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2007 H2 Mathematics Paper 1 Possible Solutions

1)Show that

22 192 + 3 + 2 1 = 2 4 212 + 3 + 2 . [1]Hence, without using a calculator, solve the inequality

22

192 + 3 + 2 > 1 [4]22 192 + 3 + 2 1 = (22 19) (2 + 3 + 2)2 + 3 + 2

=22 19 2 3 22 + 3 + 2

=2 4 212 + 3 + 2 2

2

19

2 + 3 + 2> 1

22 192 + 3 + 2 1 > 02 4 212 + 3 + 2 > 0( 7)( + 3)( + 1)( + 2) > 0

( 7)( + 3)( + 1)( + 2) ( + 1)2( + 3)2 > 0

( 7)( + 3)( + 1)( + 2) > 0

{: , < 3 or 2 < < 1 or > 7}x

2 13 7

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2) Function fand g are defined byf: 1 3 for , 3g:

2 for

i) Only one of the composite functions fg and fg exists. Give a definition (including thedomain) of the composite that exists, and explain why the other composite does not

exists. [3]

ii) Find f1() and state the domain off1. [3]i)

f() g() = (,)\{0} = [0,)Since = (,) \{0} = (,), therefore gfexist.

gf:

1

( 3)2

for

,

3

Since = [0,) = (,)\{3}, therefore fg doesnt exist.ii)

1 3 = 1 = 3

=

1

+ 3

f1: 1 + 3 for , 0

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3) a) Sketch, on an Argnd diagram,,the locus of points representing the complex number such that | + 2 3i| = 13 [3]

b) The complex number is such that + 2 = 3 + 4i, where is the complexconjugate of. Find in the form + i, where and are real. [4]

a)

The locus of| + 2 3i| = 13 is a circle centre at (2,3) and radius 13.

b) = + i = i + 2 = 3 + 4i

( + i)( i) + 2( + i) = 3 + 4i2 i + i + 2 + 2 + i2 = 3 + 4i2 + 2 + 2 + i2 = 3 + 4iComparing real and imaginary,

2 = 4 = 22 + 2 + 2 = 32 + 2 + 22 = 32 + 2 + 1 = 0( + 1)2 = 0 = 1

Using the GC to check

2

3223213 +=

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4) The current in an electric current at time satisfies the differential equation4 = 2 3.

Find in terms of, given that = 2 when = 0. [6]State what happens to the current in this circuit for large value of

[1]

4 = 2 3. 42 3 = 1 4

3 3

2 3 = 1 32 3 = 34

ln|2 3| = 34 +

|2 3

|=

34

+

|2 3| = 34 2 3 = 34 , = > 0

Since = 0 is a solution to the d.e,2 3 = 34 ,

When = 0, = 22 6 = 0

=

4

Therefore

2 3 = 434 = 231 + 234

As , 34 0, 23. Therefore the current approach the value

2

3.

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5)Show that the equation

2

72

+

+=x

xy can be written as

2++=x

BAy , where and are

constants to be found. Hence state a sequence of transformations which transform the

graph ofx

y1

= to the graph of2

72

+

+=x

xy .

[4]

Sketch the graph of2

72

+

+=x

xy , giving the equations of any asymptotes and the

coordinates of any points of intersection with the x- and y-axes. [3]

By long division, = 2 + 7 + 2 = 2 + 3 + 2Hence = 2, = 3.Let f(

) =

1

= 2 + 7 + 2 = 2 + 3 + 2 = 2 + 3f( + 2)The list of transformations are

1) translation 2 units parallel to the x-axis2) scaling 3 units parallel to the y-axis

3) translation 2 units parallel to the y-axis

2=y

2=x

5.3

5.3

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6) Referred to the origin , the position vectors of points and are + 2 and 2 + 4 +

respectively.

i) Show that

is perpendicular to

. [2]

ii) Find the position vector of the point on the line segment such that: =1: 2 [3]iii) The point has position vector 4 + + 2. Use a vector product to find theexact area of triangle . [4]

i)

= 112

= 24

1

= 112 241 = 1 2 + (1) 4 + 2 1 = 0Since = 0, is perpendicular to .ii) Using ratio theorem,

= 2 + 3

=1

32 11

2

+ 241

= 1342

5

iii)

Area of triangle =

1

2 11

2

422

= 12 1 2 2 21 2 2 (4)

1 2 (1) (4) =1

2 6102 =

3

51

= 35 units2

O

A M B1 2

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7) The polynomial () has real coefficients. The equation () = 0 has a root i , where > 0and 0 < < .i) Write down a second root in terms of and , and hence show that a quadratic

factor of() is 2 2 cos + 2. [3]ii) Solve the equation

6 =

64, expressing the solutions in the form

i , where

where > 0 and < . [4]iii) Hence, or otherwise, express 6 + 64 as the product of three quadratics factors withreal coefficients, giving each in non-trigonometrical form. [3]

i) Since the equation has real coefficients, complex roots are in conjugate pair. Therefore the other

complex root is i .The two roots are (cos + i sin ) and (cos() + isin()).() = [ (cos + i sin)][ (cos() + isin())]

= [

(cos

+ i sin

)][

(cos

isin

)]

= 2

(cos isin) (cos + isin ) + 2

(cos + isin)(cos isin)= 2 cos + i sin cos i sin + 2(cos2 + sin2 )= 2 2 cos + 2ii)6 = 646 = 26[cos + i sin]6 = 26[cos(2 + ) + i sin(2 + )] ,where 6 = 26i(2+ ) = 2i(2+)6 , = 3,2,1,0,1,2

= 2

i5

6, 2

i

2, 2

i

6, 2

i

6, 2

i

2, 2

i5

6

iii) From part i)6 + 64 = 2i56 2i56 2i2 2i2 2i6 2i6= 2 4 cos 5

6+ 4 2 4 cos

2+ 4 2 4 cos

6+ 4

= 2 + 4 cos6

+ 4 [2 + 4] 2 4 32+ 4

= 2 + 23 + 4(2 + 4)2 23 + 4Using the GC to check,

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8) The line passes through the points and with coordinates (1,2,4) and (2,3,1)respectively. The place has equation 3 + 2 = 17. Findi) the coordinates of the point of intersection of and , [5]ii) the acute angle between and , [3]iii) the perpendicular distance from

to

. [3]

i)

= 124

, = 231

, = 231

124

= 313Equation of line

= 124

+ 313 , Equation of plane

312 = 17Let the point of intersection be and = 12

4

+ 313 for some . Then12

4

+ 313 312

= 1712

4

312

+ 31

3

312

= 173

2 + 8

9

6

= 17

9 16 = 17 = 12

Therefore the position vector of intersection is 124

12313 = 12

5

311

.The point of intersection is 5

2,

3

2,

11

2.

ii) Let the angle of intersection be sin =

1

193

13 1

143

12 sin = 1266 |9 1 6|sin = 16266 = 1.38 rad

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iii)

= sin = 16266 = 16

266

12 53

11

124

= 8266 313 = 8266 32 + 12 + 32 = 8266 19 = 814 = 4

714 units

p

l

d

A

C

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9)

The diagram shows the graph of = 3. The two roots of the equation 3 = 0 aredenoted by and , where < .i) Find the values of and , each corrected to 3 decimal places. [2]A sequence of real number 1 , 2, 3, satisfies the recurrence relation+1 = 13 for 1.ii) Prove algebraically that, if the sequence converges, then it converges to or . [2]iii) Use a calculator to determine the behaviour of the sequence for each of the cases1 = 0, 1 = 1, 1 = 2. [3]iv) By considering +1 , prove that+1 < if < <

+1 >

if

[2]

v) State briefly how the results in part (iv) relate to the behaviours determined in part(iii). [2]

i) Using the GC

= 0.619 and = 1.512

O x

y

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ii)+1 = 13

Let the convergence values be

. As

,

and

+1

. Therefore

= 13 3 = 3 = 0 = ,

The series converges to or .iii) Using the GC

1 = 0

The series converges to

1 = 1

The series converges to 1 = 2

The series diverges.

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iv)+1 = 13

=1

3( 3)

If < < , from the graph of = 3, 3 < 0+1 = 13

( 3) < 0+1 < 0+1 < If < or > , from the graph of = 3, 3 > 0

+1

=

1

3(

3

) > 0

+1 > 0+1 >

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v)

When 1 = 0, 0 < and +1 > .As increases, the value of increases, therefore converges to

When

1 = 1,

< 0 +1

O nx

y

nx

01=x

nn xx >+1

O nx

y

nx

11=x

nn xx

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10) A geometric series has common ratio , and an arithmetic series has first term and commondifference , where and are non-zero. The first three terms of the geometric series areequal to the first, fourth and sixth terms respectively of the arithmetic series.

i) Show that 32 5 + 2 = 0 [4]ii) Deduce that the geometric series is convergent and find, in terms of

, the sum to

infinity. [5]iii) The sum of the first terms of the arithmetic series is denoted by . Given that > 0, find the set of possible values of for which exceeds 4. [5]

i)1 of GP = 2 of GP = + 33 of GP = + 52 of GP1 of GP = 3 of GP2 of GP = r

+ 3 = (1) + 5 + 3 = (2)From (1), = 1

3( 1)

Substitute into (2)

+ 5 13( 1)

+ 3

13

(

1)

=

13[3 + 5 5]13[3 + 3 3] =

5 23 =

32 5 + 2 = 0ii)

32 5 + 2 = 0( 1)(3 2) = 0

= 1,2

3

= 1 is rejected since , 0 =

1 23

= 3iii) =

2[2 + ( 1)]

> 4

2 [2 + ( 1)] > 4 (3)

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Since = 23, from (1) + 3 = 23

+ 3

=

2

3

= 19Substitute = 1

9 into (3)

22 + ( 1) 1

9 > 4

18[18 ( 1)] > 4

18[19 ] > 4

Using the GC

The set of values for is { +, 6 13}

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11) A curve has parametric equations = cos2 , = sin3 , for 0 12

i) Sketch the curve. [2]ii) The tangent to the curve at point (2,sin3 ), where 0 < < 1

2, meets the x-

and y-axes at

and

respectively. The origin is denoted by

. Show that the area of

triangle is 112

sin (3cos2 + 2sin2 )2 [6]iii)

Show that the area under the curve for 0 12 is 2 cos sin4 12

0, and use

the substitution sin = to find this area. [5]i) Using the GC

ii) = cos2 = 2(cos )( sin ) = 2(sin )(cos ) = sin3 = 3(sin2 )(cos )

= = 3(sin2 )(cos )2(sin )(cos ) = 32 sin Equation of tangent at = , (2,sin3 ) sin3 = 3

2sin [ cos2 ]

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When the tangent cuts the x-axis, = 00 sin3 = 3

2sin [ cos2 ]

2

3sin2 = cos2

= cos2 + 23 sin2 = 13

[3cos2 + 2sin2 ]When the tangent cuts the y-axis, = 0 sin3 = 3

2sin [0 cos2 ]

sin3 = 32

sin cos2 = sin

3

2 cos

2

+ sin2

= 12

sin [3cos2 + 2 sin2 ]

Area of triangle =

1

2()()

=1

21

2sin [3cos2 + 2sin2 ] 1

3[3cos2 + 2 sin2 ]

=1

12sin

(3cos2

+ 2sin2

)2

( ) 32 sin,cos

[ ]

+ 0,sin2cos3

3

1 22 Q

[ ]

+ 22 sin2cos3

2

1,0R

Ox

y

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iii)

When = 0, = 1. When = 12, = 0.

Area under the curve for 0 12

=

1

0

= sin3 10

= sin3 0

12

= sin3 012 [2(sin )(cos )]

= 2 cos sin4 120

= sin = cos

When = 0, = 0. When = 12, = 1.

Area under the curve

= 2 cos sin4 120

= 2 cos sin

4

1

0 = 2 cos 4 110 = 2 41

0

= 2 5

5

0

1

=2

5units2

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2007 H2 Mathematics Paper 2 Possible Solutions

1) Four friends buy three different kinds of fruits in the market. When they get home theycannot remember the individual prices per kilogram, but three of them can remember the

total amount that they each paid. The weights of fruits and the total amount paid areshown in the following table.

Suresh Fandi Cindy Lee Lian

Pineapples (kg) 1.15 1.20 2.15 1.30

Mangoes (kg) 0.60 0.45 0.90 0.25

Lychees (kg) 0.55 0.30 0.65 0.50

Total amount paid in $ 8.28 6.84 13.05

Assuming that, for each variety of fruit, the price per kilogram paid by each of the friends

is the same, calculate the total amount that Lee Lian paid. [6]

Let the cost of pineapple per kg be Let the cost of mangoes per kg beLet the cost of lychess per kg be 1.15 + 0.60 + 0.55 = 8.28 (1)1.20 + 0.45 + 0.30 = 6.84 (2)2.15 + 0.90 + 0.65 = 13.05 (3)

Therefore the cost of pineapple per kg is $3.50, cost of mangoes per kg is $2.60 and the cost oflychess per kg is $4.90.

Total Lee Lian paid = 1.30 3.50 + 0.25 2.60 + 0.50 4.90 = $7.65

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2) A sequence 1 ,2,3 , is such that 1 = 1 and+1 = 2 + 12( + 1)2 , for all 1.i) Use the method of mathematical induction to prove that = 12. [4]ii)

Hence find 2

+ 1

2( + 1)2=1 . [2]iii) Give a reason why the series in part (ii) is convergent and state the sum to infinity. [2]iv)

Use your answer to part () to find 2 12( 1)2=2 . [2]

i)

Let () be = 12 , for all 1When

= 1,

= 1 = 1 = 112

= 1 = Therefore(1) is true.Assume P() is true, = 12 , for all 1When = + 1

+1 =1

( + 1)2

= +1= 2 + 12( + 1)2=

12 2 + 12( + 1)2 =

(+ 1)2 (2 + 1)2( + 1)2 =2 + 2 + 1 2 12( + 1)2

= 2

2( + 1)2 = 1(+ 1)2

= Therefore() is true implies ( + 1) is true. Since (1) is true, by mathematical induction, ()is true.

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ii)

+1 = 2 + 12( + 1)22 + 1

2(

+ 1)2

= +1 2 + 12( + 1)2=1 = ( +1)=1

=

+1 2+2 3+3 4 +2 1+1 + +1

= 1 +1= 1

1

( + 1)2

iii)

As , 1(+1)2 0, therefore the sum will converges to 1.

iv)

Let = + 1. When = 2, = 1. When = , = 1. Therefore 2 12( 1)2=2 =

2( + 1) 1( + 1)2( + 1 1)2

1=1

= 2

+ 1

2( + 1)21

=1 = 1 12

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3) i) By successive differentiating (1 + ) , find Maclaurins series for (1 + ) , up toand including the term in 3. [4]

ii)Obtain the expansion of(4 )32 (1 + 22)32 up to and including the term in 3. [5]

iii) Find the set of values of for which the expansion in part (ii) is valid. [2]i)

Let f() = (1 + ) f() = (1 + )1f () = ( 1)(1 + )2f () = ( 1)( 2)(1 + )3f(0) = f (0) = ( 1)f (0) = ( 1)( 2)Therefore,

f(

)

= f(0

)+ f

(0

)+

2

2! f(0

)+

3

3! f(0

)+

= 1 + + ( 1)2!

2 + ( 1)( 2)3!

3 +ii)

(4 )32 = 432 1 432

= 8 1 + 32

4+ 32 32 1

2!

42 + 32 32 1 32 2

3!

43 +

= 8 1 3

8 +3

1282

+

1

10243

+= 8 3 + 3162 + 1

1283 + , for

4 < 1

(1 + 22)32 = 1 + 32

(22) += 1 + 32 + , for |22| < 1

(4 )32(1 + 22)32 = 8 3 + 3162 + 1

1283 + (1 + 32)

=8

3

+

3

16 2 +

1

1283

242 93 +

= 8 3 + 38716

2 1151128

3 +iii)

The expansion is valid for 4 < 1 and |22| < 1

4 < 1 4 < < 4

|22| < 1 1

2

< < 1

2

Therefore the expansion is valid for

: ,1

2 < 5 and = 182.4 > 5,~(240,182.4) approximately(230 260) (229.5 260.5) = 0.717

Let be the random variable number of people out of 1000 that have gene ~(1000,0.003)Since = 3 < 5, ~(3) approximately

(2

< 5) =

(

= 2) +

(

= 3) +

(

= 4) = 0.616

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7) A large number of students in a college have completed a geography project. The time, hours, taken by a student to complete the project is noted for a random sample of 150

students. The results are summarised by = 4626, 2 = 147691.Find unbiased estimates for the population mean and variance.

[2]Test, at 5% significance level, whether the population mean time for student to complete

the project exceeds 30 hours.

[4]

State, giving reason, whether any assumption about the population are needed in order

for the test to be valid. [1]

The unbiased estimate of the mean = = 4626150 = 30.84The unbiased estimate of the variance

2 =

1 2

2

=150

149147691

150 30.842

= 33.7259 = 33.7

Since sample size is large (>50), by central limit theorem

~30.84, 33.7259150

0:

= 30

1:

> 30

Since = 0.038 < 0.05, we conclude that we have sufficient evidence at 5% level of significance toreject0 and accept 1 , the mean time a student takes to complete the project exceeds 30 hours.Since the sample size is large, by central limit theorem, the population mean will be normallydistributed. Hence we can use the Z-test without any assumption.

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8) Chickens and turkeys are sold by weight. The masses, in kg, of chickens and turkeys aremodelled as having independent normal distributions with mean and standard deviations as

shown in the table.

Mean mass Standard deviation

Chickens 2.2 0.5Turkeys 10.5 2.1

Chickens are sold at $3 per kg and turkeys at $5 per kg.

i) Find the probability that a randomly chosen chicken has a selling price exceeding $7. [2]ii) Find the probability of the event that both a randomly chosen chicken has a selling

price exceeding $7 and a randomly chosen turkey has a selling price exceeding $55. [3]

iii) Find the probability that the total selling price of a randomly chosen chicken and arandomly chosen turkey is more than $62. [4]

iv) Explain why the answer to part (iii) is greater than the answer to part (ii). [1]Let be the random variable selling price of a chicken.~(2.2 3,0.52 32)~(6.6,1.52)Let be the random variable selling price of a turkey.~(10.5 2.12 52)~(52.5,10.52)i) ( > 7) = 0.395

ii) ( > 7 and > 55) = ( > 7) ( > 55) = 0.160

iii) + ~(6.6 + 52.5,1.52 + 10.52) + ~(59.1,112.5)( + > 62) = 0.392

iv) Part (iii) answer is greater since it is more likely for the price of the turkey to be higher as it has a

larger variance. This will help make up the price to exceed $62.

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9) A group of 12 people consists of 6 married couples.i) The group stand in a line.

a) Find the number of different possible orders [1]b) Find the number of different possible orders in which each man stands next to

his wife [3]

ii) The group stand in a circle.a) Find the number of different possible arrangements. [1]b) Find the number of different possible arrangements if men and women

alternate. [2]

c) Find the number of different possible arrangements if each man stands next tohis wife and men and women alternate.

ia) Ways they can stand in a line = 12! = 479001600

ib)

Ways = 6 2 5 2 4 2 3 2 2 2 1 2 = 6! 26 = 46080

iia) Ways = 11! = 39916800 iib(

Ways = 6 5 5 4 4 3 3 2 2 1 1 = 86400

Man (fixed)

Woman (6)

Woman (5)

Woman (4)Woman (3)

Woman (2)

Woman (1)

Man (5)

Man (4)

Man (3)

Man (1)

Man (1)

Ways to arrange the 6 couples

6 5 4 3 2 1

Each couple can swap position

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iic)

Case 1

Case 2

Ways = 5! 2 = 240

Man (fixed)

Wife of fixed man(1)

Wife (1)

Wife (1)Wife (1)

Wife (1)

Wife (1)

Man (5)

Man (4)

Man (3)

Man (1)

Man (1)

Man (fixed)

Wife of fixed man(1)

Wife (1)

Wife (1)Wife (1)

Wife (1)

Wife (1)

Man (5)

Man (4)

Man (3)

Man (1)

Man (1)

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10) A player throws three darts at a target. The probability that he is successful in hitting the targetwith his first throw is

1

8. For each of his second and third throws, the probability of success is

twice the probability of success on the preceding throw if that throw was successful, the same as the probability of success on the preceding throw if that throw was

unsuccessful.

Construct a probability tree showing this information. [3]Find

i) the probability that all three throws are successful, [2]ii) the probability that at least two throws are successful, [2]iii) the probability that the third throw is successful given that exactly two of the three

throws are successful. [4]

i) (all three throws successful) = 18

2

8

4

8=

1

64

ii) (at least two throws sucessful) = () + () + () + ()=

1

64+

1

8

2

8

4

8+

1

8

6

8

2

8+

7

8

1

8

2

8

=21

256

iii) (third throw successful | exactly two throws successful

)

= (third throw successful and exactly two throws successful)(exactly two throws successful) =

() + ()() + () + ()=

18

68

28

+78

18

28

18

28

48

+18

68

28

+78

18

28

=13

17

MissHit

Hit

MissHit

Miss

MissHit

Hit

MissHit

Miss

MissHit

Start

8

1

8

7

8

1

8

7

8

1

8

7

8

2

8

6

8

2

8

6

8

6

8

2

8

4

8

4

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11) Research is being carried out into how the concentration of a drug in the bloodstream varieswith time, measured from when the drug is given. Observations at successive times give the

data shown in the following table.

Time (

minutes) 15 30 60 90 120 150 180 240 300

Concentration( microgram per litre) 82 65 43 37 22 19 12 6 2

It is given that the value of the product moment correlation coefficient for this data is 0.912,correct to 3 decimal places. The scatter diagram for the data is shown below.

Calculate the equation of the regression line on . [2]Calculate the corresponding estimated value of when = 300, and comment on thesuitability of the linear model. [2]

The variable is defined by = ln . For variables and ,i) calculate the product moment correlation coefficient and comment on its value, [2]ii) Calculate the equation of the appropriate regression line [3]Use a regression line to give the best estimate that you can of the time when the drug

concentration is 15 microgram per litre. [2]

The regression line on is = 0.260 + 66.2

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When = 300 s, = 11.7 g.The linear model is not valid, since from the scatter diagram, it displayed a logarithmic relationship.

Also > 0, but the linear regression line gives < 0 when = 300.i)

product moment correlation coefficient = 0.994.The values is close to 1. There a good negative linear relationship between ln and .ii) The regression line is ln = 0.0123 + 4.62.Since we want to find the given , we need the regression line on ln

When

= 15

g,

= 155s

Alternatively, since = 0.994 is close to 1. We can also use regression line ln on to find thevalue of

ln = 0.0123434 + 4.620609When = 15 g,ln 15 = 0.0123434 + 4.620609 = 155 s