2011 Year 6 H2 Mathematics Exercises for June Holiday _Solution

Raffles Institution H2 Mathematics 9740

Year 6 2011 ____________________________________

2011 Year 6 H2 Mathematics June Holiday Revision of 82

JUNE HOLIDAY REVISION EXERCISE (Worked solutions will be posted on Discovery on 15 June (Wednesday) 2011. Please try the questions first before accessing the solutions.)

Equations, Inequalities and System of Linear Equations 1. NJC 08/Prelim/I/2

A Sumo wrestler would like to have fish fillet, salad and fries for breakfast. As he is on a special diet, he must make sure that his intake (in grams) of protein, carbohydrates and fats per meal is in the ratio of 4:8:3. The table below shows the nutritional breakdown for one serving of each item.

Protein

(in grams)

Carbohydrates

(in grams)

Fats

(in grams)

Fish Fillet 150 60 25

Salad with

dressing 15 30 5

Fries 5 250 110

Calculate the ratio of the servings of fish fillet, salad and fries that the wrestler should take. [4]

[Ratio 1:3:1]

Let x : y : z be the ratio for the servings of fish fillet, salad and fries. 150x + 15y + 5z = 4k 60x + 30y + 250z = 8k 25x + 5y + 110z = 3k where k is a constant. 150x + 15y + 5z = 4 60x + 30y + 250z = 8 25x + 5y + 110z = 3 Solving the system of linear equations with the GC, we have

x = 0.02, y = 0.06, z =0.02 Ratio is 1:3:1 (ans)

2. DHS10/CT2/1

(i) Solve the inequality 1

1( 1)( 3)x x

≥ −+ +

. [3]

(ii) Hence, or otherwise, solve 1

1(ln 1)(ln 3)x x

≥ −+ +

, leaving your answer in exact

form. [2]

[ 3 or 2 or 1x x x< − = − > − ,3 2

1 1 10 or or

e e ex x x< < = > ]

Raffles Institution 2011 Year 6 H2 Mathematics

_________________________________________________________________________________________________________


(i) (ii)

2

11

( 1)( 3)

11 0

( 1)( 3)

( 2)0

( 1)( 3)

x x

x x

x

x x

≥ −+ +

+ ≥+ +

+≥

+ +

3 or 2 or 1x x x< − = − > −

Replace x by ln x,

3 2 1

3 2

ln 3 or ln 2 or ln 1

0 e or e or e

1 1 1 0 or or

e e e

x x x

x x x

x x x

− − −

< − = − > −

∴ < < = >

< < = >

3. NJC08/Prelim/I/1

Without the use of a calculator, solve the inequality 32

5≥

−x

x. Hence, find the range of

values of x that satisfy 32

5≥

−x

x. [5]

[ 3−≤x or 2>x , 2x > or 2−<x ]

32

5≥

−x

x 0

2

63

2

5≥

−−

−−

⇒x

x

x

x 0

2

635≥

−+−

⇒x

xx

02

62≥

−+

⇒x

x

( )( ) 0262 ≥−+⇒ xx , 2≠x

⇒ 3−≤x or 2>x

Replace x by x

3−≤⇒ x or 2>x

2>⇒ x or 2−<x

-1 -3 x

+ + _ _

-2


_________________________________________________________________________________________________________


4. JJC09/Prelim/I/6b

By using an algebraic method, solve the inequality 2 18

63

xx

x

−> −

+. [3]

Hence, solve the inequality 2ln 18

ln 6ln 3

xx

x

−> −

+. [3]

[ 3x < − or 0 < x < 5, 0 <3

x e−< or

51 x e< < ]

2 18( 6) 0

3

xx

x

−− − >

+

2 18 ( 3)( 6)0

3

x x x

x

− − + −>

+

2 50

3

x x

x

− +>

+

( 5)0

3

x x

x

−<

+

3x < − or 0 < x < 5

2ln 18

ln 6ln 3

xx

x

−> −

+

2ln 18 ln 6

ln 3

xx

x

−⇒ > −

+

For 2ln 18

ln 6ln 3

xx

x

−> −

+, replace x with ln x

ln 3x < − or 0 ln 5x< <

0 <3

x e−< or

51 x e< <

Functions 1 MJC 10/JC2/CT2/I/5

The function f is defined by 1 1

f :2

x xx

+ −� for ,x a x≥ ∈� .

Given that the inverse of f exists, find the smallest value of a.

Hence, write down the

domain of -1f . [2] Using the smallest value of a found,

(i) find, in a similar form, -1ff , stating its range. [2]

(ii) solve the equation ( ) ( )f ffx x= . [3]


_________________________________________________________________________________________________________


(iii) using the result in part (ii), state the value of ( )99f 2 . [1]

[a = 1, [ )-1f 1.5,= ∞ , [ )1.5,∞ , 2x = , ( )99f 2 2∴ = ]

From the graph of 1 1

2y x

x= + −

The function 1 1

f :2

x xx

+ −� is 1-1 when 1x ≥ .

Therefore, smallest a = 1

Domain of [ )-1f 1.5,= ∞ (i) ( )-1ff x x=

And its domain follows the domain of -1f .

Therefore,

1ff : x x−� , 1.5,x x≥ ∈�

Range of [ )-1ff 1.5,= ∞

(ii)

( ) ( )f ffx x=

Method 1

( ) ( )( )

-1 -1f f f ff

f

1 1

2

2

x x

x x

x xx

x

=

=

= + −

=

Method 2

(1, 1.5)


_________________________________________________________________________________________________________


( ) ( )

( )

2

2

2

2

2

1 1 1 21

2 2 2

1 21

2 2 2

2 11

2 2 2

4 2 2

2 5 2 0

2 1 2 0

12 or NA, 1

2

xx x

x x x x

x

x x

xx

x x

x x x

x x

x x

x x

+ − = + − +− +

− = − +− +

= ≥− +

= − +

− + =

− − =

= ≥

∵

∵

(iii) From part (ii), f(2) = ff(2) and since f(2) = 2, ( )99f 2 2∴ =

2 SAJC 10/Promos/I/5

The functions f and g are defined as follows:

2f : 6,

g : 6 5,

x x x x

x x x

+ − ∈

+ ∈

� �

� �

(i) Explain why the inverse function of f does not exist. [1]

(ii) Suppose that the domain of f is restricted to x a< , where a is a constant. Find the largest value of a for which the inverse function of f exists. [1]

With the new domain of f found in (ii),

(iii) Define 1f − in a similar form; [3] (iv) Show that the composite function gf exists; [1] (v) Define gf in a similar form and find its range. [3]

[ 0.5a = − , 1 1 25 25f : , ,

2 4 4x x x x− − − + ∈ > −� � ,

2gf : 6 6 31, , 0.5x x x x x+ − ∈ < −� � , gf

65,

2R

= − ∞

]


_________________________________________________________________________________________________________


(i) (ii) (iii) (iv) (v)

f is not one-to-one as there exists a horizontal line 0y = (or any other value within the

range of the function), which cuts the graph of f at two points, i.e. 3x = − and 2x = . Therefore the inverse function of f does not exist. Alternative method : Since f(-3) = f(2) = 0, f is not one-to-one. Hence inverse function of f does not exist.

By GC, the minimum point on the graph is 1 25

,2 6

− −

, i.e. ( )0.5, 6.25− − .

Alternatively, x-coordinate of minimum point on the graph is 3 2 1

2 2

− += − .

0.5a∴ = −

Let ( ) 2f 6, 0.5y x x x x= = + − < −

2

1 25

2 4y x

= + −

21 25

2 4

1 25

2 4

x y

x y

+ = +

= − ± +

Since 0.5x < − , 1 25

2 4x y= − − +

1 1 25 25 f : , ,

2 4 4x x x x

−∴ − − + ∈ > −� �

( )f g

25, ,

4R D

= − ∞ ⊆ = −∞ ∞

, therefore gf exists.

( )2 2gf ( ) 6 6 5 6 6 31x x x x x= + − + = + −

( )gf f , 0.5D D= = −∞ −

2 gf : 6 6 31, , 0.5x x x x x∴ + − ∈ < −� �

3−

y

x

0

( )fy x=

2


_________________________________________________________________________________________________________


( ) f g25 650.5 , ,

4 2

−∞ − → − ∞ → − ∞

gf

65 ,

2R

∴ = − ∞

3 CJC 08/Prelim/I/4

Functions f and g are defined by

0,,1:

0,,1

1:

>ℜ∈−−→

>ℜ∈−

→−

xxxxg

xxe

xfx

(i) With the aid of a diagram, show that 1−f exists.

(ii) Find 1−f in a similar form.

(iii) Only one of the composite functions fg and gf exists. Give a definition (including the domain) of the composite that exists and explain why the other composite does not exist.

(iv) The function g has an inverse if its domain is restricted to kx ≥ . Find the smallest

possible value of k and define the inverse function 1−g corresponding to this domain

for g.

[2] [3] [3] [2]

[ 1,1

1ln)(1 >

−−=− xx

xf , 0,1

11

1

1)( >

−−−=

−=

−−x

eegxgf

xx,

1( ) 1 , 0g x x x− = − ≤ ]


_________________________________________________________________________________________________________


(i) (ii) (iii) (iv)

Since every horizontal line , 0y k k= > cuts the graph of f at exactly one point, therefore f

is one-one. Since f is one-one, hence 1−f exists.

−−=

−=

=−

−=

−

−

−

yx

ye

ye

ey

x

x

x

11ln

11

11

1

1

1,1

1ln)(1 >

−−=− xx

xf

Since ),0()0,( ∞=⊄−∞= fg DR , fg does not exist.

Since ),0(),1( ∞=⊂∞= gf DR , gf exists.

0,1

11

1

1)( >

−−−=

−=

−−x

eegxgf

xx

k =1

0,1)(

1

1

1 ≤−=

−=

−=

− xxxg

yx

xy


_________________________________________________________________________________________________________


4

SAJC09/Prelims/II/4

The functions f and g are defined by

2f : 5x x−� , x k<

g : 3x x+� , x∈�

(i) Given that the inverse 1f − exists, find the largest value of k and define 1f − in similar form. [3]

(ii) Sketch, on the same diagram, the graphs of f, 1f − and 1ff − .

Solve the equation 1f( ) f ( )x x−= , giving your answer(s) in exact form. [4]

(iii) Determine whether the composite functions 1gf − and 1f g− exist. [2]

[1f : 5 , 5x x x

− − − <� , 1 21

2x

− −= , -1 -1 gf exists, f g does not exist. ]

(i) (ii)

Largest k = 0

Let 25y x= −

5 ( 5 is rejected as 0)x y x y x= − − = − <

1f : 5 , 5x x x− − − <�

1f( ) f ( ) f( )x x x x−= ⇒ =

y=f(x)

y=ff-1

(x)

y=f-1

(x)

5

5


_________________________________________________________________________________________________________


(iii)

2

2

5

5 0

1 21 1 21 (Rej as 0)

2 2

1 21 (ans in exact form)

2

x x

x x

x x x

x

− =

+ − =

− ± − += = <

− −∴ =

-1

-1

ff

-1

f

R D ( ,0)

D ( , )

Since R D , gf exists.

g

g

= = −∞

= −∞ ∞

⊆

-1

-1

f

-1

f

R [0, )

D ( ,5)

Since R D , f g does not exist.

g

g

= ∞

= −∞

⊄

Graphing Techniques 1. AJC09 Prelim/I/11

The graph of

−=2

3 fx

y has 2 stationary points at (8, -2) and (14, 2) and intersects the

x-axis at 18 and 10 ,6 === xxx as shown in the diagram below.

-4

(8, -2)

−=2

3 fx

y

x

y

6 10 18

(14, 2)


_________________________________________________________________________________________________________


Sketch, on separate clearly labeled diagrams, the graphs of

(a)

−=

2

3 f

1

xy [3]

(b) 2 f 3 2

xy

= − −

[3]

(c) )f( xy = [3]

Show clearly the asymptotes, stationary points and points of intersection with the axes in your diagrams.

(a)

y = -1/4 (8, -1/2)

(14, 1/2)

6 10 18 x

y

1

f 3 2

yx

= −


_________________________________________________________________________________________________________


(b)

x

y

y = -2 6 10

)2,8( −

)2,8(

y=2

2 f 3- 2

xy

=−

18

(c)

(-1, -2) 0 -2 -6

(-4, 2)

-4

( )xy f=

3


_________________________________________________________________________________________________________


2. HCI09/Prelim/I/3

The diagram below shows the graph of f '( )y x= . The curve passes through the origin and

has turning points at (−3, 0) and (−1.5, −1). The x-axis and x = 1 are the two asymptotes of the curve.

(i) Find the range of values of x for which the graph of y = f ( )x is strictly increasing and

concave downwards. [1]

(ii) State the x-coordinates of all the stationary points of the graph of f ( )y x= and

determine the nature of each point. [2]

(iii) Given that f (0) 1= , sketch the graph of f ( )y x= for 1x < . Your sketch should

indicate clearly all stationary points, asymptotes and intersections with the axes. [2]

[ 1x > , Point of inflexion at x = −3, Minimum point at x = 0.]

x

y

x=1

O −3

(−1.5, −1)


_________________________________________________________________________________________________________


(i)

Concave downwards ⇒ f ''( ) 0x < so the gradient of f '( )x <0

Strictly increasing ⇒ f '( )x >0

1x > (ans) (ii)

Stationary points: f '( )x =0 ⇒ 3, 0x = −

x (−3)− −3 (−3)+

f '( )x -ve 0 -ve

x 0− 0 0+

f '( )x -ve 0 +ve

(iii)

x

y

1

−3 −1.5

x =1

Point of inflexion at

x = −3

Minimum point at x

= 0.


_________________________________________________________________________________________________________


3. DHS08/Prelim/I/4

The diagram shows the graph of f ( )y x= . The curve has turning points at ( 2, )A p− and

(0, )B q where p and q are positive real numbers.

Sketch, on separate clearly labelled diagrams, the graphs of:

(i) ( )

1

f 2y

x=

+, [3]

(ii) ( )fy x= − . [3]

1

2y =

y

x

( 2, )A p−

(0, )B q

f( )y x=


_________________________________________________________________________________________________________


4. RJC/II/4

The graph of f ( )y x= has a minimum turning point at (4,0) and passes through the

origin. The lines 2x = and 2y = are asymptotes to the graph, as shown in the diagram

below.

f ( )y x=

2

2

0 4

y

x


_________________________________________________________________________________________________________


(i) State the range of values of x for which the graph of 1

f( )y

x= is decreasing. [1]

(ii) State the range of values of x for which the graph of f '( )y x= is below the x-axis. [2]

(iii)Sketch the graph of 2 f ( )y x= , showing clearly the equations of all asymptotes and the

shape of the graph at the origin. [3]

(iv) Sketch the graph of f (| |) 2y x= + , showing clearly the equations of all asymptotes and

the coordinates of the stationary points. [3]

[ (4, )x∈ ∞ , ( ,2) (2,4)x∈ −∞ ∪ ]

(i) The graph of 1

f( )y

x= is decreasing for (4, )x∈ ∞ .

(ii) The graph of f '( )y x= is below the x -axis for ( ,2) (2,4)x∈ −∞ ∪ .

(iii)

(iv)

2 f ( )y x=

2

√2

0 4

y

x

−√2

x = 2

y = √2

y = −√2

f ( ) 2y x= +

2

4

0

(4,2)

y

x

(−4,2)

−2

x=−2 x=2

y=4

2


_________________________________________________________________________________________________________


Arithmetic and Geometric Progression

1 JJC 09/1/8

A geometric series has first term a, common ratio r and nth term denoted by Gn. An arithmetic series has first term a, common difference d and nth term denoted by An. It is given that a, r and d are non-zero and the two series are related by the following equations

4 2 3A A G− = and 5 55 6 9A a G− = .

Show that 4 29 10 1 0r r− + = . [4]

It is also given that r > 0, 1r ≠ . (i) Deduce that the geometric series is convergent and show that its sum to infinity is

3

2a . [3]

(ii) Find the least value of N for which

1 1

10

N

n n

n n

A G

∞

= =

>∑ ∑ where a > 0. [4]

[ans: 12]

4 2 3

2

22

( 3 ) ( )

22

A A G

a d a d ar

ard ar d

− =

+ − + =

= ⇒ =

5 5

4

4

5 6 9

5( 4 ) 6 9

9 20 0 (1)

A a G

a d a ar

ar d a

− =

+ − =

− + = − − −

Sub 2

2

ard = into (1),

24

4 2

4 2

9 20 02

9 10 0

9 10 1 0 (Shown)

arar a

ar ar a

r r

− + =

− + =

− + =

4 2 2 2 2 2 1(i) 9 10 1 0 ( 1)(9 1) 0 1 or

9

1 11 (Reject r 1) or 1 (Reject 0) or (Reject 0) or

3 3

r r r r r r

r r r r r r

− + = ⇒ − − = ⇒ = =

= ≠ = − > = − > =∵ ∵ ∵

Since 1

13

r = < , the geometric series is convergent. 3

11 213

a aS a

r∞ = = =

− −.


_________________________________________________________________________________________________________


[ ]

1 1

22

2

(ii) 10

2 ( 1) 10 .......(2)2

1( )3Sub into (2),

2 2 18

32 ( 1) 10

2 18 2

Since 0, ( 35) 540 35 540 0

N

n n

n n

A G

Na N d S

aar ad

N aa N a

a N N N N

∞

= =

∞

>

+ − >

= = =

+ − >

> + > ⇒ + − >

∑ ∑

From GC, 46.59 or 11.59N N< − > Therefore least value of N =12.

2 NYJC 09/1/9

(a) In a convergent geometric progression, the sum of the first n terms is equal to the

sum of the remaining terms. Given also that the ( )1n + th term is 1

2, determine the

value of the ( )2 1n + th term. [4]

(b) An arithmetic progression has first term a and common difference d . The eighth, third and second term of the progression are successive terms of an infinitegeometric progression.

If the first term of the geometric progression is 10, find the sum of the even-numbered terms of the progression. [4]

A sequence is formed in which the n th term is given by 1ln n nu u + , where nu is the

n th term of the geometric progression. Show that the sequence forms an arithmetic progression and state the value of the common difference. [3]

[ans: 1

4 ,

25

12, 2ln 5− ]

( ) 2n n na S S S S S∞ ∞= − ⇒ =

( )1

21 1

na r a

r r

−=

− −

( ) 1Since 1, 2 1 1

2

n nr r r≠ − = ⇒ =

Given 1

( 1) term = 2

th nn ar+ = , hence 1a =

22 1 1

(2 1) term = 2 4

th nn ar

+ = =


_________________________________________________________________________________________________________


(b)

22( 7 )( ) ( 2 )

7 2

a d a da d a d a d

a d a d

+ += ⇒ + + = +

+ +

23 4 0d ad+ =

40 (Reject 0) or

3d d d a= ≠ = −∵

413

82 5

3

a aa d

ra d

a a

−+= = =

+ −

Sum of even-numbered terms

2 2

10255121 1

15

ar

r= = =

− −

1 1

21 1

1 1

term ( 1) term= ln ln

1ln ln ln 2ln which is a constant

5

th thn n n n

n n n

n n n

n n u u u u

u u ur

u u u

+ −

+ +

− −

− − −

= = = =

Therefore the sequence forms an arithmetic progression. Common difference 2ln 5= −

3 TJC 09/1/9

(a) Find the sum of the arithmetic series

(m + 1) + (m + 3) + (m + 5) + . . . . + (3m − 3) where m is a positive integer. [4]

(b) A customer purchases a new 52-inch LCD television set from Counts Hypermarket for $7000 and decides to pay the entire amount by loan instalment. He takes the loan at the beginning of June during the Mid-Year Sale and he repays $p at the end of each month where p < 7000. The Hypermarket will then charge a 5% interest on the outstanding balance after each monthly repayment. Show that the outstanding amount

owes at the end of the nth month is ( )7000(1.05) 21 1.05 1n np− − . [3]

Deduce the least monthly repayment amount p, rounded off to the nearest dollar, required to pay off the entire loan by the end of the 12th month. [3]

[ans: ( 1)(2 1)m m− −,

$753 ]

(a) Let n be the number of terms.

nth term = 3m − 3 = (m + 1) + (n − 1)2 ⇒ n = m − 1

Sum of the arithmetic series = [ ]1 1( 1) (3 3) ( 1)(4 2) ( 1)(2 1)

2 2

mm m m m m m

−+ + − = − − = − −

(b) Month Start of month End of month 1 7000 (7000 − p)1.05 2 (7000 − p)1.05 [(7000−p)1.05 − p]1.05


_________________________________________________________________________________________________________


=(7000)1.052 − p(1.052 + 1.05) 3 (7000)1.052 − p(1.052 + 1.05) [(7000)1.052 − p(1.052 + 1.05) − p]1.05 =(7000)1.053 − p(1.053 + 1.052 + 1.05) At the end of the nth month, the outstanding amount owes = (7000)1.05n − p(1.05n + 1.05n−1 + . . . + 1.05)

=( )1.05 1.05 1

7000(1.05)1.05 1

n

np

−−

− = ( )7000(1.05) 21 1.05 1n np− − (shown)

To pay off the entire loan by the 12th month, the outstanding amount owes at the end of the

12th

month must be less than or equal to 0.

( )12 127000(1.05) 21 1.05 1p− − ≤ 0

12

12 7000(1.05 )(1.05 1)

21p − ≥

752.17p ≥

The least monthly installment is $753.

4 DHS 09/1/9b

The arithmetic progression 2, 4, 6, 8, 10, 12, ... is arranged in rows in the following way: 1st Row: 2 2nd Row: 4, 6 3rd Row: 8, 10, 12 4th Row: 14, 16, 18, 20

…

(i) Show that the first term in the nth row is 2 2n n− + . [2]

(ii) Find the sum of all the terms from the 1st row to the (n – 1)th row. [3]

[ans: ( )( )21 2

4

n n n n− − +]

(i) First term of each row: 2, 4, 8, 14…. Difference between the first terms: 2, 4, 6…. is AP with a = 2, d = 2.

Sum of the first (n – 1) differences = ( ) ( ) 212 2 2 2

2

nn n n

−+ − = −

First term of nth row = Sum of the first (n – 1) differences + 1st term = 2 2n n− +

(ii) Total no. of terms from 1st row to (n – 1)th row = 1 + 2 + 3 …+ (n – 1)

( ) ( ) ( )1 1

1 12 2

n n nn

− −= + − =

Sum of all terms from 1st row to (n – 1)th row

( )

( )2

1

2 2 2 22

n n

n n

−

= + − + − ( )( )21 2

4

n n n n− − +=


_________________________________________________________________________________________________________


Summation, Recurrence Relations and Mathematical Induction

1 AJC 09/1/12

A sequence of real positive numbers 1 2 3, , , ...u u u satisfies the recurrence relation

1 4 4 1,n n nu u u n+

+ = + + + ∈Z and 1 15u = .

(i) Prove by induction that ( )24 1 1nu n= + − for all n

+∈Z . [4]

Express 1

nu in the form

2 1 2 3

A B

n n+

+ +, where A and B are constants to be determined.

[2]

(iii) Hence find 1

1N

n nu=∑ in terms of N. [2]

(iv) Using your answer in part (iii), find 2

1

1

4 1

N

n n= −∑ . [3]

[ans: (ii) ( ) ( )

1 1

2 2 1 2 2 3n n−

+ +

(iii) ( )

1 1

6 2 2 3N−

+ (iv)

1 1

2 2(2 1)N−

+]

(i) Let Pn be the statement 1 4 4 1,n n nu u u n+

+ = + + + ∈Z

When 1n = ,

L.H.S. = 1u = 15 (given)

R.H.S. = ( )24 2 1− = 15 Since LHS = RHS ∴ 1P is true

Assume Pk is true for some k+∈Z ,

i.e. ( )24 1 1ku k= + −

Prove that 1Pk + is true,

i.e. to prove ( )2

1 4 2 1k

u k+ = + −

When 1n k= + ,

L.H.S. = 1ku + = 4 4 1 (given)k ku u+ + +

= ( ) ( )2 24 1 1 4 4 4 1 1 1k k+ − + + + − +

= ( ) ( )24 1 1 4 4 2 1k k+ − + + × +

= ( ) ( )24 1 2 1 1 1k k + + + + −

= ( )2

4 1 1 1k + + −


_________________________________________________________________________________________________________


= ( )24 2 1k + −

Since Pk is true implies 1Pk + is true, and 1P is true, by mathematical induction, Pn is true

for all n+∈Z .

(ii) ( )2

1 1

4 1 1nu n=

+ −

( )

2

1

2 1 1n=

+ −

( )( ) ( )( )

1

2 1 1 2 1 1n n=

+ − + +

( ) ( )

1

2 1 2 3n n=

+ +

( ) ( )

1 1

2 2 1 2 2 3n n= −

+ +

1 1

,2 2

A B∴ = = −

(iii) 1

1N

N

n n

Su=

=∑

( ) ( )1

1 1

2 2 1 2 2 3

N

n n n=

= − + + ∑

1 1 1

2 3 5

1 1

5 7

1 1

2 1 2 1

1 1

2 1 2 3

N N

N N

= −

+ −

+ −− +

+ − + +

�

= 1 1 1

2 3 2 3N

− + =

( )1 1

6 2 2 3N−

+

(iv) 2

1

1

4 1

N

n n= −∑1

20

1

4( 1) 1

N

t t

−

=

=+ −∑


_________________________________________________________________________________________________________


1

21

1 1

3 4( 1) 1

N

t t

−

=

= ++ −∑

1 1 1

3 6 2(2( 1)) 3)N= + −

− +

1 1

2 2(2 1)N= −

+

2 JJC 09/1/7

(a) A sequence of integers 0 1 2, , ,u u u � is defined by 0 1u = and 1 3 7n nu u+ = − .

Prove by induction that ( )17 5 3

2

n

nu = −

for all non-negative integral values

of n. [4]

(b) Prove that for all positive integers n , ( )

1 1 1

! 1 ! ! ( 1)!n n n n− =

+ + −. [2]

Hence evaluate ( )

1

1

! 1 !

N

nn n

=+ −∑ in terms of N . [2]

Deduce that

1

12

!n

n

∞

=

<∑ . [2]

[ans: (b) ( )

11

1 !N−

+]

Let Pn be the statement ( )17 5 3

2

n

nu = −

for , 0n n∈ ≥Z

when = 0n , L.H.S. = 0 1u = (given)

R.H.S. = ( )017 5 3 1

2 − =

∴ 0P is true.

Assume that Pk is true for some , 0k k∈ ≥Z ,

i.e. ( )17 5 3

2

k

ku = −

Prove that 1+kP is true,

i.e. to prove ( ) 1

1

17 5 3

2

k

ku

+

+ = −

L.H.S. = 1ku +


_________________________________________________________________________________________________________


= 3 7ku − (given)

= ( )37 5 3 7

2

k − −

( ) 117 5 3

2

k+ = −

∴ 1Pk+ is true

Since 0P is true and Pk is true 1Pk+⇒ is true,

by Mathematical Induction, Pn is true for all non-negative integral values of n.

L.H.S. = ( ) ( )

1 1 1 1

! 1 ! ! ! 1n n n n n− = −

+ +

( )

( )1 1

! 1

n

n n

+ −=

+

= ( )! 1

n

n n +

= ( )

1

1 ! ( 1)n n n

n

− +

= ( )

1

1 !( 1)n n− +

= ( ) ( )

1

1 ! 1 !n n n− + −

=1

! ( 1)!n n+ − = R.H.S. (Shown)

( )1

1

! 1 !

N

nn n= + −∑

= ( )1

1 1

! 1 !

N

n n n=

−

+ ∑

= ( )

1 1 1 1 1 1 1 1.........

1! 2! 2! 3! 3! 4! ! 1 !N N

− + − + − + + − +

= ( )

11

1 !N−

+


_________________________________________________________________________________________________________


Since ( )1 1

! 1 ! ! !n n n n>

+ − +

( )1 1

1 1

! 1 ! ! !

N N

n nn n n n= =

>+ − +∑ ∑

( )1 1

1 1

! 1 ! 2 !

N N

n nn n n= =

>+ −∑ ∑

( )1 1

1 12

! ! 1 !

N N

n nn n n= =

<+ −∑ ∑

1

1 12 1

! ( 1)!

N

n n N=

< − +

∑

As ( )

1, 0

1 !N

N→ ∞ →

+

∴1

12

!n n

∞

=

<∑ (shown)

3 TJC 09/1/7

A sequence of positive real numbers 0x , x1 , x 2 , x 3 , ... satisfies the recurrence relation

1

1

2 1

3n

n

n

xx

x

−

−

+=

+, for positive integers n.

(i) As n → ∞ , nx → α . Determine the exact value of α . [3]

(ii) Describe the behaviour of the sequence when 0 0.5x = . [1]

(iii) Show that 1n nx x+ < when n

x α> . [4]

[ans: (i) 1 5

2

− +

(ii) sequence is increasing and converges to α ]


_________________________________________________________________________________________________________


(i) As n → ∞ , nx → α ,

2 1

3

αα

α+

=+

2

3 2 1α α α∴ + = +

2

1 0α α+ − =

α⇒ =1 5

2

− + or

1 5

2

− − (rejected ∵ n

x > 0)

(ii) Using GC, the sequence is increasing and converges to α when 0 0.5x = .

(iii) 1

2 1

3n

n n n

n

xx x x

x+

+− = −

+

Sketch the graph of 2 1

3

xy x

x

+= −

+ and

observe that when x > α , 2 1

03

xx

x

+− <

+

Therefore, replace x by nx ,

1 1

2 10 0

3n

n n n n n

n

xx x x x x

x+ +

+− < ⇒ − < ⇒ <

+

4 MJC 09/1/1

Express 2

2

6 8r r+ + in partial fractions. [2]

Find 2

1

2

6 8

n

r r r= + +∑ , giving your answer in the form ( )fk n− , where k is a constant. [2]

Find also 2

21

6 88

6 8

n

r

r r

r r=

+ ++ +∑ . [2]


_________________________________________________________________________________________________________


[ans: 1 1

;2 4r r

−+ +

( )7 1 112 3 4n n+ +− + ; ( )70 1 1

3 3 440

n nn + ++ − + ]

2

2 1 1

6 8 2 4r r r r= −

+ + + +

( )

21 1

1 153

1 14 6

1 11 3

1 12 4

1 1 1 13 4 3 4

7 1 112 3 4

2 1 1

6 8 2 4

n n

r r

n n

n n

n n

n n

r r r r= =

+ +

+ +

+ +

+ +

= − + + + +

= −

+ −

+ −

+ −

= + − −

= − +

∑ ∑

�

( )( )

2

2 21 1

21

7 1 112 3 4

70 1 13 3 4

6 88 801

6 8 6 8

240

6 8

40

40

n n

r r

n

r

n n

n n

r r

r r r r

nr r

n

n

= =

=

+ +

+ +

+ + = + + + + +

= ++ +

= + − +

= + − +

∑ ∑

∑

5 SRJC 09/1/4 Prove by induction that

( )( ) ( )( ) ( )( ) ( )( ) ( ) ( ) ( )22 2 2 2 2 3 1 22 1 3 2 4 2 5 2 ........ 1 2 2 2 2n nn n

−+ + + + + + = + −

for all positive integers n. [5]


_________________________________________________________________________________________________________


Let nP be the statement ( )2 1 2

1

( 1) (2 ) 2 2 2,n

r n

r

r n n− +

=

+ = + − ∈∑ Z

When 1n = : ( )( )2LHS 2 1 4= =

( )( )2RHS 1 2 2 2 4 LHS= + − = =

Hence 1P is true.

Assume that kP is true for some k

+∈Z .

i.e. ( )2 1 2

1

( 1) (2 ) 2 2 2k

r k

r

r k−

=

+ = + −∑

Prove that 1kP + is true.

i.e. ( )1

2 1 2 1

1

( 1) (2 ) ( 1) 2 2 2k

r k

r

r k+

− +

=

+ = + + −∑

( ) ( ) ( )1

22 1 2

1

( 1) (2 ) 2 2 2 2 2k

r k k

r

r k k+

−

=

+ = + − + +∑

( )( )( )

2 2

2 1

2 1

2 2 4 4 2

2 3 2 2

1 2 2 2

k

k

k

k k k

k k

k

+

+

= + + + + −

= + + −

= + + −

kP is true ⇒ 1k

P + is true.

1P is true and kP is true ⇒ 1k

P + is true. Hence by mathematical induction, nP is true for all

positive integers n.

Binomial Expansion and Power Series 1 IJC 09/1/7

Find the expansion of

1

31

1 2

x

x

− +

in ascending powers of x, up to and including the term in

2x . State the range of values of x for which the expansion is valid. [5]

Without performing any calculations, explain why putting 1

6x = into the result gives a

better approximation to 3 5 than putting 4

11x = − . [1]

Hence by putting 1

6x = , find an approximation for 3 5 , expressing your answer as a

fraction in its lowest terms. [2]


_________________________________________________________________________________________________________


2 31[1 ..., ]

18x x− + +

( ) ( )1

2 23

1 1 1 2 1 11 1 ... 1 ...

3 2! 3 3 3 9x x x x x

− = − + ⋅ − − + = − − +

( ) ( )1

2 23

1 1 1 4 2 81 2 1 (2 ) 2 ... 1 ...

3 2! 3 3 3 9x x x x x

− + = − + ⋅ − − + = − + +

11 1

33 3

2 2 2

1(1 ) (1 2 )

1 2

1 1 2 81 ... 1 ... 1 ...

3 9 3 9

xx x

x

x x x x x x

−− = − − +

= − − + − + + = − + +

For expansion to be valid, 1 and | 2 | 1 x x< < ⇒1

1 and | |2

x x< <

1 1 1 1 1Range of value of ,

2 2 2 2 2x x x

∴ < ⇒ − < < ⇒ = −

Both values of x are within the range of values of x. 1

6x = will give a better approximation

as it is closer to zero.

Substitute 1

6x = ,

1 1

3 32

1 51

1 16 6 12 8 6 616 6

− = ≈ − + +

( )

1

3

1

3

5 1 11

8 6 36

31 315 2

36 18

≈ − +

≈ =

2 TPJC 09/1/3

Expand 2

2

4 9x + in ascending powers of x up to and including the term in 4

x .

State the range of values of x for which the series expansion is valid. [3]

Deduce the expansion of 3

2 2(4 9)x−

+ in ascending powers of x up to and including the term

in 2x . [3]

2 4 22 4 4 1 2[ ..., ...]

3 27 81 27 81x x x− + + − +


_________________________________________________________________________________________________________


2

2

4 9x + =

1

224

2 9 19

x

− +

=

1

221 4

2 13 9

x

− +

=

2

2 2

1 3

2 1 4 42 21 ...

3 2 9 2! 9x x

− − − + +

= 2 42 4 4...

3 27 81x x− + +

Expansion is valid for 24 3 3 3 31 Range of value of ,

9 2 2 2 2x x x

< ⇒ − < < ⇒ = −

.

3

2 2(4 9)x−

+ = 1

2 2 12(4 9) (4 9)x x− −+ +

=

1

2 4 21 2 4 4 1 4... 1

2 3 27 81 9 9x x x

− − + + +

= 2 41 2 4 4...

2 3 27 81x x

− + +

( ) ( ) 2

2 21 21 4 4

1 ...9 9 2! 9

x x − − − + +

= 2 21 2 8 4...

18 3 27 27x x

− − +

= 21 2...

27 81x− +

Alternatively,

Now, 3

2 2

2

d 28 (4 9)

d 4 9x x

x x

− = − +

+ .

So, 3

2 2(4 9)x−

+ = 2 41 d 2 4 4...

8 d 3 27 81x x

x x

− − + +

= 31 8 16

...8 27 81

x xx

− − + +

= 21 2...

27 81x− +

3 AJC 08/1/4

It is given that ( )1

2cosy x= .

(i) Show that

222

2

d d2 2 0

d d

y yy y

x x

+ + =

.

(ii) Find Maclaurin’s series for y in ascending powers of x, up to and including the term in x2.


_________________________________________________________________________________________________________


(iii) By choosing a suitable value for x, deduce the approximate relation 2

4

11

2kπ≈ + ,

where k is a constant to be determined.

21[1 ]

4x−

(i) ( )1

22cos cosy x y x= ⇒ =

2 22 22

2 2

222

2

2 sin

2 2 cos 2 2

2 2 0

dyy x

dx

d y dy d y dyy x y y

dx dx dx dx

d y dyy y

dx dx

= −

+ = − ⇒ + = −

∴ + + =

(ii) When x = 0, y = 1

2 2

2 2

2 0 0

12 0 1 0

2

dy dy

dx dx

d y d y

dx dx

= ⇒ =

+ + = ⇒ = −

2 2

1

121 ... 1

2 4y x x

− = + + ≈ −

(iii) ( )1

24

1 1cos cos

42 2x x x

π= ⇒ = ⇒ =

Sub 4

xπ

= :

2

2

4

1 1 11 1

4 4 642

ππ ≈ − = −

4 RJC 08/2/5

The function f ( )y x= satisfies the differential equation 2d2 1

d

yy y

x= − and its graph

passes through the point (0,3) .

Show that when 0x = , 2

2

d 20

d 27

y

x= .

Find the Maclaurin’s series of y in ascending powers of x, up to and including the term

in 3x .

2 34 10 14[3 ...]

3 27 243x x x+ + + +


_________________________________________________________________________________________________________


2d2 1

d

yy y

x= − ----- (1)

Differentiate (1) w.r.t. x , we get

22

2

d d d2 2 2

d d d

y y yy y

x x x

+ =

22

2

d d d

d d d

y y yy y

x x x

+ =

----- (2)

When 0x = , 3y = , d 4

d 3

y

x=

Subst into (2),

22

2

d 4 43 3

d 3 3

y

x

+ =

2

2

d 1 16 20 4

d 3 9 27

y

x

⇒ = − =

(shown)

Differentiate (2) w.r.t. x , we get

23 2 2 2

3 2 2 2

d d d d d d d2

d d d d d d d

y y y y y y yy y

x x x x x x x

+ + = +

23 2 2

3 2 2

d d d d d3

d d d d d

y y y y yy y

x x x x x

+ = +

When 0x = , 3y = , d 4

d 3

y

x= ,

2

2

d 20

d 27

y

x= ,

3

3

d 28

d 81

y

x=

By Maclaurin’s Thm,

2 34 20 28

3 ...3 27 2! 81 3!

x xy x

= + + + +

2 34 10 143 ...

3 27 243x x x= + + + +

5 In triangle ABC, angle 1

3A π= radians, angle

1

3B xπ = +

radians and

1

3C xπ = −

radians, where x is small. By using the sine formula, or otherwise, show that

2

3

axb c− ≈ .


_________________________________________________________________________________________________________


(sine rule)

sin( ) sin sin( )3 3 3

sin cos cos sin sin cos cos sin3 3 3 3

and

sin sin3 3

1 1cos sin and cos sin

3 3

2 2sin for small (shown)

3 3

b a c

x x

a x x a x x

b c

b a x x c a x x

a axb c x x

π π π

π π π π

π π

= =+ −

+ − = =

= + = −

∴ − = ≈

Differentiation

1. RJC/2006/Promo/RJC/P1/5

A sphere of radius 1 m is inscribed in a right circular cone of base radius r and height h as shown in Figure 1.

(i) By considering similar triangles ACD and AEB shown in Figure 2 or otherwise, show

that 2

2

hr

h=

−. Find the volume of the cone in terms of h and π. [3]

(ii) Find the values of h and r which will result in the smallest volume of the cone. [3]

[The formula for the volume of a cone is 21

3r hπ .]

[2

3m3( 2)

h

h

π−

; , 24h r= = ]

1(i)

Consider similar triangles ACD and AEB.

AD CD AC

AB EB AE= =

( )

2 2

21 1 1 1

h r r h

h h

+⇒ = =

− − −

A

B C

Figure 1

r

h

1

1

Figure 2

B

C D

E

A


_________________________________________________________________________________________________________


( )

22

22

2

22 1 1

(Shown)2

h hr r

h hh h

hr

h

= ⇒ =−− + −

⇒ =−

2

23

1

3

1( )

3 2

m3( 2)

V r h

hh

h

h

h

π

π

π

=

=−

=−

(ii)

From the GC, minimum value of V occurs when 4h = .

2 42

4 2r r= ⇒ =

−

Alternatively, obtain ( )

( )2

4d

d 3 2

h hV

h h

π −=

−.

Set d

0d

V

h= to obtain h=0 or h=4 as stationary values. Use graph or 1st derivative test

or 2nd derivative test to conclude V is minimum at h=4 and

2 4

24 2

r r= ⇒ =−

.

2. (a) Prove that 2 2d 1e (2 1) e

d 4

x xx x

x

− = . [2]

(b) Write down ( )2dtan( ) .

dx

x Hence find ( )3 2 2sec d .x x x∫ [4]

[ ( ) ( )2 2 21tan ln sec

2x x x c − +

]

2(a)

( )2

2 2

2

1 de (2 1)

4 d1

e (2) (2 1)(2e )4

e

x

x x

x

xx

x

x

−

= + −

=

2(b)

( )( ) ( )2 2 2dtan 2 sec

dx x x

x=

B

A

D

E

C

1

1 1h −

2 2h r+

r


_________________________________________________________________________________________________________


( ) ( )

( ) ( )

( ) ( )

3 2 2 2 2 2

2 2 2

2 2 2

1sec d 2 sec d

2

1 tan 2 tan d

2

1 tan ln sec

2

x x x x x x x

x x x x x

x x x c

=

= −

= − +

∫ ∫

∫

3. RJC/2008/Prelim/P1/Q11

(a) The curve C has the equation 2 yx

− = . The point A on C has x-coordinate a where a > 0.

Show that d 1

d ln 2

y

x a= − at A and find the equation of the tangent to C at A. [3]

Hence find the equation of the tangent to C which passes through the origin. [2]

The straight line y mx= intersects C at 2 distinct points. Write down the range of

values of m. [1]

[1 1 ln

ln 2 ln 2 ln 2

ay x

a= − + − ,

1 1, 0

e ln 2 e ln 2y x m= − − < < ]

(b) A movie theatre screen which is 5 m high, has its lower edge 1 m above an observer’s

eye. The visual angle θ of the observer seated x m away is as shown in the diagram below.

(i) Show that 1 16 1tan tan

x xθ − − = −

. [1]

(ii) Find the exact distance the observer should sit to obtain the largest visual angle. [4] [You need not establish that the distance gives the largest visual angle.]

θ

x

1

5


_________________________________________________________________________________________________________


(iii) Suppose that the observer is situated between 2 m and 15 m from the screen. Find, to the nearest degree, the smallest visual angle. [2]

[ 6, 18° ]

3(a)

[3]

[2]

[1]

ln2

ln 2

d 1

d ln 2

d 1When , [shown]

d ln 2

ln1ln 2ln 2

1 1 ln

ln 2 ln 2 ln 2

y xx y

y

x x

yx a

x a

ay

x a a

ay x

a

− = ⇔ = −

= −

= = −

+= −

−

= − + −

1 ln0 e

ln 2 ln 2

aa− = ⇒ =

Equation of tangent passing through origin is 1

e ln 2y x= −

From the graph, 1

0e ln 2

m− < < and so the range of m is 1

,0e ln 2

−

.

3(b)

[1]

2

1

e l n 2

yx

y x

− =

= −

θ

x

1

5

A

D

B

C


_________________________________________________________________________________________________________


[4]

[2]

(i) 1

tan ABDx

∠ = and 6

tan ABCx

∠ =

1 16 1tan tanABC ABD

x xθ − − = ∠ − ∠ = −

[shown]

(ii) 2 22 2

d 1 6 1 1

d 6 11 1

x x x

x x

θ − − = − + +

2 2

6 1

36 1x x

−= +

+ +

For stationary value, 2 2

d 6 10 0

d 36 1x x x

θ −= ⇒ + =

+ +

2 26( 1) 36 0x x⇒ − + + + = 25 30

6 or 6 (NA)

x

x

⇒ =

⇒ = −

The required distance is 6 m.

(iii)

From the graph we obtain minimum value when x=15.

The required angle is 18° .


_________________________________________________________________________________________________________


4. NJC/2010/Promo/Q9

A piece of paper is used to make a conical cup (as shown in Figure 1) with base radius r cm,

height h cm, and fixed slant height AB, of 12 cm. Let V be the volume of the cone. Show that

31(144 )

3V h hπ= − . [1]

(i) Let h1 be the height of the cone such that the volume of the cone is a maximum. Find the exact value of h1. [3]

[ 1 4 3h = ]

To produce disposable conical cups with maximum volume, a manufacturer adopts the following design: a typical cup (as shown in Figure 2) has a fixed slant height AB of 12 cm, fixed height of h1 cm (as found in (i)), and fixed radius of r1 cm. The height of the water in the cone is y cm, and the radius of the water in the cone is x cm. Let W be the volume of water in the cup.

Show that

2

31

1

1

3

rW y

hπ

=

. [1]

(ii) Water is leaking from a hole at the bottom of this cup at a constant rate of 3 120 cm s− . Find the rate of change in the height of water in the cup at the instant when the height

of water is equal to 1

2

h. [ 0.265 cm/s− ]


_________________________________________________________________________________________________________


4 2 2 2 212 144r h h= − = −

2 2 31 1(144 ) (144 )

3 3

1

3r h h h h hV π π π= − = −=

4(i) 2(14 31

3)

d

d4

V

hhπ −=

When the height is a maximum, d

0,d

V

h= so

2

1 1

144(144 3 ) 0 4 3

3

1

3h hπ − = ⇒ = =

Since 2

2

d

d02

V

hhπ= <− for all ,h this gives a maximum.

4 1

1

1

1

h y

r x

rx y

h

=

=

Let W be the volume of the water in the cone ,

2

2

1

1

2

31

1

1

3

1 =

3

1 =

3

W x y

ry y

h

ry

h

π

π

π

=

12 y

x

h1

r1


_________________________________________________________________________________________________________


4(ii) 2

21

1

d

d

W ry

y hπ

=

2 2

1 1

1

d d d 1. .( 20)

d d d

2

y y W

t W t r h

hπ

= = −

( )

( )

22

2

1 122

1 1

144 4 3 962

484 3

r r

h h

− = = = =

( ) ( )d 1

.( 20)d 2 12

20

24

50.265 cm/s

6

y

t π

π

π

= −

−=

−= = −

5. DHS/2010/Prelim/P1/Q8

The diagram shows a hexagon PQRSTU inscribed in a circle with radius 6 cm. The sides QR

and UT are parallel, and 2 cm.QR UT x= =

(i) Show that A, the area of the hexagon PQRSTU, is ( ) 2 22 6 36 cm .x x+ − [3]

(ii) Using differentiation, find the value of x when A is a maximum. (You need not verify that it gives a maximum value.) [4]


_________________________________________________________________________________________________________


Initially 6 cm,x = and the lengths of the parallel sides QR and UT are each decreasing at a

constant rate of 11cm s .

10

− Find the rate of change of A at the instant when 2 cm.x = [3]

[ Area of a trapezium = 1

sum of the lengths of the parallel sides height2

× × ]

[ 3 cmx = ; 12 cm s

5

−− ]

5(i) 236ON x= −

( )

( )

2

2

12 12 2 36

2

2 6 36

A x x

x x

= × × + −

= + −

5(ii)

( )2

2

d 1 22 36 2 6

d 2 36

A xx x

x x

= − + + − −

( )

( ) ( )

2 2

2

2

2

2

72 2 12 2

36

4 18 3

36

4 6 3

36

x x x

x

x x

x

x x

x

− − −=

−

− −=

−

+ −=

−

dFor maximum , 0 : 0 3 cm

d= > ⇒ =

AA x x

x

( ) ( ) ( )d d d 1 d 12 2

d d d 10 d 20

xQR x x

t t t t= = = − ⇒ = −

( )( )4 8 1d

When 2, 32 4 2d 32

Ax

x= = = =

• P S

Q R

U T

2x

6 6 O

N

6 6


_________________________________________________________________________________________________________


Integration

1 DHS 09/1/8

(a) Evaluate 2

2

0 4 3 dx x x− +∫ without the use of the graphic calculator. [3]

(b) Use integration by parts to find 1tan dx x−

∫ . [3]

(c) Use the substitution π

tan , 02

x t t= < < to find 2 2

1 d

1x

x x +∫ . [4]

21 21 1

[ 2, tan ln(1 ) , ]2

xx x x c c

x

− +− + + − +

Solution

(a) 2 1 2

2 2 2

0 0 1 4 3 d ( 4 3) d ( 4 3) dx x x x x x x x x− + = − + − − +∫ ∫ ∫

1 23 3

2 2

0 1

2 3 2 33 3

1 8 12 3 8 6 2 3

3 3 3

2

x xx x x x

= − + − − +

= − + − − + + − +

=

2 4 3 0 for 0 1x x x− + > < < 2 4 3 0 for 1 2x x x− + < < <

(b)

1 1

2tan d tan d

1

xx x x x x

x

− −= −+∫ ∫

1 dtan , 1

d

vu x

x

−= =

2

d 1,

d 1

uv x

x x= =

+

2 1

d d d

d d d

1 4 2

20

2 cm s

5

A A x

t x t

−

= ×

= × −

= −

A is decreasing at the rate of 12 cm s .

5

−


_________________________________________________________________________________________________________


1

2

1 2

1 2tan d

2 1

1tan ln(1 )

2

xx x x

x

x x x c

−

−

= −+

= − + +

∫

(c)

2

22 2

1 1 d sec d

tan sec1x t t

t tx x=

+∫ ∫

2

2

1

2

cos d (cos )(sin ) dsin

(sin ) 1

1 sin

1

tt t t t

t

tc c

t

xc

x

−

−

= =

= + = − +−

+= − +

∫ ∫

π

tan , 02

x t t= < <

2dsec

d

xt

t=

2 21 tan 1 secx t t+ = + =

2 (a) NJC09/1/8a

Write down the constants A and B such that, for all values of x ,

(2 4)x A x B= + + .

Hence find 2

d4 6

xx

x x+ +∫ . [5]

(b) TPJC 09/1/2

Using the substitution 2 1u x= − , evaluate the integral 2

1

2 1 d

1

xx

x

−+∫ , giving

your answer in exact form. [5]

2 1 2 3[ ln 4 6 2 tan , 2( 3 1) ]

62

xx x c

π− + + + − + − −

(a) 1

(2 4) 22

x x= + − . Hence 1

2A = and 2B = − .

2 2 2

1 2 4 1d d 2 d

4 6 2 4 6 4 6

x xx x x

x x x x x x

+= −

+ + + + + +∫ ∫ ∫


_________________________________________________________________________________________________________


2 2 2

2 1

2 1

1 2 4 1 d 2 d

2 4 6 ( 2) ( 2)

1 2 2ln( 4 6) tan

2 2 2

2ln 4 6 2 tan

2

xx x

x x x

xx x c

xx x c

−

−

+= −

+ + + +

+ = + + − +

+ = + + − +

∫ ∫

(b) 2 3

1 1 2

2 1 d d

11( 3)

2

x ux u u

xu

−=

+ +∫ ∫

[ ]

23

21

3

21

3

3 1

1

1

2 d

3

6 2 d

3

62 tan

3 3

62( 3 1)

4 63

32( 3 1)

6

uu

u

uu

uu

π π

π

−

=+

= − +

= −

= − − −

= − −

∫

∫

2 1u x= −

21( 1)

2x u= +

d 1

(2 )d 2

xu u

u= =

2 21 11 ( 1) 1 ( 3)

2 2x u u+ = + + = +

1, 1x u= =

2, 3x u= =

3 VJC 09/1/10

Use integration by parts to show that 2

2 ee d (2 1)

4

xx

x x x c= − +∫ , where c is an arbitrary

constant. [2]

x

y exy x=

O

4e

2y x=

4(4, 2e )


_________________________________________________________________________________________________________


The diagram above shows the graph of a straight line 4e

2y x= and part of the graph of a

curve C given by exy x= .

(a) The region R is bounded by the line 4e

2y x= and C between 0x = and 8x = .

Find, correct to 1 decimal place, the area of R . [2]

(b) The region S is bounded by C and the line4e

2y x= between 0x = and 4x = .Show

that the volume V of the solid formed when S is rotated 2π radians about the x -

axis is given by 8( e )V A Bπ= − , where A and B are exact constants to be

determined. [4]

2 2 21e d e e d

2 2

x x xxx x x= −∫ ∫

2 2

2

1e e

2 4

e(2 1)

4

x x

x

xc

x c

= − +

= − +

(a)

Area of R 4

8

0

e e d 7239.2 (1 d.p.)

2

xx x x= − =∫

(b)

Required volume 4

4 2 2

0

48 2

0

48 2

0

8 8

8

1(2e ) (4) (e ) d

3

16e e d

3

16e e (2 1)

3 4

16e (7e 1)

3 4

43 1e

12 4

x

x

x

x x

x x

x

π π

π π

ππ

ππ

π

= −

= −

= − −

= − +

= −

∫

∫

2d, e

d

xvu x

x= =

2d 11, e

d 2

xuv

x= =


_________________________________________________________________________________________________________


43 1 ,

12 4A B∴ = =

4 ACJC 09/2/3 (part)

The curve C is defined parametrically by 2

tx = and ty ln= , 0t > .

The region bounded by the curve C , the x -axis and the lines 2=x and 5=x is denoted by R .

(i) Sketch the graph of C for 52 ≤≤ x , indicating clearly the coordinates of the end-points. [2]

(ii) Express the area of R as an integral in t and show that it has an exact value given by

ln 5 ln 2α β γ+ + , where α , β and γ are constants to be determined. [5]

5 3

[ , 1, ]2 2

α β γ= = − = −

(i)

(ii) Area of R 2 d 2

d

xx t t

t= ⇒ =

2, 2x t= = , 5, 5x t= =

dln , 2

d

vu t t

x= =

2d 1,

d

uv t

x t= =

1ln 2

2

1ln 5

2

2 5

12, ln 2

2

15, ln 5

2


_________________________________________________________________________________________________________


5

2

5

2

5 52

2 2

52

2

d

(ln )(2 ) d

ln d

5ln 5 2 ln 22

5 3ln 5 ln 2

2 2

y x

t t t

t t t t

t

=

=

= −

= − −

= − −

∫

∫

∫

5 3

, 1, 2 2

α β γ∴ = = − = −

5 NYJC 09/2/3a

The parametric equations of a curve are (1 cos ), ( sin ),x a t y a t t= − = +

where a is a positive constant and 0 2t π≤ ≤ . (i) Sketch the curve. [2]

(ii) The region bounded by the curve and the y -axis is denoted by R . Find the volume

of solid formed when R is rotated through 2π radians about the y -axis. [4] 3[9.87 ]a

Solution

(i)

(ii) d

( sin ) (1 cos )d

yy a t t a t

t= + ⇒ = +

When 0, 0y t= = ; When 2 , 2y a tπ π= =

Required volume


_________________________________________________________________________________________________________


[ ] [ ]

22

0

22

0

23 2

0

3

d

(1 cos ) (1 cos ) d

(1 cos ) (1 cos ) d

9.87 (3 s.f.)

a

x y

a t a t t

a t t t

a

π

π

π

π

π

π

=

= − +

= − +

=

∫

∫

∫

6 HCI/10/MYE/BT2/Q6

State a geometrical transformation which would transform the graph of exy = − onto the

graph of e xy −= − . [1]

The region R is bounded by the two curves and the line 1x = − . (i) Find the exact value of the area of R. [3]

(ii) Hence find the exact value of 1

1| e e | dx x x−

−−∫ . [2]

(iii) Find the volume of the solid of revolution when R is rotated through 4 right angles

about the y-axis. [3] 1 1[ 2 e e , 4 2e 2e , 4.62 ]− −− + + − + +


_________________________________________________________________________________________________________


The graph of e xy −= − can be obtained by reflecting the graph of exy = − in the y-axis.

(i)

0

1

0

1

0

1

1

1

( e ) ( e )d

(e e )d

[ e e ]

[ 1 1] [ e e ]

2 e e

x x

x x

x x

R x

x

−

−

−

−

−−

−

−

= − − −

= −

= − −

= − − − − −

= − + +

∫

∫

(ii)

1

1

0 1

1 0

0

1

1

| e e | d

e e d e e d

2 e e d

(by the symmetry relationship of the two graphs)

= 4 2e 2e

x x

x x x x

x x

x

x x

x

−

−

− −

−

−

−

−

−

= − + −

= −

− + +

∫

∫ ∫

∫

e xy −= −

exy = −

1x = −


_________________________________________________________________________________________________________


(iii)

e ln( )

e ln( )

x

x

y x y

y x y−

= − ⇒ = −

= − ⇒ = − −

1 1when 1, f( ) e , g( ) ex x x−= − = − = − 1

1

1 e2 1 2 2

e 1

e2 1 2

e

(1) (e e ) [ ln( )] d [ln( )] d

(1) (e e ) [ln( )] d

4.62

V y y y y

y y

π π π

π π

−

−

− −−

− −

−−

−

= − − − − − −

= − − −

=

∫ ∫

∫

7 MJC/10/MYE/P1/Q11

(a) Show that 2 1x x+ + is positive for all real values of x. [1]

Hence, evaluate 1

21

2 1 d

1

xx

x x−

+

+ +∫ , giving your answer in the form lnp

q, where p

and q are integers. [4]

(b) (i) Find e cos dxx x

−∫ . [3]

(ii) Show that 2

2

2e cos d 2 e cos 2 dx xx x x x

β β

α α− −=∫ ∫ . [2]

(iii) Hence, find the volume of the solid formed when the finite region

bounded by the curve e cos 1xy x

−= + , the axes and the line πx = is

rotated through 4 right angles about the x-axis. Give your answer in an exact form. [5]

( ) 216 1 3 11[ ln , e sin cos , e e ]

3 2 8 8

xx x c

π ππ π− − − − + − + + +

(a) 2

2 1 31 0

2 4x x x

+ + = + + >

for all real x

OR

Discriminant =1 4 3 0− = − < and coefficient of 2 1 0x = >

( )11 1

212 2 21 12

2 12 1 2 1 d d d

1 1 1

xx xx x x

x x x x x x

−

− − −

− ++ += +

+ + + + + +∫ ∫ ∫

11

2 221

12

ln 1 ln 1x x x x−

− − = − + + + + +


_________________________________________________________________________________________________________


3 3

ln ln1 ln 3 ln4 4

= − − + −

3

ln 3 2ln4

= −

16ln 3

9

16ln

3

= ×

=

(b)(i)

e cos d e sin e sin dx x xx x x x x

− − −= +∫ ∫

e sin e cos e cos dx x xx x x x

− − −= − − ∫

( )

2 e cos d e sin e cos '

1e cos d e sin cos

2

x x x

x x

x x x x c

x x x x c

− − −

− −

= − +∫

= − +∫

(ii) Let 2x y=

d2

d

x

y=

When 2 , x yα α= =

2 , x yβ β= =

2 22 e cos d e cos 2 2dx y

x x y yβ βα α

− −=∫ ∫

22 e cos 2 dyy y

βα

−= ∫

22 e cos 2 dxx x

βα

−= ∫ (shown)

(iii) Volume required

( )( )

2

0

2 2

0

e cos 1 d

e cos 2e cos 1 d

x

x x

x x

x x x

π

π

π

π

−

− −

= +

= + +

∫

∫

2 2

0

1 1e cos 2 e 2e cos 1 d

2 2

x x xx x x

ππ − − − = + + +

∫

2 2

0 0 0

1e 1 d e cos 2 d 2 e cos d

2 2

x x xx x x x x

π π πππ π− − − = + + +

∫ ∫ ∫

22

0 00

1 1e . e cos d 2 e cos d

4 2 2

x x xx x x x x

ππ ππ

π π− − − = − + + + ∫ ∫

( ) ( )2

2

0 0

1 1 1 1e e sin cos 2 e sin cos

4 4 4 2 2

x xx x x x

π ππ π

π π π− − − = − + + + − + −


_________________________________________________________________________________________________________


( ) ( ) ( ) ( )2 2

2

1 1e e 1 1 e 1 1

4 4 8

3 11e e

8 8

π π π

π π

ππ π π

π π

− − −

− −

= − + + + − − − + − −

= − + + +

8 SRJC/10/MYE/Q12

(a) (i) Find ( )22 sin dx x x∫ . [1]

(ii) Hence solve ( )6 3 20

sin dx x xπ

∫ exactly. [3]

(b) The diagram below shows the shaded region B that is bounded by the curve

2 9y x= − and the line 7

4y x= − .

7

4y x= −

(i) Find the exact volume generated when region B is rotated 2π radian about

the axisy − . Leave your answers in terms of π . [3]

(ii) Hence, state the exact volume generated when region B and region C are

rotated 2π radian about the axisy − . Explain your answers clearly. [2]

( )2 6 3 272[ cos , , ]

24 3x c

ππ

−− +

y

x

0

– 3 3

2 9y x= −

B C


_________________________________________________________________________________________________________


a. (i) ( ) ( )2 22 sin d cosx x x x c= − +∫

(ii) ( ) ( )2

3 2 26 6

0 0sin d 2 sin d

2

xx x x x x x

π π

= ∫ ∫

( ) ( )2 6

2 26

00

cos cos d2

xx x x x

= − +

∫

ππ

( )26

0

1cos 2 cos d

12 6 2x x x

= − +

∫ππ π

( )2 6

0

3 1sin

24 2

3 1sin

24 2 6

6 3

24

x = − +

= − +

−=

ππ

π π

π

b. (i) ( ) ( )0 7

2

9 0

16Volume 9 d 9 d

49y y y y y

−

= + + + −

∫ ∫π π

272

3= π units3

(ii) Since curve 2 9y x= − is symmetrical about the y-axis, so the volume

generated about y-axis due to C overlap with the volume generated by B about the y-axis. Thus

272Volume generated due to B and C about y-axis

3= π units3


_________________________________________________________________________________________________________


Differential Equations

1 Solve the following differential equations:

(a) 2

2

2

d12 6 2

d

yx x

x= + + , [3]

(b) 2

2

dsin

d

Lt t

t= , given that

d0 and 0

d

LL

t= = when 0,t = [6]

(c) 2d

d

yy

x= and sketch the family of solution curves, [5]

(d) 2

1 d1

d

y

y x+ = . [4]

[(a) 4 3 2y x x x Ax B= + + + + ;(b) 2 2cos sinL t t t= − − ;(c) 1

yC x

=−

;

(d) 1 1

ln2 1

yy x C

y

−+ = +

+]

(a)2

2

2

3 2

4 3 2

d12 6 2

dd

4 3 2d

, where and are arbitrary constants.

yx x

xy

x x x Ax

y x x x Ax B A B

= + +

⇒ = + + +

⇒ = + + + +

(b)2

2

dsin

dd

cos cos dd

= cos sin where is an arbitrary constant.

Lt t

tL

t t t tt

t t t A A

=

⇒ = − − −

− + +

∫

( )

dcos sin

d

cos sin d

= sin sin d cos +

= sin cos cos +

= sin 2 cos + where is an arbitrary constant.

Lt t t A

t

L t t t A t

t t t t t At

t t t t At B

t t t At B B

= − + +

⇒ = − + +

− − − −

− − − +− − +

∫∫

Since d

0d

L

t= when 0,t = we get 0.A =

Using 0L = when 0t = gives 2,B = so 2 2cos sin .L t t t= − −


_________________________________________________________________________________________________________


(c)

2

2

2

d

d1 d

1d

1 d d

1

1 where and are an arbitrary constants

yy

xy

y x

y xy

x By

y B CC x

=

⇒ =

⇒ =

⇒ − = +

⇒ =−

∫ ∫

(d)

0C =

y

1C = 1C = −

−1 1

x

−1

1


_________________________________________________________________________________________________________


2

2

2

2

2

2

2

2

2

1 d 1

d

d 11

d

d 1

d

d1

1 d

d 1 d1

11 d 1 d

1

y

y x

y

x y

y y

x y

y y

y x

yy x

y

y xy

+ =

⇒ = −

−⇒ =

⇒ =−

⇒ = −

⇒ + = −

∫ ∫

∫ ∫

1 1ln where is an arbitrary constant

2 1

yy x C C

y

−⇒ + = +

+

2. [MI 09 Prelim P1 Q11(b)] (i) Find the general solution of the differential equation

212

=

+dx

dy

x

x.

[2] (ii) Find the particular solution of the differential equation in (i) for which y = 0 and

x = 0. [1] (iii) Sketch, on a single diagram, three graphs from the same family of curves, including

the particular solution found in part (ii). [2] Describe the gradient of every solution curve as x → +∞ . [1]

[(i) ( )2ln 1y x c= + + ;(ii) ( )2ln 1y x= + ; gradient tends towards zero as x → +∞ ]

(i)

( )

2

2

2

12

d 2

d 1

ln 1 , where is an arbitrary constant

x dy

x dx

y x

x x

y x c c

+=

=+

= + +

(ii) When 0x = and 0y = , we have 0c = .

Hence the particular solution is ( )1ln 2 += xy

(iii


_________________________________________________________________________________________________________


Gradient tends towards zero as x → +∞

(

2

2

1

d 00

d 1 1 0x

x

y

x= → =

+ +∵ as x → +∞ )

3. [PJC 09 MYE P2 Q3]

The quantities x and y are related by the differential equation ( )2 d 11

d

yx yx x

x y+ = − . By

means of the substitution v yx= , show that this differential equation can be reduced to 2d 1

d

v v

x v

−= . Given that 2y = − when 1x = − , find, in the form of ( )2 fy x= , the

particular solution of this differential equation. [7]

[ 2 2 2

2 2

1 3 xy e

x x

+= + ]

( )2 d 11

d

yx yx x

x y+ = − ----(1)

v yx= ----(2)

Differentiate (2) w.r.t. x , we have d d

d d

v yy x

x x= + ⇒

d d

d d

y vx y

x x= − ---(3)

Substitute (2) and (3) into (1):

( )d1

d

v xx y v x

x v

− + = −

d

d

v xx xy vx v

x v− + = −

1c =

0c =

1c = −

Note that since the gradient at

0x = is zero, the curve should have a minimum

turning point at 0x = .


_________________________________________________________________________________________________________


d

d

v xx v vx v

x v− = − −

2d

d

v v x xx

x v

−=

2d 1

d

v v

x v

−= (shown)

2d d

1

vv x

v=

−∫ ∫

21ln 1

2v x c− = + where c is an arbitrary constant.

2 2 2 21 c x xv e e Ae− = ± = , where 2c

A e= ± 2 21 x

v Ae= +

From (2), when 1 and 2, 2x y v= − = − =

∴ 2 2 22 1 3Ae A e−= + ⇒ =

Hence ( )2 2 21 3 xv e e= + ⇒ 2 2 2 21 3 xy x e += + ⇒ 2 2 2

2 2

1 3 xy e

x x

+= +

4 AJC 05/1/15(a)

Show that the substitution z x y= + reduces the differential equation

( ) ( )1d

d 2 2 1

x y x yy

x x y

+ − +=

+ + to

2d 1

d 2 1

z z z

x z

+ +=

+.

Hence find the general solution of the differential equation

( ) ( )1d

d 2 2 1

x y x yy

x x y

+ − +=

+ +. [7]

[ ( )2

1 exx y x y A+ + + + = ]

d d1

d d

z x y

z y

x x

= +

⇒ = +

So ( ) ( )1d

d 2 2 1

x y x yy

x x y

+ − +=

+ + becomes

d ( 1)1

d 2 1

z z z

x z

−− =

+

Hence 2 2d ( 1) 2 1 1

1d 2 1 2 1 2 1

z z z z z z z z

x z z z

− − + + + += + = =

+ + + (shown)


_________________________________________________________________________________________________________


2

2

2 1 d 1 d

1ln 1 where is an arbitrary constant.

zz x

z zz z x C C

+ ⇒ = + + ⇒ + + = +

∫ ∫

( )( )( )( )

2

2

2

2

Hence ln 1

1 e .e

1 e .e

1 e where is an arbitrary constant.

C x

C x

x

x y x y x C

x y x y

x y x y

x y x y A A

+ + + + = +

⇒ + + + + =

⇒ + + + + = ±⇒ + + + + =

The general solution is ( )2

1 e xx y x y A+ + + + = .

5 SAJC 07/Y6/CT2/11

A circular patch of weeds starts to grow at the center on the surface of a pond of radius a

at a rate that is proportional to the area of the pond not covered by the weeds.

(i) Show that the radius, r m of the circular patch of weeds satisfies the differential

equation2 2d ( )

d 2

r k a r

t r

−= . [4]

(ii) Solve the differential equation, expressing r in terms of t, a and k. [3] (iii) Given that the initial radius of the circular patch of weeds is 0 m, will the weeds

cover the whole pond? Justify your answer. [3]

[(ii) 2 e ktr a B

−= − ]

(i)

Let A m2 be the area of the circular patch of weeds and r m be the radius of the patch at time

t.

Then ( )2 2d

d

Ak a r

tπ π= − where k is a positive constant.

But A = πr2. So

d d2

d d

A rr

t tπ=

Hence d

2d

rr

tπ ( )2 2k a rπ= −

Thus 2 2d ( )

d 2

r k a r

t r

−= .

(ii)


_________________________________________________________________________________________________________


2 2

2 2

2 2

2 2 ( )

2 2

2 2

2 2

2

d ( )

d 2

2 d d

ln where is an arbitrary constant.

e e .e

e .e

e where is an arbitrary constant

Thus e

As 0, e

kt c c kt

c kt

kt

kt

k

r k a r

t r

rr k t

a r

a r kt c c

a r

a r

a r B B

r a B

r r a B

− + − −

− −

−

−

−

−=

⇒ =−

⇒ − − = +

⇒ − = =

⇒ − = ±

⇒ − =

= −

> = −

∫ ∫

t

(iii) When t = 0, r = 0, so B = a2.

The particular solution is 2 2e 1 ekt ktr a a a

− −= − = − .

Since e−kt > 0 for all t ≥ 0, where k is a positive constant, we have 1 − e−kt < 1.

Hence r < a. Thus the weeds will never cover the whole pond.

Vectors

1 AJC/09/Prelims/I/13

The points P and Q have position vectors i j− and 3 13 6i j k+ + respectively. The

plane 1π contains the point P and the line 1 , 02

xz y= − − = .

(i) Find a vector equation of the plane 1π in scalar product form. [3]

(ii) Find the position vector of the foot of the perpendicular from Q to 1π . [2]

The line l1 passes through the points P and Q.

(iii) The line l2 is the reflection of the line l1 about the plane 1π . Find a vector equation

of l2. [3]

The plane 2π has the equation 6

4

a

b

• =

r .

Find the values of a and b such that

(iv) 1π and 2π are parallel and at a distance of 224 apart. [3]

(v) 1π and 2π are intersecting. [1]


_________________________________________________________________________________________________________


[Ans: (i)

1

3 2

2

• = −

r (ii)

1

1

2

ON

− = −

�� (iii)

5 3

11 5 ,

10 5

r λ λ

− = − + ∈ −

��

(iv) 2a = ,

108 or -116b = (v) 2,a b≠ ∈� ]

(i)

0 2

1 , 0 0 02

1 1

xz y r λ

= − − = ⇒ = + − −

�

Vector parallel to 2

1 0 1

1 0 1

0 1 1

π = − − = − −

1

1 2 1

1 0 3

1 1 2

1 0 1 1

: 3 0 3 3 2

2 1 2 2

n

π

= − × = −

• = • ⇒ • = − −

r r

�

(ii) Let foot of perpendicular from Q to 2π be N.

3 1

13 3

6 2

ON λ = +

��

3 1

13 3 3 2

6 2 2

4

λλ

λ

λ

+ + • = +

⇒ = −

1

1

2

ON

− = −

��

(iii)

1 3 5'

' 2 1 13 112

2 6 10

OQ OQON OQ

− − + = ⇒ = − = −

− −

��


_________________________________________________________________________________________________________


2 2

5 1 6 3

' 11 1 10 2 5

10 0 10 5

1 3 5 3

: 1 5 OR : 11 5

0 5 10 5

PQ

l r l rλ λ

− − = − − − = − = − − −

− ∴ = − + ∴ = − + −

��

� �

(iv)

1

3

2

= 6 2

4

a

k a

⇒ =

Method 1:

1 2

1 1

: 3 2 : 32

2 2

bπ π

• = − • =

r r

Distance between the 2 planes = ( )2

2 2241 9 4

b− −

=+ +

108 or -116b⇒ =

Method 2:

Distance QN = ( )( ) ( ) ( )( )2 223 1 13 1 6 2 224− − + − + − − =

2 3 2 2 5 2

6 13 6 108 OR ' 6 11 6 116

4 6 4 4 10 4

b OQ b OQ

− = • = • = = • = − • = − −

��

Method 3

1

1 11 2

: 3 2 314 14

2 2

π • = − ⇒ • = −

r r

2

11 2

: 3 22414 14

2

1

3 2 56 54 58 108 116

2

b

π • = − ±

⇒ • = − ± = − ⇒ = −

r

r or or


_________________________________________________________________________________________________________


(v) 2,a b≠ ∈�

2 ACJC/09/Prelims/I/12

The coordinates of the points A and B are (-1, 2, 4) and (0, -2, 5) respectively. The equations of two planes p1 and p2 are 7x y z+ + = and 1x y az− + = respectively, where

a is a constant.

(i) Find the position vector of the foot of perpendicular from the point A to the plane p1. [4]

(ii) The two planes p1 and p2 intersect at the line . Verify that the point with coordinates (4, 3, 0) is a common point on the planes p1 and p2 and hence find the equation of the line . [3]

(iii) Another plane p3 has equation 2 3x y z b+ + = . Find the values of a and b if all the

three planes p1, p2 and p3 intersect at the line . [4]

(iv) If the length of projection of the line segment AB on the line is 6

1 unit, find

the values of a. [3]

[Ans: (i)

11

83

14

−

(ii) r

4 1

3 1

0 2

a

aλ

+ = + − −

(iii) 3a = − ; b = 10 (iv) 0.781a = or 1.25a = ]

(i) Let N be the foot of perpendicular from A to the plane 1p .

1

1

1

AN λ =

��

1 1 1

2 1 2

4 1 4

ON OA AN

λ

λ λλ

− − + = + = + = + +

��

Since N is a point on the plane 1p ,

1

. 1 7

1

ON

=

��

1 1

2 . 1 7

4 1

λ

λλ

− + + = +


_________________________________________________________________________________________________________


( 1 ) (2 ) (4 ) 7λ λ λ− + + + + + =

2λ =

21

31 12 1

2 2 83 3

4 142

43

ON

λ

λ

λ

− + − + − = + = + = +

+

��

(ii)

4 1

3 . 1 4 3 0 7

0 1

= + + =

(satisfies equation of plane 1p )

4 1

3 . 1 4 3 0 1

0 a

− = − + =

(satisfies equation of plane 2p )

Hence the point (4, 3, 0) is on both planes 1p and 2p .

(4, 3, 0) is then a point on the line where the 2 planes intersect.

1 1 1

1 1 1

1 2

a

a

a

+ × − = − −

is the direction vector of the line ,

Hence r

4 1

3 1

0 2

a

aλ

+ = + − −

is the equation of line .

(iii) If the planes 3p with equation r

1

. 2

3

b

=

intersects with 1p and 2p at line , then

1 1

2 . 1 0

3 2

a

a

+ − = −

(1 ) 2(1 ) 6 0a a+ + − − =

3a = −

The line lies in the plane 3p and thus

4

3

0

lies in the plane 3p also.


_________________________________________________________________________________________________________


4 1

3 . 2 4 6 0 10

0 3

b b

= ⇒ = + + =

(iv)

0 1 1

2 2 4

5 4 1

AB OB OA

− = − = − − = −

��.

Given the length of projection of line segment AB on the line is 1

6.

1

. 1

2 1

1 6

1

2

a

AB a

a

a

+ − − =+

− −

��

2 2 2

(1 ) 4(1 ) 2 1

6(1 ) (1 ) ( 2)

a a

a a

+ − − −=

+ + − + −

2

5( 1) 1

3( 3)

a

a

−=

+

2

2

25( 1) 1

( 3) 3

a

a

−=

+

2 275( 1) 3a a− = + 274 150 72 0a a− + =

150 1188 75 247

2(74) 74a

± ±= =

0.781a = or 1.25a =

3 CJC/09/Prelims/I/11

The planes p1 and p2, which meet in the line l, have vector equations

r =

+

−

+

1

1

0

1

0

1

6

4

2

11 µλ ,


_________________________________________________________________________________________________________


r =

+

+

1

0

1

0

3

2

6

4

2

22 µλ

respectively, where 121 ,, µλλ and 2µ are real constants.

(i) Show that l is parallel to the vector 5i + 6j + k. [3] (ii) Calculate the acute angle between p1 and p2. [2] (iii) Find, in exact form, the perpendicular distance from the point with coordinates

(4, 2, 2) to p2. [2] The plane p3 has equation ax – 2y + 2z = b, where a, b ℜ∈ .

(iv) Find b in terms of a such that all three planes meet at the single common point

with position vector

6

4

2

. [4]

(v) If given instead that a = 2, find the values of b, such that the distance between the

planes p1 and p3 is 3

1 units. [3]

[Ans: (ii) 75.7o (iii) 22 (iv) b = 2a + 4, where a ≠ 2 (v) b = 6 or 10]

(i) Let n1 and n2 be the normals of p1 and p2 respectively.

n1 = =

×

− 1

1

0

1

0

1

−

1

1

1

n2 =

−

−=

×

3

2

3

1

0

1

0

3

2

=

−

−×

−

1

6

5

3

2

3

1

1

1

Therefore l is parallel to 5i + 6j + k.

(ii) Acute angle between p1 and p2 = 223

3

2

3

1

1

1

cos 1

−

−⋅

−

−


_________________________________________________________________________________________________________


= 75.7o

(iii)Perpendicular distance = 22

3

2

3

6

4

2

2

2

4

−

−⋅

−

= 22

3

2

3

4

2

2

−

−⋅

−

−

= 22

22−

= 22 (iv) The point (2, 4, 6) must lie on p3

So 2a – 8 + 12 = b b = 2a + 4

Since the three planes meet at a single point, we must exclude the case where they meet along a line.

If p3 meets in the line l, the normal of p3 is perpendicular to l.

So

⋅

−

1

6

5

2

2

a

= 0

a = 2 Therefore b = 2a + 4, where a ≠ 2

(v) p1: 4

1

1-

1

. =

r and p3: b=

2

2-

2

. r

Distance between p1 and p2 = 3

1

3

1

323

4=−

b

3

1

323

4=−

b or

3

1

323

4 −=−

b

b = 6 or 10


_________________________________________________________________________________________________________


4 DHS/09/Prelims/I/5

The position vectors of points A, B, C and D are given by 2 3 4= + −a i j k , 5 2= − +b i j k ,

11 14λ= + +c i j k and 4 3= −d j k respectively. Find

(a) the position vector of P on AB if AP : PB = 2 : 1, [2] (b) the position vector of E, in terms of λ such that ABCE is a parallelogram, [2] (c) the angle DAB, giving your answer to the nearest degree, [2]

(d) the value of λ if A, B and C are collinear. [2]

[Ans: ((a)

4

13

0

(b)

8

4

8

λ +

(c) 102° (d) 9λ = − ]

(a) By ratio theorem,

2 5

3 2 1 44 2 1

330

OP

+ − − = =

��

(b)

11 3 8

4 4

14 6 8

AB EC OE λ λ = ⇒ = − − = +

��

(c)

2

1

1

AD

− =

��

Angle required = 1

3 2

4 1

6 1cos 102

61 6

−

− − • = °

(d)

11 5 6 3

1 1 4

14 2 12 6

BC k AB BC kλ λ = ⇒ = − − = + = −

��

2 & 8 1 9k λ∴ = = − − = −


_________________________________________________________________________________________________________


5 DHS/09/Prelims/I/10

The equation of the line l is

1 7

8 2

8 1

λ = + −

r and the equation of the plane 1π is

7 2 1x y z− + = − . The point A has coordinates (1, 8, 8).

(i) Verify that point A lies on plane 1π . [1]

The equation of the plane 2π is

0 1 0

1 4 2

2 1 4

µ γ = + +

r .

(ii) Find the angle line l makes with 2π . [3]

(iii) Describe, with a reason, the geometrical representation of the planes 1π and 2π .

[1]

(iv) Hence find the exact shortest distance between 1π and 2π . [3]

(v) The point P has position vector 4 +i j . Find the position vector of point P’, the

reflection of point P in the plane 1π . [4]

[Ans: (ii) 90° (iii) Since l is both perpendicular to π1 and π2, π1 and π2 must be parallel or Since

the normal vectors of π1 and π2 are parallel, π1 and π2 are also parallel. (iv) 1

54 (v)

3

3

1

− −

]

(i) At point A, ( ) ( ) ( )7 1 2 8 8 1 RHS− + = − =

(ii) Normal vector of π2 =

1 0 14

4 2 4

1 4 2

× = −

Let θ be the angle required.

14 7

4 2

2 1 108sin 1 90 or

2216 54 11664

πθ θ

− • − = = = ⇒ = °

(iii)Normal vector of π1=

7

2

1

−

is parallel to direction vector of l

Since l is both perpendicular to π1 and π2, π1 and π2 must be parallel. Alternative:


_________________________________________________________________________________________________________


Normal vector of π1=

7

2

1

−

=

141

42

2

−

Since the normal vectors of π1 and π2 are parallel, π1 and π2 are also parallel.

(iv) Observe that π2 contains origin,

Shortest distance required = ( )22 2

1 1

547 2 1

−=

+ − +

Alternative:

( )Let 0,1, 2B ≡

0 1 1

1 8 7

2 8 6

AB

− = − = − −

��

Shortest distance required = ( )22 2

1 7

7 2

6 1 1

547 2 1

− − • − − =

+ − +

(v) Let M be the foot of perpendicular of P on π1.

( )

3 7 7

7 2 27

8 1 1 1ˆ ˆ 2

254 541

PM PA

− • − − = • = = − −

n n��

' '

7 4 31

2 2 1 32

1 0 1

OP PP OP= +

− = − − + = −

��

Alternative:

Let l2 be the line passing through point P and parallel to normal vector of π1.


_________________________________________________________________________________________________________


2

4 7

: 1 2

0 1

l s

= + −

r

4 7 7

1 2 2 1

0 1 1

s

+ − • − = −

128 49 2 4 1

2s s s s+ − + + = − ⇒ = −

Let M be the foot of perpendicular of P on π1 and P’ be the point of reflection required.

12

2 & '

12

a

OM OP b

c

= =

−

��

Using midpoint theorem: 4 1 0 1 1

, , , 2,2 2 2 2 2

a b c+ + + = −

3

Thus ' 3

1

OP

− = −

��

6 IJC/09/Prelims/I/13

The diagram shows a cuboid with horizontal rectangular base ABCD, where 5AB = units,

3AD = units and 2AP = units. The edges AP, BQ, CR and DS are vertical, and PQRS is the top of the cuboid. The point A is taken as the origin and vectors i, j, k, each of length 1 unit, are taken along AB, AD, AP respectively.

(i) Show that the line DQ has equation

A B

C D

P Q

R S

i

j k


_________________________________________________________________________________________________________


3 (5 3 2 )λ= + − +r j i j k ,

where λ is a parameter. Find an equation of the line BS. [4]

(ii) Hence calculate the acute angle between the lines DQ and BS, giving your answer to

the nearest 0.1°. [3]

The point X lies on RC such that 2

,3

RX RC= and the point M is the midpoint of BD. The

plane π contains the points R, B and M.

(iii) Find, in scalar product form, an equation of the plane π. [3]

(iv) Find the exact value of the shortest distance from X to the plane π . [3]

[Ans: (i) 3 (5 3 2 )λ= + − +r j i j k 5 ( 5 3 2 )µ= + − + +r j i j k (ii) 37.9

(iii)

6

10 30

15

• = −

r (iv) 20

units19

]

(i) 5 3 2= − +i j k��DQ

Vector equation of line DQ,

0 5

3 3

0 2

3 (5 3 2 )

AD DQλ

λ

λ

= +

= + −

= + − +

r

j i j k

��

5i 3j 2kBS = − + +��

Vector equation of line BS,

5 ( 5 3 2 )

AB BSµµ

= +

= + − + +

r

j i j k

��

(ii) Let the angle between the lines DQ and BS be θ

1

cos

5 5

3 3 38 38 cos

2 2

30 cos 142.15

38

DQ BS DQ BS θ

θ

θ −

=

− − =

− = =

�� i

i

Therefore, acute angle between the lines DQ and BS

180 142.15

37.9 (to1dp)

= −

=


_________________________________________________________________________________________________________


(iii) = +�� BD BA AD

5= − i+3j

3 2

BR BC CR= +

= +j k

��

n

5 0 6

3 3 10

0 2 15

− = × = −

Equation of plane π,

6 5 6

10 0 10 30

15 0 15

• = • = − −

r

(iv) = +�� BX BC CX

2

33

= +j k

Shortest distance from X to the planeπ,

0 6

3 10

2 15

3

36 100 225

20units

19

•

− =

+ +

=

7 IJC/II/2

(a) The plane П is given by the vector equation 14= −r n⋅⋅⋅⋅ .

Given that 7

6=n units, find the shortest distance from the origin O to the plane П.

[2] (b) Relative to the origin O, the position vectors of points A and B are a and b

respectively. Given that angle AOB is 90°, show that the position vector of the foot of the perpendicular from O to AB is

( )2

2 2+ −

+

aa b a

a b [5]

[Ans: (a) 12]


_________________________________________________________________________________________________________


(a) shortest distance from the origin O to the plane П 14

127

6

−= =

r•n= units

n

(b) Let F be the foot of perpendicular from O to AB. ( )OF λ= + −a b a��

Since OF is perpendicular to the line AB,

[ ]

( )( )

2 2 2

2 2 2

2

2 2

( ) 0

( ) ( ) 0

( ) ( ) ( ) 0

2 0

0

OF

λ

λ

λ

λ

λ

• − =

+ − • − =

• − + − • − =

• − + − • + =

− + + =

=+

b a

a b a b a

a b a b a b a

a b a b a b a

a b a

a

a b

��

( )

2

2 2

( )OF λ= + −

= + −+

a b a

aa b a

a b

��

8 JJC/09/Prelims/I/9

Relative to a fixed point O, the position vectors of the points A, B and C are given as follows :

,i kOA = +��

2 ,i jOB = +��

4 2i j kOC = + +��

.

It is known that A, B and C exist in the plane 1Π .

(i) Show that the equation of plane 1Π is 2 3 1x y z− − = . [2]

(ii) The point D has position vector 5i j k+ − . Find the position vector of the foot of

perpendicular from D to plane 1Π and hence find the exact distance from D to plane

1Π . [4]

(iii) It is further known that the point D is the reflection of point C about another plane

2Π . Show that the equation of plane 2Π is 2 3x y z− − = .

Find the equation of the line of intersection between planes 1Π and 2Π . [5]


_________________________________________________________________________________________________________


[(ii) 4

5/ 2

1/ 2

−

, 14

2. (iii)

8 5

5 3 , .

0 1

λ λ = + ∈

r where � ]

(i)

1 2 4

0 1 2

1 0 1

= = =

�� OA OB OC

1 3

1 2

1 0

= = −

�� AB AC

2

3

1

= × = − −

��

�n AB AC

r.n=a.n

2 1 2

. 3 0 . 3

1 1 1

− = − − −

r

2 3 1 (Shown)− − =x y z

(ii) Eqn of perpendicular line passing thru D to plane is

5 2

1 3 (1)

1 1

λ = + − − − − − −

r

2

. 3 1 (2)

1

− = − − − −

Eqn of plane r

5 2 2

Sub (1) into (2) 1 3 . 3 1

1 1

1

2

λ

λλ

λ

+ − − = −− −

= −

5 2 41

Foot of perpendicular : 1 3 5 / 22

1 1 1/ 2

r

= − − = − − −

�


_________________________________________________________________________________________________________


4 5 114

Shortest Dist from D to plane : 5 / 2 1 3 / 22

1/ 2 1 1/ 2

− − = = − −

(iii) 2If and are reflections, is a normal vector of ΠC D CD

5 4 1

1 2 = 1

1 1 2

= − − − −

��CD

2If and are reflections, mid point of will exist in

5 4 9 / 21

Mid point of = 1 2 3 / 22

1 1 0

Π

+ + = − +

C D CD

CD

2

1 9 / 2 1

Equation of : . 1 3 / 2 . 1

2 0 2

2 3 (Shown)

r

x y z

Π − = − − −

− − =

�

9 MJC/I/9

The equations of three planes Π 1, Π 2, and Π 3 are 3 8,

3 0

11,

x y az

x y bz

x py qz

+ + =

+ + =

+ + =

and

respectively, where a, b, p and q are constants.

(i) If the point ( ), ,0α β lies on both Π 1 and Π 2, find the values of α and β . [2]

(ii) Given that the line l1 lies on both Π 1 and Π 2, find a vector equation of l1 in terms of a and b. [2]

The line l2 has equation r =

5

2

2

+

4

1

0

µ

, where µ is a parameter.

(iii) Given that l1 and l2 intersect at a point and 0a b+ = , find the values of a and b.[4]

(iv) If Π 1, Π 2, and Π 3 have no point in common, using the values of a and b found in (iii), comment on the value of p and find an equation relating p and q. [3]

[(i) 1, 3α β= − = (ii) 1

1 3

: r 3 3 ,

0 8

b a

l a bλ λ

− − = + − ∈ −

��

(iii) 2, 2a b= = − (iv) 1p q− = ]

(i) Sub ( ), ,0α β into Π 1 and Π 2.


_________________________________________________________________________________________________________


1

2

: 1( ) 3( ) (0) 8

: 3( ) 1( ) (0) 0

a

b

π α β

π α β

+ + =

+ + =

Using GC, 1, 3α β= − =

(ii)

1 3 3

3 1 3

8

b a

a b

a b

− × = − −

1

1 3

: r 3 3 ,

0 8

b a

l a bλ λ

− − = + − ∈ −

��

(iv)

1 3

3 3

0 8

b a

a bλ

− − + − = −

5

2

2

+

4

1

0

µ

1

4λ⇒ = −

Sub a b= − and 1

4λ = − , we have

( )

( )

11 4 5 4

4

13 4 2

4

a

a

µ

µ

− − − = +

− = +

2

2

a

b

∴ =

= −

(iv) Since the 3 planes have no point in common, l1 cannot intersect Π 3.

1

1 1

: r 3 1 ,

0 1

l λ λ

− − = + ∈ −

��

Condition 1:

( )1,3,0− is not on Π 3.

Hence, ( 1) (3) (0) 11

4

p q

p

− + + ≠

⇒ ≠


_________________________________________________________________________________________________________


Condition 2:

1

p

q

and

1

1

1

− −

are perpendicular.

1 1

1 0

1

1 0

1

.p

q

p q

p q

− ⇒ = −

⇒ − + − =

⇒ − =

10 PJC/II/4

The point A has position vector i + j + 3k and the plane, π , has equation

( )3 4 40.+ =r i ki

Find (i) the vector equation of the line l passing through A and normal to the plane, π ,

[1]

(ii) the position vector of the foot of perpendicular from the point A to the plane, π . [3]

A sphere of radius 2 units has its centre at A. Show that the shortest distance between the

sphere and the plane, π , is 3 units. Hence find the position vector of the nearest point on the sphere to the plane. [5]

[(ii)

4

1

7

=

q ,

111

55

23

=

p ]

1 3

: 1 0

3 4

l λ = +

r

i A(1,1,3)

P

Q

l

3

0

4


_________________________________________________________________________________________________________


1 3 3

pt : 1 0 40

3 4 4

1

4

1

7

Q

λ

λ

λ

+ = +

→ =

=

q

i

2

Let be pt on sphere nearest to

9 0 16, 5 2 & 3

shortest dist between sphere & Plane = 3

P

AQ AQ AP PQ

π

= + + = → = =

1 4 111

5 3 1 2 1 , 55

3 7 23

= + =

p p

Complex Numbers 1. [PJC 09 MY P2 Q5(b)]

The complex number w is such that * 252

3 4iww w+ =

−, where *w is the complex

conjugate of w . Find w in the form ia b+ , where a and b are real. [3] [ 1 2i− + ]

* 252

3 4iww w+ =

− * 25

2 3 4i3 4i

ww w+ = = +−

( )( ) ( )i i 2 i 3 4ia b a b a b+ − + + = +

2 2 2 2 i 3 4ia b a b+ + + = +

Comparing real and imaginary parts 2 4 2b b= ⇒ =

2 2 2 3a b a+ + = ⇒ 2 2 1 0a a+ + = ⇒ ( )21 0a + = ⇒ 1a = − ⇒ 1 2iw = − +

2 Solve the simultaneous equations 4i 3 7 5i,

2 (1 i) 10 i,

z w

z w

+ = +

− + = −

giving z and w in the form ia b+ , where a and b are real. [3]

[ 11

5 i ; 5 5i2

− − − ]


_________________________________________________________________________________________________________


4i 3 7 5i, --- (1)z w+ = +

2 (1 i) 10 i, --- (2)

(2) 2i,

4i 2i(1 i) (10 i)2i

4 i 2 2i 20i 2 --- (3)

(1) - (3),

2i 5 15i

5 15i5 5i

1 2i

z w

z w

z w w

w w

w

− + = −

×

− + = −

+ − = +

+ = −

−⇒ = = − −

+

Sub into (2),

2 (1 i)( 5 5i) 10 iz − + − − = −

2 10i 10 iz + = − 11

5 i2

z⇒ = −

3 [NJC 09 J2 CT Q10 (a) 1st part] Solve for the complex number w in the form ix y+ such that

* 10 10i2

2 i

− ++ =

−w w [2]

[ 2 2i− + ]

* 10 10i2

2 i

− ++ =

−w w

Let iw x y= + ,

( ) ( )i 2 i 6 2i 3 6 2

Comparing real and & imaginary parts,

2 and 2

2 2i

x y x y x iy i

x y

w

− + + = − + ⇒ + = − +

= − =

∴ = − +

4. [ACJC 09 Prelim P1 Q10a]

Without the use of a calculator, find the complex numbers z and w in the form

iba + that satisfy the two simultaneous equations (2 ) 9 16z i w i+ + = − + and iwz 3* =+ . [4]

[ 2 4i+ ; 2 7i− + ]

(2 ) 9 16z i w i+ + = − + (1) * 3z w i+ = (2)

Substitute *3w i z= − into equation (1) *(2 )(3 ) 9 16z i i z i+ + − = − +


_________________________________________________________________________________________________________


*( 2 ) ( 3 6 ) 9 16z i z i i+ − − + − + = − + *( 2 ) 6 10z i z i+ − − = − +

Let z x iy= +

( ) ( 2 )( ) 6 10x iy i x iy i+ + − − − = − +

( ) ( 3 ) 6 10x y i x y i− − + − + = − +

Equating real parts: 6 6x y x y− − = − ⇔ + = (3)

Equating imaginary parts: 3 10x y− + = (4)

Solving equations (3) and (4): 2x = and 4y =

2 4z i= + 3 (2 4 ) 2 7w i i i= − − = − +

☺☺☺☺ GOOD JOB! YOU DID IT!! GOOD LUCK FOR YOUR COMMON TEST!!! ☺☺☺☺

2011 Year 6 H2 Mathematics Exercises for June Holiday _Solution

Documents

Transcript of 2011 Year 6 H2 Mathematics Exercises for June Holiday _Solution