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JANUARY 2006 MATHEMATICS GENERAL PROFICIENCY (PAPER 2)
Section I
1. a. (i) Required To Calculate:
Calculation: Numerator
Denominator
Hence,
(ii) Required To Calculate:
Calculation:
b. Data: Amount of loan = $12 000 Rate = 14% per annum (i) Required To Calculate: Interest on the loan at the end of the first year. Calculation:
Interest on loan at the end of 1st year
21
53
54
412
-
´
59
54
4954
412
=
´=
´
101
105621
53
=-
-
10159
21
53
54
412
=-
´
form)exact(in18110
59
=
´=
( )211.275.18 -
( )
figurestsignifican3to3.147992.14
4521.475.1811.275.18 2
==
-=-
1200010014
´=
1680$=
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(ii) Required To Calculate: Total amount owing at the end of 1st year. Calculation: Total amount owing at the end of 1st year = Original amount borrowed + Interest after 1 year. = $12 000 + $1 680 = $13 680
Data: Repayment at start of 2nd year = $7 800
(iii) Required To Calculate: Amount outstanding at the start of 2nd year. Calculation: Amount owed at start of second year = $13 680 - $7 800 = $5 880
(iv) Required To Calculate: Interest on the outstanding amount at the end of
2nd year. Calculation: Interest on the outstanding amount at the end of 2nd year
2. a. Data: and Required To Calculate: Calculation:
b. Data: ,
Required To Calculate: The value of x and of y Calculation:
Let …(1) and …(2) Equation (2) …(3) Equation (1) + (3)
20.823$
588010014
=
´=
2-=m 4=n( )( )nmnm -+ 22
( )( ) ( )( ) ( )( )( )( )
080
444442242222
=-´=
--+-=--+-=-+ nmnm
3765 =+ yx 432 =- yx
3765 =+ yx 432 =- yx2´
864 =- yx
459
)3(864)1(3765
=
=-=+
x
yxyx
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When Substitute in equation (1)
Hence and OR Let …(1) and …(2) From (2)
Substituting in (1)
Substitute in
Hence, and OR
5945
=
=
x
x
5=x
( )
2126
25376376553765
==
-==+=+
yyyyyx
5=x 2=y
3765 =+ yx 432 =- yx
243432
432
+=
+==-
yx
yxyx
( ) ( ) ( )
254277412201537262435
3762435
===++=++
=+÷øö
çèæ +
yyyyyy
yy
2=y243 +
=yx
( )
52102423
=
=
+=x
5=x 2=y
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Obtaining 2 points on the straight line
x y -1 7 11 -3
Obtaining 2 points on the straight line
x y -1 -2 8 4
Plotting both straight lines on the same axes.
Point of intersection is (5, 2), therefore, and . OR and and
…matrix equation
Let
det
3765 =+ yx
432 =- yx
5=x 2=y
3765 =+ yx 432 =- yx
÷÷ø
öççè
æ=÷÷
ø
öççè
æ÷÷ø
öççè
æ- 4
373265
yx
÷÷ø
öççè
æ-
=3265
A
( ) ( )2635 ´--´=A
271215
-=--=
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Matrix equation
Equating corresponding entries and c. Required To Factorise: (i) , (ii) , (iii) Factorising: (i)
(ii)
( )( )
÷÷÷÷
ø
ö
çççç
è
æ
-=
÷÷ø
öççè
æ-
---=\ -
275
272
276
273
5263
2711A
1-´ A
÷÷ø
öççè
æ=
÷÷÷÷
ø
ö
çççç
è
æ
=
÷÷÷÷
ø
ö
çççç
è
æ
÷øö
çèæ ´-+÷
øö
çèæ ´
÷øö
çèæ ´+÷
øö
çèæ ´
=
÷÷ø
öççè
æ
÷÷÷÷
ø
ö
çççç
è
æ
-=÷÷
ø
öççè
æ
÷÷ø
öççè
æ=÷÷
ø
öççè
æ´
÷÷ø
öççè
æ=÷÷
ø
öççè
æ´´
-
--
25275427135
427537
272
427637
273
437
275
272
276
273
437
437
1
11
yx
Ayx
I
Ayx
AA
5=x 2=y
254 2 -x qsqpsp 6496 -+- 443 2 -+ xx
254 2 -x( ) ( )
( )( )5252squares2ofDifference
52 22
+-=
-=
xx
x
qsqpsp 6496 -+-( ) ( )
( )( )qpssqsp
2332322323
+-=-+-=
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(iii)
3. a. Data:
Required To Express: u in terms of v, s and t. Solution:
b. Data: Venn diagram illustrating the customers’ purchases at a bakery. (i) Required To Complete: The Venn diagram to represent the data given.
Solution:
(ii) Required To Find: An expression in terms of x to represent the total number of customers to visit the bakery that day. Solution: Total no. of customers who visited the bakery
443 2 -+ xx( )( )223 +- xx
( )tvus +=21
( )
( )
( )
utvts
utvtsvtutstvus
tvus
tvus
=-
=-+=+=
+=
+=
22
22
21
21
vts
tvtsu -
-=\
2or2
80702 +++= xx1503 += x
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(iii) Required To Calculate: The number or customers who bought bread only. Calculation:
(data)
No. of customers who bought bread only
4. a. Data: Line, l, with equation (i) Required To Find: The gradient of and line parallel to l.
Solution:
is of the form , where is the gradient.
(ii) Required To Find: The equation of the line parallel to l that passes
through (2, -6) Solution:
The gradient of the required line = 4 (Parallel lines have the same gradient). Hence equation of the required line is
3001503 =+x
501503
==\xx
\ x2=( )100502
==
54 += xy
54 += xy cmxy += 4=m
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b. Data: Map showing the position of three cities A, B and C with a scale of
1 : 20 000 000. (i) Required To Find: The length of line segment BC. Solution:
cm (by measurement)
(ii) Required To Calculate: The actual shortest distance from B to C. Calculation: The shortest distance between B and C cm
km
(iii) Required To Find: The bearing of B from A. Solution:
By measurement, the bearing of B from A = 050°
( )
( )
144846246
426
-=-=+-=+
=---
xyxyxy
xy
7.6=BC
000000207.6 ´=
1001000000000207.6
´´
=
km1340=
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5. a. Data: Diagram illustrating a solid glass paper weight consisting of a hemisphere
mounted on a cylinder.
(i) Required To Calculate: The curved surface area of the cylinder
Calculation: The area of the curved surface of the cylinder is
(ii) Required To Calculate: the surface area of the hemisphere.
Calculation:
Surface area of the hemisphere
(iii) Required To Calculate: total surface area of the solid paper weight
Calculation: Total surface area of the paper weight = Area of curved surface of cylinder + Area of curved surface of hemisphere + Area of the circular base.
( )( )( )2
2
cm72.150cm8314.322
=
=rhp
( )2421 rp=
( )( )2
2
cm52.56
314.3421
=
´´=
( )2
2
cm50.235314.352.5672.150
=
++=
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b. Required to Construct: Parallelogram KLMN with KL = 8 cm, KN = 6 cm and
Solution:
6. Data: Diagram illustrating PQRS and its image under a rotation.
a. Required To Find: the coordinates of and Solution:
°= 60NK̂L
SRQP ¢¢¢¢
R¢ S ¢
( )4,2-=¢R
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(from the given diagram) b. Required To Describe: the rotation completely.
Solution: Two object vertices – P and R Two image vertices - and Broken line segments join P to and R to Perpendicular bisectors of these line segments were constructed and found to intersect at the origin (0, 0).
The centre of rotation is (0, 0).
Line segments are drawn from an object vertex and the image vertex to the centre of rotation, example OP and . The angle between both line segments is the angle of rotation.
The angle of rotation is 90°.
( )1,4-=¢S
P¢ R¢P¢ R¢
\
PO ¢
\
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Quadrilateral PQRS is mapped onto under a rotation of 90° anti-
clockwise about O. This may be represented by the matrix .
c. Data: is a reflection of the image in the x – axis.
Required To Draw: Solution:
d. Required To Describe: the transformation that maps PQRS onto
Solution:
The simple transformation that maps PQRS onto is
. This is equivalent to a reflection in the line .
SRQP ¢¢¢¢
÷÷ø
öççè
æ -0110
SRQP ¢¢¢¢¢¢¢¢ SRQP ¢¢¢¢SRQP ¢¢¢¢¢¢¢¢
SRQP ¢¢¢¢¢¢¢¢
SRQPPQRS
SRQPSRQPPQRS
SRQPSRQPPQRS axisxinflectionOaboutclockwiseanti
ofRotation
¢¢¢¢¢¢¢¢¾¾¾¾¾¾¾ ®¾
¢¢¢¢¢¢¢¢¾¾¾ ®¾¢¢¢¢¾¾¾ ®¾
¢¢¢¢¢¢¢¢¾¾¾¾¾ ®¾¢¢¢¢¾¾¾¾ ®¾
÷÷ø
öççè
æ-
-=÷÷ø
öççè
æ -÷÷ø
öççè
æ-
÷÷ø
öççè
æ-÷÷
ø
öççè
æ -
--
°
0110
0110
1001
1001
0110
Re
90
\ SRQP ¢¢¢¢¢¢¢¢
÷÷ø
öççè
æ-
-0110
xy -=
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7. Data: The lengths of the right foot of 25 students. a. Required to complete: the grouped frequency table for the given data. Solution:
Modifying and completing the table of values for the continuous variable. Length of right foot,
x cm L.C.B U.C.B. Frequency Mid-class Interval
14 – 16 4
17 – 19 6
20 – 22 8
23 – 25 5
26 – 28 2
Frequency for class interval 17 – 19
b. Required To Find: The lower class boundary of the class interval 14 – 16 Solution: The lower class boundary of the class interval 14 – 16 is 13.5 (as shown in the above table).
c. Required To Find: The width of class interval 20 – 22.
Solution: The width of class interval 20 – 22 is
d. Required To Find: The class interval in which a measurement of 16.8 cm would
lie. Solution: A measurement of 16.8 cm lies in the class interval 17 – 19.
e. Required To Calculate: The probability that a randomly chosen student has a right foot measuring greater than or equal to 20 cm.
2U.C.B.L.C.B +
5.165.13 <£ x15
25.165.13=
+
5.195.16 <£ x18
25.195.16=
+
5.225.19 <£ x21
25.225.19=
+
5.255.22 <£ x24
25.255.22=
+
5.285.25 <£ x27
25.285.25=
+
25=å f( )258425 +++-=
6=
35.195.22L.C.B.U.C.B
=-=-
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Calculation:
f. Required to find: The modal length of a student’s right foot.
Solution: Modal class is 20 – 22 since this class corresponds to the greatest frequency.
Modal length of student’s foot
g. Required To Estimate: the mean length of a student’s foot using the mid-class
intervals for the data given. Solution: Mean length of student’s foot where
f = frequency x = mid-class interval of class
8. Data: Path of a ball follows the equation , h = height in m and t = time in seconds.
a. Required To Complete: the table of values of t and corresponding values of h. Solution:
t 0 0.5 1 1.5 2 2.5 3 3.5 4 h 0.0 8.8 15 18.8 (20) 18.8 (15) 8.8 0
When When
( )
532515
classtheinstudentsofnumberTotalcm20footwithstudentsofNo.cm20footsstudent'ofLength
=
=
³=³P
\2L.C.B.U.C.B +
=
cm212
5.225.19
=
+=
x=
åå=ffx
x
( ) ( ) ( ) ( ) ( )
cm4.2025
272245218186154
=
´+´+´+´+´=x
2520 tth -=
2=t ( ) ( )225220 -=h20=
3=t ( ) ( )235320 -=h15=
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b. Data: Scale of 2 cm to represent 0.5 seconds on t – axis for . Scale of 1 cm to represent 1 metre on the h – axis for .
Required To Draw: the graph of for Solution:
40 ££ t0.210.0 ££ h
2520 tth -= 40 ££ t
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c.
(i) Required To Find: the greatest height above the ground reached by the
ball. Solution: The greatest height reached by the ball is 20 m.
(ii) Required To Find: the duration for which the ball was more than 12 m
above the ground. Solution:
m from the ground to
Time
That is, m for 2.52 seconds.
12>h74.0=t 26.3=t
74.026.3 -=seconds52.2=12>h
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(iii) Required To Find: the time interval during which the ball was moving upwards. Solution: The ball is moving upwards from to , that is for 2 – 0 = 2 seconds.
Section II 9. a. Data: and
Required To Calculate: x and y Calculation: Let …(1) and …(2) From (1)
Substitute in (2)
When
When
Hence, and or and
b. Required To Express: in the form
Solution:
Half the coefficient of x is
0=t 2=t
72 =+ yx 62 =- xyx
72 =+ yx 62 =- xyx
xyyx
2772-=
=+
( )
( )( ) 0323067306270627
2
22
2
=-+=--
=-+-
=---
xxxxxxxxxx
3or32
-=x
32
-=x ÷øö
çèæ--=3227y
318
347
=
+=
3=x ( )327 -=y1=
32
-=x318=y 3=x 1=y
3124 2 -- xx ( ) khxa ++ 2
( ) 3343124 22 --=-- xxxx
( )233
21
-=-
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is of the form where
OR
Equating coefficient of Equating coefficient of x
Equating constants
12?3
1291244934
234
?2343124
2
22
22
-=-
-+-
÷øö
çèæ +-=÷
øö
çèæ -
+÷øö
çèæ -=--
and
xxand
xxx
xxx
122343124
22 -÷
øö
çèæ -º--\ xxx ( ) khxa ++ 2
ÂÎ-=
ÂÎ-=
ÂÎ=
12234
k
h
a
( )( )
kahahxaxkhhxxa
khxax
+++=
+++=
++=--
22
22
22
22
3124
2xÂÎ= 4a
( )
ÂÎ-=
-=
=-
238124212
h
h
122343124
1293
2343
22
2
-÷øö
çèæ -º--\
ÂÎ-=+=-
+÷øö
çèæ-=-
xxx
kk
k
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c. (i) Required To Find: the minimum value of Solution:
Minimum value of
(ii) Required To Find: the value of x at which the minimum point occurs. Solution: Minimum value occurs at
OR Let . The axis of symmetry passes through the minimum point and occurs at
Therefore,
Therefore,
3124 2 -- xx
xx
xxx
"³÷øö
çèæ -
-÷øö
çèæ -º--
0234
122343124
2
22
\ 3124 2 -- xx 120 -=12-=
23
023
023
0234
2
2
=
=-
=÷øö
çèæ -
=÷øö
çèæ -
x
x
x
x
3124 2 --= xxy
( )( )
238124212
=
=
--=x
12
32312
234
2
min
-=
-÷øö
çèæ-÷
øö
çèæ=y
23at12min =-= xy
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(iii) Required To Solve:
Solution:
to 3 significant figures. OR
to 3 significant figures
10. a. Data: No. of Sonix radios = x No. of Zent radios = y Space on shelf for up to 20 radios. (i) Required To Find: The inequality to represent the data given.
Solution: Therefore, total must be less than or equal to 20. Hence, …(1)
03124 2 =-- xx
( ) ( ) ( )( )( )
0232.0or223.38192128
4814412
423441212
031242
2
-=
±=
+±=
---±--=
=--
x
xx
232.0or23.3 -=
0232.0or223.3732.15.1
323
323
323
412
23
012234
03124
2
2
2
2
-=±=
±=
±=-
=÷øö
çèæ -
=÷øö
çèæ -
=-÷øö
çèæ -
=--
x
x
x
x
x
xx
232.0or23.3 -=
20£+ yx
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(ii) Data: Cost of Sonix radio = $150 Cost of Zent radio = $ 300 Shop’s owner has $4500 Required To Find: The inequality to represent the data given Solution: Cost of x Sonix radios at $150 each and y Zent radios at $300 each Total available to spent is $4500.
(iii) Data: Owner decides to stock at least 6 Sonix and at least 6 Zent radios. Required To Find: The two inequalities to represent the data given. Solution: Stock is at least 6 Sonix and at least 6 Zent radios.
b. Data: Scale of 2 cm to represent 5 Sonix radios and 2 cm to represent 5 Zent radios.
Horizontal axis - Vertical axis - Required To: Draw the boundary lines for all four above inequalities, shade the region that satisfies all four inequalities and state the vertices of the shaded region. Solution: The line is a straight vertical line. The region is represented as
The line is a straight horizontal line. The region is represented as
( ) ( )300150 yx +=
302150
4500300150
£+÷
£+\
yx
yx
6and6 ³³\ yx
300 ££ x250 ££ y
6=x6³x
6=y6³y
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Obtaining two points on the line . When The line passes through the point (0, 20).
When The line passes through the point (20, 0).
The region with the smaller angle satisfies the region. The region is
Obtaining two points on the line When
The line passes through the point (0, 15).
20=+ yx0=x 200 =+ y
20=y20=+ yx
0=y 200 =+x20=x
20=+ yx
£20£+ yx
302 =+ yx0=x 3020 =+ y
15230302
=
=
=
y
y
302 =+ yx
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When The line passes through the point (30, 0).
The region with the smaller angle satisfies the region. The region is
The region which satisfies all 4 inequalities is the area where all shaded regions overlap.
0=y ( ) 3002 =+x30=x
302 =+ yx
£302 £+ yx
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(iv) The vertices of the shaded region are (6, 6), (14, 6), (10, 10) and (6, 12).
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c. Data: The owner of the shop sells to make a profit of $80 on each Sonix radio and $100 on each Zent radio. (i) Required To Express: The total profit in terms of x and y Solution: Profit, P on x Sonix radios at $80 each and y Zent radios at $100 each is
(ii) Required To Calculate: The maximum profit Calculation: Test Test
Test
Maximum profit = $1800 when the owner sells 10 Sonix radios and 10
Zent radios.
11. Data: Circle, centre O with and . and are chords. a. (i) Required To Calculate: Solution:
(Angle subtended by chord (AB) at centre of circle is twice the angle that the chord subtends at the circumference, standing on the same arc.)
( ) ( )( )yx
yxP10080$10080
+=´+´=
126 == yx 1010 == yx( ) ( )1680$
12100680=
+=P ( ) ( )1800$
101001080=
´+´=PP
614 == xx( ) ( )1720$
61001480=
´+´=P
\
°= 130ˆBOA °= 30ˆCAD AEC BEDACBÐ
( )
°=
°=
65
13021ˆBCA
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(ii) Required To Find: Solution:
(Chord CD subtends equal angles at the circumference, standing on the same arc).
(iii) Required To Find: Solution:
(Sum of angles in a triangle = 180°). (Vertically opposite angles).
b. Required To Show: is similar to Solution:
(Sum of angles in a triangle = 180°).
are equi-angular OR similar.
CBDÐ
°= 30ˆDBC
AEDÐ
( )°=
°+°-°=85
3560180ˆBEC
°= 85ˆDEA
BCED ADED
°= 65ˆADE
DC
EAngleAB
ˆˆcommonis
ˆˆ
=
=
ADEBCE DD\ and
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c. Data: CE = 6 cm, EA = 9.1 cm and DE = 5 cm (i) Required To Calculate: length of EB.
Calculation: If , then the ratio of their corresponding sides are the same. That is,
(ii) Required To Calculate: the area of
Calculation:
Area of
12. This question is not done since it involves latitude and longitude (Earth Geometry) which has been removed from the syllabus.
13. Data: A (1, 2), B (5, 2), C (6, 4) and D (2, 4) are the vertices of a quadrilateral ABCD.
a. (i)Required To Express: The position vectors of in the
form .
Solution:
If then is of the form where and .
ADEBCE DD ~
cm92.1056.5451.96
1.956
=
=
´=
=
==
EB
EBAEBE
DECE
ADBC
AEDD
( )( ) °=D 85sin1.9521AED
placedecimal1tocm7.22cm66.222
2
=
=
ODOCOBOA and,,
÷÷ø
öççè
æyx
( )2,1=A ÷÷ø
öççè
æ=21
OA ÷÷ø
öççè
æyx
1=x 2=y
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If then is of the form where and .
If then is of the form where and .
If then is of the form where and .
(ii) Required To Find: and Solution:
b. Required To Find: and the unit vector in the direction of
Solution:
( )2,5=B ÷÷ø
öççè
æ=25
OB ÷÷ø
öççè
æyx
5=x 2=y
( )4,6=C ÷÷ø
öççè
æ=46
OC ÷÷ø
öççè
æyx
6=x 4=y
( )4,2=D ÷÷ø
öççè
æ=42
OD ÷÷ø
öççè
æyx
2=x 4=y
AB DC
÷÷ø
öççè
æ=
÷÷ø
öççè
æ+÷÷ø
öççè
æ-=
+=
04
25
21OBAOAB
÷÷ø
öççè
æ=
÷÷ø
öççè
æ+÷÷ø
öççè
æ-=
+=
04
46
42OCDODC
AB AB
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assm
aths.c
om
Any vector in the direction of , where is a scalar.
If the vector is unit then the magnitude is 1 unit.
The unit vector in the direction of
.
c. (i)Required To Find: Two geometrical relationships between the line segments AB and DC. Solution:
Hence, and is parallel to .
(ii) Required To Explain: Why ABCD is a parallelogram. Solution:
If one pair of opposite sides of a quadrilateral are both parallel and equal
then the quadrilateral is a parallelogram. Hence, ABCD is a parallelogram.
( ) ( )units4
04 22
=
+=AB
÷÷ø
öççè
æ=
04
aAB a
( ) ( )
41104
104
22
=
=+
=\
a
a
a
÷÷ø
öççè
æ=
04
41AB
÷÷ø
öççè
æ=01
÷÷ø
öççè
æ=
÷÷ø
öççè
æ=
÷÷ø
öççè
æ=
04
1
04
04
DC
AB
DCAB = AB DC
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aths.c
om
c. Required To Find: The position vector of G, the midpoint of line AC and the coordinates of the point of intersection of the diagonals AC and BD. Solution:
Position vector of G is
The diagonals of a parallelogram bisect each other. AC and BD intersect at G.
Intersection of AC and BD is G = .
÷÷
ø
ö
çç
è
æ=
÷÷
ø
ö
çç
è
æ+÷÷ø
öççè
æ=
+=
÷÷ø
öççè
æ=
÷÷ø
öççè
æ=
÷÷ø
öççè
æ+÷÷ø
öççè
æ-=
+=
3213
1212
21
25
21
25
46
21
AGOAOG
AG
OCAOAC
÷÷
ø
ö
çç
è
æ=
3213OG
\ ÷øö
çèæ 3,213
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assm
aths.c
om
14. a. Data:
(i) Required To Calculate: The determinant of L. Calculation: Det (ii) Required To Calculate: The values of x given that L is singular. Calculation: If L is singular, then det L = 0 When
b. Data:
(i) Required To Calculate: Calculation: det
(ii) Data:
Required To Calculate: The value of x and of y Calculation:
÷÷ø
öççè
æ=
xx
L14
( ) ( )14´-´= xxL42 -= x
042 =-x
242
±==xx
÷÷ø
öççè
æ=
6213
M
1-M
( ) ( )2163 ´-´=M
16218
=-=
( )( )
÷÷÷÷
ø
ö
çççç
è
æ
-
-=
÷÷ø
öççè
æ-
-=-
163
162
161
166
3216
1611M
÷÷ø
öççè
æ-
=÷÷ø
öççè
æ812
yx
M
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assm
aths.c
om
Equating corresponding entries. and
c. Data:
(i) Required To Calculate: , the image of (3, -1) under N. Calculation:
When
The image of (3, -1) is (11, -4).
÷÷ø
öççè
æ-
=÷÷ø
öççè
æ
´
÷÷ø
öççè
æ-
=÷÷ø
öççè
æ
--
-
812
812
11
1
Myx
MM
M
yx
M
÷÷ø
öççè
æ-
=
÷÷÷÷
ø
ö
çççç
è
æ
÷øö
çèæ -´+÷
øö
çèæ ´-
÷øö
çèæ -´-+÷
øö
çèæ ´
=
÷÷ø
öççè
æ-÷÷
÷÷
ø
ö
çççç
è
æ
-
-=÷÷
ø
öççè
æ
÷÷ø
öççè
æ-
=÷÷ø
öççè
æ -
35
816312
162
816112
166
812
163
162
161
166
8121
yx
Myx
I
5=x 3-=y
÷÷ø
öççè
æ-
+÷÷ø
öççè
æ÷÷ø
öççè
æ=÷÷
ø
öççè
梢
25
2002
yx
yx
( )yx ¢¢,
( ) ÷÷ø
öççè
æ-
=13
, yx
( ) ( )( ) ( )
÷÷ø
öççè
æ-
=
÷÷ø
öççè
æ-
+÷÷ø
öççè
æ-
=
÷÷ø
öççè
æ-
+÷÷ø
öççè
æ-´+´-´+´
=
÷÷ø
öççè
æ-
+÷÷ø
öççè
æ-÷÷
ø
öççè
æ=÷÷
ø
öççè
梢
411
25
26
25
12301032
25
13
2002
yx
\
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assm
aths.c
om
(ii) Required To Calculate: when is (7, 4) Calculation:
Equation corresponding entries
and
The point which is mapped onto (7, 4) is (1, 3).
( )yx, ( )yx ¢¢,
÷÷ø
öççè
æ-
+÷÷ø
öççè
æ=÷÷
ø
öççè
æ
÷÷ø
öççè
æ-
+÷÷ø
öççè
æ÷÷ø
öççè
æ=÷÷
ø
öççè
æ
25
22
47
25
2002
47
yx
yx
÷÷ø
öççè
æ-+
=÷÷ø
öççè
æ2252
47
yx
122
527
==
+=
xx
x
326
224
==
-=
yyy
\
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