2mdash2 983090
2
983092 = ndash2[( x ndash 1)2 ndash 1]
= ndash2( x ndash 1)2 + 2
Hence the maximum value is 2
(g) f ( x) = 10 + 5 x ndash 3 x2
= ndash3 x2 + 5 x + 10
= ndash3983089 x2 ndash5mdash3 x983090 + 10
= ndash3983091 x2 ndash5mdash3 x + 983089 5
mdash6 983090
2
ndash 983089 5mdash6 983090
2
983092 + 10
= ndash3983091983089 x ndash5mdash6 983090
2
ndash25
ndashndashndash36 983092 + 10
= ndash3983089 x ndash5mdash6 983090
2
+25
ndashndashndash12
+ 10
= ndash3983089 x ndash5mdash6 983090
2
+145
ndashndashndashndash12
Hence the maximum value is145
ndashndashndashndash12
(h) f ( x) = (2 x ndash 1)( x + 3)
= 2 x2 + 5 x ndash 3
= 2983089 x2
+
5
mdash2 x983090 ndash 3
= 2983091 x2 +5mdash2 x + 983089 5
mdash4 983090
2
ndash 983089 5mdash4 983090
2
983092 ndash 3
= 2983091983089 x +5mdash4 983090
2
ndash25
ndashndashndash16 983092 ndash 3
= 2983089 x +5mdash4 983090
2
ndash25
ndashndashndash8
ndash 3
= 2983089 x +5mdash4 983090
2
ndash49
ndashndashndash8
Hence the minimum value is ndash49
ndashndashndash8
(i) f ( x) = (1 ndash 4 x)( x + 2)
= x + 2 ndash 4 x2 ndash 8 x
= ndash 4 x2 ndash 7 x + 2
= ndash 4983089 x2 +7mdash4 x983090 + 2
= ndash 4983091 x2 +7mdash4 x + 983089 7
mdash8 983090
2
ndash 983089 7mdash8 983090
2
983092 + 2
= ndash 4983091983089 x + 7mdash8 983090
2
ndash49
ndashndashndash64 983092 + 2
= ndash 4983089 x + 7mdash8 983090
2
+49
ndashndashndash16
+ 2
= ndash 4983089 x + 7mdash8 983090
2
+81
ndashndashndash16
Hence the maximum value is81
ndashndashndash16
9 (a) f ( x) = x2 ndash 4
Therefore the minimum point is (0 ndash 4)
x ndash2 2 4 f ( x) 0 0 12
f (x )
x
ndash4 ndash2 2 4
12
0
(b) f ( x) = 3 x2 + 5
Therefore the minimum point is (0 5)
x ndash1 3
f ( x) 8 32
f (x )
x
ndash1 3
5
8
32
0
(c) f ( x) = 8 ndash x2
Therefore the maximum point is (0 8)
x ndash3 plusmn9831059831068 3
f ( x) ndash1 0 ndash1
f (x )
x
ndash1
8
ndash3 3 ndash 8 8
0
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(d) f ( x) = 10 ndash 2 x2
Therefore the maximum point is (0 10)
x ndash3 plusmn9831059831065 4
f ( x) ndash8 0 ndash22
f (x )
x
ndash8
10
ndash3 4 ndash 5 5 0
ndash22
(e) f ( x) = x( x + 2)
= x2 + 2 x
= x2 + 2 x + 12 ndash 12
= ( x + 1)2 ndash 1
Therefore the minimum point is (ndash1 ndash1)
x ndash 4 ndash2 0 2 f ( x) 8 0 0 8
f (x )
x
(ndash1ndash1) ndash4
8
ndash2 20
(f) f ( x) = ( x ndash 1)(2 x + 1)
= 2 x2
ndash x ndash 1 = 2983089 x2 ndash
xmdash2 983090 ndash 1
= 2983091 x2 ndash xmdash2
+ 983089 1mdash4 983090
2
ndash 983089 1mdash4 983090
2
983092 ndash 1
= 2983091983089 x ndash 1mdash4 983090
2
ndash1
ndashndashndash16 983092 ndash 1
= 2983089 x ndash 1mdash4 983090
2
ndash1mdash8
ndash 1
= 2983089 x ndash 1mdash4 983090
2
ndash9mdash8
Therefore the minimum point is (
1
mdash4 ndash
9
mdash8 )
x ndash1 ndash1mdash2
0 1 2
f ( x) 2 0 ndash1 0 5
f (x )
x
1 2ndash1 1
2ndash ndash
98ndashndash)14 ( ndash
2
0
5
ndash1
(g) f ( x) = ndash( x ndash 3)2 + 5
Therefore the maximum point is (3 5)
x ndash2 0 3 ndash 9831059831065 3 + 9831059831065 6
f ( x) ndash20 ndash 4 0 0 ndash4
f (x )
x
ndash2 ndash4
ndash20
(3 5)
63 + 53 ndash 50
(h) f ( x) = x2 + 4 x + 5
= x2 + 4 x + 22 ndash 22 + 5
= ( x + 2)2 + 1
Therefore the minimum point is (ndash2 1)
x ndash3 0 1 f ( x) 2 5 10
f (x )
x
(ndash2 1)
ndash3 1
2
5
10
0
(i) f ( x) = 2 x2 + 6 x ndash 8
= 2( x2
+ 3 x) ndash 8
= 2983091 x2 + 3 x + 983089 3mdash2 983090
2
ndash 983089 3mdash2 983090
2
983092 ndash 8
= 2983091983089 x +3mdash2 983090
2
ndash9mdash4 983092 ndash 8
= 2983089 x + 3mdash2 983090
2
ndash9mdash2
ndash 8
= 2983089 x + 3mdash2 983090
2
ndash25
ndashndashndash2
Therefore the minimum point is (ndash3mdash2
ndash25
ndashndashndash2
)
x ndash3 0 1 2 f ( x) ndash8 ndash8 0 12
ndash25
2ndashndash)3
2( ndash
f (x )
x
ndash3
12
ndash8
0 1 2
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(j) f ( x) = ( x ndash 4)2
Therefore the minimum point is (4 0)
x 0 4 5
f ( x) 16 0 1
f (x )
x
4 501
16
(k) f ( x) = ndash x2 + 6 x ndash 9
= ndash( x2 ndash 6 x) ndash 9
= ndash( x2 ndash 6 x + 32 ndash 32) ndash 9
= ndash[( x ndash 3)2 ndash 9] ndash 9
= ndash( x ndash 3)2
Therefore the maximum point is (3 0)
x 0 3 4
f ( x) ndash9 0 ndash1
f (x )
x 0
ndash1
ndash9
(3 0)
4
10 (a) x( x ndash 2) 983102 0
0 2x
f (x )
The range of values of x is x 983100 0 or x 983102 2
(b) ( x ndash 3)( x ndash 4) 983100 0
0 43
x
f (x )
The range of values of x is 3 983100 x 983100 4
(c) x2 ndash 3 x ndash 4 983086 0
( x ndash 4)( x + 1) 983086 0
4 ndash1 0 x
f (x )
The range of values of x is x 983084 ndash1 or x 983086 4
(d) 2 x2 + 5 x ndash 3 983084 0
(2 x ndash 1)( x + 3) 983084 0
1 ndash
2 ndash3 0
x
f (x )
The range of values of x is ndash3 983084 x 983084 1mdash2
(e) ( x ndash 3)( x + 2) 983100 ndash4
x2 ndash x ndash 6 983100 ndash4
x2 ndash x ndash 2 983100 0
( x ndash 2)( x + 1) 983100 0
ndash1 20 x
f (x )
The range of values of x is ndash1 983100 x 983100 2
(f) (2 x ndash 1)( x ndash 3) 983100 4( x ndash 3)
2 x2 ndash 6 x ndash x + 3 983100 4 x ndash 12
2 x2 ndash 11 x + 15 983100 0
( x ndash 3)(2 x ndash 5) 983100 0
5 ndash
2
30 x
f (x )
The range of values of x is5mdash2
983100 x 983100 3
(g) x2 + 4
ndashndashndashndashndashndash5
983100 2 x ndash 1
x2 + 4 983100 5(2 x ndash 1)
x2 + 4 983100 10 x ndash 5
x2 ndash 10 x + 9 983100 0
( x ndash 1)( x ndash 9) 983100 0
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910 x
f (x )
The range of values of x is 1 983100 x 983100 9
(h) x (1 ndash 4 x) 983084 5 x ndash 8
x ndash 4 x2 ndash 5 x + 8 983084 0
ndash 4 x2 ndash 4 x + 8 983084 0
x2 + x ndash 2 983086 0
( x + 2)( x ndash 1) 983086 0
1 ndash2 0 x
f (x )
The range of values of x is x 983084 ndash2 or x 983086 1
1 (a) x-coordinate of the maximum part =1 + 7
ndashndashndashndashndash2
= 4
Therefore the equation of the axis of symmetry
is x = 4
(b) f ( x) = p ndash ( x + q)2
= 5 ndash ( x ndash 4)2
2 f ( x) = 2 x2 ndash 16 x + k 2 + 2k + 1
= 2( x2 ndash 8 x) + k 2 + 2k + 1
= 2983091 x2 ndash 8 x + 983089 8mdash2 983090
2
ndash 983089 8mdash2 983090
2
983092 + k 2 + 2k + 1
= 2[( x ndash 4)2 ndash 16] + k 2 + 2k + 1
= 2( x ndash 4)2 ndash 32 + k 2 + 2k + 1
= 2( x ndash 4)2 + k 2 + 2k ndash 31
Given minimum value = ndash28
there4 k 2 + 2k ndash 31 = ndash28
k 2 + 2k ndash 3 = 0
(k + 3)(k ndash 1) = 0 k = ndash3 1
3 (a) f ( x) = 2( x ndash 3)2 + k
p is the x-coordinate of the minimum point
Therefore p = 3
(b) k is the minimum value of f ( x)
Therefore k = ndash4
(c) The equation of the axis of symmetry is x = 3
4 f ( x) = 3 x2 ndash 2 x + p
a = 3 b = ndash2 c = p
Since the graph does not intersect the x-axis
b2 ndash 4ac lt 0
(ndash2)2 ndash 4(3)( p) lt 0
4 ndash 12 p lt 0
4 lt 12 p
13
lt p
p gt13
5 f ( x) = 2 x2 ndash 12 x + 5
= 2( x2 ndash 6 x) + 5
= 2[( x ndash 3)2 ndash 32] + 5
= 2( x ndash 3)2 ndash 18 + 5
= 2( x ndash 3)2 ndash 13
there4 p = 2 q = ndash3 ndashr + 1 = ndash13
r = 14
6 (a) f ( x) = ndash x2 + 6 px + 1 ndash 4 p2
= ndash( x2 ndash 6 px) + 1 ndash 4 p2
= ndash983091 x2 ndash 6 px + 983089 6 p ndashndashndash
2 983090
2
ndash 983089 6 p ndashndashndash
2 983090
2
983092 + 1 ndash 4 p2
= ndash[( x ndash 3 p)2 ndash 9 p2] + 1 ndash 4 p2
= ndash ( x ndash 3 p)2 + 9 p2 + 1 ndash 4 p2
= ndash ( x ndash 3 p)2 + 1 + 5 p2
The maximum value given is q2 ndash p
Therefore q2 ndash p = 1 + 5 p2
5 p2 + p + 1 = q2
(b) x = 3 is symmetrical axis 3 p = 3
p = 1
Substitute p = 1 into 5 p2 + p + 1 = q2
5(1)2 + 1 + 1 = q2
q2 = 7
q = plusmn9831059831067 Hence p = 1 q = plusmn9831059831067
7 4t (t + 1) ndash 3t 2 + 12 983086 0
4t 2 + 4t ndash 3t 2 + 12 983086 0
t 2 + 4t + 12 983086 0
(t + 2)(t + 6) 983086 0
0 ndash2 ndash6x
f (x )
The range of values of t is t 983084 ndash6 or t 983086 ndash2
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1 x-coordinate of maximum point = ndash 4 + 0 ndashndashndashndashndashndash
2 = ndash2
Equation of the axis of symmetry is x = ndash2
2 Let x be the x-coordinate of A
0 + x
ndashndashndashndashndash2
= 3
x = 6
The coordinates of A are (6 4)
3 Let x be the x-coordinate of A
x + 6 ndashndashndashndashndash
2 = 2
x = 4 ndash 6
x = ndash2
The coordinates of A are (ndash2 0)
4 x-coordinate of A =0 + 8
ndashndashndashndashndash2
= 4
Let C be the centre of OB
4
5
A
C O
AC 2
= OA2
ndash OC 2
= 52 ndash 42
= 9
AC = 3
The coordinates of A are (4 3)
5 x-coordinate of minimum point =0 + 4
ndashndashndashndashndash2
= 2
x-coordinate of minimum point for the image is ndash2
6 (a) y = ( x ndash p)2 + q and minimum point is (2 ndash1)
Hence p = 2 and q = ndash1
(b) y = ( x ndash 2)2 ndash 1
When y = 0
( x ndash 2)2 ndash 1 = 0
( x ndash 2)2 = 1
x ndash 2 = plusmn1
x = plusmn1 + 2
= 1 3
Hence A is (1 0) and B is (3 0)
7 f ( x) = 2k + 1 ndash 983089 x + 1mdash2 p983090
2
Given (ndash1 k ) is the maximum point
Therefore 2k + 1 = k
k = ndash1
x +1mdash2 p = 0 when x = ndash1
ndash1 + 1mdash2 p = 0
1mdash2 p = 1
p = 2
8 Given ( p 2q) is the minimum point of
y = 2 x2 ndash 4 x + 5
= 2( x2 ndash 2 x) + 5
= 2( x2 ndash 2 x + 12 ndash 12) + 5
= 2[( x ndash 1)2 ndash 1] + 5
= 2( x ndash 1)2 ndash 2 + 5
= 2( x ndash 1)2 + 3
2q = 3
q =3mdash2
p ndash 1 = 0
p = 1
9 (a) Since (1 4) is the point on y = x2 ndash 2kx + 1
substitute x = 1 y = 4 into the equation
4 = 12 ndash 2k (1) + 1
2k = ndash2
k = ndash1
(b) y = x2 ndash 2(ndash1) x + 1
= x2 + 2 x + 1
= ( x + 1)2
Minimum value of y is 0
10 f ( x) = ndash x2 ndash 8 x + k ndash 1
= ndash( x2 + 8 x) + k ndash 1
= ndash( x2 + 8 x + 42 ndash 42) + k ndash 1
= ndash[( x + 4)2 ndash 16] + k ndash 1
= ndash( x + 4)2 + 16 + k ndash 1
= ndash( x + 4)2 + 15 + k
Since 13 is the maximum value
then 15 + k = 13
k = ndash2
11 f ( x) = 2 x2 ndash 6 x + 7
= 2( x2 ndash 3 x) + 7
= 2983091 x2 ndash 3 x + 983089 3mdash2 983090
2
ndash 983089 3mdash2 983090
2
983092 + 7
= 2983091983089 x ndash3mdash2 983090
2
ndash9mdash4 983092 + 7
= 2983089 x ndash3mdash2 983090
2
ndash9mdash2
+ 7
= 2983089 x ndash3mdash2 983090
2
+5mdash2
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The minimum point is (3mdash2
5mdash2
)
x ndash1 0 3
f ( x) 15 7 7
ndash ndash
3
5322
0
7
15
x
f (x )
ndash1
The range is5mdash2
983100 f ( x) 983100 15
12 f ( x) = 5 ndash 4 x ndash 2 x2
= ndash2 x2 ndash 4 x + 5
= ndash2( x2 + 2 x) + 5
= ndash2( x2 + 2 x + 12 ndash 12) + 5
= ndash2[( x + 1)2 ndash 1] + 5
= ndash2( x + 1)2 + 2 + 5
= ndash2( x + 1)2 + 7
When f ( x) = ndash1
ndash2( x + 1)2 + 7 = ndash1
ndash2( x + 1)2 = ndash8
( x + 1)2 = 4
x + 1 = plusmn2
x = plusmn2 ndash 1
= ndash3 or 1
13 y = ( x ndash 3)2 ndash 4
Minimum point is (3 ndash4)
x ndash1 0 1 5 6
y 12 5 0 0 5
x 0 1 5 6
5
12
ndash1
(3 ndash4)
y
The range is ndash 4983100 y 983100 12
14 y = ndash x2 + 4 x ndash 5
= ndash( x2 ndash 4 x) ndash 5
= ndash( x2 ndash 4 x + 22 ndash 22) ndash 5
= ndash [( x ndash 2)2 ndash 4] ndash 5
= ndash( x ndash 2)2 + 4 ndash 5
= ndash( x ndash 2)2 ndash 1
Maximum point is (2 ndash1)
x ndash1 0 3
y ndash10 ndash5 ndash2
x
0 3(2 ndash1)
ndash1
ndash5
ndash10
y
The range is ndash10 983100 y 983100 ndash1
15 y = 9 ndash ( x ndash 3)2Maximum point is (3 9)
x ndash1 0 6 7
y 7 0 0 7
x
y
0 ndash1 6
(3 9)
7
7
The range is 0 983100 y 983100 9
16 3 x2 983084 x
3 x2
ndash x 983084 0 x(3 x ndash 1)983084 0
ndash
13
0 x
f (x )
The range is 0 983084 x 983084 1mdash3
17 3 x ndash x
2
ndashndashndashndashndashndash2 983084 1
3 x ndash x2 983084 2
ndash x2 + 3 x ndash 2 983084 0
x2 ndash 3 x + 2 983086 0
( x ndash 1)( x ndash 2) 983086 0
20 x
f (x )
1
The range is x 983084
1 or x 983086
2
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18 Given that x ndash 2 y = 1
there4 x = 1 + 2 y 983089
Substitute 983089 into y + 3983102 2 xy
y + 3983102 2(1 + 2 y) y
y + 3983102 2 y + 4 y2
0983102 4 y2 + y ndash 3
0983102 (4 y ndash 3)( y + 1)
3 ndash
4
0 y
f (y )
ndash1
The range is ndash1 983100 y 983100 3mdash4
19 f ( x) 983084 0
5 x2 ndash 4 x ndash 1 983084 0
(5 x + 1)( x ndash 1) 983084 0
1 ndash ndash
5
0 x
f (x )
1
The range is ndash1mdash5
983084 x 983084 1
20 g( x) 983086 0
4 x2 ndash 9 983086 0(2 x + 3)(2 x ndash 3) 983086 0
3 ndash ndash
23 ndash
2
0 x
g (x )
The range is x 983084 ndash3mdash2
or x 983086 3mdash2
21 (a) Since y = 3 x
2
ndash 9 x + t 983086
0 for all values of x andit does not have root when y = 0
Then b2 ndash 4ac 983084 0 for 3 x2 ndash 9 x + t = 0
(ndash9)2 ndash 4(3)(t ) 983084 0
81 ndash 12t 983084 0
ndash12t 983084 ndash81
t 983086 ndash81 ndashndashndashndash ndash12
t 983086 27
ndashndashndash4
(b) Let f ( x) = a( x ndash b)2 + c
f ( x) = a( x ndash 2)2 + 0
f ( x) = a( x ndash 2)2
Substitute x = 0 f ( x) = ndash3 into the equation
ndash3 = a(0 ndash 2)2
= 4a
a = ndash3mdash4
Hence the quadratic function is
f ( x) = ndash3mdash4
( x ndash 2)2
22 (a) Given 2 x2 ndash 3 y + 2 = 0
3 y = 2 x2 + 2
y =2 x2
ndashndashndash3
+2mdash3
983089
Substitute 983089 into y 983084 10
2 x2
ndashndashndash
3
+2mdash
3
983084 10
2 x2 + 2 983084 30
2 x2 ndash 28 983084 0
x2 ndash 14 983084 0
( x + 98310598310698310614 )( x ndash 98310598310698310614 ) 983084 0
14 ndash14
0 x
f (x )
The range is ndash98310598310698310614 983084 x 983084 98310598310698310614
(b) 2 x2 ndash 8 x ndash 10 = 2( x2 ndash 4 x) ndash 10
= 2( x2 ndash 4 x + 22 ndash 22) ndash 10
= 2[( x ndash 2)2 ndash 4] ndash 10
= 2( x ndash 2)2 ndash 8 ndash 10
= 2( x ndash 2)2 ndash 18
Therefore a = 2 b = ndash2 and c = ndash18
Hence the minimum value of 2 x2 ndash 8 x ndash 10 is
ndash18
23 5 ndash 2 x 983100 0 3 x2 ndash 4 x 983086 ndash1
3 x2
ndash 4 x + 1983086
0(3 x ndash 1)( x ndash 1) 983086 0
11 ndash
3
0 x
f (x )
x 983084 1mdash3
x 983086 1
ndash2 x 983100
ndash5 x 983102
ndash5 ndashndashndash ndash2
x 983102 5mdash2
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5ndash
2
1ndash
3
1
x 5mdash2
x x 11mdash3
The range is x 983102 5mdash2
24 5 983084 f ( x) 983084 9
5 983084 5 ndash 3 x + x2 983084 9
5 983084 5 ndash 3 x + x2 5 ndash 3 x + x2 983084 9
5 ndash 3 x + x2 ndash 9 983084 0
x2 ndash 3 x ndash 4 983084 0
( x ndash 4)( x + 1) 983084 0
4 ndash1 0 x
f (x )
ndash1 983084 x 983084 4
0 983084 x2 ndash 3 x
0 983084 x( x ndash 3)
30 x
f (x )
x 983084 0 x 983086 3
0 ndash1 43
x lt 0 x gt 3
ndash1 lt x lt 4
The range is ndash1983084
x 983084
0 or 3983084
x 983084
4
25 1 983102 x2 + 3 x ndash 3 983086 ndash3
x2 + 3 x ndash 3 983086 ndash3
x2 + 3 x 983086 0
x( x + 3) 983086 0
ndash3 0 x
f (x )
x 983084 ndash3 x 983086 0
1 983102 x2 + 3 x ndash 3
0 983102 x2 + 3 x ndash 4
0 983102 ( x + 4)( x ndash 1)
1 ndash4 0 x
f (x )
ndash 4 983100 x 983100 1
ndash4
x lt ndash3 x gt 0
ndash4 x 1
ndash3 0 1
The range is ndash 4983100 x 983084 ndash3 or 0 983084 x 983100 1
26 x2 + 5 x ndash 10 983086 4
x2 + 5 x ndash 10 983084 ndash 4 x2 + 5 x ndash 10 983086 4
x2 + 5 x ndash 14 983086 0
( x + 7)( x ndash 2) 983086 0
ndash7 20
x
f (x )
x 983084 ndash7 x 983086 2
x2 + 5 x ndash 6 983084 0
( x + 6)( x ndash 1) 983084 0
ndash6 10
x
f (x )
ndash6 983084 x 983084 1
The ranges are ndash6 lt x lt 1 x lt ndash7 x gt 2
27 f ( x) = (r + 1) x2 + 2rx + r ndash 3
Given that f ( x) does not intersect the x-axis
therefore b2 ndash 4ac 983084 0
(2r)2 ndash 4(r + 1)(r ndash 3) 983084 0
4r2 ndash 4(r2 ndash 2r ndash 3) 983084 0
4r2 ndash 4r2 + 8r + 12 983084 0
2r + 3 983084 0
r 983084 ndash3mdash2
28 Given that f ( x) = 9 ndash 6 x + 2 x2 does not have real root
when f ( x) = k
there4 9 ndash 6 x + 2 x2 = k
2 x2 ndash 6 x + 9 ndash k = 0
Use b2 ndash 4ac 983084 0
(ndash 6)2 ndash 4(2)(9 ndash k ) 983084 0
36 ndash 72 + 8k 983084 0
ndash36 + 8k 983084 0 8k 983084 36
k 983084 36
ndashndashndash8
k 983084 9mdash2
29 2 x2 + 10 x ndash 20 983100 8
ndash8 983100 2 x2 + 10 x ndash 20 983100 8
ndash8983100 2 x2 + 10 x ndash 20 2 x2 + 10 x ndash 20 983100 8
2 x2 + 10 x ndash 28 983100 0
x2 + 5 x ndash 14 983100 0
( x + 7)( x ndash 2) 983100 0
ndash7 20
x
f (x )
ndash7 983100 x 983100 2
0983100 2 x2 + 10 x ndash 12
0983100 x2 + 5 x ndash 6
0983100 ( x + 6)( x ndash 1)
ndash6 10
x
f (x )
x 983100 ndash6 x 983102 1
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x ndash6 x 1
ndash7 x 2
ndash6 ndash7 1 2
The range is ndash7 983100 x 983100 ndash6 or 1 983100 x 983100 2
30 y = x2 + 5 x ndash 6
= x2 + 5 x + 983089 5mdash2 983090
2
ndash 983089 5mdash2 983090
2
ndash 6
= 983089 x + 5mdash2 983090
2
ndash25
ndashndashndash4
ndash 6
= 983089 x + 5mdash2 983090
2
ndash49
ndashndashndash4
The minimum point is (ndash5mdash2
ndash49
ndashndashndash4
)
x ndash13 ndash 6 0 1 3
y 98 0 ndash 6 0 18
x 0
ndash6 ndash13 ndash6
18
98
1 3
) ) ndash ndash ndash ndash
5
2
49
4
y
The range is ndash49
ndashndashndash4
983084 y 983084 98
31 y = ndash x2 + 2 x ndash nx + 16
= ndash x2 + (2 ndash n) x + 16
= ndash[ x2 ndash (2 ndash n) x] + 16
= ndash983091 x2 ndash (2 ndash n) x + 983089 2 ndash n ndashndashndashndashndash
2 983090
2
ndash 983089 2 ndash n ndashndashndashndashndash
2 983090
2
983092 + 16
= ndash
983091983089 x ndash
2 ndash n ndashndashndashndashndash
2 983090
2
ndash
9830892 ndash n
ndashndashndashndashndash2 983090
2
983092 + 16
= ndash983089 x ndash 2 ndash n ndashndashndashndashndash
2 983090
2
+ 983089 2 ndash n ndashndashndashndashndash
2 983090
2
+ 16
Since y = x2 ndash 3k and y = ndash x2 + 2 x ndash nx + 16 have the
same axis of symmetry that is x = 0
then ndash2 ndash n
ndashndashndashndashndash2
= 0
2 ndash n = 0
n = 2
The equation y = ndash x2 + 2 x ndash nx + 16
= ndash x2 + 2 x ndash 2 x + 16
= ndash983089 x ndash 2 ndash 2 ndashndashndashndashndash
2 983090
2
+ 983089 2 ndash 2 ndashndashndashndashndash
2 983090
2
+ 16
= ndash x2 + 16
When y = 0
ndash x2
+ 16 = 0 x2 = 16
x = plusmn98310598310698310616
= plusmn4
Therefore B = (4 0)
Substitute x = 4 y = 0 into y = x2 ndash 3k
0 = 42 ndash 3k
0 = 16 ndash 3k
k =16
ndashndashndash3
32 (a) y = ndash2[(3k ndash x)2 + n] ndash 10 has a maximum point
(4 11)
y = ndash2(3k ndash x)2 ndash 2n ndash 10
there4 3k ndash 4 = 0 and ndash2n ndash 10 = 11
k =4mdash3
ndash2n = 21
n = ndash21
ndashndashndash2
(b) Substitute k =4mdash3
and n = ndash21
ndashndashndash2
into
y = ndash2[(3k ndash x)2 + n] ndash 10
y = ndash2983091(4 ndash x)2 ndash21
ndashndashndash2 983092 ndash 10
= ndash2(4 ndash x)2 + 21 ndash 10
= ndash2(4 ndash x)2
+ 11 When y = 0
ndash2(4 ndash x)2 + 11= 0
2(4 ndash x)2 = 11
(4 ndash x)2 =11
ndashndashndash2
4 ndash x = plusmn98310598310698310698310611 ndashndashndash
2
x = 4 plusmn 98310598310698310698310611 ndashndashndash
2
= 4 ndash 98310598310698310698310611 ndashndashndash
2 4 + 98310598310698310698310611
ndashndashndash2
= 1655 6345
(c) y = ndash2(4 ndash x)2 + 11
The maximum point is (4 11)
x ndash1 0 5
y ndash39 ndash21 9
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1 y = 3(2 x ndash 1)( x + 1) ndash x(4 x ndash 5) + 2
= 3(2 x2 + x ndash 1) ndash 4 x2 + 5 x + 2
= 6 x2 + 3 x ndash 3 ndash 4 x2 + 5 x + 2
= 2 x2 + 8 x ndash 1
= 2( x2 + 4 x) ndash 1
= 2( x2 + 4 x + 22 ndash 22) ndash 1
= 2[( x + 2)2 ndash 4] ndash 1
= 2( x + 2)2 ndash 8 ndash 1
= 2( x + 2)2 ndash 9
Since a = 2 983086 0 therefore the minimum value of
y is ndash9
When y = 0 2( x + 2)2 ndash 9 = 0
( x + 2)2 =9mdash2
x + 2 = plusmn9831059831069831069mdash2 x = plusmn9831059831069831069mdash2 ndash 2
= 9831059831069831069mdash2 ndash 2 or ndash9831059831069831069mdash2 ndash 2
= 01213 or ndash 4121
When x = 0 y = 2(2)2 ndash 9
= ndash1
The minimum point is (ndash2 ndash9)
ndash1 ndash4121
(ndash2 ndash9)
012130
y
x
2 5 983084 Area of rectangle ABCD 983084 21
5 983084 ( x + 3)( x ndash 1) 983084 21
5 983084 ( x + 3)( x ndash 1)
5 983084 x2 + 2 x ndash 3
0 983084 x2 + 2 x ndash 8
0 983084 ( x + 4)( x ndash 2)
0 ndash4 2
f (x )
x
x 983084 ndash 4 x gt 2
( x + 3)( x ndash 1) 983084 21
x2 + 2 x ndash 3 983084 21
x2 + 2 x ndash 24 983084 0
( x ndash 4)( x + 6) 983084 0
0 ndash6 4
f (x )
x
ndash 6 983084 x 983084 4
ndash4
ndash6 lt x lt 4
x lt ndash4 x gt 2
2
x
ndash6 4
The range is ndash 6983084 x 983084 ndash 4 or 2 983084 x 983084 4
3 (a) p =1 + 5
ndashndashndashndashndash2
= 3
(b) y = ( x ndash 1)( x ndash 5) = x2 ndash 6 x + 5
4 y = a( x ndash 2)2 + 1
Substitute x = 0 y = 9 into the equation
9 = a(ndash2)2 + 1
8 = 4a
a = 2
Therefore the quadratic function is
f ( x) = 2( x ndash 2)2 + 1
5 x2
+ (1 + k ) x ndash k 2
+ 1 = 0For quadratic equation to have real roots
b2 ndash 4ac 983102 0
(1 + k )2 ndash 4(1)(1 ndash k 2) 983102 0
1 + 2k + k 2 ndash 4 + 4k 2 983102 0
5k 2 + 2k ndash 3 983102 0
(5k ndash 3)(k + 1) 983102 0
0 ndash1 3 ndash
5
f (k )
k
The range of values of k is k 983100 ndash1 or k 983102 3mdash5
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6 y = x2 + 7 x ndash 8 ndash 2k
For y to be positive for all real values of x there is
no roots for y = 0
Therefore b2 ndash 4ac 983084 0
72 ndash 4(1)(ndash8 ndash 2k ) 983084 0
49 + 32 + 8k 983084 0
8k 983084 ndash81
k 983084 ndash81
ndashndashndash8
Alternative
y = x2 + 7 x ndash 8 ndash 2k
= x2 + 7 x + 983089 7mdash2 983090
2
ndash 983089 7mdash2 983090
2
ndash 8 ndash 2k
= 983089 x +7mdash2 983090
2
ndash49
ndashndashndash4
ndash 8 ndash 2k
For y to be positive for all real values of x
ndash49
ndashndashndash4
ndash 8 ndash 2k 983086 0
ndash2k 983086 49
ndashndashndash4
+ 8
ndash2k 983086 81
ndashndashndash4
k 983084 ndash81
ndashndashndash8
7 Substitute x = 6 y = 0 into y = px2 + qx
0 = p(6)2 + q(6)
0 = 36 p + 6q
q + 6 p = 0 983089
y = px2 + qx
= p983089 x2 +q mdash p x983090
= p983091 x2 +q mdash p x + 983089
q ndashndashndash2 p 983090
2
ndash 983089q
ndashndashndash2 p 983090
2
983092
= p983091983089 x +q
ndashndashndash2 p 983090
2
ndashq2
ndashndashndashndash4 p2 983092
= p983089 x +q
ndashndashndash2 p 983090
2
ndashq2
ndashndashndash4 p
ndash q2
ndashndashndash4 p
= ndash12
q2 = 48 p
p =q2
ndashndashndash48
983090
Substitute 983090 into 983089
q + 6983089 q2
ndashndashndash48
983090 = 0
q +q2
mdash8
= 0
8q + q2 = 0
q(8 + q) = 0
q = 0 or q = ndash8
When q = 0 p =02
ndashndashndash48
= 0
When q = ndash8 p =(ndash8)2
ndashndashndashndashndash48
=64
ndashndashndash48
=4mdash3
Therefore the values of p =
4
mdash3 and q = ndash8
8 (2 ndash 3k ) x2 + x +3mdash4k = 0
b2 ndash 4ac = 12 ndash 4(2 ndash 3k )983089 3mdash4k 983090
= 1 ndash 6k + 9k 2
= 9k 2 ndash 6k + 1
= (3k ndash 1)2
Since (3k ndash 1)2 983102 0 for all values of k
therefore (2 ndash 3k ) x2 + x +3mdash4k = 0 has real roots for
all values of k
9 f ( x) = 3( x2 + 2mx + m2 + n)
= 3[( x + m)2 + n]
= 3( x + m)2 + 3n
The minimum point is (ndashm 3n)
Compare to A(t 3t 2)
there4 ndashm = t and 3n = 3t 2
m = ndasht n = t 2
10 (a) y = px2 + 8 x + 10 ndash p
When the graph does not intercept the x-axis
there are no roots for px2 + 8 x + 10 ndash p = 0 Therefore b2 ndash 4ac 983084 0
82 ndash 4 p(10 ndash p) 983084 0
64 ndash 40 p + 4 p2 983084 0
p2 ndash 10 p + 16 983084 0
( p ndash 2)( p ndash 8) 983084 0
2 8
Hence r = 2 and t = 8
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(b) When p = 2
y = 2 x2 + 8 x + 8
= 2( x2 + 4 x) + 8
= 2( x2 + 4 x + 22 ndash 22) + 8
= 2[( x + 2)2 ndash 4] + 8
= 2( x + 2)2 ndash 8 + 8
= 2( x + 2)2
Therefore the minimum point is (ndash2 0)
When x = 0 y = 8
When y = 0 2( x + 2)2 = 0
x = ndash2
When p = 8
y = 8 x2 + 8 x + 2
= 8( x2 + x) + 2
= 8983091 x2 + x + 983089 1mdash2 983090
2
ndash 983089 1mdash2 983090
2
983092 + 2
= 8983091983089 x +1mdash2 983090
2
ndash1mdash4 983092 + 2
= 8983089 x +1mdash2 983090
2
ndash 2 + 2
= 8983089 x +1mdash2 983090
2
Therefore the minimum point is (ndash1mdash2
0)
When x = 0 y = 2
When y = 0 0 = 8983089 x + 1mdash2 983090
2
x = ndash1mdash2
0
2
8
1 ndash ndash
2
ndash2
y p = 2 p = 8
x
11 (a) f ( x) = 24 x ndash 4 x2 + r
= ndash 4 x2 + 24 x + r
= ndash 4( x2 ndash 6 x) + r
= ndash 4( x2 ndash 6 x + 32 ndash 32) + r
= ndash 4[( x ndash 3)2 ndash 9] + r
= ndash 4( x ndash 3)2 + 36 + r
Compare to f ( x) = p( x ndash q)2 + 16
Therefore p = ndash 4 q = 3 and 36 + r = 16
r = ndash20
(b) The turning point is (3 16)
(c) f ( x) = 24 x ndash 4 x2 ndash 20
When x = 0 f ( x) = ndash20
When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0
4( x ndash 3)2 = 16
( x ndash 3)2 = 4
x ndash 3 = plusmn2
x = plusmn2 + 3
= ndash2 + 3 or 2 + 3
= 1 or 5
0 1
(3 16)
5
ndash20
y
x
12 (a) y = ndash| p( x ndash 3)2 + q|
Substitute x = 3 y = ndash5 into the equation
ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5
Substitute x = 4 y = 0 into the equation
0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0
p = 5
Therefore p = 5 q = ndash5 or p = ndash5 q = 5
(b) When x = 3 y = ndash5
For p = 5 q = ndash5
When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40
Based on the graph the range of values of y is
ndash 40 983100 y 983100 0
For p = ndash5 q = 5
When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40
Therefore the range of values of y is
ndash 40 983100 y 983100 0
13 (a) y = ndash2( x ndash 3)2 + 2k
= ndash x2 + 2 x + px ndash 8
= ndash x2 + (2 + p) x ndash 8
= ndash[ x2 ndash (2 + p) x] ndash 8
= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash
2 9830902
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983091983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
+(2 + p)2
ndashndashndashndashndashndashndash4
ndash 8
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Since the x-coordinate of the maximum point for
both the graphs are same
therefore2 + p
ndashndashndashndashndash2
= 3
p = 4
y = ndash x2 + 2 x + px ndash 8 becomes
y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8
When y = 0
ndash x2 + 6 x ndash 8 = 0
x2 ndash 6 x + 8 = 0
( x ndash 2)( x ndash 4) = 0
x = 2 or 4
Hence A(2 0) and B(4 0)
Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k
0 = ndash2(2 ndash 3)2 + 2k
2k = 2
k = 1 Hence k = 1 and p = 4
(b) For y = ndash2( x ndash 3)2 + 2k
= ndash2( x ndash 3)2 + 2(1)
= ndash2( x ndash 3)2 + 2
Maximum value of the curve is 2
For y = ndash x2 + 2 x + px ndash 8
= ndash x2 + 2 x + 4 x ndash 8
= ndash x2 + 6 x ndash 8
When x = 3
y = ndash9 + 18 ndash 8
= 1
Maximum value of the curve is 1
14 Since 3 x2 983102 0 for all values of x
therefore3 x2
ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)
983100 0
(2 x ndash 1)( x + 4) 983100 0
0 ndash4 1 ndash
2
f (x )
x
Hence ndash 4983100 x 983100 1mdash2
15 Since x2 + 1 983086 0
therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash
x2 + 1 983086 0
x2 + 3 x + 2 983086 0
( x + 1)( x + 2) 983086 0
0 ndash2 ndash1
f (x )
x
Hence x 983084 ndash2 x gt ndash1
16 ndash4
ndashndashndashndashndashndash1 ndash 3 x
983100 x
0 983100 x +4
ndashndashndashndashndashndash1 ndash 3 x
0 983100 x(1 ndash 3 x) + 4
ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0983100
x ndash 3 x2 + 4
ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash
1 ndash 3 x
0 983100 (ndash3 x + 4)( x + 1)
ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
For ndash3 x + 4 983102 0
4 983102 3 x
x 983100 4mdash3
For x + 1 983102 0
x 983102 ndash1
For 1 ndash 3 x 983086 0
ndash3 x 983086 ndash1
x 983084 1mdash3
ndash1 ndash + ndash +
1 4 ndash ndash
3
4x ndash
3
1x lt ndash
3
x ndash1
3
x
Therefore the range is ndash1 983100 x 983084 1mdash3
x 983102 4mdash3
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17 y2 ndash 9 = x
x = y2 ndash 9
When x = 0
y2 = 9
y = plusmn3
When y = 0
x = ndash9
When x = 7
7 = y2 ndash 9
y2 = 16
y = plusmn4
x ndash9 0 7
y 0 plusmn3 plusmn4
0 ndash9 7
ndash4 ndash3
34
y
x
The range of values of y is ndash 4 983100 y 983100 4
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(d) f ( x) = 6 x ndash x2
= ndash( x2 ndash 6 x)
= ndash983091 x2 ndash 6 x + 983089 6mdash2 983090
2
ndash 983089 6mdash2 983090
2
983092 = ndash[( x ndash 3)2 ndash 9]
= ndash ( x ndash 3)2 + 9
Hence the maximum value is 9
(e) f ( x) = 3 ndash 4 x ndash x2
= ndash x2 ndash 4 x + 3
= ndash( x2 + 4 x) + 3
= ndash983091 x2 + 4 x + 983089 4mdash2 983090
2
ndash 983089 4mdash2 983090
2
983092 + 3
= ndash[( x + 2)2 ndash 4] + 3
= ndash( x + 2)2 + 4 + 3
= ndash( x + 2)2 + 7
Hence the maximum value is 7
(f) f ( x) = 4 x ndash 2 x2
= ndash2 x2 + 4 x
= ndash2( x2
ndash 2 x) = ndash2983091 x2 ndash 2 x + 983089 2
mdash2 983090
2
ndash 983089 2mdash2 983090
2
983092 = ndash2[( x ndash 1)2 ndash 1]
= ndash2( x ndash 1)2 + 2
Hence the maximum value is 2
(g) f ( x) = 10 + 5 x ndash 3 x2
= ndash3 x2 + 5 x + 10
= ndash3983089 x2 ndash5mdash3 x983090 + 10
= ndash3983091 x2 ndash5mdash3 x + 983089 5
mdash6 983090
2
ndash 983089 5mdash6 983090
2
983092 + 10
= ndash3983091983089 x ndash5mdash6 983090
2
ndash25
ndashndashndash36 983092 + 10
= ndash3983089 x ndash5mdash6 983090
2
+25
ndashndashndash12
+ 10
= ndash3983089 x ndash5mdash6 983090
2
+145
ndashndashndashndash12
Hence the maximum value is145
ndashndashndashndash12
(h) f ( x) = (2 x ndash 1)( x + 3)
= 2 x2 + 5 x ndash 3
= 2983089 x2
+
5
mdash2 x983090 ndash 3
= 2983091 x2 +5mdash2 x + 983089 5
mdash4 983090
2
ndash 983089 5mdash4 983090
2
983092 ndash 3
= 2983091983089 x +5mdash4 983090
2
ndash25
ndashndashndash16 983092 ndash 3
= 2983089 x +5mdash4 983090
2
ndash25
ndashndashndash8
ndash 3
= 2983089 x +5mdash4 983090
2
ndash49
ndashndashndash8
Hence the minimum value is ndash49
ndashndashndash8
(i) f ( x) = (1 ndash 4 x)( x + 2)
= x + 2 ndash 4 x2 ndash 8 x
= ndash 4 x2 ndash 7 x + 2
= ndash 4983089 x2 +7mdash4 x983090 + 2
= ndash 4983091 x2 +7mdash4 x + 983089 7
mdash8 983090
2
ndash 983089 7mdash8 983090
2
983092 + 2
= ndash 4983091983089 x + 7mdash8 983090
2
ndash49
ndashndashndash64 983092 + 2
= ndash 4983089 x + 7mdash8 983090
2
+49
ndashndashndash16
+ 2
= ndash 4983089 x + 7mdash8 983090
2
+81
ndashndashndash16
Hence the maximum value is81
ndashndashndash16
9 (a) f ( x) = x2 ndash 4
Therefore the minimum point is (0 ndash 4)
x ndash2 2 4 f ( x) 0 0 12
f (x )
x
ndash4 ndash2 2 4
12
0
(b) f ( x) = 3 x2 + 5
Therefore the minimum point is (0 5)
x ndash1 3
f ( x) 8 32
f (x )
x
ndash1 3
5
8
32
0
(c) f ( x) = 8 ndash x2
Therefore the maximum point is (0 8)
x ndash3 plusmn9831059831068 3
f ( x) ndash1 0 ndash1
f (x )
x
ndash1
8
ndash3 3 ndash 8 8
0
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(d) f ( x) = 10 ndash 2 x2
Therefore the maximum point is (0 10)
x ndash3 plusmn9831059831065 4
f ( x) ndash8 0 ndash22
f (x )
x
ndash8
10
ndash3 4 ndash 5 5 0
ndash22
(e) f ( x) = x( x + 2)
= x2 + 2 x
= x2 + 2 x + 12 ndash 12
= ( x + 1)2 ndash 1
Therefore the minimum point is (ndash1 ndash1)
x ndash 4 ndash2 0 2 f ( x) 8 0 0 8
f (x )
x
(ndash1ndash1) ndash4
8
ndash2 20
(f) f ( x) = ( x ndash 1)(2 x + 1)
= 2 x2
ndash x ndash 1 = 2983089 x2 ndash
xmdash2 983090 ndash 1
= 2983091 x2 ndash xmdash2
+ 983089 1mdash4 983090
2
ndash 983089 1mdash4 983090
2
983092 ndash 1
= 2983091983089 x ndash 1mdash4 983090
2
ndash1
ndashndashndash16 983092 ndash 1
= 2983089 x ndash 1mdash4 983090
2
ndash1mdash8
ndash 1
= 2983089 x ndash 1mdash4 983090
2
ndash9mdash8
Therefore the minimum point is (
1
mdash4 ndash
9
mdash8 )
x ndash1 ndash1mdash2
0 1 2
f ( x) 2 0 ndash1 0 5
f (x )
x
1 2ndash1 1
2ndash ndash
98ndashndash)14 ( ndash
2
0
5
ndash1
(g) f ( x) = ndash( x ndash 3)2 + 5
Therefore the maximum point is (3 5)
x ndash2 0 3 ndash 9831059831065 3 + 9831059831065 6
f ( x) ndash20 ndash 4 0 0 ndash4
f (x )
x
ndash2 ndash4
ndash20
(3 5)
63 + 53 ndash 50
(h) f ( x) = x2 + 4 x + 5
= x2 + 4 x + 22 ndash 22 + 5
= ( x + 2)2 + 1
Therefore the minimum point is (ndash2 1)
x ndash3 0 1 f ( x) 2 5 10
f (x )
x
(ndash2 1)
ndash3 1
2
5
10
0
(i) f ( x) = 2 x2 + 6 x ndash 8
= 2( x2
+ 3 x) ndash 8
= 2983091 x2 + 3 x + 983089 3mdash2 983090
2
ndash 983089 3mdash2 983090
2
983092 ndash 8
= 2983091983089 x +3mdash2 983090
2
ndash9mdash4 983092 ndash 8
= 2983089 x + 3mdash2 983090
2
ndash9mdash2
ndash 8
= 2983089 x + 3mdash2 983090
2
ndash25
ndashndashndash2
Therefore the minimum point is (ndash3mdash2
ndash25
ndashndashndash2
)
x ndash3 0 1 2 f ( x) ndash8 ndash8 0 12
ndash25
2ndashndash)3
2( ndash
f (x )
x
ndash3
12
ndash8
0 1 2
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(j) f ( x) = ( x ndash 4)2
Therefore the minimum point is (4 0)
x 0 4 5
f ( x) 16 0 1
f (x )
x
4 501
16
(k) f ( x) = ndash x2 + 6 x ndash 9
= ndash( x2 ndash 6 x) ndash 9
= ndash( x2 ndash 6 x + 32 ndash 32) ndash 9
= ndash[( x ndash 3)2 ndash 9] ndash 9
= ndash( x ndash 3)2
Therefore the maximum point is (3 0)
x 0 3 4
f ( x) ndash9 0 ndash1
f (x )
x 0
ndash1
ndash9
(3 0)
4
10 (a) x( x ndash 2) 983102 0
0 2x
f (x )
The range of values of x is x 983100 0 or x 983102 2
(b) ( x ndash 3)( x ndash 4) 983100 0
0 43
x
f (x )
The range of values of x is 3 983100 x 983100 4
(c) x2 ndash 3 x ndash 4 983086 0
( x ndash 4)( x + 1) 983086 0
4 ndash1 0 x
f (x )
The range of values of x is x 983084 ndash1 or x 983086 4
(d) 2 x2 + 5 x ndash 3 983084 0
(2 x ndash 1)( x + 3) 983084 0
1 ndash
2 ndash3 0
x
f (x )
The range of values of x is ndash3 983084 x 983084 1mdash2
(e) ( x ndash 3)( x + 2) 983100 ndash4
x2 ndash x ndash 6 983100 ndash4
x2 ndash x ndash 2 983100 0
( x ndash 2)( x + 1) 983100 0
ndash1 20 x
f (x )
The range of values of x is ndash1 983100 x 983100 2
(f) (2 x ndash 1)( x ndash 3) 983100 4( x ndash 3)
2 x2 ndash 6 x ndash x + 3 983100 4 x ndash 12
2 x2 ndash 11 x + 15 983100 0
( x ndash 3)(2 x ndash 5) 983100 0
5 ndash
2
30 x
f (x )
The range of values of x is5mdash2
983100 x 983100 3
(g) x2 + 4
ndashndashndashndashndashndash5
983100 2 x ndash 1
x2 + 4 983100 5(2 x ndash 1)
x2 + 4 983100 10 x ndash 5
x2 ndash 10 x + 9 983100 0
( x ndash 1)( x ndash 9) 983100 0
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910 x
f (x )
The range of values of x is 1 983100 x 983100 9
(h) x (1 ndash 4 x) 983084 5 x ndash 8
x ndash 4 x2 ndash 5 x + 8 983084 0
ndash 4 x2 ndash 4 x + 8 983084 0
x2 + x ndash 2 983086 0
( x + 2)( x ndash 1) 983086 0
1 ndash2 0 x
f (x )
The range of values of x is x 983084 ndash2 or x 983086 1
1 (a) x-coordinate of the maximum part =1 + 7
ndashndashndashndashndash2
= 4
Therefore the equation of the axis of symmetry
is x = 4
(b) f ( x) = p ndash ( x + q)2
= 5 ndash ( x ndash 4)2
2 f ( x) = 2 x2 ndash 16 x + k 2 + 2k + 1
= 2( x2 ndash 8 x) + k 2 + 2k + 1
= 2983091 x2 ndash 8 x + 983089 8mdash2 983090
2
ndash 983089 8mdash2 983090
2
983092 + k 2 + 2k + 1
= 2[( x ndash 4)2 ndash 16] + k 2 + 2k + 1
= 2( x ndash 4)2 ndash 32 + k 2 + 2k + 1
= 2( x ndash 4)2 + k 2 + 2k ndash 31
Given minimum value = ndash28
there4 k 2 + 2k ndash 31 = ndash28
k 2 + 2k ndash 3 = 0
(k + 3)(k ndash 1) = 0 k = ndash3 1
3 (a) f ( x) = 2( x ndash 3)2 + k
p is the x-coordinate of the minimum point
Therefore p = 3
(b) k is the minimum value of f ( x)
Therefore k = ndash4
(c) The equation of the axis of symmetry is x = 3
4 f ( x) = 3 x2 ndash 2 x + p
a = 3 b = ndash2 c = p
Since the graph does not intersect the x-axis
b2 ndash 4ac lt 0
(ndash2)2 ndash 4(3)( p) lt 0
4 ndash 12 p lt 0
4 lt 12 p
13
lt p
p gt13
5 f ( x) = 2 x2 ndash 12 x + 5
= 2( x2 ndash 6 x) + 5
= 2[( x ndash 3)2 ndash 32] + 5
= 2( x ndash 3)2 ndash 18 + 5
= 2( x ndash 3)2 ndash 13
there4 p = 2 q = ndash3 ndashr + 1 = ndash13
r = 14
6 (a) f ( x) = ndash x2 + 6 px + 1 ndash 4 p2
= ndash( x2 ndash 6 px) + 1 ndash 4 p2
= ndash983091 x2 ndash 6 px + 983089 6 p ndashndashndash
2 983090
2
ndash 983089 6 p ndashndashndash
2 983090
2
983092 + 1 ndash 4 p2
= ndash[( x ndash 3 p)2 ndash 9 p2] + 1 ndash 4 p2
= ndash ( x ndash 3 p)2 + 9 p2 + 1 ndash 4 p2
= ndash ( x ndash 3 p)2 + 1 + 5 p2
The maximum value given is q2 ndash p
Therefore q2 ndash p = 1 + 5 p2
5 p2 + p + 1 = q2
(b) x = 3 is symmetrical axis 3 p = 3
p = 1
Substitute p = 1 into 5 p2 + p + 1 = q2
5(1)2 + 1 + 1 = q2
q2 = 7
q = plusmn9831059831067 Hence p = 1 q = plusmn9831059831067
7 4t (t + 1) ndash 3t 2 + 12 983086 0
4t 2 + 4t ndash 3t 2 + 12 983086 0
t 2 + 4t + 12 983086 0
(t + 2)(t + 6) 983086 0
0 ndash2 ndash6x
f (x )
The range of values of t is t 983084 ndash6 or t 983086 ndash2
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1 x-coordinate of maximum point = ndash 4 + 0 ndashndashndashndashndashndash
2 = ndash2
Equation of the axis of symmetry is x = ndash2
2 Let x be the x-coordinate of A
0 + x
ndashndashndashndashndash2
= 3
x = 6
The coordinates of A are (6 4)
3 Let x be the x-coordinate of A
x + 6 ndashndashndashndashndash
2 = 2
x = 4 ndash 6
x = ndash2
The coordinates of A are (ndash2 0)
4 x-coordinate of A =0 + 8
ndashndashndashndashndash2
= 4
Let C be the centre of OB
4
5
A
C O
AC 2
= OA2
ndash OC 2
= 52 ndash 42
= 9
AC = 3
The coordinates of A are (4 3)
5 x-coordinate of minimum point =0 + 4
ndashndashndashndashndash2
= 2
x-coordinate of minimum point for the image is ndash2
6 (a) y = ( x ndash p)2 + q and minimum point is (2 ndash1)
Hence p = 2 and q = ndash1
(b) y = ( x ndash 2)2 ndash 1
When y = 0
( x ndash 2)2 ndash 1 = 0
( x ndash 2)2 = 1
x ndash 2 = plusmn1
x = plusmn1 + 2
= 1 3
Hence A is (1 0) and B is (3 0)
7 f ( x) = 2k + 1 ndash 983089 x + 1mdash2 p983090
2
Given (ndash1 k ) is the maximum point
Therefore 2k + 1 = k
k = ndash1
x +1mdash2 p = 0 when x = ndash1
ndash1 + 1mdash2 p = 0
1mdash2 p = 1
p = 2
8 Given ( p 2q) is the minimum point of
y = 2 x2 ndash 4 x + 5
= 2( x2 ndash 2 x) + 5
= 2( x2 ndash 2 x + 12 ndash 12) + 5
= 2[( x ndash 1)2 ndash 1] + 5
= 2( x ndash 1)2 ndash 2 + 5
= 2( x ndash 1)2 + 3
2q = 3
q =3mdash2
p ndash 1 = 0
p = 1
9 (a) Since (1 4) is the point on y = x2 ndash 2kx + 1
substitute x = 1 y = 4 into the equation
4 = 12 ndash 2k (1) + 1
2k = ndash2
k = ndash1
(b) y = x2 ndash 2(ndash1) x + 1
= x2 + 2 x + 1
= ( x + 1)2
Minimum value of y is 0
10 f ( x) = ndash x2 ndash 8 x + k ndash 1
= ndash( x2 + 8 x) + k ndash 1
= ndash( x2 + 8 x + 42 ndash 42) + k ndash 1
= ndash[( x + 4)2 ndash 16] + k ndash 1
= ndash( x + 4)2 + 16 + k ndash 1
= ndash( x + 4)2 + 15 + k
Since 13 is the maximum value
then 15 + k = 13
k = ndash2
11 f ( x) = 2 x2 ndash 6 x + 7
= 2( x2 ndash 3 x) + 7
= 2983091 x2 ndash 3 x + 983089 3mdash2 983090
2
ndash 983089 3mdash2 983090
2
983092 + 7
= 2983091983089 x ndash3mdash2 983090
2
ndash9mdash4 983092 + 7
= 2983089 x ndash3mdash2 983090
2
ndash9mdash2
+ 7
= 2983089 x ndash3mdash2 983090
2
+5mdash2
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The minimum point is (3mdash2
5mdash2
)
x ndash1 0 3
f ( x) 15 7 7
ndash ndash
3
5322
0
7
15
x
f (x )
ndash1
The range is5mdash2
983100 f ( x) 983100 15
12 f ( x) = 5 ndash 4 x ndash 2 x2
= ndash2 x2 ndash 4 x + 5
= ndash2( x2 + 2 x) + 5
= ndash2( x2 + 2 x + 12 ndash 12) + 5
= ndash2[( x + 1)2 ndash 1] + 5
= ndash2( x + 1)2 + 2 + 5
= ndash2( x + 1)2 + 7
When f ( x) = ndash1
ndash2( x + 1)2 + 7 = ndash1
ndash2( x + 1)2 = ndash8
( x + 1)2 = 4
x + 1 = plusmn2
x = plusmn2 ndash 1
= ndash3 or 1
13 y = ( x ndash 3)2 ndash 4
Minimum point is (3 ndash4)
x ndash1 0 1 5 6
y 12 5 0 0 5
x 0 1 5 6
5
12
ndash1
(3 ndash4)
y
The range is ndash 4983100 y 983100 12
14 y = ndash x2 + 4 x ndash 5
= ndash( x2 ndash 4 x) ndash 5
= ndash( x2 ndash 4 x + 22 ndash 22) ndash 5
= ndash [( x ndash 2)2 ndash 4] ndash 5
= ndash( x ndash 2)2 + 4 ndash 5
= ndash( x ndash 2)2 ndash 1
Maximum point is (2 ndash1)
x ndash1 0 3
y ndash10 ndash5 ndash2
x
0 3(2 ndash1)
ndash1
ndash5
ndash10
y
The range is ndash10 983100 y 983100 ndash1
15 y = 9 ndash ( x ndash 3)2Maximum point is (3 9)
x ndash1 0 6 7
y 7 0 0 7
x
y
0 ndash1 6
(3 9)
7
7
The range is 0 983100 y 983100 9
16 3 x2 983084 x
3 x2
ndash x 983084 0 x(3 x ndash 1)983084 0
ndash
13
0 x
f (x )
The range is 0 983084 x 983084 1mdash3
17 3 x ndash x
2
ndashndashndashndashndashndash2 983084 1
3 x ndash x2 983084 2
ndash x2 + 3 x ndash 2 983084 0
x2 ndash 3 x + 2 983086 0
( x ndash 1)( x ndash 2) 983086 0
20 x
f (x )
1
The range is x 983084
1 or x 983086
2
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18 Given that x ndash 2 y = 1
there4 x = 1 + 2 y 983089
Substitute 983089 into y + 3983102 2 xy
y + 3983102 2(1 + 2 y) y
y + 3983102 2 y + 4 y2
0983102 4 y2 + y ndash 3
0983102 (4 y ndash 3)( y + 1)
3 ndash
4
0 y
f (y )
ndash1
The range is ndash1 983100 y 983100 3mdash4
19 f ( x) 983084 0
5 x2 ndash 4 x ndash 1 983084 0
(5 x + 1)( x ndash 1) 983084 0
1 ndash ndash
5
0 x
f (x )
1
The range is ndash1mdash5
983084 x 983084 1
20 g( x) 983086 0
4 x2 ndash 9 983086 0(2 x + 3)(2 x ndash 3) 983086 0
3 ndash ndash
23 ndash
2
0 x
g (x )
The range is x 983084 ndash3mdash2
or x 983086 3mdash2
21 (a) Since y = 3 x
2
ndash 9 x + t 983086
0 for all values of x andit does not have root when y = 0
Then b2 ndash 4ac 983084 0 for 3 x2 ndash 9 x + t = 0
(ndash9)2 ndash 4(3)(t ) 983084 0
81 ndash 12t 983084 0
ndash12t 983084 ndash81
t 983086 ndash81 ndashndashndashndash ndash12
t 983086 27
ndashndashndash4
(b) Let f ( x) = a( x ndash b)2 + c
f ( x) = a( x ndash 2)2 + 0
f ( x) = a( x ndash 2)2
Substitute x = 0 f ( x) = ndash3 into the equation
ndash3 = a(0 ndash 2)2
= 4a
a = ndash3mdash4
Hence the quadratic function is
f ( x) = ndash3mdash4
( x ndash 2)2
22 (a) Given 2 x2 ndash 3 y + 2 = 0
3 y = 2 x2 + 2
y =2 x2
ndashndashndash3
+2mdash3
983089
Substitute 983089 into y 983084 10
2 x2
ndashndashndash
3
+2mdash
3
983084 10
2 x2 + 2 983084 30
2 x2 ndash 28 983084 0
x2 ndash 14 983084 0
( x + 98310598310698310614 )( x ndash 98310598310698310614 ) 983084 0
14 ndash14
0 x
f (x )
The range is ndash98310598310698310614 983084 x 983084 98310598310698310614
(b) 2 x2 ndash 8 x ndash 10 = 2( x2 ndash 4 x) ndash 10
= 2( x2 ndash 4 x + 22 ndash 22) ndash 10
= 2[( x ndash 2)2 ndash 4] ndash 10
= 2( x ndash 2)2 ndash 8 ndash 10
= 2( x ndash 2)2 ndash 18
Therefore a = 2 b = ndash2 and c = ndash18
Hence the minimum value of 2 x2 ndash 8 x ndash 10 is
ndash18
23 5 ndash 2 x 983100 0 3 x2 ndash 4 x 983086 ndash1
3 x2
ndash 4 x + 1983086
0(3 x ndash 1)( x ndash 1) 983086 0
11 ndash
3
0 x
f (x )
x 983084 1mdash3
x 983086 1
ndash2 x 983100
ndash5 x 983102
ndash5 ndashndashndash ndash2
x 983102 5mdash2
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5ndash
2
1ndash
3
1
x 5mdash2
x x 11mdash3
The range is x 983102 5mdash2
24 5 983084 f ( x) 983084 9
5 983084 5 ndash 3 x + x2 983084 9
5 983084 5 ndash 3 x + x2 5 ndash 3 x + x2 983084 9
5 ndash 3 x + x2 ndash 9 983084 0
x2 ndash 3 x ndash 4 983084 0
( x ndash 4)( x + 1) 983084 0
4 ndash1 0 x
f (x )
ndash1 983084 x 983084 4
0 983084 x2 ndash 3 x
0 983084 x( x ndash 3)
30 x
f (x )
x 983084 0 x 983086 3
0 ndash1 43
x lt 0 x gt 3
ndash1 lt x lt 4
The range is ndash1983084
x 983084
0 or 3983084
x 983084
4
25 1 983102 x2 + 3 x ndash 3 983086 ndash3
x2 + 3 x ndash 3 983086 ndash3
x2 + 3 x 983086 0
x( x + 3) 983086 0
ndash3 0 x
f (x )
x 983084 ndash3 x 983086 0
1 983102 x2 + 3 x ndash 3
0 983102 x2 + 3 x ndash 4
0 983102 ( x + 4)( x ndash 1)
1 ndash4 0 x
f (x )
ndash 4 983100 x 983100 1
ndash4
x lt ndash3 x gt 0
ndash4 x 1
ndash3 0 1
The range is ndash 4983100 x 983084 ndash3 or 0 983084 x 983100 1
26 x2 + 5 x ndash 10 983086 4
x2 + 5 x ndash 10 983084 ndash 4 x2 + 5 x ndash 10 983086 4
x2 + 5 x ndash 14 983086 0
( x + 7)( x ndash 2) 983086 0
ndash7 20
x
f (x )
x 983084 ndash7 x 983086 2
x2 + 5 x ndash 6 983084 0
( x + 6)( x ndash 1) 983084 0
ndash6 10
x
f (x )
ndash6 983084 x 983084 1
The ranges are ndash6 lt x lt 1 x lt ndash7 x gt 2
27 f ( x) = (r + 1) x2 + 2rx + r ndash 3
Given that f ( x) does not intersect the x-axis
therefore b2 ndash 4ac 983084 0
(2r)2 ndash 4(r + 1)(r ndash 3) 983084 0
4r2 ndash 4(r2 ndash 2r ndash 3) 983084 0
4r2 ndash 4r2 + 8r + 12 983084 0
2r + 3 983084 0
r 983084 ndash3mdash2
28 Given that f ( x) = 9 ndash 6 x + 2 x2 does not have real root
when f ( x) = k
there4 9 ndash 6 x + 2 x2 = k
2 x2 ndash 6 x + 9 ndash k = 0
Use b2 ndash 4ac 983084 0
(ndash 6)2 ndash 4(2)(9 ndash k ) 983084 0
36 ndash 72 + 8k 983084 0
ndash36 + 8k 983084 0 8k 983084 36
k 983084 36
ndashndashndash8
k 983084 9mdash2
29 2 x2 + 10 x ndash 20 983100 8
ndash8 983100 2 x2 + 10 x ndash 20 983100 8
ndash8983100 2 x2 + 10 x ndash 20 2 x2 + 10 x ndash 20 983100 8
2 x2 + 10 x ndash 28 983100 0
x2 + 5 x ndash 14 983100 0
( x + 7)( x ndash 2) 983100 0
ndash7 20
x
f (x )
ndash7 983100 x 983100 2
0983100 2 x2 + 10 x ndash 12
0983100 x2 + 5 x ndash 6
0983100 ( x + 6)( x ndash 1)
ndash6 10
x
f (x )
x 983100 ndash6 x 983102 1
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x ndash6 x 1
ndash7 x 2
ndash6 ndash7 1 2
The range is ndash7 983100 x 983100 ndash6 or 1 983100 x 983100 2
30 y = x2 + 5 x ndash 6
= x2 + 5 x + 983089 5mdash2 983090
2
ndash 983089 5mdash2 983090
2
ndash 6
= 983089 x + 5mdash2 983090
2
ndash25
ndashndashndash4
ndash 6
= 983089 x + 5mdash2 983090
2
ndash49
ndashndashndash4
The minimum point is (ndash5mdash2
ndash49
ndashndashndash4
)
x ndash13 ndash 6 0 1 3
y 98 0 ndash 6 0 18
x 0
ndash6 ndash13 ndash6
18
98
1 3
) ) ndash ndash ndash ndash
5
2
49
4
y
The range is ndash49
ndashndashndash4
983084 y 983084 98
31 y = ndash x2 + 2 x ndash nx + 16
= ndash x2 + (2 ndash n) x + 16
= ndash[ x2 ndash (2 ndash n) x] + 16
= ndash983091 x2 ndash (2 ndash n) x + 983089 2 ndash n ndashndashndashndashndash
2 983090
2
ndash 983089 2 ndash n ndashndashndashndashndash
2 983090
2
983092 + 16
= ndash
983091983089 x ndash
2 ndash n ndashndashndashndashndash
2 983090
2
ndash
9830892 ndash n
ndashndashndashndashndash2 983090
2
983092 + 16
= ndash983089 x ndash 2 ndash n ndashndashndashndashndash
2 983090
2
+ 983089 2 ndash n ndashndashndashndashndash
2 983090
2
+ 16
Since y = x2 ndash 3k and y = ndash x2 + 2 x ndash nx + 16 have the
same axis of symmetry that is x = 0
then ndash2 ndash n
ndashndashndashndashndash2
= 0
2 ndash n = 0
n = 2
The equation y = ndash x2 + 2 x ndash nx + 16
= ndash x2 + 2 x ndash 2 x + 16
= ndash983089 x ndash 2 ndash 2 ndashndashndashndashndash
2 983090
2
+ 983089 2 ndash 2 ndashndashndashndashndash
2 983090
2
+ 16
= ndash x2 + 16
When y = 0
ndash x2
+ 16 = 0 x2 = 16
x = plusmn98310598310698310616
= plusmn4
Therefore B = (4 0)
Substitute x = 4 y = 0 into y = x2 ndash 3k
0 = 42 ndash 3k
0 = 16 ndash 3k
k =16
ndashndashndash3
32 (a) y = ndash2[(3k ndash x)2 + n] ndash 10 has a maximum point
(4 11)
y = ndash2(3k ndash x)2 ndash 2n ndash 10
there4 3k ndash 4 = 0 and ndash2n ndash 10 = 11
k =4mdash3
ndash2n = 21
n = ndash21
ndashndashndash2
(b) Substitute k =4mdash3
and n = ndash21
ndashndashndash2
into
y = ndash2[(3k ndash x)2 + n] ndash 10
y = ndash2983091(4 ndash x)2 ndash21
ndashndashndash2 983092 ndash 10
= ndash2(4 ndash x)2 + 21 ndash 10
= ndash2(4 ndash x)2
+ 11 When y = 0
ndash2(4 ndash x)2 + 11= 0
2(4 ndash x)2 = 11
(4 ndash x)2 =11
ndashndashndash2
4 ndash x = plusmn98310598310698310698310611 ndashndashndash
2
x = 4 plusmn 98310598310698310698310611 ndashndashndash
2
= 4 ndash 98310598310698310698310611 ndashndashndash
2 4 + 98310598310698310698310611
ndashndashndash2
= 1655 6345
(c) y = ndash2(4 ndash x)2 + 11
The maximum point is (4 11)
x ndash1 0 5
y ndash39 ndash21 9
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1 y = 3(2 x ndash 1)( x + 1) ndash x(4 x ndash 5) + 2
= 3(2 x2 + x ndash 1) ndash 4 x2 + 5 x + 2
= 6 x2 + 3 x ndash 3 ndash 4 x2 + 5 x + 2
= 2 x2 + 8 x ndash 1
= 2( x2 + 4 x) ndash 1
= 2( x2 + 4 x + 22 ndash 22) ndash 1
= 2[( x + 2)2 ndash 4] ndash 1
= 2( x + 2)2 ndash 8 ndash 1
= 2( x + 2)2 ndash 9
Since a = 2 983086 0 therefore the minimum value of
y is ndash9
When y = 0 2( x + 2)2 ndash 9 = 0
( x + 2)2 =9mdash2
x + 2 = plusmn9831059831069831069mdash2 x = plusmn9831059831069831069mdash2 ndash 2
= 9831059831069831069mdash2 ndash 2 or ndash9831059831069831069mdash2 ndash 2
= 01213 or ndash 4121
When x = 0 y = 2(2)2 ndash 9
= ndash1
The minimum point is (ndash2 ndash9)
ndash1 ndash4121
(ndash2 ndash9)
012130
y
x
2 5 983084 Area of rectangle ABCD 983084 21
5 983084 ( x + 3)( x ndash 1) 983084 21
5 983084 ( x + 3)( x ndash 1)
5 983084 x2 + 2 x ndash 3
0 983084 x2 + 2 x ndash 8
0 983084 ( x + 4)( x ndash 2)
0 ndash4 2
f (x )
x
x 983084 ndash 4 x gt 2
( x + 3)( x ndash 1) 983084 21
x2 + 2 x ndash 3 983084 21
x2 + 2 x ndash 24 983084 0
( x ndash 4)( x + 6) 983084 0
0 ndash6 4
f (x )
x
ndash 6 983084 x 983084 4
ndash4
ndash6 lt x lt 4
x lt ndash4 x gt 2
2
x
ndash6 4
The range is ndash 6983084 x 983084 ndash 4 or 2 983084 x 983084 4
3 (a) p =1 + 5
ndashndashndashndashndash2
= 3
(b) y = ( x ndash 1)( x ndash 5) = x2 ndash 6 x + 5
4 y = a( x ndash 2)2 + 1
Substitute x = 0 y = 9 into the equation
9 = a(ndash2)2 + 1
8 = 4a
a = 2
Therefore the quadratic function is
f ( x) = 2( x ndash 2)2 + 1
5 x2
+ (1 + k ) x ndash k 2
+ 1 = 0For quadratic equation to have real roots
b2 ndash 4ac 983102 0
(1 + k )2 ndash 4(1)(1 ndash k 2) 983102 0
1 + 2k + k 2 ndash 4 + 4k 2 983102 0
5k 2 + 2k ndash 3 983102 0
(5k ndash 3)(k + 1) 983102 0
0 ndash1 3 ndash
5
f (k )
k
The range of values of k is k 983100 ndash1 or k 983102 3mdash5
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6 y = x2 + 7 x ndash 8 ndash 2k
For y to be positive for all real values of x there is
no roots for y = 0
Therefore b2 ndash 4ac 983084 0
72 ndash 4(1)(ndash8 ndash 2k ) 983084 0
49 + 32 + 8k 983084 0
8k 983084 ndash81
k 983084 ndash81
ndashndashndash8
Alternative
y = x2 + 7 x ndash 8 ndash 2k
= x2 + 7 x + 983089 7mdash2 983090
2
ndash 983089 7mdash2 983090
2
ndash 8 ndash 2k
= 983089 x +7mdash2 983090
2
ndash49
ndashndashndash4
ndash 8 ndash 2k
For y to be positive for all real values of x
ndash49
ndashndashndash4
ndash 8 ndash 2k 983086 0
ndash2k 983086 49
ndashndashndash4
+ 8
ndash2k 983086 81
ndashndashndash4
k 983084 ndash81
ndashndashndash8
7 Substitute x = 6 y = 0 into y = px2 + qx
0 = p(6)2 + q(6)
0 = 36 p + 6q
q + 6 p = 0 983089
y = px2 + qx
= p983089 x2 +q mdash p x983090
= p983091 x2 +q mdash p x + 983089
q ndashndashndash2 p 983090
2
ndash 983089q
ndashndashndash2 p 983090
2
983092
= p983091983089 x +q
ndashndashndash2 p 983090
2
ndashq2
ndashndashndashndash4 p2 983092
= p983089 x +q
ndashndashndash2 p 983090
2
ndashq2
ndashndashndash4 p
ndash q2
ndashndashndash4 p
= ndash12
q2 = 48 p
p =q2
ndashndashndash48
983090
Substitute 983090 into 983089
q + 6983089 q2
ndashndashndash48
983090 = 0
q +q2
mdash8
= 0
8q + q2 = 0
q(8 + q) = 0
q = 0 or q = ndash8
When q = 0 p =02
ndashndashndash48
= 0
When q = ndash8 p =(ndash8)2
ndashndashndashndashndash48
=64
ndashndashndash48
=4mdash3
Therefore the values of p =
4
mdash3 and q = ndash8
8 (2 ndash 3k ) x2 + x +3mdash4k = 0
b2 ndash 4ac = 12 ndash 4(2 ndash 3k )983089 3mdash4k 983090
= 1 ndash 6k + 9k 2
= 9k 2 ndash 6k + 1
= (3k ndash 1)2
Since (3k ndash 1)2 983102 0 for all values of k
therefore (2 ndash 3k ) x2 + x +3mdash4k = 0 has real roots for
all values of k
9 f ( x) = 3( x2 + 2mx + m2 + n)
= 3[( x + m)2 + n]
= 3( x + m)2 + 3n
The minimum point is (ndashm 3n)
Compare to A(t 3t 2)
there4 ndashm = t and 3n = 3t 2
m = ndasht n = t 2
10 (a) y = px2 + 8 x + 10 ndash p
When the graph does not intercept the x-axis
there are no roots for px2 + 8 x + 10 ndash p = 0 Therefore b2 ndash 4ac 983084 0
82 ndash 4 p(10 ndash p) 983084 0
64 ndash 40 p + 4 p2 983084 0
p2 ndash 10 p + 16 983084 0
( p ndash 2)( p ndash 8) 983084 0
2 8
Hence r = 2 and t = 8
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(b) When p = 2
y = 2 x2 + 8 x + 8
= 2( x2 + 4 x) + 8
= 2( x2 + 4 x + 22 ndash 22) + 8
= 2[( x + 2)2 ndash 4] + 8
= 2( x + 2)2 ndash 8 + 8
= 2( x + 2)2
Therefore the minimum point is (ndash2 0)
When x = 0 y = 8
When y = 0 2( x + 2)2 = 0
x = ndash2
When p = 8
y = 8 x2 + 8 x + 2
= 8( x2 + x) + 2
= 8983091 x2 + x + 983089 1mdash2 983090
2
ndash 983089 1mdash2 983090
2
983092 + 2
= 8983091983089 x +1mdash2 983090
2
ndash1mdash4 983092 + 2
= 8983089 x +1mdash2 983090
2
ndash 2 + 2
= 8983089 x +1mdash2 983090
2
Therefore the minimum point is (ndash1mdash2
0)
When x = 0 y = 2
When y = 0 0 = 8983089 x + 1mdash2 983090
2
x = ndash1mdash2
0
2
8
1 ndash ndash
2
ndash2
y p = 2 p = 8
x
11 (a) f ( x) = 24 x ndash 4 x2 + r
= ndash 4 x2 + 24 x + r
= ndash 4( x2 ndash 6 x) + r
= ndash 4( x2 ndash 6 x + 32 ndash 32) + r
= ndash 4[( x ndash 3)2 ndash 9] + r
= ndash 4( x ndash 3)2 + 36 + r
Compare to f ( x) = p( x ndash q)2 + 16
Therefore p = ndash 4 q = 3 and 36 + r = 16
r = ndash20
(b) The turning point is (3 16)
(c) f ( x) = 24 x ndash 4 x2 ndash 20
When x = 0 f ( x) = ndash20
When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0
4( x ndash 3)2 = 16
( x ndash 3)2 = 4
x ndash 3 = plusmn2
x = plusmn2 + 3
= ndash2 + 3 or 2 + 3
= 1 or 5
0 1
(3 16)
5
ndash20
y
x
12 (a) y = ndash| p( x ndash 3)2 + q|
Substitute x = 3 y = ndash5 into the equation
ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5
Substitute x = 4 y = 0 into the equation
0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0
p = 5
Therefore p = 5 q = ndash5 or p = ndash5 q = 5
(b) When x = 3 y = ndash5
For p = 5 q = ndash5
When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40
Based on the graph the range of values of y is
ndash 40 983100 y 983100 0
For p = ndash5 q = 5
When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40
Therefore the range of values of y is
ndash 40 983100 y 983100 0
13 (a) y = ndash2( x ndash 3)2 + 2k
= ndash x2 + 2 x + px ndash 8
= ndash x2 + (2 + p) x ndash 8
= ndash[ x2 ndash (2 + p) x] ndash 8
= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash
2 9830902
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983091983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
+(2 + p)2
ndashndashndashndashndashndashndash4
ndash 8
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Since the x-coordinate of the maximum point for
both the graphs are same
therefore2 + p
ndashndashndashndashndash2
= 3
p = 4
y = ndash x2 + 2 x + px ndash 8 becomes
y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8
When y = 0
ndash x2 + 6 x ndash 8 = 0
x2 ndash 6 x + 8 = 0
( x ndash 2)( x ndash 4) = 0
x = 2 or 4
Hence A(2 0) and B(4 0)
Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k
0 = ndash2(2 ndash 3)2 + 2k
2k = 2
k = 1 Hence k = 1 and p = 4
(b) For y = ndash2( x ndash 3)2 + 2k
= ndash2( x ndash 3)2 + 2(1)
= ndash2( x ndash 3)2 + 2
Maximum value of the curve is 2
For y = ndash x2 + 2 x + px ndash 8
= ndash x2 + 2 x + 4 x ndash 8
= ndash x2 + 6 x ndash 8
When x = 3
y = ndash9 + 18 ndash 8
= 1
Maximum value of the curve is 1
14 Since 3 x2 983102 0 for all values of x
therefore3 x2
ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)
983100 0
(2 x ndash 1)( x + 4) 983100 0
0 ndash4 1 ndash
2
f (x )
x
Hence ndash 4983100 x 983100 1mdash2
15 Since x2 + 1 983086 0
therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash
x2 + 1 983086 0
x2 + 3 x + 2 983086 0
( x + 1)( x + 2) 983086 0
0 ndash2 ndash1
f (x )
x
Hence x 983084 ndash2 x gt ndash1
16 ndash4
ndashndashndashndashndashndash1 ndash 3 x
983100 x
0 983100 x +4
ndashndashndashndashndashndash1 ndash 3 x
0 983100 x(1 ndash 3 x) + 4
ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0983100
x ndash 3 x2 + 4
ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash
1 ndash 3 x
0 983100 (ndash3 x + 4)( x + 1)
ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
For ndash3 x + 4 983102 0
4 983102 3 x
x 983100 4mdash3
For x + 1 983102 0
x 983102 ndash1
For 1 ndash 3 x 983086 0
ndash3 x 983086 ndash1
x 983084 1mdash3
ndash1 ndash + ndash +
1 4 ndash ndash
3
4x ndash
3
1x lt ndash
3
x ndash1
3
x
Therefore the range is ndash1 983100 x 983084 1mdash3
x 983102 4mdash3
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17 y2 ndash 9 = x
x = y2 ndash 9
When x = 0
y2 = 9
y = plusmn3
When y = 0
x = ndash9
When x = 7
7 = y2 ndash 9
y2 = 16
y = plusmn4
x ndash9 0 7
y 0 plusmn3 plusmn4
0 ndash9 7
ndash4 ndash3
34
y
x
The range of values of y is ndash 4 983100 y 983100 4
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(d) f ( x) = 10 ndash 2 x2
Therefore the maximum point is (0 10)
x ndash3 plusmn9831059831065 4
f ( x) ndash8 0 ndash22
f (x )
x
ndash8
10
ndash3 4 ndash 5 5 0
ndash22
(e) f ( x) = x( x + 2)
= x2 + 2 x
= x2 + 2 x + 12 ndash 12
= ( x + 1)2 ndash 1
Therefore the minimum point is (ndash1 ndash1)
x ndash 4 ndash2 0 2 f ( x) 8 0 0 8
f (x )
x
(ndash1ndash1) ndash4
8
ndash2 20
(f) f ( x) = ( x ndash 1)(2 x + 1)
= 2 x2
ndash x ndash 1 = 2983089 x2 ndash
xmdash2 983090 ndash 1
= 2983091 x2 ndash xmdash2
+ 983089 1mdash4 983090
2
ndash 983089 1mdash4 983090
2
983092 ndash 1
= 2983091983089 x ndash 1mdash4 983090
2
ndash1
ndashndashndash16 983092 ndash 1
= 2983089 x ndash 1mdash4 983090
2
ndash1mdash8
ndash 1
= 2983089 x ndash 1mdash4 983090
2
ndash9mdash8
Therefore the minimum point is (
1
mdash4 ndash
9
mdash8 )
x ndash1 ndash1mdash2
0 1 2
f ( x) 2 0 ndash1 0 5
f (x )
x
1 2ndash1 1
2ndash ndash
98ndashndash)14 ( ndash
2
0
5
ndash1
(g) f ( x) = ndash( x ndash 3)2 + 5
Therefore the maximum point is (3 5)
x ndash2 0 3 ndash 9831059831065 3 + 9831059831065 6
f ( x) ndash20 ndash 4 0 0 ndash4
f (x )
x
ndash2 ndash4
ndash20
(3 5)
63 + 53 ndash 50
(h) f ( x) = x2 + 4 x + 5
= x2 + 4 x + 22 ndash 22 + 5
= ( x + 2)2 + 1
Therefore the minimum point is (ndash2 1)
x ndash3 0 1 f ( x) 2 5 10
f (x )
x
(ndash2 1)
ndash3 1
2
5
10
0
(i) f ( x) = 2 x2 + 6 x ndash 8
= 2( x2
+ 3 x) ndash 8
= 2983091 x2 + 3 x + 983089 3mdash2 983090
2
ndash 983089 3mdash2 983090
2
983092 ndash 8
= 2983091983089 x +3mdash2 983090
2
ndash9mdash4 983092 ndash 8
= 2983089 x + 3mdash2 983090
2
ndash9mdash2
ndash 8
= 2983089 x + 3mdash2 983090
2
ndash25
ndashndashndash2
Therefore the minimum point is (ndash3mdash2
ndash25
ndashndashndash2
)
x ndash3 0 1 2 f ( x) ndash8 ndash8 0 12
ndash25
2ndashndash)3
2( ndash
f (x )
x
ndash3
12
ndash8
0 1 2
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(j) f ( x) = ( x ndash 4)2
Therefore the minimum point is (4 0)
x 0 4 5
f ( x) 16 0 1
f (x )
x
4 501
16
(k) f ( x) = ndash x2 + 6 x ndash 9
= ndash( x2 ndash 6 x) ndash 9
= ndash( x2 ndash 6 x + 32 ndash 32) ndash 9
= ndash[( x ndash 3)2 ndash 9] ndash 9
= ndash( x ndash 3)2
Therefore the maximum point is (3 0)
x 0 3 4
f ( x) ndash9 0 ndash1
f (x )
x 0
ndash1
ndash9
(3 0)
4
10 (a) x( x ndash 2) 983102 0
0 2x
f (x )
The range of values of x is x 983100 0 or x 983102 2
(b) ( x ndash 3)( x ndash 4) 983100 0
0 43
x
f (x )
The range of values of x is 3 983100 x 983100 4
(c) x2 ndash 3 x ndash 4 983086 0
( x ndash 4)( x + 1) 983086 0
4 ndash1 0 x
f (x )
The range of values of x is x 983084 ndash1 or x 983086 4
(d) 2 x2 + 5 x ndash 3 983084 0
(2 x ndash 1)( x + 3) 983084 0
1 ndash
2 ndash3 0
x
f (x )
The range of values of x is ndash3 983084 x 983084 1mdash2
(e) ( x ndash 3)( x + 2) 983100 ndash4
x2 ndash x ndash 6 983100 ndash4
x2 ndash x ndash 2 983100 0
( x ndash 2)( x + 1) 983100 0
ndash1 20 x
f (x )
The range of values of x is ndash1 983100 x 983100 2
(f) (2 x ndash 1)( x ndash 3) 983100 4( x ndash 3)
2 x2 ndash 6 x ndash x + 3 983100 4 x ndash 12
2 x2 ndash 11 x + 15 983100 0
( x ndash 3)(2 x ndash 5) 983100 0
5 ndash
2
30 x
f (x )
The range of values of x is5mdash2
983100 x 983100 3
(g) x2 + 4
ndashndashndashndashndashndash5
983100 2 x ndash 1
x2 + 4 983100 5(2 x ndash 1)
x2 + 4 983100 10 x ndash 5
x2 ndash 10 x + 9 983100 0
( x ndash 1)( x ndash 9) 983100 0
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910 x
f (x )
The range of values of x is 1 983100 x 983100 9
(h) x (1 ndash 4 x) 983084 5 x ndash 8
x ndash 4 x2 ndash 5 x + 8 983084 0
ndash 4 x2 ndash 4 x + 8 983084 0
x2 + x ndash 2 983086 0
( x + 2)( x ndash 1) 983086 0
1 ndash2 0 x
f (x )
The range of values of x is x 983084 ndash2 or x 983086 1
1 (a) x-coordinate of the maximum part =1 + 7
ndashndashndashndashndash2
= 4
Therefore the equation of the axis of symmetry
is x = 4
(b) f ( x) = p ndash ( x + q)2
= 5 ndash ( x ndash 4)2
2 f ( x) = 2 x2 ndash 16 x + k 2 + 2k + 1
= 2( x2 ndash 8 x) + k 2 + 2k + 1
= 2983091 x2 ndash 8 x + 983089 8mdash2 983090
2
ndash 983089 8mdash2 983090
2
983092 + k 2 + 2k + 1
= 2[( x ndash 4)2 ndash 16] + k 2 + 2k + 1
= 2( x ndash 4)2 ndash 32 + k 2 + 2k + 1
= 2( x ndash 4)2 + k 2 + 2k ndash 31
Given minimum value = ndash28
there4 k 2 + 2k ndash 31 = ndash28
k 2 + 2k ndash 3 = 0
(k + 3)(k ndash 1) = 0 k = ndash3 1
3 (a) f ( x) = 2( x ndash 3)2 + k
p is the x-coordinate of the minimum point
Therefore p = 3
(b) k is the minimum value of f ( x)
Therefore k = ndash4
(c) The equation of the axis of symmetry is x = 3
4 f ( x) = 3 x2 ndash 2 x + p
a = 3 b = ndash2 c = p
Since the graph does not intersect the x-axis
b2 ndash 4ac lt 0
(ndash2)2 ndash 4(3)( p) lt 0
4 ndash 12 p lt 0
4 lt 12 p
13
lt p
p gt13
5 f ( x) = 2 x2 ndash 12 x + 5
= 2( x2 ndash 6 x) + 5
= 2[( x ndash 3)2 ndash 32] + 5
= 2( x ndash 3)2 ndash 18 + 5
= 2( x ndash 3)2 ndash 13
there4 p = 2 q = ndash3 ndashr + 1 = ndash13
r = 14
6 (a) f ( x) = ndash x2 + 6 px + 1 ndash 4 p2
= ndash( x2 ndash 6 px) + 1 ndash 4 p2
= ndash983091 x2 ndash 6 px + 983089 6 p ndashndashndash
2 983090
2
ndash 983089 6 p ndashndashndash
2 983090
2
983092 + 1 ndash 4 p2
= ndash[( x ndash 3 p)2 ndash 9 p2] + 1 ndash 4 p2
= ndash ( x ndash 3 p)2 + 9 p2 + 1 ndash 4 p2
= ndash ( x ndash 3 p)2 + 1 + 5 p2
The maximum value given is q2 ndash p
Therefore q2 ndash p = 1 + 5 p2
5 p2 + p + 1 = q2
(b) x = 3 is symmetrical axis 3 p = 3
p = 1
Substitute p = 1 into 5 p2 + p + 1 = q2
5(1)2 + 1 + 1 = q2
q2 = 7
q = plusmn9831059831067 Hence p = 1 q = plusmn9831059831067
7 4t (t + 1) ndash 3t 2 + 12 983086 0
4t 2 + 4t ndash 3t 2 + 12 983086 0
t 2 + 4t + 12 983086 0
(t + 2)(t + 6) 983086 0
0 ndash2 ndash6x
f (x )
The range of values of t is t 983084 ndash6 or t 983086 ndash2
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1 x-coordinate of maximum point = ndash 4 + 0 ndashndashndashndashndashndash
2 = ndash2
Equation of the axis of symmetry is x = ndash2
2 Let x be the x-coordinate of A
0 + x
ndashndashndashndashndash2
= 3
x = 6
The coordinates of A are (6 4)
3 Let x be the x-coordinate of A
x + 6 ndashndashndashndashndash
2 = 2
x = 4 ndash 6
x = ndash2
The coordinates of A are (ndash2 0)
4 x-coordinate of A =0 + 8
ndashndashndashndashndash2
= 4
Let C be the centre of OB
4
5
A
C O
AC 2
= OA2
ndash OC 2
= 52 ndash 42
= 9
AC = 3
The coordinates of A are (4 3)
5 x-coordinate of minimum point =0 + 4
ndashndashndashndashndash2
= 2
x-coordinate of minimum point for the image is ndash2
6 (a) y = ( x ndash p)2 + q and minimum point is (2 ndash1)
Hence p = 2 and q = ndash1
(b) y = ( x ndash 2)2 ndash 1
When y = 0
( x ndash 2)2 ndash 1 = 0
( x ndash 2)2 = 1
x ndash 2 = plusmn1
x = plusmn1 + 2
= 1 3
Hence A is (1 0) and B is (3 0)
7 f ( x) = 2k + 1 ndash 983089 x + 1mdash2 p983090
2
Given (ndash1 k ) is the maximum point
Therefore 2k + 1 = k
k = ndash1
x +1mdash2 p = 0 when x = ndash1
ndash1 + 1mdash2 p = 0
1mdash2 p = 1
p = 2
8 Given ( p 2q) is the minimum point of
y = 2 x2 ndash 4 x + 5
= 2( x2 ndash 2 x) + 5
= 2( x2 ndash 2 x + 12 ndash 12) + 5
= 2[( x ndash 1)2 ndash 1] + 5
= 2( x ndash 1)2 ndash 2 + 5
= 2( x ndash 1)2 + 3
2q = 3
q =3mdash2
p ndash 1 = 0
p = 1
9 (a) Since (1 4) is the point on y = x2 ndash 2kx + 1
substitute x = 1 y = 4 into the equation
4 = 12 ndash 2k (1) + 1
2k = ndash2
k = ndash1
(b) y = x2 ndash 2(ndash1) x + 1
= x2 + 2 x + 1
= ( x + 1)2
Minimum value of y is 0
10 f ( x) = ndash x2 ndash 8 x + k ndash 1
= ndash( x2 + 8 x) + k ndash 1
= ndash( x2 + 8 x + 42 ndash 42) + k ndash 1
= ndash[( x + 4)2 ndash 16] + k ndash 1
= ndash( x + 4)2 + 16 + k ndash 1
= ndash( x + 4)2 + 15 + k
Since 13 is the maximum value
then 15 + k = 13
k = ndash2
11 f ( x) = 2 x2 ndash 6 x + 7
= 2( x2 ndash 3 x) + 7
= 2983091 x2 ndash 3 x + 983089 3mdash2 983090
2
ndash 983089 3mdash2 983090
2
983092 + 7
= 2983091983089 x ndash3mdash2 983090
2
ndash9mdash4 983092 + 7
= 2983089 x ndash3mdash2 983090
2
ndash9mdash2
+ 7
= 2983089 x ndash3mdash2 983090
2
+5mdash2
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The minimum point is (3mdash2
5mdash2
)
x ndash1 0 3
f ( x) 15 7 7
ndash ndash
3
5322
0
7
15
x
f (x )
ndash1
The range is5mdash2
983100 f ( x) 983100 15
12 f ( x) = 5 ndash 4 x ndash 2 x2
= ndash2 x2 ndash 4 x + 5
= ndash2( x2 + 2 x) + 5
= ndash2( x2 + 2 x + 12 ndash 12) + 5
= ndash2[( x + 1)2 ndash 1] + 5
= ndash2( x + 1)2 + 2 + 5
= ndash2( x + 1)2 + 7
When f ( x) = ndash1
ndash2( x + 1)2 + 7 = ndash1
ndash2( x + 1)2 = ndash8
( x + 1)2 = 4
x + 1 = plusmn2
x = plusmn2 ndash 1
= ndash3 or 1
13 y = ( x ndash 3)2 ndash 4
Minimum point is (3 ndash4)
x ndash1 0 1 5 6
y 12 5 0 0 5
x 0 1 5 6
5
12
ndash1
(3 ndash4)
y
The range is ndash 4983100 y 983100 12
14 y = ndash x2 + 4 x ndash 5
= ndash( x2 ndash 4 x) ndash 5
= ndash( x2 ndash 4 x + 22 ndash 22) ndash 5
= ndash [( x ndash 2)2 ndash 4] ndash 5
= ndash( x ndash 2)2 + 4 ndash 5
= ndash( x ndash 2)2 ndash 1
Maximum point is (2 ndash1)
x ndash1 0 3
y ndash10 ndash5 ndash2
x
0 3(2 ndash1)
ndash1
ndash5
ndash10
y
The range is ndash10 983100 y 983100 ndash1
15 y = 9 ndash ( x ndash 3)2Maximum point is (3 9)
x ndash1 0 6 7
y 7 0 0 7
x
y
0 ndash1 6
(3 9)
7
7
The range is 0 983100 y 983100 9
16 3 x2 983084 x
3 x2
ndash x 983084 0 x(3 x ndash 1)983084 0
ndash
13
0 x
f (x )
The range is 0 983084 x 983084 1mdash3
17 3 x ndash x
2
ndashndashndashndashndashndash2 983084 1
3 x ndash x2 983084 2
ndash x2 + 3 x ndash 2 983084 0
x2 ndash 3 x + 2 983086 0
( x ndash 1)( x ndash 2) 983086 0
20 x
f (x )
1
The range is x 983084
1 or x 983086
2
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18 Given that x ndash 2 y = 1
there4 x = 1 + 2 y 983089
Substitute 983089 into y + 3983102 2 xy
y + 3983102 2(1 + 2 y) y
y + 3983102 2 y + 4 y2
0983102 4 y2 + y ndash 3
0983102 (4 y ndash 3)( y + 1)
3 ndash
4
0 y
f (y )
ndash1
The range is ndash1 983100 y 983100 3mdash4
19 f ( x) 983084 0
5 x2 ndash 4 x ndash 1 983084 0
(5 x + 1)( x ndash 1) 983084 0
1 ndash ndash
5
0 x
f (x )
1
The range is ndash1mdash5
983084 x 983084 1
20 g( x) 983086 0
4 x2 ndash 9 983086 0(2 x + 3)(2 x ndash 3) 983086 0
3 ndash ndash
23 ndash
2
0 x
g (x )
The range is x 983084 ndash3mdash2
or x 983086 3mdash2
21 (a) Since y = 3 x
2
ndash 9 x + t 983086
0 for all values of x andit does not have root when y = 0
Then b2 ndash 4ac 983084 0 for 3 x2 ndash 9 x + t = 0
(ndash9)2 ndash 4(3)(t ) 983084 0
81 ndash 12t 983084 0
ndash12t 983084 ndash81
t 983086 ndash81 ndashndashndashndash ndash12
t 983086 27
ndashndashndash4
(b) Let f ( x) = a( x ndash b)2 + c
f ( x) = a( x ndash 2)2 + 0
f ( x) = a( x ndash 2)2
Substitute x = 0 f ( x) = ndash3 into the equation
ndash3 = a(0 ndash 2)2
= 4a
a = ndash3mdash4
Hence the quadratic function is
f ( x) = ndash3mdash4
( x ndash 2)2
22 (a) Given 2 x2 ndash 3 y + 2 = 0
3 y = 2 x2 + 2
y =2 x2
ndashndashndash3
+2mdash3
983089
Substitute 983089 into y 983084 10
2 x2
ndashndashndash
3
+2mdash
3
983084 10
2 x2 + 2 983084 30
2 x2 ndash 28 983084 0
x2 ndash 14 983084 0
( x + 98310598310698310614 )( x ndash 98310598310698310614 ) 983084 0
14 ndash14
0 x
f (x )
The range is ndash98310598310698310614 983084 x 983084 98310598310698310614
(b) 2 x2 ndash 8 x ndash 10 = 2( x2 ndash 4 x) ndash 10
= 2( x2 ndash 4 x + 22 ndash 22) ndash 10
= 2[( x ndash 2)2 ndash 4] ndash 10
= 2( x ndash 2)2 ndash 8 ndash 10
= 2( x ndash 2)2 ndash 18
Therefore a = 2 b = ndash2 and c = ndash18
Hence the minimum value of 2 x2 ndash 8 x ndash 10 is
ndash18
23 5 ndash 2 x 983100 0 3 x2 ndash 4 x 983086 ndash1
3 x2
ndash 4 x + 1983086
0(3 x ndash 1)( x ndash 1) 983086 0
11 ndash
3
0 x
f (x )
x 983084 1mdash3
x 983086 1
ndash2 x 983100
ndash5 x 983102
ndash5 ndashndashndash ndash2
x 983102 5mdash2
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5ndash
2
1ndash
3
1
x 5mdash2
x x 11mdash3
The range is x 983102 5mdash2
24 5 983084 f ( x) 983084 9
5 983084 5 ndash 3 x + x2 983084 9
5 983084 5 ndash 3 x + x2 5 ndash 3 x + x2 983084 9
5 ndash 3 x + x2 ndash 9 983084 0
x2 ndash 3 x ndash 4 983084 0
( x ndash 4)( x + 1) 983084 0
4 ndash1 0 x
f (x )
ndash1 983084 x 983084 4
0 983084 x2 ndash 3 x
0 983084 x( x ndash 3)
30 x
f (x )
x 983084 0 x 983086 3
0 ndash1 43
x lt 0 x gt 3
ndash1 lt x lt 4
The range is ndash1983084
x 983084
0 or 3983084
x 983084
4
25 1 983102 x2 + 3 x ndash 3 983086 ndash3
x2 + 3 x ndash 3 983086 ndash3
x2 + 3 x 983086 0
x( x + 3) 983086 0
ndash3 0 x
f (x )
x 983084 ndash3 x 983086 0
1 983102 x2 + 3 x ndash 3
0 983102 x2 + 3 x ndash 4
0 983102 ( x + 4)( x ndash 1)
1 ndash4 0 x
f (x )
ndash 4 983100 x 983100 1
ndash4
x lt ndash3 x gt 0
ndash4 x 1
ndash3 0 1
The range is ndash 4983100 x 983084 ndash3 or 0 983084 x 983100 1
26 x2 + 5 x ndash 10 983086 4
x2 + 5 x ndash 10 983084 ndash 4 x2 + 5 x ndash 10 983086 4
x2 + 5 x ndash 14 983086 0
( x + 7)( x ndash 2) 983086 0
ndash7 20
x
f (x )
x 983084 ndash7 x 983086 2
x2 + 5 x ndash 6 983084 0
( x + 6)( x ndash 1) 983084 0
ndash6 10
x
f (x )
ndash6 983084 x 983084 1
The ranges are ndash6 lt x lt 1 x lt ndash7 x gt 2
27 f ( x) = (r + 1) x2 + 2rx + r ndash 3
Given that f ( x) does not intersect the x-axis
therefore b2 ndash 4ac 983084 0
(2r)2 ndash 4(r + 1)(r ndash 3) 983084 0
4r2 ndash 4(r2 ndash 2r ndash 3) 983084 0
4r2 ndash 4r2 + 8r + 12 983084 0
2r + 3 983084 0
r 983084 ndash3mdash2
28 Given that f ( x) = 9 ndash 6 x + 2 x2 does not have real root
when f ( x) = k
there4 9 ndash 6 x + 2 x2 = k
2 x2 ndash 6 x + 9 ndash k = 0
Use b2 ndash 4ac 983084 0
(ndash 6)2 ndash 4(2)(9 ndash k ) 983084 0
36 ndash 72 + 8k 983084 0
ndash36 + 8k 983084 0 8k 983084 36
k 983084 36
ndashndashndash8
k 983084 9mdash2
29 2 x2 + 10 x ndash 20 983100 8
ndash8 983100 2 x2 + 10 x ndash 20 983100 8
ndash8983100 2 x2 + 10 x ndash 20 2 x2 + 10 x ndash 20 983100 8
2 x2 + 10 x ndash 28 983100 0
x2 + 5 x ndash 14 983100 0
( x + 7)( x ndash 2) 983100 0
ndash7 20
x
f (x )
ndash7 983100 x 983100 2
0983100 2 x2 + 10 x ndash 12
0983100 x2 + 5 x ndash 6
0983100 ( x + 6)( x ndash 1)
ndash6 10
x
f (x )
x 983100 ndash6 x 983102 1
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x ndash6 x 1
ndash7 x 2
ndash6 ndash7 1 2
The range is ndash7 983100 x 983100 ndash6 or 1 983100 x 983100 2
30 y = x2 + 5 x ndash 6
= x2 + 5 x + 983089 5mdash2 983090
2
ndash 983089 5mdash2 983090
2
ndash 6
= 983089 x + 5mdash2 983090
2
ndash25
ndashndashndash4
ndash 6
= 983089 x + 5mdash2 983090
2
ndash49
ndashndashndash4
The minimum point is (ndash5mdash2
ndash49
ndashndashndash4
)
x ndash13 ndash 6 0 1 3
y 98 0 ndash 6 0 18
x 0
ndash6 ndash13 ndash6
18
98
1 3
) ) ndash ndash ndash ndash
5
2
49
4
y
The range is ndash49
ndashndashndash4
983084 y 983084 98
31 y = ndash x2 + 2 x ndash nx + 16
= ndash x2 + (2 ndash n) x + 16
= ndash[ x2 ndash (2 ndash n) x] + 16
= ndash983091 x2 ndash (2 ndash n) x + 983089 2 ndash n ndashndashndashndashndash
2 983090
2
ndash 983089 2 ndash n ndashndashndashndashndash
2 983090
2
983092 + 16
= ndash
983091983089 x ndash
2 ndash n ndashndashndashndashndash
2 983090
2
ndash
9830892 ndash n
ndashndashndashndashndash2 983090
2
983092 + 16
= ndash983089 x ndash 2 ndash n ndashndashndashndashndash
2 983090
2
+ 983089 2 ndash n ndashndashndashndashndash
2 983090
2
+ 16
Since y = x2 ndash 3k and y = ndash x2 + 2 x ndash nx + 16 have the
same axis of symmetry that is x = 0
then ndash2 ndash n
ndashndashndashndashndash2
= 0
2 ndash n = 0
n = 2
The equation y = ndash x2 + 2 x ndash nx + 16
= ndash x2 + 2 x ndash 2 x + 16
= ndash983089 x ndash 2 ndash 2 ndashndashndashndashndash
2 983090
2
+ 983089 2 ndash 2 ndashndashndashndashndash
2 983090
2
+ 16
= ndash x2 + 16
When y = 0
ndash x2
+ 16 = 0 x2 = 16
x = plusmn98310598310698310616
= plusmn4
Therefore B = (4 0)
Substitute x = 4 y = 0 into y = x2 ndash 3k
0 = 42 ndash 3k
0 = 16 ndash 3k
k =16
ndashndashndash3
32 (a) y = ndash2[(3k ndash x)2 + n] ndash 10 has a maximum point
(4 11)
y = ndash2(3k ndash x)2 ndash 2n ndash 10
there4 3k ndash 4 = 0 and ndash2n ndash 10 = 11
k =4mdash3
ndash2n = 21
n = ndash21
ndashndashndash2
(b) Substitute k =4mdash3
and n = ndash21
ndashndashndash2
into
y = ndash2[(3k ndash x)2 + n] ndash 10
y = ndash2983091(4 ndash x)2 ndash21
ndashndashndash2 983092 ndash 10
= ndash2(4 ndash x)2 + 21 ndash 10
= ndash2(4 ndash x)2
+ 11 When y = 0
ndash2(4 ndash x)2 + 11= 0
2(4 ndash x)2 = 11
(4 ndash x)2 =11
ndashndashndash2
4 ndash x = plusmn98310598310698310698310611 ndashndashndash
2
x = 4 plusmn 98310598310698310698310611 ndashndashndash
2
= 4 ndash 98310598310698310698310611 ndashndashndash
2 4 + 98310598310698310698310611
ndashndashndash2
= 1655 6345
(c) y = ndash2(4 ndash x)2 + 11
The maximum point is (4 11)
x ndash1 0 5
y ndash39 ndash21 9
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1 y = 3(2 x ndash 1)( x + 1) ndash x(4 x ndash 5) + 2
= 3(2 x2 + x ndash 1) ndash 4 x2 + 5 x + 2
= 6 x2 + 3 x ndash 3 ndash 4 x2 + 5 x + 2
= 2 x2 + 8 x ndash 1
= 2( x2 + 4 x) ndash 1
= 2( x2 + 4 x + 22 ndash 22) ndash 1
= 2[( x + 2)2 ndash 4] ndash 1
= 2( x + 2)2 ndash 8 ndash 1
= 2( x + 2)2 ndash 9
Since a = 2 983086 0 therefore the minimum value of
y is ndash9
When y = 0 2( x + 2)2 ndash 9 = 0
( x + 2)2 =9mdash2
x + 2 = plusmn9831059831069831069mdash2 x = plusmn9831059831069831069mdash2 ndash 2
= 9831059831069831069mdash2 ndash 2 or ndash9831059831069831069mdash2 ndash 2
= 01213 or ndash 4121
When x = 0 y = 2(2)2 ndash 9
= ndash1
The minimum point is (ndash2 ndash9)
ndash1 ndash4121
(ndash2 ndash9)
012130
y
x
2 5 983084 Area of rectangle ABCD 983084 21
5 983084 ( x + 3)( x ndash 1) 983084 21
5 983084 ( x + 3)( x ndash 1)
5 983084 x2 + 2 x ndash 3
0 983084 x2 + 2 x ndash 8
0 983084 ( x + 4)( x ndash 2)
0 ndash4 2
f (x )
x
x 983084 ndash 4 x gt 2
( x + 3)( x ndash 1) 983084 21
x2 + 2 x ndash 3 983084 21
x2 + 2 x ndash 24 983084 0
( x ndash 4)( x + 6) 983084 0
0 ndash6 4
f (x )
x
ndash 6 983084 x 983084 4
ndash4
ndash6 lt x lt 4
x lt ndash4 x gt 2
2
x
ndash6 4
The range is ndash 6983084 x 983084 ndash 4 or 2 983084 x 983084 4
3 (a) p =1 + 5
ndashndashndashndashndash2
= 3
(b) y = ( x ndash 1)( x ndash 5) = x2 ndash 6 x + 5
4 y = a( x ndash 2)2 + 1
Substitute x = 0 y = 9 into the equation
9 = a(ndash2)2 + 1
8 = 4a
a = 2
Therefore the quadratic function is
f ( x) = 2( x ndash 2)2 + 1
5 x2
+ (1 + k ) x ndash k 2
+ 1 = 0For quadratic equation to have real roots
b2 ndash 4ac 983102 0
(1 + k )2 ndash 4(1)(1 ndash k 2) 983102 0
1 + 2k + k 2 ndash 4 + 4k 2 983102 0
5k 2 + 2k ndash 3 983102 0
(5k ndash 3)(k + 1) 983102 0
0 ndash1 3 ndash
5
f (k )
k
The range of values of k is k 983100 ndash1 or k 983102 3mdash5
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6 y = x2 + 7 x ndash 8 ndash 2k
For y to be positive for all real values of x there is
no roots for y = 0
Therefore b2 ndash 4ac 983084 0
72 ndash 4(1)(ndash8 ndash 2k ) 983084 0
49 + 32 + 8k 983084 0
8k 983084 ndash81
k 983084 ndash81
ndashndashndash8
Alternative
y = x2 + 7 x ndash 8 ndash 2k
= x2 + 7 x + 983089 7mdash2 983090
2
ndash 983089 7mdash2 983090
2
ndash 8 ndash 2k
= 983089 x +7mdash2 983090
2
ndash49
ndashndashndash4
ndash 8 ndash 2k
For y to be positive for all real values of x
ndash49
ndashndashndash4
ndash 8 ndash 2k 983086 0
ndash2k 983086 49
ndashndashndash4
+ 8
ndash2k 983086 81
ndashndashndash4
k 983084 ndash81
ndashndashndash8
7 Substitute x = 6 y = 0 into y = px2 + qx
0 = p(6)2 + q(6)
0 = 36 p + 6q
q + 6 p = 0 983089
y = px2 + qx
= p983089 x2 +q mdash p x983090
= p983091 x2 +q mdash p x + 983089
q ndashndashndash2 p 983090
2
ndash 983089q
ndashndashndash2 p 983090
2
983092
= p983091983089 x +q
ndashndashndash2 p 983090
2
ndashq2
ndashndashndashndash4 p2 983092
= p983089 x +q
ndashndashndash2 p 983090
2
ndashq2
ndashndashndash4 p
ndash q2
ndashndashndash4 p
= ndash12
q2 = 48 p
p =q2
ndashndashndash48
983090
Substitute 983090 into 983089
q + 6983089 q2
ndashndashndash48
983090 = 0
q +q2
mdash8
= 0
8q + q2 = 0
q(8 + q) = 0
q = 0 or q = ndash8
When q = 0 p =02
ndashndashndash48
= 0
When q = ndash8 p =(ndash8)2
ndashndashndashndashndash48
=64
ndashndashndash48
=4mdash3
Therefore the values of p =
4
mdash3 and q = ndash8
8 (2 ndash 3k ) x2 + x +3mdash4k = 0
b2 ndash 4ac = 12 ndash 4(2 ndash 3k )983089 3mdash4k 983090
= 1 ndash 6k + 9k 2
= 9k 2 ndash 6k + 1
= (3k ndash 1)2
Since (3k ndash 1)2 983102 0 for all values of k
therefore (2 ndash 3k ) x2 + x +3mdash4k = 0 has real roots for
all values of k
9 f ( x) = 3( x2 + 2mx + m2 + n)
= 3[( x + m)2 + n]
= 3( x + m)2 + 3n
The minimum point is (ndashm 3n)
Compare to A(t 3t 2)
there4 ndashm = t and 3n = 3t 2
m = ndasht n = t 2
10 (a) y = px2 + 8 x + 10 ndash p
When the graph does not intercept the x-axis
there are no roots for px2 + 8 x + 10 ndash p = 0 Therefore b2 ndash 4ac 983084 0
82 ndash 4 p(10 ndash p) 983084 0
64 ndash 40 p + 4 p2 983084 0
p2 ndash 10 p + 16 983084 0
( p ndash 2)( p ndash 8) 983084 0
2 8
Hence r = 2 and t = 8
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(b) When p = 2
y = 2 x2 + 8 x + 8
= 2( x2 + 4 x) + 8
= 2( x2 + 4 x + 22 ndash 22) + 8
= 2[( x + 2)2 ndash 4] + 8
= 2( x + 2)2 ndash 8 + 8
= 2( x + 2)2
Therefore the minimum point is (ndash2 0)
When x = 0 y = 8
When y = 0 2( x + 2)2 = 0
x = ndash2
When p = 8
y = 8 x2 + 8 x + 2
= 8( x2 + x) + 2
= 8983091 x2 + x + 983089 1mdash2 983090
2
ndash 983089 1mdash2 983090
2
983092 + 2
= 8983091983089 x +1mdash2 983090
2
ndash1mdash4 983092 + 2
= 8983089 x +1mdash2 983090
2
ndash 2 + 2
= 8983089 x +1mdash2 983090
2
Therefore the minimum point is (ndash1mdash2
0)
When x = 0 y = 2
When y = 0 0 = 8983089 x + 1mdash2 983090
2
x = ndash1mdash2
0
2
8
1 ndash ndash
2
ndash2
y p = 2 p = 8
x
11 (a) f ( x) = 24 x ndash 4 x2 + r
= ndash 4 x2 + 24 x + r
= ndash 4( x2 ndash 6 x) + r
= ndash 4( x2 ndash 6 x + 32 ndash 32) + r
= ndash 4[( x ndash 3)2 ndash 9] + r
= ndash 4( x ndash 3)2 + 36 + r
Compare to f ( x) = p( x ndash q)2 + 16
Therefore p = ndash 4 q = 3 and 36 + r = 16
r = ndash20
(b) The turning point is (3 16)
(c) f ( x) = 24 x ndash 4 x2 ndash 20
When x = 0 f ( x) = ndash20
When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0
4( x ndash 3)2 = 16
( x ndash 3)2 = 4
x ndash 3 = plusmn2
x = plusmn2 + 3
= ndash2 + 3 or 2 + 3
= 1 or 5
0 1
(3 16)
5
ndash20
y
x
12 (a) y = ndash| p( x ndash 3)2 + q|
Substitute x = 3 y = ndash5 into the equation
ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5
Substitute x = 4 y = 0 into the equation
0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0
p = 5
Therefore p = 5 q = ndash5 or p = ndash5 q = 5
(b) When x = 3 y = ndash5
For p = 5 q = ndash5
When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40
Based on the graph the range of values of y is
ndash 40 983100 y 983100 0
For p = ndash5 q = 5
When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40
Therefore the range of values of y is
ndash 40 983100 y 983100 0
13 (a) y = ndash2( x ndash 3)2 + 2k
= ndash x2 + 2 x + px ndash 8
= ndash x2 + (2 + p) x ndash 8
= ndash[ x2 ndash (2 + p) x] ndash 8
= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash
2 9830902
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983091983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
+(2 + p)2
ndashndashndashndashndashndashndash4
ndash 8
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Since the x-coordinate of the maximum point for
both the graphs are same
therefore2 + p
ndashndashndashndashndash2
= 3
p = 4
y = ndash x2 + 2 x + px ndash 8 becomes
y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8
When y = 0
ndash x2 + 6 x ndash 8 = 0
x2 ndash 6 x + 8 = 0
( x ndash 2)( x ndash 4) = 0
x = 2 or 4
Hence A(2 0) and B(4 0)
Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k
0 = ndash2(2 ndash 3)2 + 2k
2k = 2
k = 1 Hence k = 1 and p = 4
(b) For y = ndash2( x ndash 3)2 + 2k
= ndash2( x ndash 3)2 + 2(1)
= ndash2( x ndash 3)2 + 2
Maximum value of the curve is 2
For y = ndash x2 + 2 x + px ndash 8
= ndash x2 + 2 x + 4 x ndash 8
= ndash x2 + 6 x ndash 8
When x = 3
y = ndash9 + 18 ndash 8
= 1
Maximum value of the curve is 1
14 Since 3 x2 983102 0 for all values of x
therefore3 x2
ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)
983100 0
(2 x ndash 1)( x + 4) 983100 0
0 ndash4 1 ndash
2
f (x )
x
Hence ndash 4983100 x 983100 1mdash2
15 Since x2 + 1 983086 0
therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash
x2 + 1 983086 0
x2 + 3 x + 2 983086 0
( x + 1)( x + 2) 983086 0
0 ndash2 ndash1
f (x )
x
Hence x 983084 ndash2 x gt ndash1
16 ndash4
ndashndashndashndashndashndash1 ndash 3 x
983100 x
0 983100 x +4
ndashndashndashndashndashndash1 ndash 3 x
0 983100 x(1 ndash 3 x) + 4
ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0983100
x ndash 3 x2 + 4
ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash
1 ndash 3 x
0 983100 (ndash3 x + 4)( x + 1)
ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
For ndash3 x + 4 983102 0
4 983102 3 x
x 983100 4mdash3
For x + 1 983102 0
x 983102 ndash1
For 1 ndash 3 x 983086 0
ndash3 x 983086 ndash1
x 983084 1mdash3
ndash1 ndash + ndash +
1 4 ndash ndash
3
4x ndash
3
1x lt ndash
3
x ndash1
3
x
Therefore the range is ndash1 983100 x 983084 1mdash3
x 983102 4mdash3
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17 y2 ndash 9 = x
x = y2 ndash 9
When x = 0
y2 = 9
y = plusmn3
When y = 0
x = ndash9
When x = 7
7 = y2 ndash 9
y2 = 16
y = plusmn4
x ndash9 0 7
y 0 plusmn3 plusmn4
0 ndash9 7
ndash4 ndash3
34
y
x
The range of values of y is ndash 4 983100 y 983100 4
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(j) f ( x) = ( x ndash 4)2
Therefore the minimum point is (4 0)
x 0 4 5
f ( x) 16 0 1
f (x )
x
4 501
16
(k) f ( x) = ndash x2 + 6 x ndash 9
= ndash( x2 ndash 6 x) ndash 9
= ndash( x2 ndash 6 x + 32 ndash 32) ndash 9
= ndash[( x ndash 3)2 ndash 9] ndash 9
= ndash( x ndash 3)2
Therefore the maximum point is (3 0)
x 0 3 4
f ( x) ndash9 0 ndash1
f (x )
x 0
ndash1
ndash9
(3 0)
4
10 (a) x( x ndash 2) 983102 0
0 2x
f (x )
The range of values of x is x 983100 0 or x 983102 2
(b) ( x ndash 3)( x ndash 4) 983100 0
0 43
x
f (x )
The range of values of x is 3 983100 x 983100 4
(c) x2 ndash 3 x ndash 4 983086 0
( x ndash 4)( x + 1) 983086 0
4 ndash1 0 x
f (x )
The range of values of x is x 983084 ndash1 or x 983086 4
(d) 2 x2 + 5 x ndash 3 983084 0
(2 x ndash 1)( x + 3) 983084 0
1 ndash
2 ndash3 0
x
f (x )
The range of values of x is ndash3 983084 x 983084 1mdash2
(e) ( x ndash 3)( x + 2) 983100 ndash4
x2 ndash x ndash 6 983100 ndash4
x2 ndash x ndash 2 983100 0
( x ndash 2)( x + 1) 983100 0
ndash1 20 x
f (x )
The range of values of x is ndash1 983100 x 983100 2
(f) (2 x ndash 1)( x ndash 3) 983100 4( x ndash 3)
2 x2 ndash 6 x ndash x + 3 983100 4 x ndash 12
2 x2 ndash 11 x + 15 983100 0
( x ndash 3)(2 x ndash 5) 983100 0
5 ndash
2
30 x
f (x )
The range of values of x is5mdash2
983100 x 983100 3
(g) x2 + 4
ndashndashndashndashndashndash5
983100 2 x ndash 1
x2 + 4 983100 5(2 x ndash 1)
x2 + 4 983100 10 x ndash 5
x2 ndash 10 x + 9 983100 0
( x ndash 1)( x ndash 9) 983100 0
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910 x
f (x )
The range of values of x is 1 983100 x 983100 9
(h) x (1 ndash 4 x) 983084 5 x ndash 8
x ndash 4 x2 ndash 5 x + 8 983084 0
ndash 4 x2 ndash 4 x + 8 983084 0
x2 + x ndash 2 983086 0
( x + 2)( x ndash 1) 983086 0
1 ndash2 0 x
f (x )
The range of values of x is x 983084 ndash2 or x 983086 1
1 (a) x-coordinate of the maximum part =1 + 7
ndashndashndashndashndash2
= 4
Therefore the equation of the axis of symmetry
is x = 4
(b) f ( x) = p ndash ( x + q)2
= 5 ndash ( x ndash 4)2
2 f ( x) = 2 x2 ndash 16 x + k 2 + 2k + 1
= 2( x2 ndash 8 x) + k 2 + 2k + 1
= 2983091 x2 ndash 8 x + 983089 8mdash2 983090
2
ndash 983089 8mdash2 983090
2
983092 + k 2 + 2k + 1
= 2[( x ndash 4)2 ndash 16] + k 2 + 2k + 1
= 2( x ndash 4)2 ndash 32 + k 2 + 2k + 1
= 2( x ndash 4)2 + k 2 + 2k ndash 31
Given minimum value = ndash28
there4 k 2 + 2k ndash 31 = ndash28
k 2 + 2k ndash 3 = 0
(k + 3)(k ndash 1) = 0 k = ndash3 1
3 (a) f ( x) = 2( x ndash 3)2 + k
p is the x-coordinate of the minimum point
Therefore p = 3
(b) k is the minimum value of f ( x)
Therefore k = ndash4
(c) The equation of the axis of symmetry is x = 3
4 f ( x) = 3 x2 ndash 2 x + p
a = 3 b = ndash2 c = p
Since the graph does not intersect the x-axis
b2 ndash 4ac lt 0
(ndash2)2 ndash 4(3)( p) lt 0
4 ndash 12 p lt 0
4 lt 12 p
13
lt p
p gt13
5 f ( x) = 2 x2 ndash 12 x + 5
= 2( x2 ndash 6 x) + 5
= 2[( x ndash 3)2 ndash 32] + 5
= 2( x ndash 3)2 ndash 18 + 5
= 2( x ndash 3)2 ndash 13
there4 p = 2 q = ndash3 ndashr + 1 = ndash13
r = 14
6 (a) f ( x) = ndash x2 + 6 px + 1 ndash 4 p2
= ndash( x2 ndash 6 px) + 1 ndash 4 p2
= ndash983091 x2 ndash 6 px + 983089 6 p ndashndashndash
2 983090
2
ndash 983089 6 p ndashndashndash
2 983090
2
983092 + 1 ndash 4 p2
= ndash[( x ndash 3 p)2 ndash 9 p2] + 1 ndash 4 p2
= ndash ( x ndash 3 p)2 + 9 p2 + 1 ndash 4 p2
= ndash ( x ndash 3 p)2 + 1 + 5 p2
The maximum value given is q2 ndash p
Therefore q2 ndash p = 1 + 5 p2
5 p2 + p + 1 = q2
(b) x = 3 is symmetrical axis 3 p = 3
p = 1
Substitute p = 1 into 5 p2 + p + 1 = q2
5(1)2 + 1 + 1 = q2
q2 = 7
q = plusmn9831059831067 Hence p = 1 q = plusmn9831059831067
7 4t (t + 1) ndash 3t 2 + 12 983086 0
4t 2 + 4t ndash 3t 2 + 12 983086 0
t 2 + 4t + 12 983086 0
(t + 2)(t + 6) 983086 0
0 ndash2 ndash6x
f (x )
The range of values of t is t 983084 ndash6 or t 983086 ndash2
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1 x-coordinate of maximum point = ndash 4 + 0 ndashndashndashndashndashndash
2 = ndash2
Equation of the axis of symmetry is x = ndash2
2 Let x be the x-coordinate of A
0 + x
ndashndashndashndashndash2
= 3
x = 6
The coordinates of A are (6 4)
3 Let x be the x-coordinate of A
x + 6 ndashndashndashndashndash
2 = 2
x = 4 ndash 6
x = ndash2
The coordinates of A are (ndash2 0)
4 x-coordinate of A =0 + 8
ndashndashndashndashndash2
= 4
Let C be the centre of OB
4
5
A
C O
AC 2
= OA2
ndash OC 2
= 52 ndash 42
= 9
AC = 3
The coordinates of A are (4 3)
5 x-coordinate of minimum point =0 + 4
ndashndashndashndashndash2
= 2
x-coordinate of minimum point for the image is ndash2
6 (a) y = ( x ndash p)2 + q and minimum point is (2 ndash1)
Hence p = 2 and q = ndash1
(b) y = ( x ndash 2)2 ndash 1
When y = 0
( x ndash 2)2 ndash 1 = 0
( x ndash 2)2 = 1
x ndash 2 = plusmn1
x = plusmn1 + 2
= 1 3
Hence A is (1 0) and B is (3 0)
7 f ( x) = 2k + 1 ndash 983089 x + 1mdash2 p983090
2
Given (ndash1 k ) is the maximum point
Therefore 2k + 1 = k
k = ndash1
x +1mdash2 p = 0 when x = ndash1
ndash1 + 1mdash2 p = 0
1mdash2 p = 1
p = 2
8 Given ( p 2q) is the minimum point of
y = 2 x2 ndash 4 x + 5
= 2( x2 ndash 2 x) + 5
= 2( x2 ndash 2 x + 12 ndash 12) + 5
= 2[( x ndash 1)2 ndash 1] + 5
= 2( x ndash 1)2 ndash 2 + 5
= 2( x ndash 1)2 + 3
2q = 3
q =3mdash2
p ndash 1 = 0
p = 1
9 (a) Since (1 4) is the point on y = x2 ndash 2kx + 1
substitute x = 1 y = 4 into the equation
4 = 12 ndash 2k (1) + 1
2k = ndash2
k = ndash1
(b) y = x2 ndash 2(ndash1) x + 1
= x2 + 2 x + 1
= ( x + 1)2
Minimum value of y is 0
10 f ( x) = ndash x2 ndash 8 x + k ndash 1
= ndash( x2 + 8 x) + k ndash 1
= ndash( x2 + 8 x + 42 ndash 42) + k ndash 1
= ndash[( x + 4)2 ndash 16] + k ndash 1
= ndash( x + 4)2 + 16 + k ndash 1
= ndash( x + 4)2 + 15 + k
Since 13 is the maximum value
then 15 + k = 13
k = ndash2
11 f ( x) = 2 x2 ndash 6 x + 7
= 2( x2 ndash 3 x) + 7
= 2983091 x2 ndash 3 x + 983089 3mdash2 983090
2
ndash 983089 3mdash2 983090
2
983092 + 7
= 2983091983089 x ndash3mdash2 983090
2
ndash9mdash4 983092 + 7
= 2983089 x ndash3mdash2 983090
2
ndash9mdash2
+ 7
= 2983089 x ndash3mdash2 983090
2
+5mdash2
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The minimum point is (3mdash2
5mdash2
)
x ndash1 0 3
f ( x) 15 7 7
ndash ndash
3
5322
0
7
15
x
f (x )
ndash1
The range is5mdash2
983100 f ( x) 983100 15
12 f ( x) = 5 ndash 4 x ndash 2 x2
= ndash2 x2 ndash 4 x + 5
= ndash2( x2 + 2 x) + 5
= ndash2( x2 + 2 x + 12 ndash 12) + 5
= ndash2[( x + 1)2 ndash 1] + 5
= ndash2( x + 1)2 + 2 + 5
= ndash2( x + 1)2 + 7
When f ( x) = ndash1
ndash2( x + 1)2 + 7 = ndash1
ndash2( x + 1)2 = ndash8
( x + 1)2 = 4
x + 1 = plusmn2
x = plusmn2 ndash 1
= ndash3 or 1
13 y = ( x ndash 3)2 ndash 4
Minimum point is (3 ndash4)
x ndash1 0 1 5 6
y 12 5 0 0 5
x 0 1 5 6
5
12
ndash1
(3 ndash4)
y
The range is ndash 4983100 y 983100 12
14 y = ndash x2 + 4 x ndash 5
= ndash( x2 ndash 4 x) ndash 5
= ndash( x2 ndash 4 x + 22 ndash 22) ndash 5
= ndash [( x ndash 2)2 ndash 4] ndash 5
= ndash( x ndash 2)2 + 4 ndash 5
= ndash( x ndash 2)2 ndash 1
Maximum point is (2 ndash1)
x ndash1 0 3
y ndash10 ndash5 ndash2
x
0 3(2 ndash1)
ndash1
ndash5
ndash10
y
The range is ndash10 983100 y 983100 ndash1
15 y = 9 ndash ( x ndash 3)2Maximum point is (3 9)
x ndash1 0 6 7
y 7 0 0 7
x
y
0 ndash1 6
(3 9)
7
7
The range is 0 983100 y 983100 9
16 3 x2 983084 x
3 x2
ndash x 983084 0 x(3 x ndash 1)983084 0
ndash
13
0 x
f (x )
The range is 0 983084 x 983084 1mdash3
17 3 x ndash x
2
ndashndashndashndashndashndash2 983084 1
3 x ndash x2 983084 2
ndash x2 + 3 x ndash 2 983084 0
x2 ndash 3 x + 2 983086 0
( x ndash 1)( x ndash 2) 983086 0
20 x
f (x )
1
The range is x 983084
1 or x 983086
2
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18 Given that x ndash 2 y = 1
there4 x = 1 + 2 y 983089
Substitute 983089 into y + 3983102 2 xy
y + 3983102 2(1 + 2 y) y
y + 3983102 2 y + 4 y2
0983102 4 y2 + y ndash 3
0983102 (4 y ndash 3)( y + 1)
3 ndash
4
0 y
f (y )
ndash1
The range is ndash1 983100 y 983100 3mdash4
19 f ( x) 983084 0
5 x2 ndash 4 x ndash 1 983084 0
(5 x + 1)( x ndash 1) 983084 0
1 ndash ndash
5
0 x
f (x )
1
The range is ndash1mdash5
983084 x 983084 1
20 g( x) 983086 0
4 x2 ndash 9 983086 0(2 x + 3)(2 x ndash 3) 983086 0
3 ndash ndash
23 ndash
2
0 x
g (x )
The range is x 983084 ndash3mdash2
or x 983086 3mdash2
21 (a) Since y = 3 x
2
ndash 9 x + t 983086
0 for all values of x andit does not have root when y = 0
Then b2 ndash 4ac 983084 0 for 3 x2 ndash 9 x + t = 0
(ndash9)2 ndash 4(3)(t ) 983084 0
81 ndash 12t 983084 0
ndash12t 983084 ndash81
t 983086 ndash81 ndashndashndashndash ndash12
t 983086 27
ndashndashndash4
(b) Let f ( x) = a( x ndash b)2 + c
f ( x) = a( x ndash 2)2 + 0
f ( x) = a( x ndash 2)2
Substitute x = 0 f ( x) = ndash3 into the equation
ndash3 = a(0 ndash 2)2
= 4a
a = ndash3mdash4
Hence the quadratic function is
f ( x) = ndash3mdash4
( x ndash 2)2
22 (a) Given 2 x2 ndash 3 y + 2 = 0
3 y = 2 x2 + 2
y =2 x2
ndashndashndash3
+2mdash3
983089
Substitute 983089 into y 983084 10
2 x2
ndashndashndash
3
+2mdash
3
983084 10
2 x2 + 2 983084 30
2 x2 ndash 28 983084 0
x2 ndash 14 983084 0
( x + 98310598310698310614 )( x ndash 98310598310698310614 ) 983084 0
14 ndash14
0 x
f (x )
The range is ndash98310598310698310614 983084 x 983084 98310598310698310614
(b) 2 x2 ndash 8 x ndash 10 = 2( x2 ndash 4 x) ndash 10
= 2( x2 ndash 4 x + 22 ndash 22) ndash 10
= 2[( x ndash 2)2 ndash 4] ndash 10
= 2( x ndash 2)2 ndash 8 ndash 10
= 2( x ndash 2)2 ndash 18
Therefore a = 2 b = ndash2 and c = ndash18
Hence the minimum value of 2 x2 ndash 8 x ndash 10 is
ndash18
23 5 ndash 2 x 983100 0 3 x2 ndash 4 x 983086 ndash1
3 x2
ndash 4 x + 1983086
0(3 x ndash 1)( x ndash 1) 983086 0
11 ndash
3
0 x
f (x )
x 983084 1mdash3
x 983086 1
ndash2 x 983100
ndash5 x 983102
ndash5 ndashndashndash ndash2
x 983102 5mdash2
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5ndash
2
1ndash
3
1
x 5mdash2
x x 11mdash3
The range is x 983102 5mdash2
24 5 983084 f ( x) 983084 9
5 983084 5 ndash 3 x + x2 983084 9
5 983084 5 ndash 3 x + x2 5 ndash 3 x + x2 983084 9
5 ndash 3 x + x2 ndash 9 983084 0
x2 ndash 3 x ndash 4 983084 0
( x ndash 4)( x + 1) 983084 0
4 ndash1 0 x
f (x )
ndash1 983084 x 983084 4
0 983084 x2 ndash 3 x
0 983084 x( x ndash 3)
30 x
f (x )
x 983084 0 x 983086 3
0 ndash1 43
x lt 0 x gt 3
ndash1 lt x lt 4
The range is ndash1983084
x 983084
0 or 3983084
x 983084
4
25 1 983102 x2 + 3 x ndash 3 983086 ndash3
x2 + 3 x ndash 3 983086 ndash3
x2 + 3 x 983086 0
x( x + 3) 983086 0
ndash3 0 x
f (x )
x 983084 ndash3 x 983086 0
1 983102 x2 + 3 x ndash 3
0 983102 x2 + 3 x ndash 4
0 983102 ( x + 4)( x ndash 1)
1 ndash4 0 x
f (x )
ndash 4 983100 x 983100 1
ndash4
x lt ndash3 x gt 0
ndash4 x 1
ndash3 0 1
The range is ndash 4983100 x 983084 ndash3 or 0 983084 x 983100 1
26 x2 + 5 x ndash 10 983086 4
x2 + 5 x ndash 10 983084 ndash 4 x2 + 5 x ndash 10 983086 4
x2 + 5 x ndash 14 983086 0
( x + 7)( x ndash 2) 983086 0
ndash7 20
x
f (x )
x 983084 ndash7 x 983086 2
x2 + 5 x ndash 6 983084 0
( x + 6)( x ndash 1) 983084 0
ndash6 10
x
f (x )
ndash6 983084 x 983084 1
The ranges are ndash6 lt x lt 1 x lt ndash7 x gt 2
27 f ( x) = (r + 1) x2 + 2rx + r ndash 3
Given that f ( x) does not intersect the x-axis
therefore b2 ndash 4ac 983084 0
(2r)2 ndash 4(r + 1)(r ndash 3) 983084 0
4r2 ndash 4(r2 ndash 2r ndash 3) 983084 0
4r2 ndash 4r2 + 8r + 12 983084 0
2r + 3 983084 0
r 983084 ndash3mdash2
28 Given that f ( x) = 9 ndash 6 x + 2 x2 does not have real root
when f ( x) = k
there4 9 ndash 6 x + 2 x2 = k
2 x2 ndash 6 x + 9 ndash k = 0
Use b2 ndash 4ac 983084 0
(ndash 6)2 ndash 4(2)(9 ndash k ) 983084 0
36 ndash 72 + 8k 983084 0
ndash36 + 8k 983084 0 8k 983084 36
k 983084 36
ndashndashndash8
k 983084 9mdash2
29 2 x2 + 10 x ndash 20 983100 8
ndash8 983100 2 x2 + 10 x ndash 20 983100 8
ndash8983100 2 x2 + 10 x ndash 20 2 x2 + 10 x ndash 20 983100 8
2 x2 + 10 x ndash 28 983100 0
x2 + 5 x ndash 14 983100 0
( x + 7)( x ndash 2) 983100 0
ndash7 20
x
f (x )
ndash7 983100 x 983100 2
0983100 2 x2 + 10 x ndash 12
0983100 x2 + 5 x ndash 6
0983100 ( x + 6)( x ndash 1)
ndash6 10
x
f (x )
x 983100 ndash6 x 983102 1
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x ndash6 x 1
ndash7 x 2
ndash6 ndash7 1 2
The range is ndash7 983100 x 983100 ndash6 or 1 983100 x 983100 2
30 y = x2 + 5 x ndash 6
= x2 + 5 x + 983089 5mdash2 983090
2
ndash 983089 5mdash2 983090
2
ndash 6
= 983089 x + 5mdash2 983090
2
ndash25
ndashndashndash4
ndash 6
= 983089 x + 5mdash2 983090
2
ndash49
ndashndashndash4
The minimum point is (ndash5mdash2
ndash49
ndashndashndash4
)
x ndash13 ndash 6 0 1 3
y 98 0 ndash 6 0 18
x 0
ndash6 ndash13 ndash6
18
98
1 3
) ) ndash ndash ndash ndash
5
2
49
4
y
The range is ndash49
ndashndashndash4
983084 y 983084 98
31 y = ndash x2 + 2 x ndash nx + 16
= ndash x2 + (2 ndash n) x + 16
= ndash[ x2 ndash (2 ndash n) x] + 16
= ndash983091 x2 ndash (2 ndash n) x + 983089 2 ndash n ndashndashndashndashndash
2 983090
2
ndash 983089 2 ndash n ndashndashndashndashndash
2 983090
2
983092 + 16
= ndash
983091983089 x ndash
2 ndash n ndashndashndashndashndash
2 983090
2
ndash
9830892 ndash n
ndashndashndashndashndash2 983090
2
983092 + 16
= ndash983089 x ndash 2 ndash n ndashndashndashndashndash
2 983090
2
+ 983089 2 ndash n ndashndashndashndashndash
2 983090
2
+ 16
Since y = x2 ndash 3k and y = ndash x2 + 2 x ndash nx + 16 have the
same axis of symmetry that is x = 0
then ndash2 ndash n
ndashndashndashndashndash2
= 0
2 ndash n = 0
n = 2
The equation y = ndash x2 + 2 x ndash nx + 16
= ndash x2 + 2 x ndash 2 x + 16
= ndash983089 x ndash 2 ndash 2 ndashndashndashndashndash
2 983090
2
+ 983089 2 ndash 2 ndashndashndashndashndash
2 983090
2
+ 16
= ndash x2 + 16
When y = 0
ndash x2
+ 16 = 0 x2 = 16
x = plusmn98310598310698310616
= plusmn4
Therefore B = (4 0)
Substitute x = 4 y = 0 into y = x2 ndash 3k
0 = 42 ndash 3k
0 = 16 ndash 3k
k =16
ndashndashndash3
32 (a) y = ndash2[(3k ndash x)2 + n] ndash 10 has a maximum point
(4 11)
y = ndash2(3k ndash x)2 ndash 2n ndash 10
there4 3k ndash 4 = 0 and ndash2n ndash 10 = 11
k =4mdash3
ndash2n = 21
n = ndash21
ndashndashndash2
(b) Substitute k =4mdash3
and n = ndash21
ndashndashndash2
into
y = ndash2[(3k ndash x)2 + n] ndash 10
y = ndash2983091(4 ndash x)2 ndash21
ndashndashndash2 983092 ndash 10
= ndash2(4 ndash x)2 + 21 ndash 10
= ndash2(4 ndash x)2
+ 11 When y = 0
ndash2(4 ndash x)2 + 11= 0
2(4 ndash x)2 = 11
(4 ndash x)2 =11
ndashndashndash2
4 ndash x = plusmn98310598310698310698310611 ndashndashndash
2
x = 4 plusmn 98310598310698310698310611 ndashndashndash
2
= 4 ndash 98310598310698310698310611 ndashndashndash
2 4 + 98310598310698310698310611
ndashndashndash2
= 1655 6345
(c) y = ndash2(4 ndash x)2 + 11
The maximum point is (4 11)
x ndash1 0 5
y ndash39 ndash21 9
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1 y = 3(2 x ndash 1)( x + 1) ndash x(4 x ndash 5) + 2
= 3(2 x2 + x ndash 1) ndash 4 x2 + 5 x + 2
= 6 x2 + 3 x ndash 3 ndash 4 x2 + 5 x + 2
= 2 x2 + 8 x ndash 1
= 2( x2 + 4 x) ndash 1
= 2( x2 + 4 x + 22 ndash 22) ndash 1
= 2[( x + 2)2 ndash 4] ndash 1
= 2( x + 2)2 ndash 8 ndash 1
= 2( x + 2)2 ndash 9
Since a = 2 983086 0 therefore the minimum value of
y is ndash9
When y = 0 2( x + 2)2 ndash 9 = 0
( x + 2)2 =9mdash2
x + 2 = plusmn9831059831069831069mdash2 x = plusmn9831059831069831069mdash2 ndash 2
= 9831059831069831069mdash2 ndash 2 or ndash9831059831069831069mdash2 ndash 2
= 01213 or ndash 4121
When x = 0 y = 2(2)2 ndash 9
= ndash1
The minimum point is (ndash2 ndash9)
ndash1 ndash4121
(ndash2 ndash9)
012130
y
x
2 5 983084 Area of rectangle ABCD 983084 21
5 983084 ( x + 3)( x ndash 1) 983084 21
5 983084 ( x + 3)( x ndash 1)
5 983084 x2 + 2 x ndash 3
0 983084 x2 + 2 x ndash 8
0 983084 ( x + 4)( x ndash 2)
0 ndash4 2
f (x )
x
x 983084 ndash 4 x gt 2
( x + 3)( x ndash 1) 983084 21
x2 + 2 x ndash 3 983084 21
x2 + 2 x ndash 24 983084 0
( x ndash 4)( x + 6) 983084 0
0 ndash6 4
f (x )
x
ndash 6 983084 x 983084 4
ndash4
ndash6 lt x lt 4
x lt ndash4 x gt 2
2
x
ndash6 4
The range is ndash 6983084 x 983084 ndash 4 or 2 983084 x 983084 4
3 (a) p =1 + 5
ndashndashndashndashndash2
= 3
(b) y = ( x ndash 1)( x ndash 5) = x2 ndash 6 x + 5
4 y = a( x ndash 2)2 + 1
Substitute x = 0 y = 9 into the equation
9 = a(ndash2)2 + 1
8 = 4a
a = 2
Therefore the quadratic function is
f ( x) = 2( x ndash 2)2 + 1
5 x2
+ (1 + k ) x ndash k 2
+ 1 = 0For quadratic equation to have real roots
b2 ndash 4ac 983102 0
(1 + k )2 ndash 4(1)(1 ndash k 2) 983102 0
1 + 2k + k 2 ndash 4 + 4k 2 983102 0
5k 2 + 2k ndash 3 983102 0
(5k ndash 3)(k + 1) 983102 0
0 ndash1 3 ndash
5
f (k )
k
The range of values of k is k 983100 ndash1 or k 983102 3mdash5
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6 y = x2 + 7 x ndash 8 ndash 2k
For y to be positive for all real values of x there is
no roots for y = 0
Therefore b2 ndash 4ac 983084 0
72 ndash 4(1)(ndash8 ndash 2k ) 983084 0
49 + 32 + 8k 983084 0
8k 983084 ndash81
k 983084 ndash81
ndashndashndash8
Alternative
y = x2 + 7 x ndash 8 ndash 2k
= x2 + 7 x + 983089 7mdash2 983090
2
ndash 983089 7mdash2 983090
2
ndash 8 ndash 2k
= 983089 x +7mdash2 983090
2
ndash49
ndashndashndash4
ndash 8 ndash 2k
For y to be positive for all real values of x
ndash49
ndashndashndash4
ndash 8 ndash 2k 983086 0
ndash2k 983086 49
ndashndashndash4
+ 8
ndash2k 983086 81
ndashndashndash4
k 983084 ndash81
ndashndashndash8
7 Substitute x = 6 y = 0 into y = px2 + qx
0 = p(6)2 + q(6)
0 = 36 p + 6q
q + 6 p = 0 983089
y = px2 + qx
= p983089 x2 +q mdash p x983090
= p983091 x2 +q mdash p x + 983089
q ndashndashndash2 p 983090
2
ndash 983089q
ndashndashndash2 p 983090
2
983092
= p983091983089 x +q
ndashndashndash2 p 983090
2
ndashq2
ndashndashndashndash4 p2 983092
= p983089 x +q
ndashndashndash2 p 983090
2
ndashq2
ndashndashndash4 p
ndash q2
ndashndashndash4 p
= ndash12
q2 = 48 p
p =q2
ndashndashndash48
983090
Substitute 983090 into 983089
q + 6983089 q2
ndashndashndash48
983090 = 0
q +q2
mdash8
= 0
8q + q2 = 0
q(8 + q) = 0
q = 0 or q = ndash8
When q = 0 p =02
ndashndashndash48
= 0
When q = ndash8 p =(ndash8)2
ndashndashndashndashndash48
=64
ndashndashndash48
=4mdash3
Therefore the values of p =
4
mdash3 and q = ndash8
8 (2 ndash 3k ) x2 + x +3mdash4k = 0
b2 ndash 4ac = 12 ndash 4(2 ndash 3k )983089 3mdash4k 983090
= 1 ndash 6k + 9k 2
= 9k 2 ndash 6k + 1
= (3k ndash 1)2
Since (3k ndash 1)2 983102 0 for all values of k
therefore (2 ndash 3k ) x2 + x +3mdash4k = 0 has real roots for
all values of k
9 f ( x) = 3( x2 + 2mx + m2 + n)
= 3[( x + m)2 + n]
= 3( x + m)2 + 3n
The minimum point is (ndashm 3n)
Compare to A(t 3t 2)
there4 ndashm = t and 3n = 3t 2
m = ndasht n = t 2
10 (a) y = px2 + 8 x + 10 ndash p
When the graph does not intercept the x-axis
there are no roots for px2 + 8 x + 10 ndash p = 0 Therefore b2 ndash 4ac 983084 0
82 ndash 4 p(10 ndash p) 983084 0
64 ndash 40 p + 4 p2 983084 0
p2 ndash 10 p + 16 983084 0
( p ndash 2)( p ndash 8) 983084 0
2 8
Hence r = 2 and t = 8
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(b) When p = 2
y = 2 x2 + 8 x + 8
= 2( x2 + 4 x) + 8
= 2( x2 + 4 x + 22 ndash 22) + 8
= 2[( x + 2)2 ndash 4] + 8
= 2( x + 2)2 ndash 8 + 8
= 2( x + 2)2
Therefore the minimum point is (ndash2 0)
When x = 0 y = 8
When y = 0 2( x + 2)2 = 0
x = ndash2
When p = 8
y = 8 x2 + 8 x + 2
= 8( x2 + x) + 2
= 8983091 x2 + x + 983089 1mdash2 983090
2
ndash 983089 1mdash2 983090
2
983092 + 2
= 8983091983089 x +1mdash2 983090
2
ndash1mdash4 983092 + 2
= 8983089 x +1mdash2 983090
2
ndash 2 + 2
= 8983089 x +1mdash2 983090
2
Therefore the minimum point is (ndash1mdash2
0)
When x = 0 y = 2
When y = 0 0 = 8983089 x + 1mdash2 983090
2
x = ndash1mdash2
0
2
8
1 ndash ndash
2
ndash2
y p = 2 p = 8
x
11 (a) f ( x) = 24 x ndash 4 x2 + r
= ndash 4 x2 + 24 x + r
= ndash 4( x2 ndash 6 x) + r
= ndash 4( x2 ndash 6 x + 32 ndash 32) + r
= ndash 4[( x ndash 3)2 ndash 9] + r
= ndash 4( x ndash 3)2 + 36 + r
Compare to f ( x) = p( x ndash q)2 + 16
Therefore p = ndash 4 q = 3 and 36 + r = 16
r = ndash20
(b) The turning point is (3 16)
(c) f ( x) = 24 x ndash 4 x2 ndash 20
When x = 0 f ( x) = ndash20
When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0
4( x ndash 3)2 = 16
( x ndash 3)2 = 4
x ndash 3 = plusmn2
x = plusmn2 + 3
= ndash2 + 3 or 2 + 3
= 1 or 5
0 1
(3 16)
5
ndash20
y
x
12 (a) y = ndash| p( x ndash 3)2 + q|
Substitute x = 3 y = ndash5 into the equation
ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5
Substitute x = 4 y = 0 into the equation
0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0
p = 5
Therefore p = 5 q = ndash5 or p = ndash5 q = 5
(b) When x = 3 y = ndash5
For p = 5 q = ndash5
When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40
Based on the graph the range of values of y is
ndash 40 983100 y 983100 0
For p = ndash5 q = 5
When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40
Therefore the range of values of y is
ndash 40 983100 y 983100 0
13 (a) y = ndash2( x ndash 3)2 + 2k
= ndash x2 + 2 x + px ndash 8
= ndash x2 + (2 + p) x ndash 8
= ndash[ x2 ndash (2 + p) x] ndash 8
= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash
2 9830902
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983091983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
+(2 + p)2
ndashndashndashndashndashndashndash4
ndash 8
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Since the x-coordinate of the maximum point for
both the graphs are same
therefore2 + p
ndashndashndashndashndash2
= 3
p = 4
y = ndash x2 + 2 x + px ndash 8 becomes
y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8
When y = 0
ndash x2 + 6 x ndash 8 = 0
x2 ndash 6 x + 8 = 0
( x ndash 2)( x ndash 4) = 0
x = 2 or 4
Hence A(2 0) and B(4 0)
Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k
0 = ndash2(2 ndash 3)2 + 2k
2k = 2
k = 1 Hence k = 1 and p = 4
(b) For y = ndash2( x ndash 3)2 + 2k
= ndash2( x ndash 3)2 + 2(1)
= ndash2( x ndash 3)2 + 2
Maximum value of the curve is 2
For y = ndash x2 + 2 x + px ndash 8
= ndash x2 + 2 x + 4 x ndash 8
= ndash x2 + 6 x ndash 8
When x = 3
y = ndash9 + 18 ndash 8
= 1
Maximum value of the curve is 1
14 Since 3 x2 983102 0 for all values of x
therefore3 x2
ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)
983100 0
(2 x ndash 1)( x + 4) 983100 0
0 ndash4 1 ndash
2
f (x )
x
Hence ndash 4983100 x 983100 1mdash2
15 Since x2 + 1 983086 0
therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash
x2 + 1 983086 0
x2 + 3 x + 2 983086 0
( x + 1)( x + 2) 983086 0
0 ndash2 ndash1
f (x )
x
Hence x 983084 ndash2 x gt ndash1
16 ndash4
ndashndashndashndashndashndash1 ndash 3 x
983100 x
0 983100 x +4
ndashndashndashndashndashndash1 ndash 3 x
0 983100 x(1 ndash 3 x) + 4
ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0983100
x ndash 3 x2 + 4
ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash
1 ndash 3 x
0 983100 (ndash3 x + 4)( x + 1)
ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
For ndash3 x + 4 983102 0
4 983102 3 x
x 983100 4mdash3
For x + 1 983102 0
x 983102 ndash1
For 1 ndash 3 x 983086 0
ndash3 x 983086 ndash1
x 983084 1mdash3
ndash1 ndash + ndash +
1 4 ndash ndash
3
4x ndash
3
1x lt ndash
3
x ndash1
3
x
Therefore the range is ndash1 983100 x 983084 1mdash3
x 983102 4mdash3
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17 y2 ndash 9 = x
x = y2 ndash 9
When x = 0
y2 = 9
y = plusmn3
When y = 0
x = ndash9
When x = 7
7 = y2 ndash 9
y2 = 16
y = plusmn4
x ndash9 0 7
y 0 plusmn3 plusmn4
0 ndash9 7
ndash4 ndash3
34
y
x
The range of values of y is ndash 4 983100 y 983100 4
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910 x
f (x )
The range of values of x is 1 983100 x 983100 9
(h) x (1 ndash 4 x) 983084 5 x ndash 8
x ndash 4 x2 ndash 5 x + 8 983084 0
ndash 4 x2 ndash 4 x + 8 983084 0
x2 + x ndash 2 983086 0
( x + 2)( x ndash 1) 983086 0
1 ndash2 0 x
f (x )
The range of values of x is x 983084 ndash2 or x 983086 1
1 (a) x-coordinate of the maximum part =1 + 7
ndashndashndashndashndash2
= 4
Therefore the equation of the axis of symmetry
is x = 4
(b) f ( x) = p ndash ( x + q)2
= 5 ndash ( x ndash 4)2
2 f ( x) = 2 x2 ndash 16 x + k 2 + 2k + 1
= 2( x2 ndash 8 x) + k 2 + 2k + 1
= 2983091 x2 ndash 8 x + 983089 8mdash2 983090
2
ndash 983089 8mdash2 983090
2
983092 + k 2 + 2k + 1
= 2[( x ndash 4)2 ndash 16] + k 2 + 2k + 1
= 2( x ndash 4)2 ndash 32 + k 2 + 2k + 1
= 2( x ndash 4)2 + k 2 + 2k ndash 31
Given minimum value = ndash28
there4 k 2 + 2k ndash 31 = ndash28
k 2 + 2k ndash 3 = 0
(k + 3)(k ndash 1) = 0 k = ndash3 1
3 (a) f ( x) = 2( x ndash 3)2 + k
p is the x-coordinate of the minimum point
Therefore p = 3
(b) k is the minimum value of f ( x)
Therefore k = ndash4
(c) The equation of the axis of symmetry is x = 3
4 f ( x) = 3 x2 ndash 2 x + p
a = 3 b = ndash2 c = p
Since the graph does not intersect the x-axis
b2 ndash 4ac lt 0
(ndash2)2 ndash 4(3)( p) lt 0
4 ndash 12 p lt 0
4 lt 12 p
13
lt p
p gt13
5 f ( x) = 2 x2 ndash 12 x + 5
= 2( x2 ndash 6 x) + 5
= 2[( x ndash 3)2 ndash 32] + 5
= 2( x ndash 3)2 ndash 18 + 5
= 2( x ndash 3)2 ndash 13
there4 p = 2 q = ndash3 ndashr + 1 = ndash13
r = 14
6 (a) f ( x) = ndash x2 + 6 px + 1 ndash 4 p2
= ndash( x2 ndash 6 px) + 1 ndash 4 p2
= ndash983091 x2 ndash 6 px + 983089 6 p ndashndashndash
2 983090
2
ndash 983089 6 p ndashndashndash
2 983090
2
983092 + 1 ndash 4 p2
= ndash[( x ndash 3 p)2 ndash 9 p2] + 1 ndash 4 p2
= ndash ( x ndash 3 p)2 + 9 p2 + 1 ndash 4 p2
= ndash ( x ndash 3 p)2 + 1 + 5 p2
The maximum value given is q2 ndash p
Therefore q2 ndash p = 1 + 5 p2
5 p2 + p + 1 = q2
(b) x = 3 is symmetrical axis 3 p = 3
p = 1
Substitute p = 1 into 5 p2 + p + 1 = q2
5(1)2 + 1 + 1 = q2
q2 = 7
q = plusmn9831059831067 Hence p = 1 q = plusmn9831059831067
7 4t (t + 1) ndash 3t 2 + 12 983086 0
4t 2 + 4t ndash 3t 2 + 12 983086 0
t 2 + 4t + 12 983086 0
(t + 2)(t + 6) 983086 0
0 ndash2 ndash6x
f (x )
The range of values of t is t 983084 ndash6 or t 983086 ndash2
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1 x-coordinate of maximum point = ndash 4 + 0 ndashndashndashndashndashndash
2 = ndash2
Equation of the axis of symmetry is x = ndash2
2 Let x be the x-coordinate of A
0 + x
ndashndashndashndashndash2
= 3
x = 6
The coordinates of A are (6 4)
3 Let x be the x-coordinate of A
x + 6 ndashndashndashndashndash
2 = 2
x = 4 ndash 6
x = ndash2
The coordinates of A are (ndash2 0)
4 x-coordinate of A =0 + 8
ndashndashndashndashndash2
= 4
Let C be the centre of OB
4
5
A
C O
AC 2
= OA2
ndash OC 2
= 52 ndash 42
= 9
AC = 3
The coordinates of A are (4 3)
5 x-coordinate of minimum point =0 + 4
ndashndashndashndashndash2
= 2
x-coordinate of minimum point for the image is ndash2
6 (a) y = ( x ndash p)2 + q and minimum point is (2 ndash1)
Hence p = 2 and q = ndash1
(b) y = ( x ndash 2)2 ndash 1
When y = 0
( x ndash 2)2 ndash 1 = 0
( x ndash 2)2 = 1
x ndash 2 = plusmn1
x = plusmn1 + 2
= 1 3
Hence A is (1 0) and B is (3 0)
7 f ( x) = 2k + 1 ndash 983089 x + 1mdash2 p983090
2
Given (ndash1 k ) is the maximum point
Therefore 2k + 1 = k
k = ndash1
x +1mdash2 p = 0 when x = ndash1
ndash1 + 1mdash2 p = 0
1mdash2 p = 1
p = 2
8 Given ( p 2q) is the minimum point of
y = 2 x2 ndash 4 x + 5
= 2( x2 ndash 2 x) + 5
= 2( x2 ndash 2 x + 12 ndash 12) + 5
= 2[( x ndash 1)2 ndash 1] + 5
= 2( x ndash 1)2 ndash 2 + 5
= 2( x ndash 1)2 + 3
2q = 3
q =3mdash2
p ndash 1 = 0
p = 1
9 (a) Since (1 4) is the point on y = x2 ndash 2kx + 1
substitute x = 1 y = 4 into the equation
4 = 12 ndash 2k (1) + 1
2k = ndash2
k = ndash1
(b) y = x2 ndash 2(ndash1) x + 1
= x2 + 2 x + 1
= ( x + 1)2
Minimum value of y is 0
10 f ( x) = ndash x2 ndash 8 x + k ndash 1
= ndash( x2 + 8 x) + k ndash 1
= ndash( x2 + 8 x + 42 ndash 42) + k ndash 1
= ndash[( x + 4)2 ndash 16] + k ndash 1
= ndash( x + 4)2 + 16 + k ndash 1
= ndash( x + 4)2 + 15 + k
Since 13 is the maximum value
then 15 + k = 13
k = ndash2
11 f ( x) = 2 x2 ndash 6 x + 7
= 2( x2 ndash 3 x) + 7
= 2983091 x2 ndash 3 x + 983089 3mdash2 983090
2
ndash 983089 3mdash2 983090
2
983092 + 7
= 2983091983089 x ndash3mdash2 983090
2
ndash9mdash4 983092 + 7
= 2983089 x ndash3mdash2 983090
2
ndash9mdash2
+ 7
= 2983089 x ndash3mdash2 983090
2
+5mdash2
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The minimum point is (3mdash2
5mdash2
)
x ndash1 0 3
f ( x) 15 7 7
ndash ndash
3
5322
0
7
15
x
f (x )
ndash1
The range is5mdash2
983100 f ( x) 983100 15
12 f ( x) = 5 ndash 4 x ndash 2 x2
= ndash2 x2 ndash 4 x + 5
= ndash2( x2 + 2 x) + 5
= ndash2( x2 + 2 x + 12 ndash 12) + 5
= ndash2[( x + 1)2 ndash 1] + 5
= ndash2( x + 1)2 + 2 + 5
= ndash2( x + 1)2 + 7
When f ( x) = ndash1
ndash2( x + 1)2 + 7 = ndash1
ndash2( x + 1)2 = ndash8
( x + 1)2 = 4
x + 1 = plusmn2
x = plusmn2 ndash 1
= ndash3 or 1
13 y = ( x ndash 3)2 ndash 4
Minimum point is (3 ndash4)
x ndash1 0 1 5 6
y 12 5 0 0 5
x 0 1 5 6
5
12
ndash1
(3 ndash4)
y
The range is ndash 4983100 y 983100 12
14 y = ndash x2 + 4 x ndash 5
= ndash( x2 ndash 4 x) ndash 5
= ndash( x2 ndash 4 x + 22 ndash 22) ndash 5
= ndash [( x ndash 2)2 ndash 4] ndash 5
= ndash( x ndash 2)2 + 4 ndash 5
= ndash( x ndash 2)2 ndash 1
Maximum point is (2 ndash1)
x ndash1 0 3
y ndash10 ndash5 ndash2
x
0 3(2 ndash1)
ndash1
ndash5
ndash10
y
The range is ndash10 983100 y 983100 ndash1
15 y = 9 ndash ( x ndash 3)2Maximum point is (3 9)
x ndash1 0 6 7
y 7 0 0 7
x
y
0 ndash1 6
(3 9)
7
7
The range is 0 983100 y 983100 9
16 3 x2 983084 x
3 x2
ndash x 983084 0 x(3 x ndash 1)983084 0
ndash
13
0 x
f (x )
The range is 0 983084 x 983084 1mdash3
17 3 x ndash x
2
ndashndashndashndashndashndash2 983084 1
3 x ndash x2 983084 2
ndash x2 + 3 x ndash 2 983084 0
x2 ndash 3 x + 2 983086 0
( x ndash 1)( x ndash 2) 983086 0
20 x
f (x )
1
The range is x 983084
1 or x 983086
2
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18 Given that x ndash 2 y = 1
there4 x = 1 + 2 y 983089
Substitute 983089 into y + 3983102 2 xy
y + 3983102 2(1 + 2 y) y
y + 3983102 2 y + 4 y2
0983102 4 y2 + y ndash 3
0983102 (4 y ndash 3)( y + 1)
3 ndash
4
0 y
f (y )
ndash1
The range is ndash1 983100 y 983100 3mdash4
19 f ( x) 983084 0
5 x2 ndash 4 x ndash 1 983084 0
(5 x + 1)( x ndash 1) 983084 0
1 ndash ndash
5
0 x
f (x )
1
The range is ndash1mdash5
983084 x 983084 1
20 g( x) 983086 0
4 x2 ndash 9 983086 0(2 x + 3)(2 x ndash 3) 983086 0
3 ndash ndash
23 ndash
2
0 x
g (x )
The range is x 983084 ndash3mdash2
or x 983086 3mdash2
21 (a) Since y = 3 x
2
ndash 9 x + t 983086
0 for all values of x andit does not have root when y = 0
Then b2 ndash 4ac 983084 0 for 3 x2 ndash 9 x + t = 0
(ndash9)2 ndash 4(3)(t ) 983084 0
81 ndash 12t 983084 0
ndash12t 983084 ndash81
t 983086 ndash81 ndashndashndashndash ndash12
t 983086 27
ndashndashndash4
(b) Let f ( x) = a( x ndash b)2 + c
f ( x) = a( x ndash 2)2 + 0
f ( x) = a( x ndash 2)2
Substitute x = 0 f ( x) = ndash3 into the equation
ndash3 = a(0 ndash 2)2
= 4a
a = ndash3mdash4
Hence the quadratic function is
f ( x) = ndash3mdash4
( x ndash 2)2
22 (a) Given 2 x2 ndash 3 y + 2 = 0
3 y = 2 x2 + 2
y =2 x2
ndashndashndash3
+2mdash3
983089
Substitute 983089 into y 983084 10
2 x2
ndashndashndash
3
+2mdash
3
983084 10
2 x2 + 2 983084 30
2 x2 ndash 28 983084 0
x2 ndash 14 983084 0
( x + 98310598310698310614 )( x ndash 98310598310698310614 ) 983084 0
14 ndash14
0 x
f (x )
The range is ndash98310598310698310614 983084 x 983084 98310598310698310614
(b) 2 x2 ndash 8 x ndash 10 = 2( x2 ndash 4 x) ndash 10
= 2( x2 ndash 4 x + 22 ndash 22) ndash 10
= 2[( x ndash 2)2 ndash 4] ndash 10
= 2( x ndash 2)2 ndash 8 ndash 10
= 2( x ndash 2)2 ndash 18
Therefore a = 2 b = ndash2 and c = ndash18
Hence the minimum value of 2 x2 ndash 8 x ndash 10 is
ndash18
23 5 ndash 2 x 983100 0 3 x2 ndash 4 x 983086 ndash1
3 x2
ndash 4 x + 1983086
0(3 x ndash 1)( x ndash 1) 983086 0
11 ndash
3
0 x
f (x )
x 983084 1mdash3
x 983086 1
ndash2 x 983100
ndash5 x 983102
ndash5 ndashndashndash ndash2
x 983102 5mdash2
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5ndash
2
1ndash
3
1
x 5mdash2
x x 11mdash3
The range is x 983102 5mdash2
24 5 983084 f ( x) 983084 9
5 983084 5 ndash 3 x + x2 983084 9
5 983084 5 ndash 3 x + x2 5 ndash 3 x + x2 983084 9
5 ndash 3 x + x2 ndash 9 983084 0
x2 ndash 3 x ndash 4 983084 0
( x ndash 4)( x + 1) 983084 0
4 ndash1 0 x
f (x )
ndash1 983084 x 983084 4
0 983084 x2 ndash 3 x
0 983084 x( x ndash 3)
30 x
f (x )
x 983084 0 x 983086 3
0 ndash1 43
x lt 0 x gt 3
ndash1 lt x lt 4
The range is ndash1983084
x 983084
0 or 3983084
x 983084
4
25 1 983102 x2 + 3 x ndash 3 983086 ndash3
x2 + 3 x ndash 3 983086 ndash3
x2 + 3 x 983086 0
x( x + 3) 983086 0
ndash3 0 x
f (x )
x 983084 ndash3 x 983086 0
1 983102 x2 + 3 x ndash 3
0 983102 x2 + 3 x ndash 4
0 983102 ( x + 4)( x ndash 1)
1 ndash4 0 x
f (x )
ndash 4 983100 x 983100 1
ndash4
x lt ndash3 x gt 0
ndash4 x 1
ndash3 0 1
The range is ndash 4983100 x 983084 ndash3 or 0 983084 x 983100 1
26 x2 + 5 x ndash 10 983086 4
x2 + 5 x ndash 10 983084 ndash 4 x2 + 5 x ndash 10 983086 4
x2 + 5 x ndash 14 983086 0
( x + 7)( x ndash 2) 983086 0
ndash7 20
x
f (x )
x 983084 ndash7 x 983086 2
x2 + 5 x ndash 6 983084 0
( x + 6)( x ndash 1) 983084 0
ndash6 10
x
f (x )
ndash6 983084 x 983084 1
The ranges are ndash6 lt x lt 1 x lt ndash7 x gt 2
27 f ( x) = (r + 1) x2 + 2rx + r ndash 3
Given that f ( x) does not intersect the x-axis
therefore b2 ndash 4ac 983084 0
(2r)2 ndash 4(r + 1)(r ndash 3) 983084 0
4r2 ndash 4(r2 ndash 2r ndash 3) 983084 0
4r2 ndash 4r2 + 8r + 12 983084 0
2r + 3 983084 0
r 983084 ndash3mdash2
28 Given that f ( x) = 9 ndash 6 x + 2 x2 does not have real root
when f ( x) = k
there4 9 ndash 6 x + 2 x2 = k
2 x2 ndash 6 x + 9 ndash k = 0
Use b2 ndash 4ac 983084 0
(ndash 6)2 ndash 4(2)(9 ndash k ) 983084 0
36 ndash 72 + 8k 983084 0
ndash36 + 8k 983084 0 8k 983084 36
k 983084 36
ndashndashndash8
k 983084 9mdash2
29 2 x2 + 10 x ndash 20 983100 8
ndash8 983100 2 x2 + 10 x ndash 20 983100 8
ndash8983100 2 x2 + 10 x ndash 20 2 x2 + 10 x ndash 20 983100 8
2 x2 + 10 x ndash 28 983100 0
x2 + 5 x ndash 14 983100 0
( x + 7)( x ndash 2) 983100 0
ndash7 20
x
f (x )
ndash7 983100 x 983100 2
0983100 2 x2 + 10 x ndash 12
0983100 x2 + 5 x ndash 6
0983100 ( x + 6)( x ndash 1)
ndash6 10
x
f (x )
x 983100 ndash6 x 983102 1
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x ndash6 x 1
ndash7 x 2
ndash6 ndash7 1 2
The range is ndash7 983100 x 983100 ndash6 or 1 983100 x 983100 2
30 y = x2 + 5 x ndash 6
= x2 + 5 x + 983089 5mdash2 983090
2
ndash 983089 5mdash2 983090
2
ndash 6
= 983089 x + 5mdash2 983090
2
ndash25
ndashndashndash4
ndash 6
= 983089 x + 5mdash2 983090
2
ndash49
ndashndashndash4
The minimum point is (ndash5mdash2
ndash49
ndashndashndash4
)
x ndash13 ndash 6 0 1 3
y 98 0 ndash 6 0 18
x 0
ndash6 ndash13 ndash6
18
98
1 3
) ) ndash ndash ndash ndash
5
2
49
4
y
The range is ndash49
ndashndashndash4
983084 y 983084 98
31 y = ndash x2 + 2 x ndash nx + 16
= ndash x2 + (2 ndash n) x + 16
= ndash[ x2 ndash (2 ndash n) x] + 16
= ndash983091 x2 ndash (2 ndash n) x + 983089 2 ndash n ndashndashndashndashndash
2 983090
2
ndash 983089 2 ndash n ndashndashndashndashndash
2 983090
2
983092 + 16
= ndash
983091983089 x ndash
2 ndash n ndashndashndashndashndash
2 983090
2
ndash
9830892 ndash n
ndashndashndashndashndash2 983090
2
983092 + 16
= ndash983089 x ndash 2 ndash n ndashndashndashndashndash
2 983090
2
+ 983089 2 ndash n ndashndashndashndashndash
2 983090
2
+ 16
Since y = x2 ndash 3k and y = ndash x2 + 2 x ndash nx + 16 have the
same axis of symmetry that is x = 0
then ndash2 ndash n
ndashndashndashndashndash2
= 0
2 ndash n = 0
n = 2
The equation y = ndash x2 + 2 x ndash nx + 16
= ndash x2 + 2 x ndash 2 x + 16
= ndash983089 x ndash 2 ndash 2 ndashndashndashndashndash
2 983090
2
+ 983089 2 ndash 2 ndashndashndashndashndash
2 983090
2
+ 16
= ndash x2 + 16
When y = 0
ndash x2
+ 16 = 0 x2 = 16
x = plusmn98310598310698310616
= plusmn4
Therefore B = (4 0)
Substitute x = 4 y = 0 into y = x2 ndash 3k
0 = 42 ndash 3k
0 = 16 ndash 3k
k =16
ndashndashndash3
32 (a) y = ndash2[(3k ndash x)2 + n] ndash 10 has a maximum point
(4 11)
y = ndash2(3k ndash x)2 ndash 2n ndash 10
there4 3k ndash 4 = 0 and ndash2n ndash 10 = 11
k =4mdash3
ndash2n = 21
n = ndash21
ndashndashndash2
(b) Substitute k =4mdash3
and n = ndash21
ndashndashndash2
into
y = ndash2[(3k ndash x)2 + n] ndash 10
y = ndash2983091(4 ndash x)2 ndash21
ndashndashndash2 983092 ndash 10
= ndash2(4 ndash x)2 + 21 ndash 10
= ndash2(4 ndash x)2
+ 11 When y = 0
ndash2(4 ndash x)2 + 11= 0
2(4 ndash x)2 = 11
(4 ndash x)2 =11
ndashndashndash2
4 ndash x = plusmn98310598310698310698310611 ndashndashndash
2
x = 4 plusmn 98310598310698310698310611 ndashndashndash
2
= 4 ndash 98310598310698310698310611 ndashndashndash
2 4 + 98310598310698310698310611
ndashndashndash2
= 1655 6345
(c) y = ndash2(4 ndash x)2 + 11
The maximum point is (4 11)
x ndash1 0 5
y ndash39 ndash21 9
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1 y = 3(2 x ndash 1)( x + 1) ndash x(4 x ndash 5) + 2
= 3(2 x2 + x ndash 1) ndash 4 x2 + 5 x + 2
= 6 x2 + 3 x ndash 3 ndash 4 x2 + 5 x + 2
= 2 x2 + 8 x ndash 1
= 2( x2 + 4 x) ndash 1
= 2( x2 + 4 x + 22 ndash 22) ndash 1
= 2[( x + 2)2 ndash 4] ndash 1
= 2( x + 2)2 ndash 8 ndash 1
= 2( x + 2)2 ndash 9
Since a = 2 983086 0 therefore the minimum value of
y is ndash9
When y = 0 2( x + 2)2 ndash 9 = 0
( x + 2)2 =9mdash2
x + 2 = plusmn9831059831069831069mdash2 x = plusmn9831059831069831069mdash2 ndash 2
= 9831059831069831069mdash2 ndash 2 or ndash9831059831069831069mdash2 ndash 2
= 01213 or ndash 4121
When x = 0 y = 2(2)2 ndash 9
= ndash1
The minimum point is (ndash2 ndash9)
ndash1 ndash4121
(ndash2 ndash9)
012130
y
x
2 5 983084 Area of rectangle ABCD 983084 21
5 983084 ( x + 3)( x ndash 1) 983084 21
5 983084 ( x + 3)( x ndash 1)
5 983084 x2 + 2 x ndash 3
0 983084 x2 + 2 x ndash 8
0 983084 ( x + 4)( x ndash 2)
0 ndash4 2
f (x )
x
x 983084 ndash 4 x gt 2
( x + 3)( x ndash 1) 983084 21
x2 + 2 x ndash 3 983084 21
x2 + 2 x ndash 24 983084 0
( x ndash 4)( x + 6) 983084 0
0 ndash6 4
f (x )
x
ndash 6 983084 x 983084 4
ndash4
ndash6 lt x lt 4
x lt ndash4 x gt 2
2
x
ndash6 4
The range is ndash 6983084 x 983084 ndash 4 or 2 983084 x 983084 4
3 (a) p =1 + 5
ndashndashndashndashndash2
= 3
(b) y = ( x ndash 1)( x ndash 5) = x2 ndash 6 x + 5
4 y = a( x ndash 2)2 + 1
Substitute x = 0 y = 9 into the equation
9 = a(ndash2)2 + 1
8 = 4a
a = 2
Therefore the quadratic function is
f ( x) = 2( x ndash 2)2 + 1
5 x2
+ (1 + k ) x ndash k 2
+ 1 = 0For quadratic equation to have real roots
b2 ndash 4ac 983102 0
(1 + k )2 ndash 4(1)(1 ndash k 2) 983102 0
1 + 2k + k 2 ndash 4 + 4k 2 983102 0
5k 2 + 2k ndash 3 983102 0
(5k ndash 3)(k + 1) 983102 0
0 ndash1 3 ndash
5
f (k )
k
The range of values of k is k 983100 ndash1 or k 983102 3mdash5
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6 y = x2 + 7 x ndash 8 ndash 2k
For y to be positive for all real values of x there is
no roots for y = 0
Therefore b2 ndash 4ac 983084 0
72 ndash 4(1)(ndash8 ndash 2k ) 983084 0
49 + 32 + 8k 983084 0
8k 983084 ndash81
k 983084 ndash81
ndashndashndash8
Alternative
y = x2 + 7 x ndash 8 ndash 2k
= x2 + 7 x + 983089 7mdash2 983090
2
ndash 983089 7mdash2 983090
2
ndash 8 ndash 2k
= 983089 x +7mdash2 983090
2
ndash49
ndashndashndash4
ndash 8 ndash 2k
For y to be positive for all real values of x
ndash49
ndashndashndash4
ndash 8 ndash 2k 983086 0
ndash2k 983086 49
ndashndashndash4
+ 8
ndash2k 983086 81
ndashndashndash4
k 983084 ndash81
ndashndashndash8
7 Substitute x = 6 y = 0 into y = px2 + qx
0 = p(6)2 + q(6)
0 = 36 p + 6q
q + 6 p = 0 983089
y = px2 + qx
= p983089 x2 +q mdash p x983090
= p983091 x2 +q mdash p x + 983089
q ndashndashndash2 p 983090
2
ndash 983089q
ndashndashndash2 p 983090
2
983092
= p983091983089 x +q
ndashndashndash2 p 983090
2
ndashq2
ndashndashndashndash4 p2 983092
= p983089 x +q
ndashndashndash2 p 983090
2
ndashq2
ndashndashndash4 p
ndash q2
ndashndashndash4 p
= ndash12
q2 = 48 p
p =q2
ndashndashndash48
983090
Substitute 983090 into 983089
q + 6983089 q2
ndashndashndash48
983090 = 0
q +q2
mdash8
= 0
8q + q2 = 0
q(8 + q) = 0
q = 0 or q = ndash8
When q = 0 p =02
ndashndashndash48
= 0
When q = ndash8 p =(ndash8)2
ndashndashndashndashndash48
=64
ndashndashndash48
=4mdash3
Therefore the values of p =
4
mdash3 and q = ndash8
8 (2 ndash 3k ) x2 + x +3mdash4k = 0
b2 ndash 4ac = 12 ndash 4(2 ndash 3k )983089 3mdash4k 983090
= 1 ndash 6k + 9k 2
= 9k 2 ndash 6k + 1
= (3k ndash 1)2
Since (3k ndash 1)2 983102 0 for all values of k
therefore (2 ndash 3k ) x2 + x +3mdash4k = 0 has real roots for
all values of k
9 f ( x) = 3( x2 + 2mx + m2 + n)
= 3[( x + m)2 + n]
= 3( x + m)2 + 3n
The minimum point is (ndashm 3n)
Compare to A(t 3t 2)
there4 ndashm = t and 3n = 3t 2
m = ndasht n = t 2
10 (a) y = px2 + 8 x + 10 ndash p
When the graph does not intercept the x-axis
there are no roots for px2 + 8 x + 10 ndash p = 0 Therefore b2 ndash 4ac 983084 0
82 ndash 4 p(10 ndash p) 983084 0
64 ndash 40 p + 4 p2 983084 0
p2 ndash 10 p + 16 983084 0
( p ndash 2)( p ndash 8) 983084 0
2 8
Hence r = 2 and t = 8
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(b) When p = 2
y = 2 x2 + 8 x + 8
= 2( x2 + 4 x) + 8
= 2( x2 + 4 x + 22 ndash 22) + 8
= 2[( x + 2)2 ndash 4] + 8
= 2( x + 2)2 ndash 8 + 8
= 2( x + 2)2
Therefore the minimum point is (ndash2 0)
When x = 0 y = 8
When y = 0 2( x + 2)2 = 0
x = ndash2
When p = 8
y = 8 x2 + 8 x + 2
= 8( x2 + x) + 2
= 8983091 x2 + x + 983089 1mdash2 983090
2
ndash 983089 1mdash2 983090
2
983092 + 2
= 8983091983089 x +1mdash2 983090
2
ndash1mdash4 983092 + 2
= 8983089 x +1mdash2 983090
2
ndash 2 + 2
= 8983089 x +1mdash2 983090
2
Therefore the minimum point is (ndash1mdash2
0)
When x = 0 y = 2
When y = 0 0 = 8983089 x + 1mdash2 983090
2
x = ndash1mdash2
0
2
8
1 ndash ndash
2
ndash2
y p = 2 p = 8
x
11 (a) f ( x) = 24 x ndash 4 x2 + r
= ndash 4 x2 + 24 x + r
= ndash 4( x2 ndash 6 x) + r
= ndash 4( x2 ndash 6 x + 32 ndash 32) + r
= ndash 4[( x ndash 3)2 ndash 9] + r
= ndash 4( x ndash 3)2 + 36 + r
Compare to f ( x) = p( x ndash q)2 + 16
Therefore p = ndash 4 q = 3 and 36 + r = 16
r = ndash20
(b) The turning point is (3 16)
(c) f ( x) = 24 x ndash 4 x2 ndash 20
When x = 0 f ( x) = ndash20
When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0
4( x ndash 3)2 = 16
( x ndash 3)2 = 4
x ndash 3 = plusmn2
x = plusmn2 + 3
= ndash2 + 3 or 2 + 3
= 1 or 5
0 1
(3 16)
5
ndash20
y
x
12 (a) y = ndash| p( x ndash 3)2 + q|
Substitute x = 3 y = ndash5 into the equation
ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5
Substitute x = 4 y = 0 into the equation
0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0
p = 5
Therefore p = 5 q = ndash5 or p = ndash5 q = 5
(b) When x = 3 y = ndash5
For p = 5 q = ndash5
When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40
Based on the graph the range of values of y is
ndash 40 983100 y 983100 0
For p = ndash5 q = 5
When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40
Therefore the range of values of y is
ndash 40 983100 y 983100 0
13 (a) y = ndash2( x ndash 3)2 + 2k
= ndash x2 + 2 x + px ndash 8
= ndash x2 + (2 + p) x ndash 8
= ndash[ x2 ndash (2 + p) x] ndash 8
= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash
2 9830902
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983091983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
+(2 + p)2
ndashndashndashndashndashndashndash4
ndash 8
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Since the x-coordinate of the maximum point for
both the graphs are same
therefore2 + p
ndashndashndashndashndash2
= 3
p = 4
y = ndash x2 + 2 x + px ndash 8 becomes
y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8
When y = 0
ndash x2 + 6 x ndash 8 = 0
x2 ndash 6 x + 8 = 0
( x ndash 2)( x ndash 4) = 0
x = 2 or 4
Hence A(2 0) and B(4 0)
Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k
0 = ndash2(2 ndash 3)2 + 2k
2k = 2
k = 1 Hence k = 1 and p = 4
(b) For y = ndash2( x ndash 3)2 + 2k
= ndash2( x ndash 3)2 + 2(1)
= ndash2( x ndash 3)2 + 2
Maximum value of the curve is 2
For y = ndash x2 + 2 x + px ndash 8
= ndash x2 + 2 x + 4 x ndash 8
= ndash x2 + 6 x ndash 8
When x = 3
y = ndash9 + 18 ndash 8
= 1
Maximum value of the curve is 1
14 Since 3 x2 983102 0 for all values of x
therefore3 x2
ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)
983100 0
(2 x ndash 1)( x + 4) 983100 0
0 ndash4 1 ndash
2
f (x )
x
Hence ndash 4983100 x 983100 1mdash2
15 Since x2 + 1 983086 0
therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash
x2 + 1 983086 0
x2 + 3 x + 2 983086 0
( x + 1)( x + 2) 983086 0
0 ndash2 ndash1
f (x )
x
Hence x 983084 ndash2 x gt ndash1
16 ndash4
ndashndashndashndashndashndash1 ndash 3 x
983100 x
0 983100 x +4
ndashndashndashndashndashndash1 ndash 3 x
0 983100 x(1 ndash 3 x) + 4
ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0983100
x ndash 3 x2 + 4
ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash
1 ndash 3 x
0 983100 (ndash3 x + 4)( x + 1)
ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
For ndash3 x + 4 983102 0
4 983102 3 x
x 983100 4mdash3
For x + 1 983102 0
x 983102 ndash1
For 1 ndash 3 x 983086 0
ndash3 x 983086 ndash1
x 983084 1mdash3
ndash1 ndash + ndash +
1 4 ndash ndash
3
4x ndash
3
1x lt ndash
3
x ndash1
3
x
Therefore the range is ndash1 983100 x 983084 1mdash3
x 983102 4mdash3
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17 y2 ndash 9 = x
x = y2 ndash 9
When x = 0
y2 = 9
y = plusmn3
When y = 0
x = ndash9
When x = 7
7 = y2 ndash 9
y2 = 16
y = plusmn4
x ndash9 0 7
y 0 plusmn3 plusmn4
0 ndash9 7
ndash4 ndash3
34
y
x
The range of values of y is ndash 4 983100 y 983100 4
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1 x-coordinate of maximum point = ndash 4 + 0 ndashndashndashndashndashndash
2 = ndash2
Equation of the axis of symmetry is x = ndash2
2 Let x be the x-coordinate of A
0 + x
ndashndashndashndashndash2
= 3
x = 6
The coordinates of A are (6 4)
3 Let x be the x-coordinate of A
x + 6 ndashndashndashndashndash
2 = 2
x = 4 ndash 6
x = ndash2
The coordinates of A are (ndash2 0)
4 x-coordinate of A =0 + 8
ndashndashndashndashndash2
= 4
Let C be the centre of OB
4
5
A
C O
AC 2
= OA2
ndash OC 2
= 52 ndash 42
= 9
AC = 3
The coordinates of A are (4 3)
5 x-coordinate of minimum point =0 + 4
ndashndashndashndashndash2
= 2
x-coordinate of minimum point for the image is ndash2
6 (a) y = ( x ndash p)2 + q and minimum point is (2 ndash1)
Hence p = 2 and q = ndash1
(b) y = ( x ndash 2)2 ndash 1
When y = 0
( x ndash 2)2 ndash 1 = 0
( x ndash 2)2 = 1
x ndash 2 = plusmn1
x = plusmn1 + 2
= 1 3
Hence A is (1 0) and B is (3 0)
7 f ( x) = 2k + 1 ndash 983089 x + 1mdash2 p983090
2
Given (ndash1 k ) is the maximum point
Therefore 2k + 1 = k
k = ndash1
x +1mdash2 p = 0 when x = ndash1
ndash1 + 1mdash2 p = 0
1mdash2 p = 1
p = 2
8 Given ( p 2q) is the minimum point of
y = 2 x2 ndash 4 x + 5
= 2( x2 ndash 2 x) + 5
= 2( x2 ndash 2 x + 12 ndash 12) + 5
= 2[( x ndash 1)2 ndash 1] + 5
= 2( x ndash 1)2 ndash 2 + 5
= 2( x ndash 1)2 + 3
2q = 3
q =3mdash2
p ndash 1 = 0
p = 1
9 (a) Since (1 4) is the point on y = x2 ndash 2kx + 1
substitute x = 1 y = 4 into the equation
4 = 12 ndash 2k (1) + 1
2k = ndash2
k = ndash1
(b) y = x2 ndash 2(ndash1) x + 1
= x2 + 2 x + 1
= ( x + 1)2
Minimum value of y is 0
10 f ( x) = ndash x2 ndash 8 x + k ndash 1
= ndash( x2 + 8 x) + k ndash 1
= ndash( x2 + 8 x + 42 ndash 42) + k ndash 1
= ndash[( x + 4)2 ndash 16] + k ndash 1
= ndash( x + 4)2 + 16 + k ndash 1
= ndash( x + 4)2 + 15 + k
Since 13 is the maximum value
then 15 + k = 13
k = ndash2
11 f ( x) = 2 x2 ndash 6 x + 7
= 2( x2 ndash 3 x) + 7
= 2983091 x2 ndash 3 x + 983089 3mdash2 983090
2
ndash 983089 3mdash2 983090
2
983092 + 7
= 2983091983089 x ndash3mdash2 983090
2
ndash9mdash4 983092 + 7
= 2983089 x ndash3mdash2 983090
2
ndash9mdash2
+ 7
= 2983089 x ndash3mdash2 983090
2
+5mdash2
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The minimum point is (3mdash2
5mdash2
)
x ndash1 0 3
f ( x) 15 7 7
ndash ndash
3
5322
0
7
15
x
f (x )
ndash1
The range is5mdash2
983100 f ( x) 983100 15
12 f ( x) = 5 ndash 4 x ndash 2 x2
= ndash2 x2 ndash 4 x + 5
= ndash2( x2 + 2 x) + 5
= ndash2( x2 + 2 x + 12 ndash 12) + 5
= ndash2[( x + 1)2 ndash 1] + 5
= ndash2( x + 1)2 + 2 + 5
= ndash2( x + 1)2 + 7
When f ( x) = ndash1
ndash2( x + 1)2 + 7 = ndash1
ndash2( x + 1)2 = ndash8
( x + 1)2 = 4
x + 1 = plusmn2
x = plusmn2 ndash 1
= ndash3 or 1
13 y = ( x ndash 3)2 ndash 4
Minimum point is (3 ndash4)
x ndash1 0 1 5 6
y 12 5 0 0 5
x 0 1 5 6
5
12
ndash1
(3 ndash4)
y
The range is ndash 4983100 y 983100 12
14 y = ndash x2 + 4 x ndash 5
= ndash( x2 ndash 4 x) ndash 5
= ndash( x2 ndash 4 x + 22 ndash 22) ndash 5
= ndash [( x ndash 2)2 ndash 4] ndash 5
= ndash( x ndash 2)2 + 4 ndash 5
= ndash( x ndash 2)2 ndash 1
Maximum point is (2 ndash1)
x ndash1 0 3
y ndash10 ndash5 ndash2
x
0 3(2 ndash1)
ndash1
ndash5
ndash10
y
The range is ndash10 983100 y 983100 ndash1
15 y = 9 ndash ( x ndash 3)2Maximum point is (3 9)
x ndash1 0 6 7
y 7 0 0 7
x
y
0 ndash1 6
(3 9)
7
7
The range is 0 983100 y 983100 9
16 3 x2 983084 x
3 x2
ndash x 983084 0 x(3 x ndash 1)983084 0
ndash
13
0 x
f (x )
The range is 0 983084 x 983084 1mdash3
17 3 x ndash x
2
ndashndashndashndashndashndash2 983084 1
3 x ndash x2 983084 2
ndash x2 + 3 x ndash 2 983084 0
x2 ndash 3 x + 2 983086 0
( x ndash 1)( x ndash 2) 983086 0
20 x
f (x )
1
The range is x 983084
1 or x 983086
2
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18 Given that x ndash 2 y = 1
there4 x = 1 + 2 y 983089
Substitute 983089 into y + 3983102 2 xy
y + 3983102 2(1 + 2 y) y
y + 3983102 2 y + 4 y2
0983102 4 y2 + y ndash 3
0983102 (4 y ndash 3)( y + 1)
3 ndash
4
0 y
f (y )
ndash1
The range is ndash1 983100 y 983100 3mdash4
19 f ( x) 983084 0
5 x2 ndash 4 x ndash 1 983084 0
(5 x + 1)( x ndash 1) 983084 0
1 ndash ndash
5
0 x
f (x )
1
The range is ndash1mdash5
983084 x 983084 1
20 g( x) 983086 0
4 x2 ndash 9 983086 0(2 x + 3)(2 x ndash 3) 983086 0
3 ndash ndash
23 ndash
2
0 x
g (x )
The range is x 983084 ndash3mdash2
or x 983086 3mdash2
21 (a) Since y = 3 x
2
ndash 9 x + t 983086
0 for all values of x andit does not have root when y = 0
Then b2 ndash 4ac 983084 0 for 3 x2 ndash 9 x + t = 0
(ndash9)2 ndash 4(3)(t ) 983084 0
81 ndash 12t 983084 0
ndash12t 983084 ndash81
t 983086 ndash81 ndashndashndashndash ndash12
t 983086 27
ndashndashndash4
(b) Let f ( x) = a( x ndash b)2 + c
f ( x) = a( x ndash 2)2 + 0
f ( x) = a( x ndash 2)2
Substitute x = 0 f ( x) = ndash3 into the equation
ndash3 = a(0 ndash 2)2
= 4a
a = ndash3mdash4
Hence the quadratic function is
f ( x) = ndash3mdash4
( x ndash 2)2
22 (a) Given 2 x2 ndash 3 y + 2 = 0
3 y = 2 x2 + 2
y =2 x2
ndashndashndash3
+2mdash3
983089
Substitute 983089 into y 983084 10
2 x2
ndashndashndash
3
+2mdash
3
983084 10
2 x2 + 2 983084 30
2 x2 ndash 28 983084 0
x2 ndash 14 983084 0
( x + 98310598310698310614 )( x ndash 98310598310698310614 ) 983084 0
14 ndash14
0 x
f (x )
The range is ndash98310598310698310614 983084 x 983084 98310598310698310614
(b) 2 x2 ndash 8 x ndash 10 = 2( x2 ndash 4 x) ndash 10
= 2( x2 ndash 4 x + 22 ndash 22) ndash 10
= 2[( x ndash 2)2 ndash 4] ndash 10
= 2( x ndash 2)2 ndash 8 ndash 10
= 2( x ndash 2)2 ndash 18
Therefore a = 2 b = ndash2 and c = ndash18
Hence the minimum value of 2 x2 ndash 8 x ndash 10 is
ndash18
23 5 ndash 2 x 983100 0 3 x2 ndash 4 x 983086 ndash1
3 x2
ndash 4 x + 1983086
0(3 x ndash 1)( x ndash 1) 983086 0
11 ndash
3
0 x
f (x )
x 983084 1mdash3
x 983086 1
ndash2 x 983100
ndash5 x 983102
ndash5 ndashndashndash ndash2
x 983102 5mdash2
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5ndash
2
1ndash
3
1
x 5mdash2
x x 11mdash3
The range is x 983102 5mdash2
24 5 983084 f ( x) 983084 9
5 983084 5 ndash 3 x + x2 983084 9
5 983084 5 ndash 3 x + x2 5 ndash 3 x + x2 983084 9
5 ndash 3 x + x2 ndash 9 983084 0
x2 ndash 3 x ndash 4 983084 0
( x ndash 4)( x + 1) 983084 0
4 ndash1 0 x
f (x )
ndash1 983084 x 983084 4
0 983084 x2 ndash 3 x
0 983084 x( x ndash 3)
30 x
f (x )
x 983084 0 x 983086 3
0 ndash1 43
x lt 0 x gt 3
ndash1 lt x lt 4
The range is ndash1983084
x 983084
0 or 3983084
x 983084
4
25 1 983102 x2 + 3 x ndash 3 983086 ndash3
x2 + 3 x ndash 3 983086 ndash3
x2 + 3 x 983086 0
x( x + 3) 983086 0
ndash3 0 x
f (x )
x 983084 ndash3 x 983086 0
1 983102 x2 + 3 x ndash 3
0 983102 x2 + 3 x ndash 4
0 983102 ( x + 4)( x ndash 1)
1 ndash4 0 x
f (x )
ndash 4 983100 x 983100 1
ndash4
x lt ndash3 x gt 0
ndash4 x 1
ndash3 0 1
The range is ndash 4983100 x 983084 ndash3 or 0 983084 x 983100 1
26 x2 + 5 x ndash 10 983086 4
x2 + 5 x ndash 10 983084 ndash 4 x2 + 5 x ndash 10 983086 4
x2 + 5 x ndash 14 983086 0
( x + 7)( x ndash 2) 983086 0
ndash7 20
x
f (x )
x 983084 ndash7 x 983086 2
x2 + 5 x ndash 6 983084 0
( x + 6)( x ndash 1) 983084 0
ndash6 10
x
f (x )
ndash6 983084 x 983084 1
The ranges are ndash6 lt x lt 1 x lt ndash7 x gt 2
27 f ( x) = (r + 1) x2 + 2rx + r ndash 3
Given that f ( x) does not intersect the x-axis
therefore b2 ndash 4ac 983084 0
(2r)2 ndash 4(r + 1)(r ndash 3) 983084 0
4r2 ndash 4(r2 ndash 2r ndash 3) 983084 0
4r2 ndash 4r2 + 8r + 12 983084 0
2r + 3 983084 0
r 983084 ndash3mdash2
28 Given that f ( x) = 9 ndash 6 x + 2 x2 does not have real root
when f ( x) = k
there4 9 ndash 6 x + 2 x2 = k
2 x2 ndash 6 x + 9 ndash k = 0
Use b2 ndash 4ac 983084 0
(ndash 6)2 ndash 4(2)(9 ndash k ) 983084 0
36 ndash 72 + 8k 983084 0
ndash36 + 8k 983084 0 8k 983084 36
k 983084 36
ndashndashndash8
k 983084 9mdash2
29 2 x2 + 10 x ndash 20 983100 8
ndash8 983100 2 x2 + 10 x ndash 20 983100 8
ndash8983100 2 x2 + 10 x ndash 20 2 x2 + 10 x ndash 20 983100 8
2 x2 + 10 x ndash 28 983100 0
x2 + 5 x ndash 14 983100 0
( x + 7)( x ndash 2) 983100 0
ndash7 20
x
f (x )
ndash7 983100 x 983100 2
0983100 2 x2 + 10 x ndash 12
0983100 x2 + 5 x ndash 6
0983100 ( x + 6)( x ndash 1)
ndash6 10
x
f (x )
x 983100 ndash6 x 983102 1
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x ndash6 x 1
ndash7 x 2
ndash6 ndash7 1 2
The range is ndash7 983100 x 983100 ndash6 or 1 983100 x 983100 2
30 y = x2 + 5 x ndash 6
= x2 + 5 x + 983089 5mdash2 983090
2
ndash 983089 5mdash2 983090
2
ndash 6
= 983089 x + 5mdash2 983090
2
ndash25
ndashndashndash4
ndash 6
= 983089 x + 5mdash2 983090
2
ndash49
ndashndashndash4
The minimum point is (ndash5mdash2
ndash49
ndashndashndash4
)
x ndash13 ndash 6 0 1 3
y 98 0 ndash 6 0 18
x 0
ndash6 ndash13 ndash6
18
98
1 3
) ) ndash ndash ndash ndash
5
2
49
4
y
The range is ndash49
ndashndashndash4
983084 y 983084 98
31 y = ndash x2 + 2 x ndash nx + 16
= ndash x2 + (2 ndash n) x + 16
= ndash[ x2 ndash (2 ndash n) x] + 16
= ndash983091 x2 ndash (2 ndash n) x + 983089 2 ndash n ndashndashndashndashndash
2 983090
2
ndash 983089 2 ndash n ndashndashndashndashndash
2 983090
2
983092 + 16
= ndash
983091983089 x ndash
2 ndash n ndashndashndashndashndash
2 983090
2
ndash
9830892 ndash n
ndashndashndashndashndash2 983090
2
983092 + 16
= ndash983089 x ndash 2 ndash n ndashndashndashndashndash
2 983090
2
+ 983089 2 ndash n ndashndashndashndashndash
2 983090
2
+ 16
Since y = x2 ndash 3k and y = ndash x2 + 2 x ndash nx + 16 have the
same axis of symmetry that is x = 0
then ndash2 ndash n
ndashndashndashndashndash2
= 0
2 ndash n = 0
n = 2
The equation y = ndash x2 + 2 x ndash nx + 16
= ndash x2 + 2 x ndash 2 x + 16
= ndash983089 x ndash 2 ndash 2 ndashndashndashndashndash
2 983090
2
+ 983089 2 ndash 2 ndashndashndashndashndash
2 983090
2
+ 16
= ndash x2 + 16
When y = 0
ndash x2
+ 16 = 0 x2 = 16
x = plusmn98310598310698310616
= plusmn4
Therefore B = (4 0)
Substitute x = 4 y = 0 into y = x2 ndash 3k
0 = 42 ndash 3k
0 = 16 ndash 3k
k =16
ndashndashndash3
32 (a) y = ndash2[(3k ndash x)2 + n] ndash 10 has a maximum point
(4 11)
y = ndash2(3k ndash x)2 ndash 2n ndash 10
there4 3k ndash 4 = 0 and ndash2n ndash 10 = 11
k =4mdash3
ndash2n = 21
n = ndash21
ndashndashndash2
(b) Substitute k =4mdash3
and n = ndash21
ndashndashndash2
into
y = ndash2[(3k ndash x)2 + n] ndash 10
y = ndash2983091(4 ndash x)2 ndash21
ndashndashndash2 983092 ndash 10
= ndash2(4 ndash x)2 + 21 ndash 10
= ndash2(4 ndash x)2
+ 11 When y = 0
ndash2(4 ndash x)2 + 11= 0
2(4 ndash x)2 = 11
(4 ndash x)2 =11
ndashndashndash2
4 ndash x = plusmn98310598310698310698310611 ndashndashndash
2
x = 4 plusmn 98310598310698310698310611 ndashndashndash
2
= 4 ndash 98310598310698310698310611 ndashndashndash
2 4 + 98310598310698310698310611
ndashndashndash2
= 1655 6345
(c) y = ndash2(4 ndash x)2 + 11
The maximum point is (4 11)
x ndash1 0 5
y ndash39 ndash21 9
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1 y = 3(2 x ndash 1)( x + 1) ndash x(4 x ndash 5) + 2
= 3(2 x2 + x ndash 1) ndash 4 x2 + 5 x + 2
= 6 x2 + 3 x ndash 3 ndash 4 x2 + 5 x + 2
= 2 x2 + 8 x ndash 1
= 2( x2 + 4 x) ndash 1
= 2( x2 + 4 x + 22 ndash 22) ndash 1
= 2[( x + 2)2 ndash 4] ndash 1
= 2( x + 2)2 ndash 8 ndash 1
= 2( x + 2)2 ndash 9
Since a = 2 983086 0 therefore the minimum value of
y is ndash9
When y = 0 2( x + 2)2 ndash 9 = 0
( x + 2)2 =9mdash2
x + 2 = plusmn9831059831069831069mdash2 x = plusmn9831059831069831069mdash2 ndash 2
= 9831059831069831069mdash2 ndash 2 or ndash9831059831069831069mdash2 ndash 2
= 01213 or ndash 4121
When x = 0 y = 2(2)2 ndash 9
= ndash1
The minimum point is (ndash2 ndash9)
ndash1 ndash4121
(ndash2 ndash9)
012130
y
x
2 5 983084 Area of rectangle ABCD 983084 21
5 983084 ( x + 3)( x ndash 1) 983084 21
5 983084 ( x + 3)( x ndash 1)
5 983084 x2 + 2 x ndash 3
0 983084 x2 + 2 x ndash 8
0 983084 ( x + 4)( x ndash 2)
0 ndash4 2
f (x )
x
x 983084 ndash 4 x gt 2
( x + 3)( x ndash 1) 983084 21
x2 + 2 x ndash 3 983084 21
x2 + 2 x ndash 24 983084 0
( x ndash 4)( x + 6) 983084 0
0 ndash6 4
f (x )
x
ndash 6 983084 x 983084 4
ndash4
ndash6 lt x lt 4
x lt ndash4 x gt 2
2
x
ndash6 4
The range is ndash 6983084 x 983084 ndash 4 or 2 983084 x 983084 4
3 (a) p =1 + 5
ndashndashndashndashndash2
= 3
(b) y = ( x ndash 1)( x ndash 5) = x2 ndash 6 x + 5
4 y = a( x ndash 2)2 + 1
Substitute x = 0 y = 9 into the equation
9 = a(ndash2)2 + 1
8 = 4a
a = 2
Therefore the quadratic function is
f ( x) = 2( x ndash 2)2 + 1
5 x2
+ (1 + k ) x ndash k 2
+ 1 = 0For quadratic equation to have real roots
b2 ndash 4ac 983102 0
(1 + k )2 ndash 4(1)(1 ndash k 2) 983102 0
1 + 2k + k 2 ndash 4 + 4k 2 983102 0
5k 2 + 2k ndash 3 983102 0
(5k ndash 3)(k + 1) 983102 0
0 ndash1 3 ndash
5
f (k )
k
The range of values of k is k 983100 ndash1 or k 983102 3mdash5
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6 y = x2 + 7 x ndash 8 ndash 2k
For y to be positive for all real values of x there is
no roots for y = 0
Therefore b2 ndash 4ac 983084 0
72 ndash 4(1)(ndash8 ndash 2k ) 983084 0
49 + 32 + 8k 983084 0
8k 983084 ndash81
k 983084 ndash81
ndashndashndash8
Alternative
y = x2 + 7 x ndash 8 ndash 2k
= x2 + 7 x + 983089 7mdash2 983090
2
ndash 983089 7mdash2 983090
2
ndash 8 ndash 2k
= 983089 x +7mdash2 983090
2
ndash49
ndashndashndash4
ndash 8 ndash 2k
For y to be positive for all real values of x
ndash49
ndashndashndash4
ndash 8 ndash 2k 983086 0
ndash2k 983086 49
ndashndashndash4
+ 8
ndash2k 983086 81
ndashndashndash4
k 983084 ndash81
ndashndashndash8
7 Substitute x = 6 y = 0 into y = px2 + qx
0 = p(6)2 + q(6)
0 = 36 p + 6q
q + 6 p = 0 983089
y = px2 + qx
= p983089 x2 +q mdash p x983090
= p983091 x2 +q mdash p x + 983089
q ndashndashndash2 p 983090
2
ndash 983089q
ndashndashndash2 p 983090
2
983092
= p983091983089 x +q
ndashndashndash2 p 983090
2
ndashq2
ndashndashndashndash4 p2 983092
= p983089 x +q
ndashndashndash2 p 983090
2
ndashq2
ndashndashndash4 p
ndash q2
ndashndashndash4 p
= ndash12
q2 = 48 p
p =q2
ndashndashndash48
983090
Substitute 983090 into 983089
q + 6983089 q2
ndashndashndash48
983090 = 0
q +q2
mdash8
= 0
8q + q2 = 0
q(8 + q) = 0
q = 0 or q = ndash8
When q = 0 p =02
ndashndashndash48
= 0
When q = ndash8 p =(ndash8)2
ndashndashndashndashndash48
=64
ndashndashndash48
=4mdash3
Therefore the values of p =
4
mdash3 and q = ndash8
8 (2 ndash 3k ) x2 + x +3mdash4k = 0
b2 ndash 4ac = 12 ndash 4(2 ndash 3k )983089 3mdash4k 983090
= 1 ndash 6k + 9k 2
= 9k 2 ndash 6k + 1
= (3k ndash 1)2
Since (3k ndash 1)2 983102 0 for all values of k
therefore (2 ndash 3k ) x2 + x +3mdash4k = 0 has real roots for
all values of k
9 f ( x) = 3( x2 + 2mx + m2 + n)
= 3[( x + m)2 + n]
= 3( x + m)2 + 3n
The minimum point is (ndashm 3n)
Compare to A(t 3t 2)
there4 ndashm = t and 3n = 3t 2
m = ndasht n = t 2
10 (a) y = px2 + 8 x + 10 ndash p
When the graph does not intercept the x-axis
there are no roots for px2 + 8 x + 10 ndash p = 0 Therefore b2 ndash 4ac 983084 0
82 ndash 4 p(10 ndash p) 983084 0
64 ndash 40 p + 4 p2 983084 0
p2 ndash 10 p + 16 983084 0
( p ndash 2)( p ndash 8) 983084 0
2 8
Hence r = 2 and t = 8
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(b) When p = 2
y = 2 x2 + 8 x + 8
= 2( x2 + 4 x) + 8
= 2( x2 + 4 x + 22 ndash 22) + 8
= 2[( x + 2)2 ndash 4] + 8
= 2( x + 2)2 ndash 8 + 8
= 2( x + 2)2
Therefore the minimum point is (ndash2 0)
When x = 0 y = 8
When y = 0 2( x + 2)2 = 0
x = ndash2
When p = 8
y = 8 x2 + 8 x + 2
= 8( x2 + x) + 2
= 8983091 x2 + x + 983089 1mdash2 983090
2
ndash 983089 1mdash2 983090
2
983092 + 2
= 8983091983089 x +1mdash2 983090
2
ndash1mdash4 983092 + 2
= 8983089 x +1mdash2 983090
2
ndash 2 + 2
= 8983089 x +1mdash2 983090
2
Therefore the minimum point is (ndash1mdash2
0)
When x = 0 y = 2
When y = 0 0 = 8983089 x + 1mdash2 983090
2
x = ndash1mdash2
0
2
8
1 ndash ndash
2
ndash2
y p = 2 p = 8
x
11 (a) f ( x) = 24 x ndash 4 x2 + r
= ndash 4 x2 + 24 x + r
= ndash 4( x2 ndash 6 x) + r
= ndash 4( x2 ndash 6 x + 32 ndash 32) + r
= ndash 4[( x ndash 3)2 ndash 9] + r
= ndash 4( x ndash 3)2 + 36 + r
Compare to f ( x) = p( x ndash q)2 + 16
Therefore p = ndash 4 q = 3 and 36 + r = 16
r = ndash20
(b) The turning point is (3 16)
(c) f ( x) = 24 x ndash 4 x2 ndash 20
When x = 0 f ( x) = ndash20
When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0
4( x ndash 3)2 = 16
( x ndash 3)2 = 4
x ndash 3 = plusmn2
x = plusmn2 + 3
= ndash2 + 3 or 2 + 3
= 1 or 5
0 1
(3 16)
5
ndash20
y
x
12 (a) y = ndash| p( x ndash 3)2 + q|
Substitute x = 3 y = ndash5 into the equation
ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5
Substitute x = 4 y = 0 into the equation
0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0
p = 5
Therefore p = 5 q = ndash5 or p = ndash5 q = 5
(b) When x = 3 y = ndash5
For p = 5 q = ndash5
When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40
Based on the graph the range of values of y is
ndash 40 983100 y 983100 0
For p = ndash5 q = 5
When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40
Therefore the range of values of y is
ndash 40 983100 y 983100 0
13 (a) y = ndash2( x ndash 3)2 + 2k
= ndash x2 + 2 x + px ndash 8
= ndash x2 + (2 + p) x ndash 8
= ndash[ x2 ndash (2 + p) x] ndash 8
= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash
2 9830902
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983091983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
+(2 + p)2
ndashndashndashndashndashndashndash4
ndash 8
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Since the x-coordinate of the maximum point for
both the graphs are same
therefore2 + p
ndashndashndashndashndash2
= 3
p = 4
y = ndash x2 + 2 x + px ndash 8 becomes
y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8
When y = 0
ndash x2 + 6 x ndash 8 = 0
x2 ndash 6 x + 8 = 0
( x ndash 2)( x ndash 4) = 0
x = 2 or 4
Hence A(2 0) and B(4 0)
Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k
0 = ndash2(2 ndash 3)2 + 2k
2k = 2
k = 1 Hence k = 1 and p = 4
(b) For y = ndash2( x ndash 3)2 + 2k
= ndash2( x ndash 3)2 + 2(1)
= ndash2( x ndash 3)2 + 2
Maximum value of the curve is 2
For y = ndash x2 + 2 x + px ndash 8
= ndash x2 + 2 x + 4 x ndash 8
= ndash x2 + 6 x ndash 8
When x = 3
y = ndash9 + 18 ndash 8
= 1
Maximum value of the curve is 1
14 Since 3 x2 983102 0 for all values of x
therefore3 x2
ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)
983100 0
(2 x ndash 1)( x + 4) 983100 0
0 ndash4 1 ndash
2
f (x )
x
Hence ndash 4983100 x 983100 1mdash2
15 Since x2 + 1 983086 0
therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash
x2 + 1 983086 0
x2 + 3 x + 2 983086 0
( x + 1)( x + 2) 983086 0
0 ndash2 ndash1
f (x )
x
Hence x 983084 ndash2 x gt ndash1
16 ndash4
ndashndashndashndashndashndash1 ndash 3 x
983100 x
0 983100 x +4
ndashndashndashndashndashndash1 ndash 3 x
0 983100 x(1 ndash 3 x) + 4
ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0983100
x ndash 3 x2 + 4
ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash
1 ndash 3 x
0 983100 (ndash3 x + 4)( x + 1)
ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
For ndash3 x + 4 983102 0
4 983102 3 x
x 983100 4mdash3
For x + 1 983102 0
x 983102 ndash1
For 1 ndash 3 x 983086 0
ndash3 x 983086 ndash1
x 983084 1mdash3
ndash1 ndash + ndash +
1 4 ndash ndash
3
4x ndash
3
1x lt ndash
3
x ndash1
3
x
Therefore the range is ndash1 983100 x 983084 1mdash3
x 983102 4mdash3
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17 y2 ndash 9 = x
x = y2 ndash 9
When x = 0
y2 = 9
y = plusmn3
When y = 0
x = ndash9
When x = 7
7 = y2 ndash 9
y2 = 16
y = plusmn4
x ndash9 0 7
y 0 plusmn3 plusmn4
0 ndash9 7
ndash4 ndash3
34
y
x
The range of values of y is ndash 4 983100 y 983100 4
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The minimum point is (3mdash2
5mdash2
)
x ndash1 0 3
f ( x) 15 7 7
ndash ndash
3
5322
0
7
15
x
f (x )
ndash1
The range is5mdash2
983100 f ( x) 983100 15
12 f ( x) = 5 ndash 4 x ndash 2 x2
= ndash2 x2 ndash 4 x + 5
= ndash2( x2 + 2 x) + 5
= ndash2( x2 + 2 x + 12 ndash 12) + 5
= ndash2[( x + 1)2 ndash 1] + 5
= ndash2( x + 1)2 + 2 + 5
= ndash2( x + 1)2 + 7
When f ( x) = ndash1
ndash2( x + 1)2 + 7 = ndash1
ndash2( x + 1)2 = ndash8
( x + 1)2 = 4
x + 1 = plusmn2
x = plusmn2 ndash 1
= ndash3 or 1
13 y = ( x ndash 3)2 ndash 4
Minimum point is (3 ndash4)
x ndash1 0 1 5 6
y 12 5 0 0 5
x 0 1 5 6
5
12
ndash1
(3 ndash4)
y
The range is ndash 4983100 y 983100 12
14 y = ndash x2 + 4 x ndash 5
= ndash( x2 ndash 4 x) ndash 5
= ndash( x2 ndash 4 x + 22 ndash 22) ndash 5
= ndash [( x ndash 2)2 ndash 4] ndash 5
= ndash( x ndash 2)2 + 4 ndash 5
= ndash( x ndash 2)2 ndash 1
Maximum point is (2 ndash1)
x ndash1 0 3
y ndash10 ndash5 ndash2
x
0 3(2 ndash1)
ndash1
ndash5
ndash10
y
The range is ndash10 983100 y 983100 ndash1
15 y = 9 ndash ( x ndash 3)2Maximum point is (3 9)
x ndash1 0 6 7
y 7 0 0 7
x
y
0 ndash1 6
(3 9)
7
7
The range is 0 983100 y 983100 9
16 3 x2 983084 x
3 x2
ndash x 983084 0 x(3 x ndash 1)983084 0
ndash
13
0 x
f (x )
The range is 0 983084 x 983084 1mdash3
17 3 x ndash x
2
ndashndashndashndashndashndash2 983084 1
3 x ndash x2 983084 2
ndash x2 + 3 x ndash 2 983084 0
x2 ndash 3 x + 2 983086 0
( x ndash 1)( x ndash 2) 983086 0
20 x
f (x )
1
The range is x 983084
1 or x 983086
2
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18 Given that x ndash 2 y = 1
there4 x = 1 + 2 y 983089
Substitute 983089 into y + 3983102 2 xy
y + 3983102 2(1 + 2 y) y
y + 3983102 2 y + 4 y2
0983102 4 y2 + y ndash 3
0983102 (4 y ndash 3)( y + 1)
3 ndash
4
0 y
f (y )
ndash1
The range is ndash1 983100 y 983100 3mdash4
19 f ( x) 983084 0
5 x2 ndash 4 x ndash 1 983084 0
(5 x + 1)( x ndash 1) 983084 0
1 ndash ndash
5
0 x
f (x )
1
The range is ndash1mdash5
983084 x 983084 1
20 g( x) 983086 0
4 x2 ndash 9 983086 0(2 x + 3)(2 x ndash 3) 983086 0
3 ndash ndash
23 ndash
2
0 x
g (x )
The range is x 983084 ndash3mdash2
or x 983086 3mdash2
21 (a) Since y = 3 x
2
ndash 9 x + t 983086
0 for all values of x andit does not have root when y = 0
Then b2 ndash 4ac 983084 0 for 3 x2 ndash 9 x + t = 0
(ndash9)2 ndash 4(3)(t ) 983084 0
81 ndash 12t 983084 0
ndash12t 983084 ndash81
t 983086 ndash81 ndashndashndashndash ndash12
t 983086 27
ndashndashndash4
(b) Let f ( x) = a( x ndash b)2 + c
f ( x) = a( x ndash 2)2 + 0
f ( x) = a( x ndash 2)2
Substitute x = 0 f ( x) = ndash3 into the equation
ndash3 = a(0 ndash 2)2
= 4a
a = ndash3mdash4
Hence the quadratic function is
f ( x) = ndash3mdash4
( x ndash 2)2
22 (a) Given 2 x2 ndash 3 y + 2 = 0
3 y = 2 x2 + 2
y =2 x2
ndashndashndash3
+2mdash3
983089
Substitute 983089 into y 983084 10
2 x2
ndashndashndash
3
+2mdash
3
983084 10
2 x2 + 2 983084 30
2 x2 ndash 28 983084 0
x2 ndash 14 983084 0
( x + 98310598310698310614 )( x ndash 98310598310698310614 ) 983084 0
14 ndash14
0 x
f (x )
The range is ndash98310598310698310614 983084 x 983084 98310598310698310614
(b) 2 x2 ndash 8 x ndash 10 = 2( x2 ndash 4 x) ndash 10
= 2( x2 ndash 4 x + 22 ndash 22) ndash 10
= 2[( x ndash 2)2 ndash 4] ndash 10
= 2( x ndash 2)2 ndash 8 ndash 10
= 2( x ndash 2)2 ndash 18
Therefore a = 2 b = ndash2 and c = ndash18
Hence the minimum value of 2 x2 ndash 8 x ndash 10 is
ndash18
23 5 ndash 2 x 983100 0 3 x2 ndash 4 x 983086 ndash1
3 x2
ndash 4 x + 1983086
0(3 x ndash 1)( x ndash 1) 983086 0
11 ndash
3
0 x
f (x )
x 983084 1mdash3
x 983086 1
ndash2 x 983100
ndash5 x 983102
ndash5 ndashndashndash ndash2
x 983102 5mdash2
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5ndash
2
1ndash
3
1
x 5mdash2
x x 11mdash3
The range is x 983102 5mdash2
24 5 983084 f ( x) 983084 9
5 983084 5 ndash 3 x + x2 983084 9
5 983084 5 ndash 3 x + x2 5 ndash 3 x + x2 983084 9
5 ndash 3 x + x2 ndash 9 983084 0
x2 ndash 3 x ndash 4 983084 0
( x ndash 4)( x + 1) 983084 0
4 ndash1 0 x
f (x )
ndash1 983084 x 983084 4
0 983084 x2 ndash 3 x
0 983084 x( x ndash 3)
30 x
f (x )
x 983084 0 x 983086 3
0 ndash1 43
x lt 0 x gt 3
ndash1 lt x lt 4
The range is ndash1983084
x 983084
0 or 3983084
x 983084
4
25 1 983102 x2 + 3 x ndash 3 983086 ndash3
x2 + 3 x ndash 3 983086 ndash3
x2 + 3 x 983086 0
x( x + 3) 983086 0
ndash3 0 x
f (x )
x 983084 ndash3 x 983086 0
1 983102 x2 + 3 x ndash 3
0 983102 x2 + 3 x ndash 4
0 983102 ( x + 4)( x ndash 1)
1 ndash4 0 x
f (x )
ndash 4 983100 x 983100 1
ndash4
x lt ndash3 x gt 0
ndash4 x 1
ndash3 0 1
The range is ndash 4983100 x 983084 ndash3 or 0 983084 x 983100 1
26 x2 + 5 x ndash 10 983086 4
x2 + 5 x ndash 10 983084 ndash 4 x2 + 5 x ndash 10 983086 4
x2 + 5 x ndash 14 983086 0
( x + 7)( x ndash 2) 983086 0
ndash7 20
x
f (x )
x 983084 ndash7 x 983086 2
x2 + 5 x ndash 6 983084 0
( x + 6)( x ndash 1) 983084 0
ndash6 10
x
f (x )
ndash6 983084 x 983084 1
The ranges are ndash6 lt x lt 1 x lt ndash7 x gt 2
27 f ( x) = (r + 1) x2 + 2rx + r ndash 3
Given that f ( x) does not intersect the x-axis
therefore b2 ndash 4ac 983084 0
(2r)2 ndash 4(r + 1)(r ndash 3) 983084 0
4r2 ndash 4(r2 ndash 2r ndash 3) 983084 0
4r2 ndash 4r2 + 8r + 12 983084 0
2r + 3 983084 0
r 983084 ndash3mdash2
28 Given that f ( x) = 9 ndash 6 x + 2 x2 does not have real root
when f ( x) = k
there4 9 ndash 6 x + 2 x2 = k
2 x2 ndash 6 x + 9 ndash k = 0
Use b2 ndash 4ac 983084 0
(ndash 6)2 ndash 4(2)(9 ndash k ) 983084 0
36 ndash 72 + 8k 983084 0
ndash36 + 8k 983084 0 8k 983084 36
k 983084 36
ndashndashndash8
k 983084 9mdash2
29 2 x2 + 10 x ndash 20 983100 8
ndash8 983100 2 x2 + 10 x ndash 20 983100 8
ndash8983100 2 x2 + 10 x ndash 20 2 x2 + 10 x ndash 20 983100 8
2 x2 + 10 x ndash 28 983100 0
x2 + 5 x ndash 14 983100 0
( x + 7)( x ndash 2) 983100 0
ndash7 20
x
f (x )
ndash7 983100 x 983100 2
0983100 2 x2 + 10 x ndash 12
0983100 x2 + 5 x ndash 6
0983100 ( x + 6)( x ndash 1)
ndash6 10
x
f (x )
x 983100 ndash6 x 983102 1
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x ndash6 x 1
ndash7 x 2
ndash6 ndash7 1 2
The range is ndash7 983100 x 983100 ndash6 or 1 983100 x 983100 2
30 y = x2 + 5 x ndash 6
= x2 + 5 x + 983089 5mdash2 983090
2
ndash 983089 5mdash2 983090
2
ndash 6
= 983089 x + 5mdash2 983090
2
ndash25
ndashndashndash4
ndash 6
= 983089 x + 5mdash2 983090
2
ndash49
ndashndashndash4
The minimum point is (ndash5mdash2
ndash49
ndashndashndash4
)
x ndash13 ndash 6 0 1 3
y 98 0 ndash 6 0 18
x 0
ndash6 ndash13 ndash6
18
98
1 3
) ) ndash ndash ndash ndash
5
2
49
4
y
The range is ndash49
ndashndashndash4
983084 y 983084 98
31 y = ndash x2 + 2 x ndash nx + 16
= ndash x2 + (2 ndash n) x + 16
= ndash[ x2 ndash (2 ndash n) x] + 16
= ndash983091 x2 ndash (2 ndash n) x + 983089 2 ndash n ndashndashndashndashndash
2 983090
2
ndash 983089 2 ndash n ndashndashndashndashndash
2 983090
2
983092 + 16
= ndash
983091983089 x ndash
2 ndash n ndashndashndashndashndash
2 983090
2
ndash
9830892 ndash n
ndashndashndashndashndash2 983090
2
983092 + 16
= ndash983089 x ndash 2 ndash n ndashndashndashndashndash
2 983090
2
+ 983089 2 ndash n ndashndashndashndashndash
2 983090
2
+ 16
Since y = x2 ndash 3k and y = ndash x2 + 2 x ndash nx + 16 have the
same axis of symmetry that is x = 0
then ndash2 ndash n
ndashndashndashndashndash2
= 0
2 ndash n = 0
n = 2
The equation y = ndash x2 + 2 x ndash nx + 16
= ndash x2 + 2 x ndash 2 x + 16
= ndash983089 x ndash 2 ndash 2 ndashndashndashndashndash
2 983090
2
+ 983089 2 ndash 2 ndashndashndashndashndash
2 983090
2
+ 16
= ndash x2 + 16
When y = 0
ndash x2
+ 16 = 0 x2 = 16
x = plusmn98310598310698310616
= plusmn4
Therefore B = (4 0)
Substitute x = 4 y = 0 into y = x2 ndash 3k
0 = 42 ndash 3k
0 = 16 ndash 3k
k =16
ndashndashndash3
32 (a) y = ndash2[(3k ndash x)2 + n] ndash 10 has a maximum point
(4 11)
y = ndash2(3k ndash x)2 ndash 2n ndash 10
there4 3k ndash 4 = 0 and ndash2n ndash 10 = 11
k =4mdash3
ndash2n = 21
n = ndash21
ndashndashndash2
(b) Substitute k =4mdash3
and n = ndash21
ndashndashndash2
into
y = ndash2[(3k ndash x)2 + n] ndash 10
y = ndash2983091(4 ndash x)2 ndash21
ndashndashndash2 983092 ndash 10
= ndash2(4 ndash x)2 + 21 ndash 10
= ndash2(4 ndash x)2
+ 11 When y = 0
ndash2(4 ndash x)2 + 11= 0
2(4 ndash x)2 = 11
(4 ndash x)2 =11
ndashndashndash2
4 ndash x = plusmn98310598310698310698310611 ndashndashndash
2
x = 4 plusmn 98310598310698310698310611 ndashndashndash
2
= 4 ndash 98310598310698310698310611 ndashndashndash
2 4 + 98310598310698310698310611
ndashndashndash2
= 1655 6345
(c) y = ndash2(4 ndash x)2 + 11
The maximum point is (4 11)
x ndash1 0 5
y ndash39 ndash21 9
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1 y = 3(2 x ndash 1)( x + 1) ndash x(4 x ndash 5) + 2
= 3(2 x2 + x ndash 1) ndash 4 x2 + 5 x + 2
= 6 x2 + 3 x ndash 3 ndash 4 x2 + 5 x + 2
= 2 x2 + 8 x ndash 1
= 2( x2 + 4 x) ndash 1
= 2( x2 + 4 x + 22 ndash 22) ndash 1
= 2[( x + 2)2 ndash 4] ndash 1
= 2( x + 2)2 ndash 8 ndash 1
= 2( x + 2)2 ndash 9
Since a = 2 983086 0 therefore the minimum value of
y is ndash9
When y = 0 2( x + 2)2 ndash 9 = 0
( x + 2)2 =9mdash2
x + 2 = plusmn9831059831069831069mdash2 x = plusmn9831059831069831069mdash2 ndash 2
= 9831059831069831069mdash2 ndash 2 or ndash9831059831069831069mdash2 ndash 2
= 01213 or ndash 4121
When x = 0 y = 2(2)2 ndash 9
= ndash1
The minimum point is (ndash2 ndash9)
ndash1 ndash4121
(ndash2 ndash9)
012130
y
x
2 5 983084 Area of rectangle ABCD 983084 21
5 983084 ( x + 3)( x ndash 1) 983084 21
5 983084 ( x + 3)( x ndash 1)
5 983084 x2 + 2 x ndash 3
0 983084 x2 + 2 x ndash 8
0 983084 ( x + 4)( x ndash 2)
0 ndash4 2
f (x )
x
x 983084 ndash 4 x gt 2
( x + 3)( x ndash 1) 983084 21
x2 + 2 x ndash 3 983084 21
x2 + 2 x ndash 24 983084 0
( x ndash 4)( x + 6) 983084 0
0 ndash6 4
f (x )
x
ndash 6 983084 x 983084 4
ndash4
ndash6 lt x lt 4
x lt ndash4 x gt 2
2
x
ndash6 4
The range is ndash 6983084 x 983084 ndash 4 or 2 983084 x 983084 4
3 (a) p =1 + 5
ndashndashndashndashndash2
= 3
(b) y = ( x ndash 1)( x ndash 5) = x2 ndash 6 x + 5
4 y = a( x ndash 2)2 + 1
Substitute x = 0 y = 9 into the equation
9 = a(ndash2)2 + 1
8 = 4a
a = 2
Therefore the quadratic function is
f ( x) = 2( x ndash 2)2 + 1
5 x2
+ (1 + k ) x ndash k 2
+ 1 = 0For quadratic equation to have real roots
b2 ndash 4ac 983102 0
(1 + k )2 ndash 4(1)(1 ndash k 2) 983102 0
1 + 2k + k 2 ndash 4 + 4k 2 983102 0
5k 2 + 2k ndash 3 983102 0
(5k ndash 3)(k + 1) 983102 0
0 ndash1 3 ndash
5
f (k )
k
The range of values of k is k 983100 ndash1 or k 983102 3mdash5
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6 y = x2 + 7 x ndash 8 ndash 2k
For y to be positive for all real values of x there is
no roots for y = 0
Therefore b2 ndash 4ac 983084 0
72 ndash 4(1)(ndash8 ndash 2k ) 983084 0
49 + 32 + 8k 983084 0
8k 983084 ndash81
k 983084 ndash81
ndashndashndash8
Alternative
y = x2 + 7 x ndash 8 ndash 2k
= x2 + 7 x + 983089 7mdash2 983090
2
ndash 983089 7mdash2 983090
2
ndash 8 ndash 2k
= 983089 x +7mdash2 983090
2
ndash49
ndashndashndash4
ndash 8 ndash 2k
For y to be positive for all real values of x
ndash49
ndashndashndash4
ndash 8 ndash 2k 983086 0
ndash2k 983086 49
ndashndashndash4
+ 8
ndash2k 983086 81
ndashndashndash4
k 983084 ndash81
ndashndashndash8
7 Substitute x = 6 y = 0 into y = px2 + qx
0 = p(6)2 + q(6)
0 = 36 p + 6q
q + 6 p = 0 983089
y = px2 + qx
= p983089 x2 +q mdash p x983090
= p983091 x2 +q mdash p x + 983089
q ndashndashndash2 p 983090
2
ndash 983089q
ndashndashndash2 p 983090
2
983092
= p983091983089 x +q
ndashndashndash2 p 983090
2
ndashq2
ndashndashndashndash4 p2 983092
= p983089 x +q
ndashndashndash2 p 983090
2
ndashq2
ndashndashndash4 p
ndash q2
ndashndashndash4 p
= ndash12
q2 = 48 p
p =q2
ndashndashndash48
983090
Substitute 983090 into 983089
q + 6983089 q2
ndashndashndash48
983090 = 0
q +q2
mdash8
= 0
8q + q2 = 0
q(8 + q) = 0
q = 0 or q = ndash8
When q = 0 p =02
ndashndashndash48
= 0
When q = ndash8 p =(ndash8)2
ndashndashndashndashndash48
=64
ndashndashndash48
=4mdash3
Therefore the values of p =
4
mdash3 and q = ndash8
8 (2 ndash 3k ) x2 + x +3mdash4k = 0
b2 ndash 4ac = 12 ndash 4(2 ndash 3k )983089 3mdash4k 983090
= 1 ndash 6k + 9k 2
= 9k 2 ndash 6k + 1
= (3k ndash 1)2
Since (3k ndash 1)2 983102 0 for all values of k
therefore (2 ndash 3k ) x2 + x +3mdash4k = 0 has real roots for
all values of k
9 f ( x) = 3( x2 + 2mx + m2 + n)
= 3[( x + m)2 + n]
= 3( x + m)2 + 3n
The minimum point is (ndashm 3n)
Compare to A(t 3t 2)
there4 ndashm = t and 3n = 3t 2
m = ndasht n = t 2
10 (a) y = px2 + 8 x + 10 ndash p
When the graph does not intercept the x-axis
there are no roots for px2 + 8 x + 10 ndash p = 0 Therefore b2 ndash 4ac 983084 0
82 ndash 4 p(10 ndash p) 983084 0
64 ndash 40 p + 4 p2 983084 0
p2 ndash 10 p + 16 983084 0
( p ndash 2)( p ndash 8) 983084 0
2 8
Hence r = 2 and t = 8
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(b) When p = 2
y = 2 x2 + 8 x + 8
= 2( x2 + 4 x) + 8
= 2( x2 + 4 x + 22 ndash 22) + 8
= 2[( x + 2)2 ndash 4] + 8
= 2( x + 2)2 ndash 8 + 8
= 2( x + 2)2
Therefore the minimum point is (ndash2 0)
When x = 0 y = 8
When y = 0 2( x + 2)2 = 0
x = ndash2
When p = 8
y = 8 x2 + 8 x + 2
= 8( x2 + x) + 2
= 8983091 x2 + x + 983089 1mdash2 983090
2
ndash 983089 1mdash2 983090
2
983092 + 2
= 8983091983089 x +1mdash2 983090
2
ndash1mdash4 983092 + 2
= 8983089 x +1mdash2 983090
2
ndash 2 + 2
= 8983089 x +1mdash2 983090
2
Therefore the minimum point is (ndash1mdash2
0)
When x = 0 y = 2
When y = 0 0 = 8983089 x + 1mdash2 983090
2
x = ndash1mdash2
0
2
8
1 ndash ndash
2
ndash2
y p = 2 p = 8
x
11 (a) f ( x) = 24 x ndash 4 x2 + r
= ndash 4 x2 + 24 x + r
= ndash 4( x2 ndash 6 x) + r
= ndash 4( x2 ndash 6 x + 32 ndash 32) + r
= ndash 4[( x ndash 3)2 ndash 9] + r
= ndash 4( x ndash 3)2 + 36 + r
Compare to f ( x) = p( x ndash q)2 + 16
Therefore p = ndash 4 q = 3 and 36 + r = 16
r = ndash20
(b) The turning point is (3 16)
(c) f ( x) = 24 x ndash 4 x2 ndash 20
When x = 0 f ( x) = ndash20
When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0
4( x ndash 3)2 = 16
( x ndash 3)2 = 4
x ndash 3 = plusmn2
x = plusmn2 + 3
= ndash2 + 3 or 2 + 3
= 1 or 5
0 1
(3 16)
5
ndash20
y
x
12 (a) y = ndash| p( x ndash 3)2 + q|
Substitute x = 3 y = ndash5 into the equation
ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5
Substitute x = 4 y = 0 into the equation
0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0
p = 5
Therefore p = 5 q = ndash5 or p = ndash5 q = 5
(b) When x = 3 y = ndash5
For p = 5 q = ndash5
When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40
Based on the graph the range of values of y is
ndash 40 983100 y 983100 0
For p = ndash5 q = 5
When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40
Therefore the range of values of y is
ndash 40 983100 y 983100 0
13 (a) y = ndash2( x ndash 3)2 + 2k
= ndash x2 + 2 x + px ndash 8
= ndash x2 + (2 + p) x ndash 8
= ndash[ x2 ndash (2 + p) x] ndash 8
= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash
2 9830902
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983091983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
+(2 + p)2
ndashndashndashndashndashndashndash4
ndash 8
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Since the x-coordinate of the maximum point for
both the graphs are same
therefore2 + p
ndashndashndashndashndash2
= 3
p = 4
y = ndash x2 + 2 x + px ndash 8 becomes
y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8
When y = 0
ndash x2 + 6 x ndash 8 = 0
x2 ndash 6 x + 8 = 0
( x ndash 2)( x ndash 4) = 0
x = 2 or 4
Hence A(2 0) and B(4 0)
Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k
0 = ndash2(2 ndash 3)2 + 2k
2k = 2
k = 1 Hence k = 1 and p = 4
(b) For y = ndash2( x ndash 3)2 + 2k
= ndash2( x ndash 3)2 + 2(1)
= ndash2( x ndash 3)2 + 2
Maximum value of the curve is 2
For y = ndash x2 + 2 x + px ndash 8
= ndash x2 + 2 x + 4 x ndash 8
= ndash x2 + 6 x ndash 8
When x = 3
y = ndash9 + 18 ndash 8
= 1
Maximum value of the curve is 1
14 Since 3 x2 983102 0 for all values of x
therefore3 x2
ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)
983100 0
(2 x ndash 1)( x + 4) 983100 0
0 ndash4 1 ndash
2
f (x )
x
Hence ndash 4983100 x 983100 1mdash2
15 Since x2 + 1 983086 0
therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash
x2 + 1 983086 0
x2 + 3 x + 2 983086 0
( x + 1)( x + 2) 983086 0
0 ndash2 ndash1
f (x )
x
Hence x 983084 ndash2 x gt ndash1
16 ndash4
ndashndashndashndashndashndash1 ndash 3 x
983100 x
0 983100 x +4
ndashndashndashndashndashndash1 ndash 3 x
0 983100 x(1 ndash 3 x) + 4
ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0983100
x ndash 3 x2 + 4
ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash
1 ndash 3 x
0 983100 (ndash3 x + 4)( x + 1)
ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
For ndash3 x + 4 983102 0
4 983102 3 x
x 983100 4mdash3
For x + 1 983102 0
x 983102 ndash1
For 1 ndash 3 x 983086 0
ndash3 x 983086 ndash1
x 983084 1mdash3
ndash1 ndash + ndash +
1 4 ndash ndash
3
4x ndash
3
1x lt ndash
3
x ndash1
3
x
Therefore the range is ndash1 983100 x 983084 1mdash3
x 983102 4mdash3
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17 y2 ndash 9 = x
x = y2 ndash 9
When x = 0
y2 = 9
y = plusmn3
When y = 0
x = ndash9
When x = 7
7 = y2 ndash 9
y2 = 16
y = plusmn4
x ndash9 0 7
y 0 plusmn3 plusmn4
0 ndash9 7
ndash4 ndash3
34
y
x
The range of values of y is ndash 4 983100 y 983100 4
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18 Given that x ndash 2 y = 1
there4 x = 1 + 2 y 983089
Substitute 983089 into y + 3983102 2 xy
y + 3983102 2(1 + 2 y) y
y + 3983102 2 y + 4 y2
0983102 4 y2 + y ndash 3
0983102 (4 y ndash 3)( y + 1)
3 ndash
4
0 y
f (y )
ndash1
The range is ndash1 983100 y 983100 3mdash4
19 f ( x) 983084 0
5 x2 ndash 4 x ndash 1 983084 0
(5 x + 1)( x ndash 1) 983084 0
1 ndash ndash
5
0 x
f (x )
1
The range is ndash1mdash5
983084 x 983084 1
20 g( x) 983086 0
4 x2 ndash 9 983086 0(2 x + 3)(2 x ndash 3) 983086 0
3 ndash ndash
23 ndash
2
0 x
g (x )
The range is x 983084 ndash3mdash2
or x 983086 3mdash2
21 (a) Since y = 3 x
2
ndash 9 x + t 983086
0 for all values of x andit does not have root when y = 0
Then b2 ndash 4ac 983084 0 for 3 x2 ndash 9 x + t = 0
(ndash9)2 ndash 4(3)(t ) 983084 0
81 ndash 12t 983084 0
ndash12t 983084 ndash81
t 983086 ndash81 ndashndashndashndash ndash12
t 983086 27
ndashndashndash4
(b) Let f ( x) = a( x ndash b)2 + c
f ( x) = a( x ndash 2)2 + 0
f ( x) = a( x ndash 2)2
Substitute x = 0 f ( x) = ndash3 into the equation
ndash3 = a(0 ndash 2)2
= 4a
a = ndash3mdash4
Hence the quadratic function is
f ( x) = ndash3mdash4
( x ndash 2)2
22 (a) Given 2 x2 ndash 3 y + 2 = 0
3 y = 2 x2 + 2
y =2 x2
ndashndashndash3
+2mdash3
983089
Substitute 983089 into y 983084 10
2 x2
ndashndashndash
3
+2mdash
3
983084 10
2 x2 + 2 983084 30
2 x2 ndash 28 983084 0
x2 ndash 14 983084 0
( x + 98310598310698310614 )( x ndash 98310598310698310614 ) 983084 0
14 ndash14
0 x
f (x )
The range is ndash98310598310698310614 983084 x 983084 98310598310698310614
(b) 2 x2 ndash 8 x ndash 10 = 2( x2 ndash 4 x) ndash 10
= 2( x2 ndash 4 x + 22 ndash 22) ndash 10
= 2[( x ndash 2)2 ndash 4] ndash 10
= 2( x ndash 2)2 ndash 8 ndash 10
= 2( x ndash 2)2 ndash 18
Therefore a = 2 b = ndash2 and c = ndash18
Hence the minimum value of 2 x2 ndash 8 x ndash 10 is
ndash18
23 5 ndash 2 x 983100 0 3 x2 ndash 4 x 983086 ndash1
3 x2
ndash 4 x + 1983086
0(3 x ndash 1)( x ndash 1) 983086 0
11 ndash
3
0 x
f (x )
x 983084 1mdash3
x 983086 1
ndash2 x 983100
ndash5 x 983102
ndash5 ndashndashndash ndash2
x 983102 5mdash2
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5ndash
2
1ndash
3
1
x 5mdash2
x x 11mdash3
The range is x 983102 5mdash2
24 5 983084 f ( x) 983084 9
5 983084 5 ndash 3 x + x2 983084 9
5 983084 5 ndash 3 x + x2 5 ndash 3 x + x2 983084 9
5 ndash 3 x + x2 ndash 9 983084 0
x2 ndash 3 x ndash 4 983084 0
( x ndash 4)( x + 1) 983084 0
4 ndash1 0 x
f (x )
ndash1 983084 x 983084 4
0 983084 x2 ndash 3 x
0 983084 x( x ndash 3)
30 x
f (x )
x 983084 0 x 983086 3
0 ndash1 43
x lt 0 x gt 3
ndash1 lt x lt 4
The range is ndash1983084
x 983084
0 or 3983084
x 983084
4
25 1 983102 x2 + 3 x ndash 3 983086 ndash3
x2 + 3 x ndash 3 983086 ndash3
x2 + 3 x 983086 0
x( x + 3) 983086 0
ndash3 0 x
f (x )
x 983084 ndash3 x 983086 0
1 983102 x2 + 3 x ndash 3
0 983102 x2 + 3 x ndash 4
0 983102 ( x + 4)( x ndash 1)
1 ndash4 0 x
f (x )
ndash 4 983100 x 983100 1
ndash4
x lt ndash3 x gt 0
ndash4 x 1
ndash3 0 1
The range is ndash 4983100 x 983084 ndash3 or 0 983084 x 983100 1
26 x2 + 5 x ndash 10 983086 4
x2 + 5 x ndash 10 983084 ndash 4 x2 + 5 x ndash 10 983086 4
x2 + 5 x ndash 14 983086 0
( x + 7)( x ndash 2) 983086 0
ndash7 20
x
f (x )
x 983084 ndash7 x 983086 2
x2 + 5 x ndash 6 983084 0
( x + 6)( x ndash 1) 983084 0
ndash6 10
x
f (x )
ndash6 983084 x 983084 1
The ranges are ndash6 lt x lt 1 x lt ndash7 x gt 2
27 f ( x) = (r + 1) x2 + 2rx + r ndash 3
Given that f ( x) does not intersect the x-axis
therefore b2 ndash 4ac 983084 0
(2r)2 ndash 4(r + 1)(r ndash 3) 983084 0
4r2 ndash 4(r2 ndash 2r ndash 3) 983084 0
4r2 ndash 4r2 + 8r + 12 983084 0
2r + 3 983084 0
r 983084 ndash3mdash2
28 Given that f ( x) = 9 ndash 6 x + 2 x2 does not have real root
when f ( x) = k
there4 9 ndash 6 x + 2 x2 = k
2 x2 ndash 6 x + 9 ndash k = 0
Use b2 ndash 4ac 983084 0
(ndash 6)2 ndash 4(2)(9 ndash k ) 983084 0
36 ndash 72 + 8k 983084 0
ndash36 + 8k 983084 0 8k 983084 36
k 983084 36
ndashndashndash8
k 983084 9mdash2
29 2 x2 + 10 x ndash 20 983100 8
ndash8 983100 2 x2 + 10 x ndash 20 983100 8
ndash8983100 2 x2 + 10 x ndash 20 2 x2 + 10 x ndash 20 983100 8
2 x2 + 10 x ndash 28 983100 0
x2 + 5 x ndash 14 983100 0
( x + 7)( x ndash 2) 983100 0
ndash7 20
x
f (x )
ndash7 983100 x 983100 2
0983100 2 x2 + 10 x ndash 12
0983100 x2 + 5 x ndash 6
0983100 ( x + 6)( x ndash 1)
ndash6 10
x
f (x )
x 983100 ndash6 x 983102 1
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x ndash6 x 1
ndash7 x 2
ndash6 ndash7 1 2
The range is ndash7 983100 x 983100 ndash6 or 1 983100 x 983100 2
30 y = x2 + 5 x ndash 6
= x2 + 5 x + 983089 5mdash2 983090
2
ndash 983089 5mdash2 983090
2
ndash 6
= 983089 x + 5mdash2 983090
2
ndash25
ndashndashndash4
ndash 6
= 983089 x + 5mdash2 983090
2
ndash49
ndashndashndash4
The minimum point is (ndash5mdash2
ndash49
ndashndashndash4
)
x ndash13 ndash 6 0 1 3
y 98 0 ndash 6 0 18
x 0
ndash6 ndash13 ndash6
18
98
1 3
) ) ndash ndash ndash ndash
5
2
49
4
y
The range is ndash49
ndashndashndash4
983084 y 983084 98
31 y = ndash x2 + 2 x ndash nx + 16
= ndash x2 + (2 ndash n) x + 16
= ndash[ x2 ndash (2 ndash n) x] + 16
= ndash983091 x2 ndash (2 ndash n) x + 983089 2 ndash n ndashndashndashndashndash
2 983090
2
ndash 983089 2 ndash n ndashndashndashndashndash
2 983090
2
983092 + 16
= ndash
983091983089 x ndash
2 ndash n ndashndashndashndashndash
2 983090
2
ndash
9830892 ndash n
ndashndashndashndashndash2 983090
2
983092 + 16
= ndash983089 x ndash 2 ndash n ndashndashndashndashndash
2 983090
2
+ 983089 2 ndash n ndashndashndashndashndash
2 983090
2
+ 16
Since y = x2 ndash 3k and y = ndash x2 + 2 x ndash nx + 16 have the
same axis of symmetry that is x = 0
then ndash2 ndash n
ndashndashndashndashndash2
= 0
2 ndash n = 0
n = 2
The equation y = ndash x2 + 2 x ndash nx + 16
= ndash x2 + 2 x ndash 2 x + 16
= ndash983089 x ndash 2 ndash 2 ndashndashndashndashndash
2 983090
2
+ 983089 2 ndash 2 ndashndashndashndashndash
2 983090
2
+ 16
= ndash x2 + 16
When y = 0
ndash x2
+ 16 = 0 x2 = 16
x = plusmn98310598310698310616
= plusmn4
Therefore B = (4 0)
Substitute x = 4 y = 0 into y = x2 ndash 3k
0 = 42 ndash 3k
0 = 16 ndash 3k
k =16
ndashndashndash3
32 (a) y = ndash2[(3k ndash x)2 + n] ndash 10 has a maximum point
(4 11)
y = ndash2(3k ndash x)2 ndash 2n ndash 10
there4 3k ndash 4 = 0 and ndash2n ndash 10 = 11
k =4mdash3
ndash2n = 21
n = ndash21
ndashndashndash2
(b) Substitute k =4mdash3
and n = ndash21
ndashndashndash2
into
y = ndash2[(3k ndash x)2 + n] ndash 10
y = ndash2983091(4 ndash x)2 ndash21
ndashndashndash2 983092 ndash 10
= ndash2(4 ndash x)2 + 21 ndash 10
= ndash2(4 ndash x)2
+ 11 When y = 0
ndash2(4 ndash x)2 + 11= 0
2(4 ndash x)2 = 11
(4 ndash x)2 =11
ndashndashndash2
4 ndash x = plusmn98310598310698310698310611 ndashndashndash
2
x = 4 plusmn 98310598310698310698310611 ndashndashndash
2
= 4 ndash 98310598310698310698310611 ndashndashndash
2 4 + 98310598310698310698310611
ndashndashndash2
= 1655 6345
(c) y = ndash2(4 ndash x)2 + 11
The maximum point is (4 11)
x ndash1 0 5
y ndash39 ndash21 9
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1 y = 3(2 x ndash 1)( x + 1) ndash x(4 x ndash 5) + 2
= 3(2 x2 + x ndash 1) ndash 4 x2 + 5 x + 2
= 6 x2 + 3 x ndash 3 ndash 4 x2 + 5 x + 2
= 2 x2 + 8 x ndash 1
= 2( x2 + 4 x) ndash 1
= 2( x2 + 4 x + 22 ndash 22) ndash 1
= 2[( x + 2)2 ndash 4] ndash 1
= 2( x + 2)2 ndash 8 ndash 1
= 2( x + 2)2 ndash 9
Since a = 2 983086 0 therefore the minimum value of
y is ndash9
When y = 0 2( x + 2)2 ndash 9 = 0
( x + 2)2 =9mdash2
x + 2 = plusmn9831059831069831069mdash2 x = plusmn9831059831069831069mdash2 ndash 2
= 9831059831069831069mdash2 ndash 2 or ndash9831059831069831069mdash2 ndash 2
= 01213 or ndash 4121
When x = 0 y = 2(2)2 ndash 9
= ndash1
The minimum point is (ndash2 ndash9)
ndash1 ndash4121
(ndash2 ndash9)
012130
y
x
2 5 983084 Area of rectangle ABCD 983084 21
5 983084 ( x + 3)( x ndash 1) 983084 21
5 983084 ( x + 3)( x ndash 1)
5 983084 x2 + 2 x ndash 3
0 983084 x2 + 2 x ndash 8
0 983084 ( x + 4)( x ndash 2)
0 ndash4 2
f (x )
x
x 983084 ndash 4 x gt 2
( x + 3)( x ndash 1) 983084 21
x2 + 2 x ndash 3 983084 21
x2 + 2 x ndash 24 983084 0
( x ndash 4)( x + 6) 983084 0
0 ndash6 4
f (x )
x
ndash 6 983084 x 983084 4
ndash4
ndash6 lt x lt 4
x lt ndash4 x gt 2
2
x
ndash6 4
The range is ndash 6983084 x 983084 ndash 4 or 2 983084 x 983084 4
3 (a) p =1 + 5
ndashndashndashndashndash2
= 3
(b) y = ( x ndash 1)( x ndash 5) = x2 ndash 6 x + 5
4 y = a( x ndash 2)2 + 1
Substitute x = 0 y = 9 into the equation
9 = a(ndash2)2 + 1
8 = 4a
a = 2
Therefore the quadratic function is
f ( x) = 2( x ndash 2)2 + 1
5 x2
+ (1 + k ) x ndash k 2
+ 1 = 0For quadratic equation to have real roots
b2 ndash 4ac 983102 0
(1 + k )2 ndash 4(1)(1 ndash k 2) 983102 0
1 + 2k + k 2 ndash 4 + 4k 2 983102 0
5k 2 + 2k ndash 3 983102 0
(5k ndash 3)(k + 1) 983102 0
0 ndash1 3 ndash
5
f (k )
k
The range of values of k is k 983100 ndash1 or k 983102 3mdash5
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6 y = x2 + 7 x ndash 8 ndash 2k
For y to be positive for all real values of x there is
no roots for y = 0
Therefore b2 ndash 4ac 983084 0
72 ndash 4(1)(ndash8 ndash 2k ) 983084 0
49 + 32 + 8k 983084 0
8k 983084 ndash81
k 983084 ndash81
ndashndashndash8
Alternative
y = x2 + 7 x ndash 8 ndash 2k
= x2 + 7 x + 983089 7mdash2 983090
2
ndash 983089 7mdash2 983090
2
ndash 8 ndash 2k
= 983089 x +7mdash2 983090
2
ndash49
ndashndashndash4
ndash 8 ndash 2k
For y to be positive for all real values of x
ndash49
ndashndashndash4
ndash 8 ndash 2k 983086 0
ndash2k 983086 49
ndashndashndash4
+ 8
ndash2k 983086 81
ndashndashndash4
k 983084 ndash81
ndashndashndash8
7 Substitute x = 6 y = 0 into y = px2 + qx
0 = p(6)2 + q(6)
0 = 36 p + 6q
q + 6 p = 0 983089
y = px2 + qx
= p983089 x2 +q mdash p x983090
= p983091 x2 +q mdash p x + 983089
q ndashndashndash2 p 983090
2
ndash 983089q
ndashndashndash2 p 983090
2
983092
= p983091983089 x +q
ndashndashndash2 p 983090
2
ndashq2
ndashndashndashndash4 p2 983092
= p983089 x +q
ndashndashndash2 p 983090
2
ndashq2
ndashndashndash4 p
ndash q2
ndashndashndash4 p
= ndash12
q2 = 48 p
p =q2
ndashndashndash48
983090
Substitute 983090 into 983089
q + 6983089 q2
ndashndashndash48
983090 = 0
q +q2
mdash8
= 0
8q + q2 = 0
q(8 + q) = 0
q = 0 or q = ndash8
When q = 0 p =02
ndashndashndash48
= 0
When q = ndash8 p =(ndash8)2
ndashndashndashndashndash48
=64
ndashndashndash48
=4mdash3
Therefore the values of p =
4
mdash3 and q = ndash8
8 (2 ndash 3k ) x2 + x +3mdash4k = 0
b2 ndash 4ac = 12 ndash 4(2 ndash 3k )983089 3mdash4k 983090
= 1 ndash 6k + 9k 2
= 9k 2 ndash 6k + 1
= (3k ndash 1)2
Since (3k ndash 1)2 983102 0 for all values of k
therefore (2 ndash 3k ) x2 + x +3mdash4k = 0 has real roots for
all values of k
9 f ( x) = 3( x2 + 2mx + m2 + n)
= 3[( x + m)2 + n]
= 3( x + m)2 + 3n
The minimum point is (ndashm 3n)
Compare to A(t 3t 2)
there4 ndashm = t and 3n = 3t 2
m = ndasht n = t 2
10 (a) y = px2 + 8 x + 10 ndash p
When the graph does not intercept the x-axis
there are no roots for px2 + 8 x + 10 ndash p = 0 Therefore b2 ndash 4ac 983084 0
82 ndash 4 p(10 ndash p) 983084 0
64 ndash 40 p + 4 p2 983084 0
p2 ndash 10 p + 16 983084 0
( p ndash 2)( p ndash 8) 983084 0
2 8
Hence r = 2 and t = 8
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(b) When p = 2
y = 2 x2 + 8 x + 8
= 2( x2 + 4 x) + 8
= 2( x2 + 4 x + 22 ndash 22) + 8
= 2[( x + 2)2 ndash 4] + 8
= 2( x + 2)2 ndash 8 + 8
= 2( x + 2)2
Therefore the minimum point is (ndash2 0)
When x = 0 y = 8
When y = 0 2( x + 2)2 = 0
x = ndash2
When p = 8
y = 8 x2 + 8 x + 2
= 8( x2 + x) + 2
= 8983091 x2 + x + 983089 1mdash2 983090
2
ndash 983089 1mdash2 983090
2
983092 + 2
= 8983091983089 x +1mdash2 983090
2
ndash1mdash4 983092 + 2
= 8983089 x +1mdash2 983090
2
ndash 2 + 2
= 8983089 x +1mdash2 983090
2
Therefore the minimum point is (ndash1mdash2
0)
When x = 0 y = 2
When y = 0 0 = 8983089 x + 1mdash2 983090
2
x = ndash1mdash2
0
2
8
1 ndash ndash
2
ndash2
y p = 2 p = 8
x
11 (a) f ( x) = 24 x ndash 4 x2 + r
= ndash 4 x2 + 24 x + r
= ndash 4( x2 ndash 6 x) + r
= ndash 4( x2 ndash 6 x + 32 ndash 32) + r
= ndash 4[( x ndash 3)2 ndash 9] + r
= ndash 4( x ndash 3)2 + 36 + r
Compare to f ( x) = p( x ndash q)2 + 16
Therefore p = ndash 4 q = 3 and 36 + r = 16
r = ndash20
(b) The turning point is (3 16)
(c) f ( x) = 24 x ndash 4 x2 ndash 20
When x = 0 f ( x) = ndash20
When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0
4( x ndash 3)2 = 16
( x ndash 3)2 = 4
x ndash 3 = plusmn2
x = plusmn2 + 3
= ndash2 + 3 or 2 + 3
= 1 or 5
0 1
(3 16)
5
ndash20
y
x
12 (a) y = ndash| p( x ndash 3)2 + q|
Substitute x = 3 y = ndash5 into the equation
ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5
Substitute x = 4 y = 0 into the equation
0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0
p = 5
Therefore p = 5 q = ndash5 or p = ndash5 q = 5
(b) When x = 3 y = ndash5
For p = 5 q = ndash5
When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40
Based on the graph the range of values of y is
ndash 40 983100 y 983100 0
For p = ndash5 q = 5
When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40
Therefore the range of values of y is
ndash 40 983100 y 983100 0
13 (a) y = ndash2( x ndash 3)2 + 2k
= ndash x2 + 2 x + px ndash 8
= ndash x2 + (2 + p) x ndash 8
= ndash[ x2 ndash (2 + p) x] ndash 8
= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash
2 9830902
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983091983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
+(2 + p)2
ndashndashndashndashndashndashndash4
ndash 8
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Since the x-coordinate of the maximum point for
both the graphs are same
therefore2 + p
ndashndashndashndashndash2
= 3
p = 4
y = ndash x2 + 2 x + px ndash 8 becomes
y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8
When y = 0
ndash x2 + 6 x ndash 8 = 0
x2 ndash 6 x + 8 = 0
( x ndash 2)( x ndash 4) = 0
x = 2 or 4
Hence A(2 0) and B(4 0)
Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k
0 = ndash2(2 ndash 3)2 + 2k
2k = 2
k = 1 Hence k = 1 and p = 4
(b) For y = ndash2( x ndash 3)2 + 2k
= ndash2( x ndash 3)2 + 2(1)
= ndash2( x ndash 3)2 + 2
Maximum value of the curve is 2
For y = ndash x2 + 2 x + px ndash 8
= ndash x2 + 2 x + 4 x ndash 8
= ndash x2 + 6 x ndash 8
When x = 3
y = ndash9 + 18 ndash 8
= 1
Maximum value of the curve is 1
14 Since 3 x2 983102 0 for all values of x
therefore3 x2
ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)
983100 0
(2 x ndash 1)( x + 4) 983100 0
0 ndash4 1 ndash
2
f (x )
x
Hence ndash 4983100 x 983100 1mdash2
15 Since x2 + 1 983086 0
therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash
x2 + 1 983086 0
x2 + 3 x + 2 983086 0
( x + 1)( x + 2) 983086 0
0 ndash2 ndash1
f (x )
x
Hence x 983084 ndash2 x gt ndash1
16 ndash4
ndashndashndashndashndashndash1 ndash 3 x
983100 x
0 983100 x +4
ndashndashndashndashndashndash1 ndash 3 x
0 983100 x(1 ndash 3 x) + 4
ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0983100
x ndash 3 x2 + 4
ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash
1 ndash 3 x
0 983100 (ndash3 x + 4)( x + 1)
ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
For ndash3 x + 4 983102 0
4 983102 3 x
x 983100 4mdash3
For x + 1 983102 0
x 983102 ndash1
For 1 ndash 3 x 983086 0
ndash3 x 983086 ndash1
x 983084 1mdash3
ndash1 ndash + ndash +
1 4 ndash ndash
3
4x ndash
3
1x lt ndash
3
x ndash1
3
x
Therefore the range is ndash1 983100 x 983084 1mdash3
x 983102 4mdash3
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17 y2 ndash 9 = x
x = y2 ndash 9
When x = 0
y2 = 9
y = plusmn3
When y = 0
x = ndash9
When x = 7
7 = y2 ndash 9
y2 = 16
y = plusmn4
x ndash9 0 7
y 0 plusmn3 plusmn4
0 ndash9 7
ndash4 ndash3
34
y
x
The range of values of y is ndash 4 983100 y 983100 4
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5ndash
2
1ndash
3
1
x 5mdash2
x x 11mdash3
The range is x 983102 5mdash2
24 5 983084 f ( x) 983084 9
5 983084 5 ndash 3 x + x2 983084 9
5 983084 5 ndash 3 x + x2 5 ndash 3 x + x2 983084 9
5 ndash 3 x + x2 ndash 9 983084 0
x2 ndash 3 x ndash 4 983084 0
( x ndash 4)( x + 1) 983084 0
4 ndash1 0 x
f (x )
ndash1 983084 x 983084 4
0 983084 x2 ndash 3 x
0 983084 x( x ndash 3)
30 x
f (x )
x 983084 0 x 983086 3
0 ndash1 43
x lt 0 x gt 3
ndash1 lt x lt 4
The range is ndash1983084
x 983084
0 or 3983084
x 983084
4
25 1 983102 x2 + 3 x ndash 3 983086 ndash3
x2 + 3 x ndash 3 983086 ndash3
x2 + 3 x 983086 0
x( x + 3) 983086 0
ndash3 0 x
f (x )
x 983084 ndash3 x 983086 0
1 983102 x2 + 3 x ndash 3
0 983102 x2 + 3 x ndash 4
0 983102 ( x + 4)( x ndash 1)
1 ndash4 0 x
f (x )
ndash 4 983100 x 983100 1
ndash4
x lt ndash3 x gt 0
ndash4 x 1
ndash3 0 1
The range is ndash 4983100 x 983084 ndash3 or 0 983084 x 983100 1
26 x2 + 5 x ndash 10 983086 4
x2 + 5 x ndash 10 983084 ndash 4 x2 + 5 x ndash 10 983086 4
x2 + 5 x ndash 14 983086 0
( x + 7)( x ndash 2) 983086 0
ndash7 20
x
f (x )
x 983084 ndash7 x 983086 2
x2 + 5 x ndash 6 983084 0
( x + 6)( x ndash 1) 983084 0
ndash6 10
x
f (x )
ndash6 983084 x 983084 1
The ranges are ndash6 lt x lt 1 x lt ndash7 x gt 2
27 f ( x) = (r + 1) x2 + 2rx + r ndash 3
Given that f ( x) does not intersect the x-axis
therefore b2 ndash 4ac 983084 0
(2r)2 ndash 4(r + 1)(r ndash 3) 983084 0
4r2 ndash 4(r2 ndash 2r ndash 3) 983084 0
4r2 ndash 4r2 + 8r + 12 983084 0
2r + 3 983084 0
r 983084 ndash3mdash2
28 Given that f ( x) = 9 ndash 6 x + 2 x2 does not have real root
when f ( x) = k
there4 9 ndash 6 x + 2 x2 = k
2 x2 ndash 6 x + 9 ndash k = 0
Use b2 ndash 4ac 983084 0
(ndash 6)2 ndash 4(2)(9 ndash k ) 983084 0
36 ndash 72 + 8k 983084 0
ndash36 + 8k 983084 0 8k 983084 36
k 983084 36
ndashndashndash8
k 983084 9mdash2
29 2 x2 + 10 x ndash 20 983100 8
ndash8 983100 2 x2 + 10 x ndash 20 983100 8
ndash8983100 2 x2 + 10 x ndash 20 2 x2 + 10 x ndash 20 983100 8
2 x2 + 10 x ndash 28 983100 0
x2 + 5 x ndash 14 983100 0
( x + 7)( x ndash 2) 983100 0
ndash7 20
x
f (x )
ndash7 983100 x 983100 2
0983100 2 x2 + 10 x ndash 12
0983100 x2 + 5 x ndash 6
0983100 ( x + 6)( x ndash 1)
ndash6 10
x
f (x )
x 983100 ndash6 x 983102 1
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x ndash6 x 1
ndash7 x 2
ndash6 ndash7 1 2
The range is ndash7 983100 x 983100 ndash6 or 1 983100 x 983100 2
30 y = x2 + 5 x ndash 6
= x2 + 5 x + 983089 5mdash2 983090
2
ndash 983089 5mdash2 983090
2
ndash 6
= 983089 x + 5mdash2 983090
2
ndash25
ndashndashndash4
ndash 6
= 983089 x + 5mdash2 983090
2
ndash49
ndashndashndash4
The minimum point is (ndash5mdash2
ndash49
ndashndashndash4
)
x ndash13 ndash 6 0 1 3
y 98 0 ndash 6 0 18
x 0
ndash6 ndash13 ndash6
18
98
1 3
) ) ndash ndash ndash ndash
5
2
49
4
y
The range is ndash49
ndashndashndash4
983084 y 983084 98
31 y = ndash x2 + 2 x ndash nx + 16
= ndash x2 + (2 ndash n) x + 16
= ndash[ x2 ndash (2 ndash n) x] + 16
= ndash983091 x2 ndash (2 ndash n) x + 983089 2 ndash n ndashndashndashndashndash
2 983090
2
ndash 983089 2 ndash n ndashndashndashndashndash
2 983090
2
983092 + 16
= ndash
983091983089 x ndash
2 ndash n ndashndashndashndashndash
2 983090
2
ndash
9830892 ndash n
ndashndashndashndashndash2 983090
2
983092 + 16
= ndash983089 x ndash 2 ndash n ndashndashndashndashndash
2 983090
2
+ 983089 2 ndash n ndashndashndashndashndash
2 983090
2
+ 16
Since y = x2 ndash 3k and y = ndash x2 + 2 x ndash nx + 16 have the
same axis of symmetry that is x = 0
then ndash2 ndash n
ndashndashndashndashndash2
= 0
2 ndash n = 0
n = 2
The equation y = ndash x2 + 2 x ndash nx + 16
= ndash x2 + 2 x ndash 2 x + 16
= ndash983089 x ndash 2 ndash 2 ndashndashndashndashndash
2 983090
2
+ 983089 2 ndash 2 ndashndashndashndashndash
2 983090
2
+ 16
= ndash x2 + 16
When y = 0
ndash x2
+ 16 = 0 x2 = 16
x = plusmn98310598310698310616
= plusmn4
Therefore B = (4 0)
Substitute x = 4 y = 0 into y = x2 ndash 3k
0 = 42 ndash 3k
0 = 16 ndash 3k
k =16
ndashndashndash3
32 (a) y = ndash2[(3k ndash x)2 + n] ndash 10 has a maximum point
(4 11)
y = ndash2(3k ndash x)2 ndash 2n ndash 10
there4 3k ndash 4 = 0 and ndash2n ndash 10 = 11
k =4mdash3
ndash2n = 21
n = ndash21
ndashndashndash2
(b) Substitute k =4mdash3
and n = ndash21
ndashndashndash2
into
y = ndash2[(3k ndash x)2 + n] ndash 10
y = ndash2983091(4 ndash x)2 ndash21
ndashndashndash2 983092 ndash 10
= ndash2(4 ndash x)2 + 21 ndash 10
= ndash2(4 ndash x)2
+ 11 When y = 0
ndash2(4 ndash x)2 + 11= 0
2(4 ndash x)2 = 11
(4 ndash x)2 =11
ndashndashndash2
4 ndash x = plusmn98310598310698310698310611 ndashndashndash
2
x = 4 plusmn 98310598310698310698310611 ndashndashndash
2
= 4 ndash 98310598310698310698310611 ndashndashndash
2 4 + 98310598310698310698310611
ndashndashndash2
= 1655 6345
(c) y = ndash2(4 ndash x)2 + 11
The maximum point is (4 11)
x ndash1 0 5
y ndash39 ndash21 9
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1 y = 3(2 x ndash 1)( x + 1) ndash x(4 x ndash 5) + 2
= 3(2 x2 + x ndash 1) ndash 4 x2 + 5 x + 2
= 6 x2 + 3 x ndash 3 ndash 4 x2 + 5 x + 2
= 2 x2 + 8 x ndash 1
= 2( x2 + 4 x) ndash 1
= 2( x2 + 4 x + 22 ndash 22) ndash 1
= 2[( x + 2)2 ndash 4] ndash 1
= 2( x + 2)2 ndash 8 ndash 1
= 2( x + 2)2 ndash 9
Since a = 2 983086 0 therefore the minimum value of
y is ndash9
When y = 0 2( x + 2)2 ndash 9 = 0
( x + 2)2 =9mdash2
x + 2 = plusmn9831059831069831069mdash2 x = plusmn9831059831069831069mdash2 ndash 2
= 9831059831069831069mdash2 ndash 2 or ndash9831059831069831069mdash2 ndash 2
= 01213 or ndash 4121
When x = 0 y = 2(2)2 ndash 9
= ndash1
The minimum point is (ndash2 ndash9)
ndash1 ndash4121
(ndash2 ndash9)
012130
y
x
2 5 983084 Area of rectangle ABCD 983084 21
5 983084 ( x + 3)( x ndash 1) 983084 21
5 983084 ( x + 3)( x ndash 1)
5 983084 x2 + 2 x ndash 3
0 983084 x2 + 2 x ndash 8
0 983084 ( x + 4)( x ndash 2)
0 ndash4 2
f (x )
x
x 983084 ndash 4 x gt 2
( x + 3)( x ndash 1) 983084 21
x2 + 2 x ndash 3 983084 21
x2 + 2 x ndash 24 983084 0
( x ndash 4)( x + 6) 983084 0
0 ndash6 4
f (x )
x
ndash 6 983084 x 983084 4
ndash4
ndash6 lt x lt 4
x lt ndash4 x gt 2
2
x
ndash6 4
The range is ndash 6983084 x 983084 ndash 4 or 2 983084 x 983084 4
3 (a) p =1 + 5
ndashndashndashndashndash2
= 3
(b) y = ( x ndash 1)( x ndash 5) = x2 ndash 6 x + 5
4 y = a( x ndash 2)2 + 1
Substitute x = 0 y = 9 into the equation
9 = a(ndash2)2 + 1
8 = 4a
a = 2
Therefore the quadratic function is
f ( x) = 2( x ndash 2)2 + 1
5 x2
+ (1 + k ) x ndash k 2
+ 1 = 0For quadratic equation to have real roots
b2 ndash 4ac 983102 0
(1 + k )2 ndash 4(1)(1 ndash k 2) 983102 0
1 + 2k + k 2 ndash 4 + 4k 2 983102 0
5k 2 + 2k ndash 3 983102 0
(5k ndash 3)(k + 1) 983102 0
0 ndash1 3 ndash
5
f (k )
k
The range of values of k is k 983100 ndash1 or k 983102 3mdash5
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6 y = x2 + 7 x ndash 8 ndash 2k
For y to be positive for all real values of x there is
no roots for y = 0
Therefore b2 ndash 4ac 983084 0
72 ndash 4(1)(ndash8 ndash 2k ) 983084 0
49 + 32 + 8k 983084 0
8k 983084 ndash81
k 983084 ndash81
ndashndashndash8
Alternative
y = x2 + 7 x ndash 8 ndash 2k
= x2 + 7 x + 983089 7mdash2 983090
2
ndash 983089 7mdash2 983090
2
ndash 8 ndash 2k
= 983089 x +7mdash2 983090
2
ndash49
ndashndashndash4
ndash 8 ndash 2k
For y to be positive for all real values of x
ndash49
ndashndashndash4
ndash 8 ndash 2k 983086 0
ndash2k 983086 49
ndashndashndash4
+ 8
ndash2k 983086 81
ndashndashndash4
k 983084 ndash81
ndashndashndash8
7 Substitute x = 6 y = 0 into y = px2 + qx
0 = p(6)2 + q(6)
0 = 36 p + 6q
q + 6 p = 0 983089
y = px2 + qx
= p983089 x2 +q mdash p x983090
= p983091 x2 +q mdash p x + 983089
q ndashndashndash2 p 983090
2
ndash 983089q
ndashndashndash2 p 983090
2
983092
= p983091983089 x +q
ndashndashndash2 p 983090
2
ndashq2
ndashndashndashndash4 p2 983092
= p983089 x +q
ndashndashndash2 p 983090
2
ndashq2
ndashndashndash4 p
ndash q2
ndashndashndash4 p
= ndash12
q2 = 48 p
p =q2
ndashndashndash48
983090
Substitute 983090 into 983089
q + 6983089 q2
ndashndashndash48
983090 = 0
q +q2
mdash8
= 0
8q + q2 = 0
q(8 + q) = 0
q = 0 or q = ndash8
When q = 0 p =02
ndashndashndash48
= 0
When q = ndash8 p =(ndash8)2
ndashndashndashndashndash48
=64
ndashndashndash48
=4mdash3
Therefore the values of p =
4
mdash3 and q = ndash8
8 (2 ndash 3k ) x2 + x +3mdash4k = 0
b2 ndash 4ac = 12 ndash 4(2 ndash 3k )983089 3mdash4k 983090
= 1 ndash 6k + 9k 2
= 9k 2 ndash 6k + 1
= (3k ndash 1)2
Since (3k ndash 1)2 983102 0 for all values of k
therefore (2 ndash 3k ) x2 + x +3mdash4k = 0 has real roots for
all values of k
9 f ( x) = 3( x2 + 2mx + m2 + n)
= 3[( x + m)2 + n]
= 3( x + m)2 + 3n
The minimum point is (ndashm 3n)
Compare to A(t 3t 2)
there4 ndashm = t and 3n = 3t 2
m = ndasht n = t 2
10 (a) y = px2 + 8 x + 10 ndash p
When the graph does not intercept the x-axis
there are no roots for px2 + 8 x + 10 ndash p = 0 Therefore b2 ndash 4ac 983084 0
82 ndash 4 p(10 ndash p) 983084 0
64 ndash 40 p + 4 p2 983084 0
p2 ndash 10 p + 16 983084 0
( p ndash 2)( p ndash 8) 983084 0
2 8
Hence r = 2 and t = 8
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(b) When p = 2
y = 2 x2 + 8 x + 8
= 2( x2 + 4 x) + 8
= 2( x2 + 4 x + 22 ndash 22) + 8
= 2[( x + 2)2 ndash 4] + 8
= 2( x + 2)2 ndash 8 + 8
= 2( x + 2)2
Therefore the minimum point is (ndash2 0)
When x = 0 y = 8
When y = 0 2( x + 2)2 = 0
x = ndash2
When p = 8
y = 8 x2 + 8 x + 2
= 8( x2 + x) + 2
= 8983091 x2 + x + 983089 1mdash2 983090
2
ndash 983089 1mdash2 983090
2
983092 + 2
= 8983091983089 x +1mdash2 983090
2
ndash1mdash4 983092 + 2
= 8983089 x +1mdash2 983090
2
ndash 2 + 2
= 8983089 x +1mdash2 983090
2
Therefore the minimum point is (ndash1mdash2
0)
When x = 0 y = 2
When y = 0 0 = 8983089 x + 1mdash2 983090
2
x = ndash1mdash2
0
2
8
1 ndash ndash
2
ndash2
y p = 2 p = 8
x
11 (a) f ( x) = 24 x ndash 4 x2 + r
= ndash 4 x2 + 24 x + r
= ndash 4( x2 ndash 6 x) + r
= ndash 4( x2 ndash 6 x + 32 ndash 32) + r
= ndash 4[( x ndash 3)2 ndash 9] + r
= ndash 4( x ndash 3)2 + 36 + r
Compare to f ( x) = p( x ndash q)2 + 16
Therefore p = ndash 4 q = 3 and 36 + r = 16
r = ndash20
(b) The turning point is (3 16)
(c) f ( x) = 24 x ndash 4 x2 ndash 20
When x = 0 f ( x) = ndash20
When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0
4( x ndash 3)2 = 16
( x ndash 3)2 = 4
x ndash 3 = plusmn2
x = plusmn2 + 3
= ndash2 + 3 or 2 + 3
= 1 or 5
0 1
(3 16)
5
ndash20
y
x
12 (a) y = ndash| p( x ndash 3)2 + q|
Substitute x = 3 y = ndash5 into the equation
ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5
Substitute x = 4 y = 0 into the equation
0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0
p = 5
Therefore p = 5 q = ndash5 or p = ndash5 q = 5
(b) When x = 3 y = ndash5
For p = 5 q = ndash5
When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40
Based on the graph the range of values of y is
ndash 40 983100 y 983100 0
For p = ndash5 q = 5
When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40
Therefore the range of values of y is
ndash 40 983100 y 983100 0
13 (a) y = ndash2( x ndash 3)2 + 2k
= ndash x2 + 2 x + px ndash 8
= ndash x2 + (2 + p) x ndash 8
= ndash[ x2 ndash (2 + p) x] ndash 8
= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash
2 9830902
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983091983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
+(2 + p)2
ndashndashndashndashndashndashndash4
ndash 8
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Since the x-coordinate of the maximum point for
both the graphs are same
therefore2 + p
ndashndashndashndashndash2
= 3
p = 4
y = ndash x2 + 2 x + px ndash 8 becomes
y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8
When y = 0
ndash x2 + 6 x ndash 8 = 0
x2 ndash 6 x + 8 = 0
( x ndash 2)( x ndash 4) = 0
x = 2 or 4
Hence A(2 0) and B(4 0)
Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k
0 = ndash2(2 ndash 3)2 + 2k
2k = 2
k = 1 Hence k = 1 and p = 4
(b) For y = ndash2( x ndash 3)2 + 2k
= ndash2( x ndash 3)2 + 2(1)
= ndash2( x ndash 3)2 + 2
Maximum value of the curve is 2
For y = ndash x2 + 2 x + px ndash 8
= ndash x2 + 2 x + 4 x ndash 8
= ndash x2 + 6 x ndash 8
When x = 3
y = ndash9 + 18 ndash 8
= 1
Maximum value of the curve is 1
14 Since 3 x2 983102 0 for all values of x
therefore3 x2
ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)
983100 0
(2 x ndash 1)( x + 4) 983100 0
0 ndash4 1 ndash
2
f (x )
x
Hence ndash 4983100 x 983100 1mdash2
15 Since x2 + 1 983086 0
therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash
x2 + 1 983086 0
x2 + 3 x + 2 983086 0
( x + 1)( x + 2) 983086 0
0 ndash2 ndash1
f (x )
x
Hence x 983084 ndash2 x gt ndash1
16 ndash4
ndashndashndashndashndashndash1 ndash 3 x
983100 x
0 983100 x +4
ndashndashndashndashndashndash1 ndash 3 x
0 983100 x(1 ndash 3 x) + 4
ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0983100
x ndash 3 x2 + 4
ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash
1 ndash 3 x
0 983100 (ndash3 x + 4)( x + 1)
ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
For ndash3 x + 4 983102 0
4 983102 3 x
x 983100 4mdash3
For x + 1 983102 0
x 983102 ndash1
For 1 ndash 3 x 983086 0
ndash3 x 983086 ndash1
x 983084 1mdash3
ndash1 ndash + ndash +
1 4 ndash ndash
3
4x ndash
3
1x lt ndash
3
x ndash1
3
x
Therefore the range is ndash1 983100 x 983084 1mdash3
x 983102 4mdash3
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17 y2 ndash 9 = x
x = y2 ndash 9
When x = 0
y2 = 9
y = plusmn3
When y = 0
x = ndash9
When x = 7
7 = y2 ndash 9
y2 = 16
y = plusmn4
x ndash9 0 7
y 0 plusmn3 plusmn4
0 ndash9 7
ndash4 ndash3
34
y
x
The range of values of y is ndash 4 983100 y 983100 4
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x ndash6 x 1
ndash7 x 2
ndash6 ndash7 1 2
The range is ndash7 983100 x 983100 ndash6 or 1 983100 x 983100 2
30 y = x2 + 5 x ndash 6
= x2 + 5 x + 983089 5mdash2 983090
2
ndash 983089 5mdash2 983090
2
ndash 6
= 983089 x + 5mdash2 983090
2
ndash25
ndashndashndash4
ndash 6
= 983089 x + 5mdash2 983090
2
ndash49
ndashndashndash4
The minimum point is (ndash5mdash2
ndash49
ndashndashndash4
)
x ndash13 ndash 6 0 1 3
y 98 0 ndash 6 0 18
x 0
ndash6 ndash13 ndash6
18
98
1 3
) ) ndash ndash ndash ndash
5
2
49
4
y
The range is ndash49
ndashndashndash4
983084 y 983084 98
31 y = ndash x2 + 2 x ndash nx + 16
= ndash x2 + (2 ndash n) x + 16
= ndash[ x2 ndash (2 ndash n) x] + 16
= ndash983091 x2 ndash (2 ndash n) x + 983089 2 ndash n ndashndashndashndashndash
2 983090
2
ndash 983089 2 ndash n ndashndashndashndashndash
2 983090
2
983092 + 16
= ndash
983091983089 x ndash
2 ndash n ndashndashndashndashndash
2 983090
2
ndash
9830892 ndash n
ndashndashndashndashndash2 983090
2
983092 + 16
= ndash983089 x ndash 2 ndash n ndashndashndashndashndash
2 983090
2
+ 983089 2 ndash n ndashndashndashndashndash
2 983090
2
+ 16
Since y = x2 ndash 3k and y = ndash x2 + 2 x ndash nx + 16 have the
same axis of symmetry that is x = 0
then ndash2 ndash n
ndashndashndashndashndash2
= 0
2 ndash n = 0
n = 2
The equation y = ndash x2 + 2 x ndash nx + 16
= ndash x2 + 2 x ndash 2 x + 16
= ndash983089 x ndash 2 ndash 2 ndashndashndashndashndash
2 983090
2
+ 983089 2 ndash 2 ndashndashndashndashndash
2 983090
2
+ 16
= ndash x2 + 16
When y = 0
ndash x2
+ 16 = 0 x2 = 16
x = plusmn98310598310698310616
= plusmn4
Therefore B = (4 0)
Substitute x = 4 y = 0 into y = x2 ndash 3k
0 = 42 ndash 3k
0 = 16 ndash 3k
k =16
ndashndashndash3
32 (a) y = ndash2[(3k ndash x)2 + n] ndash 10 has a maximum point
(4 11)
y = ndash2(3k ndash x)2 ndash 2n ndash 10
there4 3k ndash 4 = 0 and ndash2n ndash 10 = 11
k =4mdash3
ndash2n = 21
n = ndash21
ndashndashndash2
(b) Substitute k =4mdash3
and n = ndash21
ndashndashndash2
into
y = ndash2[(3k ndash x)2 + n] ndash 10
y = ndash2983091(4 ndash x)2 ndash21
ndashndashndash2 983092 ndash 10
= ndash2(4 ndash x)2 + 21 ndash 10
= ndash2(4 ndash x)2
+ 11 When y = 0
ndash2(4 ndash x)2 + 11= 0
2(4 ndash x)2 = 11
(4 ndash x)2 =11
ndashndashndash2
4 ndash x = plusmn98310598310698310698310611 ndashndashndash
2
x = 4 plusmn 98310598310698310698310611 ndashndashndash
2
= 4 ndash 98310598310698310698310611 ndashndashndash
2 4 + 98310598310698310698310611
ndashndashndash2
= 1655 6345
(c) y = ndash2(4 ndash x)2 + 11
The maximum point is (4 11)
x ndash1 0 5
y ndash39 ndash21 9
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1 y = 3(2 x ndash 1)( x + 1) ndash x(4 x ndash 5) + 2
= 3(2 x2 + x ndash 1) ndash 4 x2 + 5 x + 2
= 6 x2 + 3 x ndash 3 ndash 4 x2 + 5 x + 2
= 2 x2 + 8 x ndash 1
= 2( x2 + 4 x) ndash 1
= 2( x2 + 4 x + 22 ndash 22) ndash 1
= 2[( x + 2)2 ndash 4] ndash 1
= 2( x + 2)2 ndash 8 ndash 1
= 2( x + 2)2 ndash 9
Since a = 2 983086 0 therefore the minimum value of
y is ndash9
When y = 0 2( x + 2)2 ndash 9 = 0
( x + 2)2 =9mdash2
x + 2 = plusmn9831059831069831069mdash2 x = plusmn9831059831069831069mdash2 ndash 2
= 9831059831069831069mdash2 ndash 2 or ndash9831059831069831069mdash2 ndash 2
= 01213 or ndash 4121
When x = 0 y = 2(2)2 ndash 9
= ndash1
The minimum point is (ndash2 ndash9)
ndash1 ndash4121
(ndash2 ndash9)
012130
y
x
2 5 983084 Area of rectangle ABCD 983084 21
5 983084 ( x + 3)( x ndash 1) 983084 21
5 983084 ( x + 3)( x ndash 1)
5 983084 x2 + 2 x ndash 3
0 983084 x2 + 2 x ndash 8
0 983084 ( x + 4)( x ndash 2)
0 ndash4 2
f (x )
x
x 983084 ndash 4 x gt 2
( x + 3)( x ndash 1) 983084 21
x2 + 2 x ndash 3 983084 21
x2 + 2 x ndash 24 983084 0
( x ndash 4)( x + 6) 983084 0
0 ndash6 4
f (x )
x
ndash 6 983084 x 983084 4
ndash4
ndash6 lt x lt 4
x lt ndash4 x gt 2
2
x
ndash6 4
The range is ndash 6983084 x 983084 ndash 4 or 2 983084 x 983084 4
3 (a) p =1 + 5
ndashndashndashndashndash2
= 3
(b) y = ( x ndash 1)( x ndash 5) = x2 ndash 6 x + 5
4 y = a( x ndash 2)2 + 1
Substitute x = 0 y = 9 into the equation
9 = a(ndash2)2 + 1
8 = 4a
a = 2
Therefore the quadratic function is
f ( x) = 2( x ndash 2)2 + 1
5 x2
+ (1 + k ) x ndash k 2
+ 1 = 0For quadratic equation to have real roots
b2 ndash 4ac 983102 0
(1 + k )2 ndash 4(1)(1 ndash k 2) 983102 0
1 + 2k + k 2 ndash 4 + 4k 2 983102 0
5k 2 + 2k ndash 3 983102 0
(5k ndash 3)(k + 1) 983102 0
0 ndash1 3 ndash
5
f (k )
k
The range of values of k is k 983100 ndash1 or k 983102 3mdash5
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6 y = x2 + 7 x ndash 8 ndash 2k
For y to be positive for all real values of x there is
no roots for y = 0
Therefore b2 ndash 4ac 983084 0
72 ndash 4(1)(ndash8 ndash 2k ) 983084 0
49 + 32 + 8k 983084 0
8k 983084 ndash81
k 983084 ndash81
ndashndashndash8
Alternative
y = x2 + 7 x ndash 8 ndash 2k
= x2 + 7 x + 983089 7mdash2 983090
2
ndash 983089 7mdash2 983090
2
ndash 8 ndash 2k
= 983089 x +7mdash2 983090
2
ndash49
ndashndashndash4
ndash 8 ndash 2k
For y to be positive for all real values of x
ndash49
ndashndashndash4
ndash 8 ndash 2k 983086 0
ndash2k 983086 49
ndashndashndash4
+ 8
ndash2k 983086 81
ndashndashndash4
k 983084 ndash81
ndashndashndash8
7 Substitute x = 6 y = 0 into y = px2 + qx
0 = p(6)2 + q(6)
0 = 36 p + 6q
q + 6 p = 0 983089
y = px2 + qx
= p983089 x2 +q mdash p x983090
= p983091 x2 +q mdash p x + 983089
q ndashndashndash2 p 983090
2
ndash 983089q
ndashndashndash2 p 983090
2
983092
= p983091983089 x +q
ndashndashndash2 p 983090
2
ndashq2
ndashndashndashndash4 p2 983092
= p983089 x +q
ndashndashndash2 p 983090
2
ndashq2
ndashndashndash4 p
ndash q2
ndashndashndash4 p
= ndash12
q2 = 48 p
p =q2
ndashndashndash48
983090
Substitute 983090 into 983089
q + 6983089 q2
ndashndashndash48
983090 = 0
q +q2
mdash8
= 0
8q + q2 = 0
q(8 + q) = 0
q = 0 or q = ndash8
When q = 0 p =02
ndashndashndash48
= 0
When q = ndash8 p =(ndash8)2
ndashndashndashndashndash48
=64
ndashndashndash48
=4mdash3
Therefore the values of p =
4
mdash3 and q = ndash8
8 (2 ndash 3k ) x2 + x +3mdash4k = 0
b2 ndash 4ac = 12 ndash 4(2 ndash 3k )983089 3mdash4k 983090
= 1 ndash 6k + 9k 2
= 9k 2 ndash 6k + 1
= (3k ndash 1)2
Since (3k ndash 1)2 983102 0 for all values of k
therefore (2 ndash 3k ) x2 + x +3mdash4k = 0 has real roots for
all values of k
9 f ( x) = 3( x2 + 2mx + m2 + n)
= 3[( x + m)2 + n]
= 3( x + m)2 + 3n
The minimum point is (ndashm 3n)
Compare to A(t 3t 2)
there4 ndashm = t and 3n = 3t 2
m = ndasht n = t 2
10 (a) y = px2 + 8 x + 10 ndash p
When the graph does not intercept the x-axis
there are no roots for px2 + 8 x + 10 ndash p = 0 Therefore b2 ndash 4ac 983084 0
82 ndash 4 p(10 ndash p) 983084 0
64 ndash 40 p + 4 p2 983084 0
p2 ndash 10 p + 16 983084 0
( p ndash 2)( p ndash 8) 983084 0
2 8
Hence r = 2 and t = 8
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(b) When p = 2
y = 2 x2 + 8 x + 8
= 2( x2 + 4 x) + 8
= 2( x2 + 4 x + 22 ndash 22) + 8
= 2[( x + 2)2 ndash 4] + 8
= 2( x + 2)2 ndash 8 + 8
= 2( x + 2)2
Therefore the minimum point is (ndash2 0)
When x = 0 y = 8
When y = 0 2( x + 2)2 = 0
x = ndash2
When p = 8
y = 8 x2 + 8 x + 2
= 8( x2 + x) + 2
= 8983091 x2 + x + 983089 1mdash2 983090
2
ndash 983089 1mdash2 983090
2
983092 + 2
= 8983091983089 x +1mdash2 983090
2
ndash1mdash4 983092 + 2
= 8983089 x +1mdash2 983090
2
ndash 2 + 2
= 8983089 x +1mdash2 983090
2
Therefore the minimum point is (ndash1mdash2
0)
When x = 0 y = 2
When y = 0 0 = 8983089 x + 1mdash2 983090
2
x = ndash1mdash2
0
2
8
1 ndash ndash
2
ndash2
y p = 2 p = 8
x
11 (a) f ( x) = 24 x ndash 4 x2 + r
= ndash 4 x2 + 24 x + r
= ndash 4( x2 ndash 6 x) + r
= ndash 4( x2 ndash 6 x + 32 ndash 32) + r
= ndash 4[( x ndash 3)2 ndash 9] + r
= ndash 4( x ndash 3)2 + 36 + r
Compare to f ( x) = p( x ndash q)2 + 16
Therefore p = ndash 4 q = 3 and 36 + r = 16
r = ndash20
(b) The turning point is (3 16)
(c) f ( x) = 24 x ndash 4 x2 ndash 20
When x = 0 f ( x) = ndash20
When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0
4( x ndash 3)2 = 16
( x ndash 3)2 = 4
x ndash 3 = plusmn2
x = plusmn2 + 3
= ndash2 + 3 or 2 + 3
= 1 or 5
0 1
(3 16)
5
ndash20
y
x
12 (a) y = ndash| p( x ndash 3)2 + q|
Substitute x = 3 y = ndash5 into the equation
ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5
Substitute x = 4 y = 0 into the equation
0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0
p = 5
Therefore p = 5 q = ndash5 or p = ndash5 q = 5
(b) When x = 3 y = ndash5
For p = 5 q = ndash5
When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40
Based on the graph the range of values of y is
ndash 40 983100 y 983100 0
For p = ndash5 q = 5
When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40
Therefore the range of values of y is
ndash 40 983100 y 983100 0
13 (a) y = ndash2( x ndash 3)2 + 2k
= ndash x2 + 2 x + px ndash 8
= ndash x2 + (2 + p) x ndash 8
= ndash[ x2 ndash (2 + p) x] ndash 8
= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash
2 9830902
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983091983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
+(2 + p)2
ndashndashndashndashndashndashndash4
ndash 8
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Since the x-coordinate of the maximum point for
both the graphs are same
therefore2 + p
ndashndashndashndashndash2
= 3
p = 4
y = ndash x2 + 2 x + px ndash 8 becomes
y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8
When y = 0
ndash x2 + 6 x ndash 8 = 0
x2 ndash 6 x + 8 = 0
( x ndash 2)( x ndash 4) = 0
x = 2 or 4
Hence A(2 0) and B(4 0)
Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k
0 = ndash2(2 ndash 3)2 + 2k
2k = 2
k = 1 Hence k = 1 and p = 4
(b) For y = ndash2( x ndash 3)2 + 2k
= ndash2( x ndash 3)2 + 2(1)
= ndash2( x ndash 3)2 + 2
Maximum value of the curve is 2
For y = ndash x2 + 2 x + px ndash 8
= ndash x2 + 2 x + 4 x ndash 8
= ndash x2 + 6 x ndash 8
When x = 3
y = ndash9 + 18 ndash 8
= 1
Maximum value of the curve is 1
14 Since 3 x2 983102 0 for all values of x
therefore3 x2
ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)
983100 0
(2 x ndash 1)( x + 4) 983100 0
0 ndash4 1 ndash
2
f (x )
x
Hence ndash 4983100 x 983100 1mdash2
15 Since x2 + 1 983086 0
therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash
x2 + 1 983086 0
x2 + 3 x + 2 983086 0
( x + 1)( x + 2) 983086 0
0 ndash2 ndash1
f (x )
x
Hence x 983084 ndash2 x gt ndash1
16 ndash4
ndashndashndashndashndashndash1 ndash 3 x
983100 x
0 983100 x +4
ndashndashndashndashndashndash1 ndash 3 x
0 983100 x(1 ndash 3 x) + 4
ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0983100
x ndash 3 x2 + 4
ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash
1 ndash 3 x
0 983100 (ndash3 x + 4)( x + 1)
ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
For ndash3 x + 4 983102 0
4 983102 3 x
x 983100 4mdash3
For x + 1 983102 0
x 983102 ndash1
For 1 ndash 3 x 983086 0
ndash3 x 983086 ndash1
x 983084 1mdash3
ndash1 ndash + ndash +
1 4 ndash ndash
3
4x ndash
3
1x lt ndash
3
x ndash1
3
x
Therefore the range is ndash1 983100 x 983084 1mdash3
x 983102 4mdash3
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17 y2 ndash 9 = x
x = y2 ndash 9
When x = 0
y2 = 9
y = plusmn3
When y = 0
x = ndash9
When x = 7
7 = y2 ndash 9
y2 = 16
y = plusmn4
x ndash9 0 7
y 0 plusmn3 plusmn4
0 ndash9 7
ndash4 ndash3
34
y
x
The range of values of y is ndash 4 983100 y 983100 4
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1 y = 3(2 x ndash 1)( x + 1) ndash x(4 x ndash 5) + 2
= 3(2 x2 + x ndash 1) ndash 4 x2 + 5 x + 2
= 6 x2 + 3 x ndash 3 ndash 4 x2 + 5 x + 2
= 2 x2 + 8 x ndash 1
= 2( x2 + 4 x) ndash 1
= 2( x2 + 4 x + 22 ndash 22) ndash 1
= 2[( x + 2)2 ndash 4] ndash 1
= 2( x + 2)2 ndash 8 ndash 1
= 2( x + 2)2 ndash 9
Since a = 2 983086 0 therefore the minimum value of
y is ndash9
When y = 0 2( x + 2)2 ndash 9 = 0
( x + 2)2 =9mdash2
x + 2 = plusmn9831059831069831069mdash2 x = plusmn9831059831069831069mdash2 ndash 2
= 9831059831069831069mdash2 ndash 2 or ndash9831059831069831069mdash2 ndash 2
= 01213 or ndash 4121
When x = 0 y = 2(2)2 ndash 9
= ndash1
The minimum point is (ndash2 ndash9)
ndash1 ndash4121
(ndash2 ndash9)
012130
y
x
2 5 983084 Area of rectangle ABCD 983084 21
5 983084 ( x + 3)( x ndash 1) 983084 21
5 983084 ( x + 3)( x ndash 1)
5 983084 x2 + 2 x ndash 3
0 983084 x2 + 2 x ndash 8
0 983084 ( x + 4)( x ndash 2)
0 ndash4 2
f (x )
x
x 983084 ndash 4 x gt 2
( x + 3)( x ndash 1) 983084 21
x2 + 2 x ndash 3 983084 21
x2 + 2 x ndash 24 983084 0
( x ndash 4)( x + 6) 983084 0
0 ndash6 4
f (x )
x
ndash 6 983084 x 983084 4
ndash4
ndash6 lt x lt 4
x lt ndash4 x gt 2
2
x
ndash6 4
The range is ndash 6983084 x 983084 ndash 4 or 2 983084 x 983084 4
3 (a) p =1 + 5
ndashndashndashndashndash2
= 3
(b) y = ( x ndash 1)( x ndash 5) = x2 ndash 6 x + 5
4 y = a( x ndash 2)2 + 1
Substitute x = 0 y = 9 into the equation
9 = a(ndash2)2 + 1
8 = 4a
a = 2
Therefore the quadratic function is
f ( x) = 2( x ndash 2)2 + 1
5 x2
+ (1 + k ) x ndash k 2
+ 1 = 0For quadratic equation to have real roots
b2 ndash 4ac 983102 0
(1 + k )2 ndash 4(1)(1 ndash k 2) 983102 0
1 + 2k + k 2 ndash 4 + 4k 2 983102 0
5k 2 + 2k ndash 3 983102 0
(5k ndash 3)(k + 1) 983102 0
0 ndash1 3 ndash
5
f (k )
k
The range of values of k is k 983100 ndash1 or k 983102 3mdash5
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6 y = x2 + 7 x ndash 8 ndash 2k
For y to be positive for all real values of x there is
no roots for y = 0
Therefore b2 ndash 4ac 983084 0
72 ndash 4(1)(ndash8 ndash 2k ) 983084 0
49 + 32 + 8k 983084 0
8k 983084 ndash81
k 983084 ndash81
ndashndashndash8
Alternative
y = x2 + 7 x ndash 8 ndash 2k
= x2 + 7 x + 983089 7mdash2 983090
2
ndash 983089 7mdash2 983090
2
ndash 8 ndash 2k
= 983089 x +7mdash2 983090
2
ndash49
ndashndashndash4
ndash 8 ndash 2k
For y to be positive for all real values of x
ndash49
ndashndashndash4
ndash 8 ndash 2k 983086 0
ndash2k 983086 49
ndashndashndash4
+ 8
ndash2k 983086 81
ndashndashndash4
k 983084 ndash81
ndashndashndash8
7 Substitute x = 6 y = 0 into y = px2 + qx
0 = p(6)2 + q(6)
0 = 36 p + 6q
q + 6 p = 0 983089
y = px2 + qx
= p983089 x2 +q mdash p x983090
= p983091 x2 +q mdash p x + 983089
q ndashndashndash2 p 983090
2
ndash 983089q
ndashndashndash2 p 983090
2
983092
= p983091983089 x +q
ndashndashndash2 p 983090
2
ndashq2
ndashndashndashndash4 p2 983092
= p983089 x +q
ndashndashndash2 p 983090
2
ndashq2
ndashndashndash4 p
ndash q2
ndashndashndash4 p
= ndash12
q2 = 48 p
p =q2
ndashndashndash48
983090
Substitute 983090 into 983089
q + 6983089 q2
ndashndashndash48
983090 = 0
q +q2
mdash8
= 0
8q + q2 = 0
q(8 + q) = 0
q = 0 or q = ndash8
When q = 0 p =02
ndashndashndash48
= 0
When q = ndash8 p =(ndash8)2
ndashndashndashndashndash48
=64
ndashndashndash48
=4mdash3
Therefore the values of p =
4
mdash3 and q = ndash8
8 (2 ndash 3k ) x2 + x +3mdash4k = 0
b2 ndash 4ac = 12 ndash 4(2 ndash 3k )983089 3mdash4k 983090
= 1 ndash 6k + 9k 2
= 9k 2 ndash 6k + 1
= (3k ndash 1)2
Since (3k ndash 1)2 983102 0 for all values of k
therefore (2 ndash 3k ) x2 + x +3mdash4k = 0 has real roots for
all values of k
9 f ( x) = 3( x2 + 2mx + m2 + n)
= 3[( x + m)2 + n]
= 3( x + m)2 + 3n
The minimum point is (ndashm 3n)
Compare to A(t 3t 2)
there4 ndashm = t and 3n = 3t 2
m = ndasht n = t 2
10 (a) y = px2 + 8 x + 10 ndash p
When the graph does not intercept the x-axis
there are no roots for px2 + 8 x + 10 ndash p = 0 Therefore b2 ndash 4ac 983084 0
82 ndash 4 p(10 ndash p) 983084 0
64 ndash 40 p + 4 p2 983084 0
p2 ndash 10 p + 16 983084 0
( p ndash 2)( p ndash 8) 983084 0
2 8
Hence r = 2 and t = 8
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(b) When p = 2
y = 2 x2 + 8 x + 8
= 2( x2 + 4 x) + 8
= 2( x2 + 4 x + 22 ndash 22) + 8
= 2[( x + 2)2 ndash 4] + 8
= 2( x + 2)2 ndash 8 + 8
= 2( x + 2)2
Therefore the minimum point is (ndash2 0)
When x = 0 y = 8
When y = 0 2( x + 2)2 = 0
x = ndash2
When p = 8
y = 8 x2 + 8 x + 2
= 8( x2 + x) + 2
= 8983091 x2 + x + 983089 1mdash2 983090
2
ndash 983089 1mdash2 983090
2
983092 + 2
= 8983091983089 x +1mdash2 983090
2
ndash1mdash4 983092 + 2
= 8983089 x +1mdash2 983090
2
ndash 2 + 2
= 8983089 x +1mdash2 983090
2
Therefore the minimum point is (ndash1mdash2
0)
When x = 0 y = 2
When y = 0 0 = 8983089 x + 1mdash2 983090
2
x = ndash1mdash2
0
2
8
1 ndash ndash
2
ndash2
y p = 2 p = 8
x
11 (a) f ( x) = 24 x ndash 4 x2 + r
= ndash 4 x2 + 24 x + r
= ndash 4( x2 ndash 6 x) + r
= ndash 4( x2 ndash 6 x + 32 ndash 32) + r
= ndash 4[( x ndash 3)2 ndash 9] + r
= ndash 4( x ndash 3)2 + 36 + r
Compare to f ( x) = p( x ndash q)2 + 16
Therefore p = ndash 4 q = 3 and 36 + r = 16
r = ndash20
(b) The turning point is (3 16)
(c) f ( x) = 24 x ndash 4 x2 ndash 20
When x = 0 f ( x) = ndash20
When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0
4( x ndash 3)2 = 16
( x ndash 3)2 = 4
x ndash 3 = plusmn2
x = plusmn2 + 3
= ndash2 + 3 or 2 + 3
= 1 or 5
0 1
(3 16)
5
ndash20
y
x
12 (a) y = ndash| p( x ndash 3)2 + q|
Substitute x = 3 y = ndash5 into the equation
ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5
Substitute x = 4 y = 0 into the equation
0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0
p = 5
Therefore p = 5 q = ndash5 or p = ndash5 q = 5
(b) When x = 3 y = ndash5
For p = 5 q = ndash5
When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40
Based on the graph the range of values of y is
ndash 40 983100 y 983100 0
For p = ndash5 q = 5
When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40
Therefore the range of values of y is
ndash 40 983100 y 983100 0
13 (a) y = ndash2( x ndash 3)2 + 2k
= ndash x2 + 2 x + px ndash 8
= ndash x2 + (2 + p) x ndash 8
= ndash[ x2 ndash (2 + p) x] ndash 8
= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash
2 9830902
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983091983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
+(2 + p)2
ndashndashndashndashndashndashndash4
ndash 8
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Since the x-coordinate of the maximum point for
both the graphs are same
therefore2 + p
ndashndashndashndashndash2
= 3
p = 4
y = ndash x2 + 2 x + px ndash 8 becomes
y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8
When y = 0
ndash x2 + 6 x ndash 8 = 0
x2 ndash 6 x + 8 = 0
( x ndash 2)( x ndash 4) = 0
x = 2 or 4
Hence A(2 0) and B(4 0)
Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k
0 = ndash2(2 ndash 3)2 + 2k
2k = 2
k = 1 Hence k = 1 and p = 4
(b) For y = ndash2( x ndash 3)2 + 2k
= ndash2( x ndash 3)2 + 2(1)
= ndash2( x ndash 3)2 + 2
Maximum value of the curve is 2
For y = ndash x2 + 2 x + px ndash 8
= ndash x2 + 2 x + 4 x ndash 8
= ndash x2 + 6 x ndash 8
When x = 3
y = ndash9 + 18 ndash 8
= 1
Maximum value of the curve is 1
14 Since 3 x2 983102 0 for all values of x
therefore3 x2
ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)
983100 0
(2 x ndash 1)( x + 4) 983100 0
0 ndash4 1 ndash
2
f (x )
x
Hence ndash 4983100 x 983100 1mdash2
15 Since x2 + 1 983086 0
therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash
x2 + 1 983086 0
x2 + 3 x + 2 983086 0
( x + 1)( x + 2) 983086 0
0 ndash2 ndash1
f (x )
x
Hence x 983084 ndash2 x gt ndash1
16 ndash4
ndashndashndashndashndashndash1 ndash 3 x
983100 x
0 983100 x +4
ndashndashndashndashndashndash1 ndash 3 x
0 983100 x(1 ndash 3 x) + 4
ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0983100
x ndash 3 x2 + 4
ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash
1 ndash 3 x
0 983100 (ndash3 x + 4)( x + 1)
ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
For ndash3 x + 4 983102 0
4 983102 3 x
x 983100 4mdash3
For x + 1 983102 0
x 983102 ndash1
For 1 ndash 3 x 983086 0
ndash3 x 983086 ndash1
x 983084 1mdash3
ndash1 ndash + ndash +
1 4 ndash ndash
3
4x ndash
3
1x lt ndash
3
x ndash1
3
x
Therefore the range is ndash1 983100 x 983084 1mdash3
x 983102 4mdash3
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17 y2 ndash 9 = x
x = y2 ndash 9
When x = 0
y2 = 9
y = plusmn3
When y = 0
x = ndash9
When x = 7
7 = y2 ndash 9
y2 = 16
y = plusmn4
x ndash9 0 7
y 0 plusmn3 plusmn4
0 ndash9 7
ndash4 ndash3
34
y
x
The range of values of y is ndash 4 983100 y 983100 4
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1 y = 3(2 x ndash 1)( x + 1) ndash x(4 x ndash 5) + 2
= 3(2 x2 + x ndash 1) ndash 4 x2 + 5 x + 2
= 6 x2 + 3 x ndash 3 ndash 4 x2 + 5 x + 2
= 2 x2 + 8 x ndash 1
= 2( x2 + 4 x) ndash 1
= 2( x2 + 4 x + 22 ndash 22) ndash 1
= 2[( x + 2)2 ndash 4] ndash 1
= 2( x + 2)2 ndash 8 ndash 1
= 2( x + 2)2 ndash 9
Since a = 2 983086 0 therefore the minimum value of
y is ndash9
When y = 0 2( x + 2)2 ndash 9 = 0
( x + 2)2 =9mdash2
x + 2 = plusmn9831059831069831069mdash2 x = plusmn9831059831069831069mdash2 ndash 2
= 9831059831069831069mdash2 ndash 2 or ndash9831059831069831069mdash2 ndash 2
= 01213 or ndash 4121
When x = 0 y = 2(2)2 ndash 9
= ndash1
The minimum point is (ndash2 ndash9)
ndash1 ndash4121
(ndash2 ndash9)
012130
y
x
2 5 983084 Area of rectangle ABCD 983084 21
5 983084 ( x + 3)( x ndash 1) 983084 21
5 983084 ( x + 3)( x ndash 1)
5 983084 x2 + 2 x ndash 3
0 983084 x2 + 2 x ndash 8
0 983084 ( x + 4)( x ndash 2)
0 ndash4 2
f (x )
x
x 983084 ndash 4 x gt 2
( x + 3)( x ndash 1) 983084 21
x2 + 2 x ndash 3 983084 21
x2 + 2 x ndash 24 983084 0
( x ndash 4)( x + 6) 983084 0
0 ndash6 4
f (x )
x
ndash 6 983084 x 983084 4
ndash4
ndash6 lt x lt 4
x lt ndash4 x gt 2
2
x
ndash6 4
The range is ndash 6983084 x 983084 ndash 4 or 2 983084 x 983084 4
3 (a) p =1 + 5
ndashndashndashndashndash2
= 3
(b) y = ( x ndash 1)( x ndash 5) = x2 ndash 6 x + 5
4 y = a( x ndash 2)2 + 1
Substitute x = 0 y = 9 into the equation
9 = a(ndash2)2 + 1
8 = 4a
a = 2
Therefore the quadratic function is
f ( x) = 2( x ndash 2)2 + 1
5 x2
+ (1 + k ) x ndash k 2
+ 1 = 0For quadratic equation to have real roots
b2 ndash 4ac 983102 0
(1 + k )2 ndash 4(1)(1 ndash k 2) 983102 0
1 + 2k + k 2 ndash 4 + 4k 2 983102 0
5k 2 + 2k ndash 3 983102 0
(5k ndash 3)(k + 1) 983102 0
0 ndash1 3 ndash
5
f (k )
k
The range of values of k is k 983100 ndash1 or k 983102 3mdash5
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6 y = x2 + 7 x ndash 8 ndash 2k
For y to be positive for all real values of x there is
no roots for y = 0
Therefore b2 ndash 4ac 983084 0
72 ndash 4(1)(ndash8 ndash 2k ) 983084 0
49 + 32 + 8k 983084 0
8k 983084 ndash81
k 983084 ndash81
ndashndashndash8
Alternative
y = x2 + 7 x ndash 8 ndash 2k
= x2 + 7 x + 983089 7mdash2 983090
2
ndash 983089 7mdash2 983090
2
ndash 8 ndash 2k
= 983089 x +7mdash2 983090
2
ndash49
ndashndashndash4
ndash 8 ndash 2k
For y to be positive for all real values of x
ndash49
ndashndashndash4
ndash 8 ndash 2k 983086 0
ndash2k 983086 49
ndashndashndash4
+ 8
ndash2k 983086 81
ndashndashndash4
k 983084 ndash81
ndashndashndash8
7 Substitute x = 6 y = 0 into y = px2 + qx
0 = p(6)2 + q(6)
0 = 36 p + 6q
q + 6 p = 0 983089
y = px2 + qx
= p983089 x2 +q mdash p x983090
= p983091 x2 +q mdash p x + 983089
q ndashndashndash2 p 983090
2
ndash 983089q
ndashndashndash2 p 983090
2
983092
= p983091983089 x +q
ndashndashndash2 p 983090
2
ndashq2
ndashndashndashndash4 p2 983092
= p983089 x +q
ndashndashndash2 p 983090
2
ndashq2
ndashndashndash4 p
ndash q2
ndashndashndash4 p
= ndash12
q2 = 48 p
p =q2
ndashndashndash48
983090
Substitute 983090 into 983089
q + 6983089 q2
ndashndashndash48
983090 = 0
q +q2
mdash8
= 0
8q + q2 = 0
q(8 + q) = 0
q = 0 or q = ndash8
When q = 0 p =02
ndashndashndash48
= 0
When q = ndash8 p =(ndash8)2
ndashndashndashndashndash48
=64
ndashndashndash48
=4mdash3
Therefore the values of p =
4
mdash3 and q = ndash8
8 (2 ndash 3k ) x2 + x +3mdash4k = 0
b2 ndash 4ac = 12 ndash 4(2 ndash 3k )983089 3mdash4k 983090
= 1 ndash 6k + 9k 2
= 9k 2 ndash 6k + 1
= (3k ndash 1)2
Since (3k ndash 1)2 983102 0 for all values of k
therefore (2 ndash 3k ) x2 + x +3mdash4k = 0 has real roots for
all values of k
9 f ( x) = 3( x2 + 2mx + m2 + n)
= 3[( x + m)2 + n]
= 3( x + m)2 + 3n
The minimum point is (ndashm 3n)
Compare to A(t 3t 2)
there4 ndashm = t and 3n = 3t 2
m = ndasht n = t 2
10 (a) y = px2 + 8 x + 10 ndash p
When the graph does not intercept the x-axis
there are no roots for px2 + 8 x + 10 ndash p = 0 Therefore b2 ndash 4ac 983084 0
82 ndash 4 p(10 ndash p) 983084 0
64 ndash 40 p + 4 p2 983084 0
p2 ndash 10 p + 16 983084 0
( p ndash 2)( p ndash 8) 983084 0
2 8
Hence r = 2 and t = 8
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(b) When p = 2
y = 2 x2 + 8 x + 8
= 2( x2 + 4 x) + 8
= 2( x2 + 4 x + 22 ndash 22) + 8
= 2[( x + 2)2 ndash 4] + 8
= 2( x + 2)2 ndash 8 + 8
= 2( x + 2)2
Therefore the minimum point is (ndash2 0)
When x = 0 y = 8
When y = 0 2( x + 2)2 = 0
x = ndash2
When p = 8
y = 8 x2 + 8 x + 2
= 8( x2 + x) + 2
= 8983091 x2 + x + 983089 1mdash2 983090
2
ndash 983089 1mdash2 983090
2
983092 + 2
= 8983091983089 x +1mdash2 983090
2
ndash1mdash4 983092 + 2
= 8983089 x +1mdash2 983090
2
ndash 2 + 2
= 8983089 x +1mdash2 983090
2
Therefore the minimum point is (ndash1mdash2
0)
When x = 0 y = 2
When y = 0 0 = 8983089 x + 1mdash2 983090
2
x = ndash1mdash2
0
2
8
1 ndash ndash
2
ndash2
y p = 2 p = 8
x
11 (a) f ( x) = 24 x ndash 4 x2 + r
= ndash 4 x2 + 24 x + r
= ndash 4( x2 ndash 6 x) + r
= ndash 4( x2 ndash 6 x + 32 ndash 32) + r
= ndash 4[( x ndash 3)2 ndash 9] + r
= ndash 4( x ndash 3)2 + 36 + r
Compare to f ( x) = p( x ndash q)2 + 16
Therefore p = ndash 4 q = 3 and 36 + r = 16
r = ndash20
(b) The turning point is (3 16)
(c) f ( x) = 24 x ndash 4 x2 ndash 20
When x = 0 f ( x) = ndash20
When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0
4( x ndash 3)2 = 16
( x ndash 3)2 = 4
x ndash 3 = plusmn2
x = plusmn2 + 3
= ndash2 + 3 or 2 + 3
= 1 or 5
0 1
(3 16)
5
ndash20
y
x
12 (a) y = ndash| p( x ndash 3)2 + q|
Substitute x = 3 y = ndash5 into the equation
ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5
Substitute x = 4 y = 0 into the equation
0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0
p = 5
Therefore p = 5 q = ndash5 or p = ndash5 q = 5
(b) When x = 3 y = ndash5
For p = 5 q = ndash5
When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40
Based on the graph the range of values of y is
ndash 40 983100 y 983100 0
For p = ndash5 q = 5
When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40
Therefore the range of values of y is
ndash 40 983100 y 983100 0
13 (a) y = ndash2( x ndash 3)2 + 2k
= ndash x2 + 2 x + px ndash 8
= ndash x2 + (2 + p) x ndash 8
= ndash[ x2 ndash (2 + p) x] ndash 8
= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash
2 9830902
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983091983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
+(2 + p)2
ndashndashndashndashndashndashndash4
ndash 8
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Since the x-coordinate of the maximum point for
both the graphs are same
therefore2 + p
ndashndashndashndashndash2
= 3
p = 4
y = ndash x2 + 2 x + px ndash 8 becomes
y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8
When y = 0
ndash x2 + 6 x ndash 8 = 0
x2 ndash 6 x + 8 = 0
( x ndash 2)( x ndash 4) = 0
x = 2 or 4
Hence A(2 0) and B(4 0)
Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k
0 = ndash2(2 ndash 3)2 + 2k
2k = 2
k = 1 Hence k = 1 and p = 4
(b) For y = ndash2( x ndash 3)2 + 2k
= ndash2( x ndash 3)2 + 2(1)
= ndash2( x ndash 3)2 + 2
Maximum value of the curve is 2
For y = ndash x2 + 2 x + px ndash 8
= ndash x2 + 2 x + 4 x ndash 8
= ndash x2 + 6 x ndash 8
When x = 3
y = ndash9 + 18 ndash 8
= 1
Maximum value of the curve is 1
14 Since 3 x2 983102 0 for all values of x
therefore3 x2
ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)
983100 0
(2 x ndash 1)( x + 4) 983100 0
0 ndash4 1 ndash
2
f (x )
x
Hence ndash 4983100 x 983100 1mdash2
15 Since x2 + 1 983086 0
therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash
x2 + 1 983086 0
x2 + 3 x + 2 983086 0
( x + 1)( x + 2) 983086 0
0 ndash2 ndash1
f (x )
x
Hence x 983084 ndash2 x gt ndash1
16 ndash4
ndashndashndashndashndashndash1 ndash 3 x
983100 x
0 983100 x +4
ndashndashndashndashndashndash1 ndash 3 x
0 983100 x(1 ndash 3 x) + 4
ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0983100
x ndash 3 x2 + 4
ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash
1 ndash 3 x
0 983100 (ndash3 x + 4)( x + 1)
ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
For ndash3 x + 4 983102 0
4 983102 3 x
x 983100 4mdash3
For x + 1 983102 0
x 983102 ndash1
For 1 ndash 3 x 983086 0
ndash3 x 983086 ndash1
x 983084 1mdash3
ndash1 ndash + ndash +
1 4 ndash ndash
3
4x ndash
3
1x lt ndash
3
x ndash1
3
x
Therefore the range is ndash1 983100 x 983084 1mdash3
x 983102 4mdash3
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17 y2 ndash 9 = x
x = y2 ndash 9
When x = 0
y2 = 9
y = plusmn3
When y = 0
x = ndash9
When x = 7
7 = y2 ndash 9
y2 = 16
y = plusmn4
x ndash9 0 7
y 0 plusmn3 plusmn4
0 ndash9 7
ndash4 ndash3
34
y
x
The range of values of y is ndash 4 983100 y 983100 4
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6 y = x2 + 7 x ndash 8 ndash 2k
For y to be positive for all real values of x there is
no roots for y = 0
Therefore b2 ndash 4ac 983084 0
72 ndash 4(1)(ndash8 ndash 2k ) 983084 0
49 + 32 + 8k 983084 0
8k 983084 ndash81
k 983084 ndash81
ndashndashndash8
Alternative
y = x2 + 7 x ndash 8 ndash 2k
= x2 + 7 x + 983089 7mdash2 983090
2
ndash 983089 7mdash2 983090
2
ndash 8 ndash 2k
= 983089 x +7mdash2 983090
2
ndash49
ndashndashndash4
ndash 8 ndash 2k
For y to be positive for all real values of x
ndash49
ndashndashndash4
ndash 8 ndash 2k 983086 0
ndash2k 983086 49
ndashndashndash4
+ 8
ndash2k 983086 81
ndashndashndash4
k 983084 ndash81
ndashndashndash8
7 Substitute x = 6 y = 0 into y = px2 + qx
0 = p(6)2 + q(6)
0 = 36 p + 6q
q + 6 p = 0 983089
y = px2 + qx
= p983089 x2 +q mdash p x983090
= p983091 x2 +q mdash p x + 983089
q ndashndashndash2 p 983090
2
ndash 983089q
ndashndashndash2 p 983090
2
983092
= p983091983089 x +q
ndashndashndash2 p 983090
2
ndashq2
ndashndashndashndash4 p2 983092
= p983089 x +q
ndashndashndash2 p 983090
2
ndashq2
ndashndashndash4 p
ndash q2
ndashndashndash4 p
= ndash12
q2 = 48 p
p =q2
ndashndashndash48
983090
Substitute 983090 into 983089
q + 6983089 q2
ndashndashndash48
983090 = 0
q +q2
mdash8
= 0
8q + q2 = 0
q(8 + q) = 0
q = 0 or q = ndash8
When q = 0 p =02
ndashndashndash48
= 0
When q = ndash8 p =(ndash8)2
ndashndashndashndashndash48
=64
ndashndashndash48
=4mdash3
Therefore the values of p =
4
mdash3 and q = ndash8
8 (2 ndash 3k ) x2 + x +3mdash4k = 0
b2 ndash 4ac = 12 ndash 4(2 ndash 3k )983089 3mdash4k 983090
= 1 ndash 6k + 9k 2
= 9k 2 ndash 6k + 1
= (3k ndash 1)2
Since (3k ndash 1)2 983102 0 for all values of k
therefore (2 ndash 3k ) x2 + x +3mdash4k = 0 has real roots for
all values of k
9 f ( x) = 3( x2 + 2mx + m2 + n)
= 3[( x + m)2 + n]
= 3( x + m)2 + 3n
The minimum point is (ndashm 3n)
Compare to A(t 3t 2)
there4 ndashm = t and 3n = 3t 2
m = ndasht n = t 2
10 (a) y = px2 + 8 x + 10 ndash p
When the graph does not intercept the x-axis
there are no roots for px2 + 8 x + 10 ndash p = 0 Therefore b2 ndash 4ac 983084 0
82 ndash 4 p(10 ndash p) 983084 0
64 ndash 40 p + 4 p2 983084 0
p2 ndash 10 p + 16 983084 0
( p ndash 2)( p ndash 8) 983084 0
2 8
Hence r = 2 and t = 8
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(b) When p = 2
y = 2 x2 + 8 x + 8
= 2( x2 + 4 x) + 8
= 2( x2 + 4 x + 22 ndash 22) + 8
= 2[( x + 2)2 ndash 4] + 8
= 2( x + 2)2 ndash 8 + 8
= 2( x + 2)2
Therefore the minimum point is (ndash2 0)
When x = 0 y = 8
When y = 0 2( x + 2)2 = 0
x = ndash2
When p = 8
y = 8 x2 + 8 x + 2
= 8( x2 + x) + 2
= 8983091 x2 + x + 983089 1mdash2 983090
2
ndash 983089 1mdash2 983090
2
983092 + 2
= 8983091983089 x +1mdash2 983090
2
ndash1mdash4 983092 + 2
= 8983089 x +1mdash2 983090
2
ndash 2 + 2
= 8983089 x +1mdash2 983090
2
Therefore the minimum point is (ndash1mdash2
0)
When x = 0 y = 2
When y = 0 0 = 8983089 x + 1mdash2 983090
2
x = ndash1mdash2
0
2
8
1 ndash ndash
2
ndash2
y p = 2 p = 8
x
11 (a) f ( x) = 24 x ndash 4 x2 + r
= ndash 4 x2 + 24 x + r
= ndash 4( x2 ndash 6 x) + r
= ndash 4( x2 ndash 6 x + 32 ndash 32) + r
= ndash 4[( x ndash 3)2 ndash 9] + r
= ndash 4( x ndash 3)2 + 36 + r
Compare to f ( x) = p( x ndash q)2 + 16
Therefore p = ndash 4 q = 3 and 36 + r = 16
r = ndash20
(b) The turning point is (3 16)
(c) f ( x) = 24 x ndash 4 x2 ndash 20
When x = 0 f ( x) = ndash20
When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0
4( x ndash 3)2 = 16
( x ndash 3)2 = 4
x ndash 3 = plusmn2
x = plusmn2 + 3
= ndash2 + 3 or 2 + 3
= 1 or 5
0 1
(3 16)
5
ndash20
y
x
12 (a) y = ndash| p( x ndash 3)2 + q|
Substitute x = 3 y = ndash5 into the equation
ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5
Substitute x = 4 y = 0 into the equation
0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0
p = 5
Therefore p = 5 q = ndash5 or p = ndash5 q = 5
(b) When x = 3 y = ndash5
For p = 5 q = ndash5
When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40
Based on the graph the range of values of y is
ndash 40 983100 y 983100 0
For p = ndash5 q = 5
When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40
Therefore the range of values of y is
ndash 40 983100 y 983100 0
13 (a) y = ndash2( x ndash 3)2 + 2k
= ndash x2 + 2 x + px ndash 8
= ndash x2 + (2 + p) x ndash 8
= ndash[ x2 ndash (2 + p) x] ndash 8
= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash
2 9830902
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983091983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
+(2 + p)2
ndashndashndashndashndashndashndash4
ndash 8
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Since the x-coordinate of the maximum point for
both the graphs are same
therefore2 + p
ndashndashndashndashndash2
= 3
p = 4
y = ndash x2 + 2 x + px ndash 8 becomes
y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8
When y = 0
ndash x2 + 6 x ndash 8 = 0
x2 ndash 6 x + 8 = 0
( x ndash 2)( x ndash 4) = 0
x = 2 or 4
Hence A(2 0) and B(4 0)
Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k
0 = ndash2(2 ndash 3)2 + 2k
2k = 2
k = 1 Hence k = 1 and p = 4
(b) For y = ndash2( x ndash 3)2 + 2k
= ndash2( x ndash 3)2 + 2(1)
= ndash2( x ndash 3)2 + 2
Maximum value of the curve is 2
For y = ndash x2 + 2 x + px ndash 8
= ndash x2 + 2 x + 4 x ndash 8
= ndash x2 + 6 x ndash 8
When x = 3
y = ndash9 + 18 ndash 8
= 1
Maximum value of the curve is 1
14 Since 3 x2 983102 0 for all values of x
therefore3 x2
ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)
983100 0
(2 x ndash 1)( x + 4) 983100 0
0 ndash4 1 ndash
2
f (x )
x
Hence ndash 4983100 x 983100 1mdash2
15 Since x2 + 1 983086 0
therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash
x2 + 1 983086 0
x2 + 3 x + 2 983086 0
( x + 1)( x + 2) 983086 0
0 ndash2 ndash1
f (x )
x
Hence x 983084 ndash2 x gt ndash1
16 ndash4
ndashndashndashndashndashndash1 ndash 3 x
983100 x
0 983100 x +4
ndashndashndashndashndashndash1 ndash 3 x
0 983100 x(1 ndash 3 x) + 4
ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0983100
x ndash 3 x2 + 4
ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash
1 ndash 3 x
0 983100 (ndash3 x + 4)( x + 1)
ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
For ndash3 x + 4 983102 0
4 983102 3 x
x 983100 4mdash3
For x + 1 983102 0
x 983102 ndash1
For 1 ndash 3 x 983086 0
ndash3 x 983086 ndash1
x 983084 1mdash3
ndash1 ndash + ndash +
1 4 ndash ndash
3
4x ndash
3
1x lt ndash
3
x ndash1
3
x
Therefore the range is ndash1 983100 x 983084 1mdash3
x 983102 4mdash3
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17 y2 ndash 9 = x
x = y2 ndash 9
When x = 0
y2 = 9
y = plusmn3
When y = 0
x = ndash9
When x = 7
7 = y2 ndash 9
y2 = 16
y = plusmn4
x ndash9 0 7
y 0 plusmn3 plusmn4
0 ndash9 7
ndash4 ndash3
34
y
x
The range of values of y is ndash 4 983100 y 983100 4
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(b) When p = 2
y = 2 x2 + 8 x + 8
= 2( x2 + 4 x) + 8
= 2( x2 + 4 x + 22 ndash 22) + 8
= 2[( x + 2)2 ndash 4] + 8
= 2( x + 2)2 ndash 8 + 8
= 2( x + 2)2
Therefore the minimum point is (ndash2 0)
When x = 0 y = 8
When y = 0 2( x + 2)2 = 0
x = ndash2
When p = 8
y = 8 x2 + 8 x + 2
= 8( x2 + x) + 2
= 8983091 x2 + x + 983089 1mdash2 983090
2
ndash 983089 1mdash2 983090
2
983092 + 2
= 8983091983089 x +1mdash2 983090
2
ndash1mdash4 983092 + 2
= 8983089 x +1mdash2 983090
2
ndash 2 + 2
= 8983089 x +1mdash2 983090
2
Therefore the minimum point is (ndash1mdash2
0)
When x = 0 y = 2
When y = 0 0 = 8983089 x + 1mdash2 983090
2
x = ndash1mdash2
0
2
8
1 ndash ndash
2
ndash2
y p = 2 p = 8
x
11 (a) f ( x) = 24 x ndash 4 x2 + r
= ndash 4 x2 + 24 x + r
= ndash 4( x2 ndash 6 x) + r
= ndash 4( x2 ndash 6 x + 32 ndash 32) + r
= ndash 4[( x ndash 3)2 ndash 9] + r
= ndash 4( x ndash 3)2 + 36 + r
Compare to f ( x) = p( x ndash q)2 + 16
Therefore p = ndash 4 q = 3 and 36 + r = 16
r = ndash20
(b) The turning point is (3 16)
(c) f ( x) = 24 x ndash 4 x2 ndash 20
When x = 0 f ( x) = ndash20
When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0
4( x ndash 3)2 = 16
( x ndash 3)2 = 4
x ndash 3 = plusmn2
x = plusmn2 + 3
= ndash2 + 3 or 2 + 3
= 1 or 5
0 1
(3 16)
5
ndash20
y
x
12 (a) y = ndash| p( x ndash 3)2 + q|
Substitute x = 3 y = ndash5 into the equation
ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5
Substitute x = 4 y = 0 into the equation
0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0
p = 5
Therefore p = 5 q = ndash5 or p = ndash5 q = 5
(b) When x = 3 y = ndash5
For p = 5 q = ndash5
When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40
Based on the graph the range of values of y is
ndash 40 983100 y 983100 0
For p = ndash5 q = 5
When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40
Therefore the range of values of y is
ndash 40 983100 y 983100 0
13 (a) y = ndash2( x ndash 3)2 + 2k
= ndash x2 + 2 x + px ndash 8
= ndash x2 + (2 + p) x ndash 8
= ndash[ x2 ndash (2 + p) x] ndash 8
= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash
2 9830902
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983091983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
ndash 983089 2 + p ndashndashndashndashndash
2 9830902
983092 ndash 8
= ndash983089 x ndash2 + p
ndashndashndashndashndash2 983090
2
+(2 + p)2
ndashndashndashndashndashndashndash4
ndash 8
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Since the x-coordinate of the maximum point for
both the graphs are same
therefore2 + p
ndashndashndashndashndash2
= 3
p = 4
y = ndash x2 + 2 x + px ndash 8 becomes
y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8
When y = 0
ndash x2 + 6 x ndash 8 = 0
x2 ndash 6 x + 8 = 0
( x ndash 2)( x ndash 4) = 0
x = 2 or 4
Hence A(2 0) and B(4 0)
Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k
0 = ndash2(2 ndash 3)2 + 2k
2k = 2
k = 1 Hence k = 1 and p = 4
(b) For y = ndash2( x ndash 3)2 + 2k
= ndash2( x ndash 3)2 + 2(1)
= ndash2( x ndash 3)2 + 2
Maximum value of the curve is 2
For y = ndash x2 + 2 x + px ndash 8
= ndash x2 + 2 x + 4 x ndash 8
= ndash x2 + 6 x ndash 8
When x = 3
y = ndash9 + 18 ndash 8
= 1
Maximum value of the curve is 1
14 Since 3 x2 983102 0 for all values of x
therefore3 x2
ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)
983100 0
(2 x ndash 1)( x + 4) 983100 0
0 ndash4 1 ndash
2
f (x )
x
Hence ndash 4983100 x 983100 1mdash2
15 Since x2 + 1 983086 0
therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash
x2 + 1 983086 0
x2 + 3 x + 2 983086 0
( x + 1)( x + 2) 983086 0
0 ndash2 ndash1
f (x )
x
Hence x 983084 ndash2 x gt ndash1
16 ndash4
ndashndashndashndashndashndash1 ndash 3 x
983100 x
0 983100 x +4
ndashndashndashndashndashndash1 ndash 3 x
0 983100 x(1 ndash 3 x) + 4
ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0983100
x ndash 3 x2 + 4
ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash
1 ndash 3 x
0 983100 (ndash3 x + 4)( x + 1)
ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
For ndash3 x + 4 983102 0
4 983102 3 x
x 983100 4mdash3
For x + 1 983102 0
x 983102 ndash1
For 1 ndash 3 x 983086 0
ndash3 x 983086 ndash1
x 983084 1mdash3
ndash1 ndash + ndash +
1 4 ndash ndash
3
4x ndash
3
1x lt ndash
3
x ndash1
3
x
Therefore the range is ndash1 983100 x 983084 1mdash3
x 983102 4mdash3
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17 y2 ndash 9 = x
x = y2 ndash 9
When x = 0
y2 = 9
y = plusmn3
When y = 0
x = ndash9
When x = 7
7 = y2 ndash 9
y2 = 16
y = plusmn4
x ndash9 0 7
y 0 plusmn3 plusmn4
0 ndash9 7
ndash4 ndash3
34
y
x
The range of values of y is ndash 4 983100 y 983100 4
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Since the x-coordinate of the maximum point for
both the graphs are same
therefore2 + p
ndashndashndashndashndash2
= 3
p = 4
y = ndash x2 + 2 x + px ndash 8 becomes
y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8
When y = 0
ndash x2 + 6 x ndash 8 = 0
x2 ndash 6 x + 8 = 0
( x ndash 2)( x ndash 4) = 0
x = 2 or 4
Hence A(2 0) and B(4 0)
Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k
0 = ndash2(2 ndash 3)2 + 2k
2k = 2
k = 1 Hence k = 1 and p = 4
(b) For y = ndash2( x ndash 3)2 + 2k
= ndash2( x ndash 3)2 + 2(1)
= ndash2( x ndash 3)2 + 2
Maximum value of the curve is 2
For y = ndash x2 + 2 x + px ndash 8
= ndash x2 + 2 x + 4 x ndash 8
= ndash x2 + 6 x ndash 8
When x = 3
y = ndash9 + 18 ndash 8
= 1
Maximum value of the curve is 1
14 Since 3 x2 983102 0 for all values of x
therefore3 x2
ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)
983100 0
(2 x ndash 1)( x + 4) 983100 0
0 ndash4 1 ndash
2
f (x )
x
Hence ndash 4983100 x 983100 1mdash2
15 Since x2 + 1 983086 0
therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash
x2 + 1 983086 0
x2 + 3 x + 2 983086 0
( x + 1)( x + 2) 983086 0
0 ndash2 ndash1
f (x )
x
Hence x 983084 ndash2 x gt ndash1
16 ndash4
ndashndashndashndashndashndash1 ndash 3 x
983100 x
0 983100 x +4
ndashndashndashndashndashndash1 ndash 3 x
0 983100 x(1 ndash 3 x) + 4
ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0983100
x ndash 3 x2 + 4
ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash
1 ndash 3 x
0 983100 (ndash3 x + 4)( x + 1)
ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x
For ndash3 x + 4 983102 0
4 983102 3 x
x 983100 4mdash3
For x + 1 983102 0
x 983102 ndash1
For 1 ndash 3 x 983086 0
ndash3 x 983086 ndash1
x 983084 1mdash3
ndash1 ndash + ndash +
1 4 ndash ndash
3
4x ndash
3
1x lt ndash
3
x ndash1
3
x
Therefore the range is ndash1 983100 x 983084 1mdash3
x 983102 4mdash3
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17 y2 ndash 9 = x
x = y2 ndash 9
When x = 0
y2 = 9
y = plusmn3
When y = 0
x = ndash9
When x = 7
7 = y2 ndash 9
y2 = 16
y = plusmn4
x ndash9 0 7
y 0 plusmn3 plusmn4
0 ndash9 7
ndash4 ndash3
34
y
x
The range of values of y is ndash 4 983100 y 983100 4
8202019 03[Anal Add Math CD]
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Additional Mathematics SPM Chapter 3
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17 y2 ndash 9 = x
x = y2 ndash 9
When x = 0
y2 = 9
y = plusmn3
When y = 0
x = ndash9
When x = 7
7 = y2 ndash 9
y2 = 16
y = plusmn4
x ndash9 0 7
y 0 plusmn3 plusmn4
0 ndash9 7
ndash4 ndash3
34
y
x
The range of values of y is ndash 4 983100 y 983100 4