2011 set b mock paper 1 solutions

8/4/2019 2011 Set B Mock Paper 1 Solutions

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2011 JC2 H2 Mathematics Set B, Mock Paper 1

Specimen Paper for Examination from 2007 of 15

Set B, Paper 1

1. Use the method of mathematical induction to prove that the sum of the first n terms of the

series

...)753()642()531( +××+××+××

is )5)(4)(1(41 +++ nnnn . [5]

- - - - - -

Let Pn be the statement

...)753()642()531( +××+××+×× + )4)(2( ++ nnn = )5)(4)(1(4

1+++ nnnn , ∀

+∈ Z n .

When n = 1,

L.H.S. = 15)531( =××

R.H.S. = 15)6)(5)(2)(1(41 = = L.H.S.

Therefore, P1 is true.

Assume that Pk is true for some+

∈ Z k .

That is, ...)753()642()531( +××+××+×× + )4)(2( ++ k k k = )5)(4)(1(4

1+++ k k k k .

We want to show Pk +1 is true.

That is,

...)753()642()531( +××+××+×× + )5)(3)(1( +++ k k k = )6)(5)(2)(1(4

1++++ k k k k .

L.H.S. = ...)753()642()531( +××+××+×× + )4)(2( ++ k k k + )5)(3)(1( +++ k k k

= )5)(4)(1(4

1+++ k k k k + )5)(3)(1( +++ k k k

= )5)(1(4

1++ k k )]3(4)4([ +++ k k k

= )5)(1(4

1++ k k ]128[

2++ k k = )6)(2)(5)(1(

4

1++++ k k k k = R.H.S.

That is, Pk is true ⇒ Pk +1 is true.

Since, P1 is true and Pk is true ⇒ Pk +1 is true, by mathematical induction, Pn is true for all+

∈ Z n .





2. A piece of wire of length d units is cut into two pieces. One piece is bent into the form of

a square and the other piece is bent into the form of a circle. Prove that, when the

combined area of the two shapes is a minimum, then the radius of the circle is

approximately 0.07d units. [7]

- - - - - -

Let r = radius of circle and A = combined area of the two shapes.

Hence,

A =

22

4

2

−+

r d r

π π = ( )22

216

1r d r π π −+

)2)(2(8

12

d

dπ π π −−+= r d r

r

A= r

d r

2

2

1

42 π

π π +− =

42

12

2 d r

π π π −

+

Let 0dd =

r A .

⇒ 42

12

2 d r

π π π −

+ = 0

⇒

+÷

= π

2

12

4

d r ≈ 0.07 d

Since =r

A

d

d

42

12

2 d r

π π π −

+ ⇒ =

2

2

d

d

r

A

+

2

2

12 π π > 0.

Hence, r ≈ 0.07 d gives minimum A.

r

Circumference = 2 π r Perimeter = d − 2 π r





3. (i) Show that)32)(12)(12(

1616

32

1

12

2

12

3

++−

+=

+−

+−

− r r r

r

r r r . [2]

(ii) Hence find ∑=

++−

+n

r r r r

r

1 )32)(12)(12(

1,

giving your answer in the form k – f(n), where k is a constant. [4]

(iii) State the sum to infinity of the series in which the r th term is

)32)(12)(12(

1

++−

+

r r r

r . [1]

- - - - -

(i)

)32)(12)(12(

)12)(12()32)(12(2)32)(12(3

32

1

12

2

12

3

++−

+−−+−−++=

+−

+−

− r r r

r r r r r r

r r r

=)32)(12)(12(

)14()344(2)384(3 222

++−

−−−+−++

r r r

r r r r r

= )32)(12)(12(

1616

++−

+

r r r

r

(ii) ∑=

++−

+n

r r r r

r

1 )32)(12)(12(

1

= ∑=

++−

+n

r r r r

r

1 )32)(12)(12(

1616

16

1= ∑

=

+

−+

−−

n

r r r r 1 32

1

12

2

12

3

16

1

= +

−−

5

1

3

2

1

3

16

1

+

−−

7

1

5

2

3

3

+

−−

9

1

7

2

5

3…

+

−−

−−

− 12

1

32

2

52

3

nnn…

+

+−

−−

− 12

1

12

2

32

3

nnn…

+−

+−

− 32

1

12

2

12

3

nnn =

−−

+−

+−+−

32

1

12

2

12

1

3

3

3

2

1

3

16

1

nnn

=

++

+−

32

1

12

3

16

1

24

5

nn.

(iii) As ∞→n , 012

3→

+n, 0

32

1→

+n.

Hence ∑∞

=∞→ ++−

+

1 )32)(12)(12(

1lim

r n r r r

r =

24

5.





4. On the same Argand diagram, sketch the loci

(i) ,5i68 =−− z [2]

(ii) arg( z – 4 – 3i) = 2α , where α =

−

4

3tan

1. [2]

The complex number w is represented by the point of intersection of the loci in parts (i)

and (ii). Find w, in the form x + i y, giving the exact values of x and y. [5]- - - - - -

Note that

1. ∆ BCD is an isosceles triangle with =∠=∠ DBC BCD and α π 2−=∠ BDC

2. α π 2cos)2cos( −=−

3. 4

3tan =α ⇒

5

4cos =α and

5

3sin =α

Applying cosine rule on ∆ BCD : 2)( BC = )2cos()5)(5(255

22α π −−+

2)( BC = 2cos5050 +

2)( BC = )1cos2(5050

2−+ α

2)( BC = 6415

425050

2

=

−

+ ⇒ BC = 8

BC

AB BCA =∠sin ⇒

82sin

AB=α ⇒

25

192

5

4

5

316cossin162sin8 =

=== α α α AB .

CA =

22

25

1928

− =

625

3136=

25

56.

Hence x = 4 +25

56=

25

156and = y

25

1923+ =

25

267. Implies w =

25

156+

25

267i .

5

D(8,6)•

α

2α

C (4,3)

•

• A

B

Re( z)

Im( z)





5. A curve has parametric equations2

1

t x = , t y 2= , where t is a non-zero parameter.

(i) Sketch the curve. [2]

(ii) Find the equation of the tangent to the curve at the point

t

t 2,

12

and show that

the equation of the normal to the curve at the same point can be written in the form

12 625−=− t xt yt . [6]

(iii) The tangent at the point P

4,

4

1meets the y-axis at T and the normal at P meets

the y-axis at N . Prove that the area of the triangle PTN is256

65. [4]

- - - - - -

(i)

(ii) 2

1

t x = ⇒

3

2

d

d

t t

x −= ; t y 2= ⇒ 2

d

d=

t

y.

Hence,3

3

22

d

d

d

d

d

dt

t

t

x

t

y

x

y−=

−×=÷= .

Equation of tangent at the point

t

t 2,

12

is

−−=−

2

3 12

t

xt t y

⇒ 033=+− xt t y

x

y





Equation of normal at the point

t

t 2,

12

is

−=−

23

112

t x

t t y

⇒

−=−

243 12

t xt yt

⇒ 12265−=− xt t yt

⇒ 12625−=− t xt yt

(iii) At the point P

4,

4

1, t = 2.

Hence equation of tangent at P is 68 +−= x y … (1)

And equation of normal at P is . 127432 =− x y . … (2)

Substitute x = 0 into (1) : 6= y .

Hence T is )6,0(

Substitute x = 0 into (2) :32

127= y .

Hence N is )32

127,0(

Area of ∆PTN =256

65

4

1

32

1276

2

1=

− .

x

y

P

T

N





6. (i) Show that ⌡

⌠ +1

dln

n

n

x x = 1ln)1ln()1( −−++ nnnn . [4]

(ii)

The diagram shows the graph of x y ln= between x = n and x = 1+n . The area of

the shaded region represents the error when the value of the integral in part (i) is

approximated by using a single trapezium. Show that the area of the shaded region

is

11

1ln2

1−

+

+

nn . [4]

(iii) Use a series expansion to show that if n is large enough for3

1

nand higher

powers of n

1to be neglected, then the area in part (ii) is approximately equal to

2n

k ,

where k is a constant to be determined. [3]

- - - - - -

(i) xu ln= ⇒ xdx

du 1= ; 1=

dx

dv ⇒ xv =

⌡

⌠ +1

dln

n

n

x x = x x

x x x

n

n

n

n

d1

))((ln

11

⌡

⌠

−

++

=

1

ln)1ln()1(

+

−−++

n

n

xnnnn

= ( )nnnnnn −+−−++ )1(ln)1ln()1(

= 1ln)1ln()1( −−++ nnnn .

x

y

n n + 1O

y = ln x





(ii) Area of the shaded region = ⌡

⌠ +1

dln

n

n

x x − ( ))1ln(ln)1(2

1++ nn

=

−−++ 1ln)1ln()1( nnnn − )1ln(

2

1ln

2

1+− nn

= 1ln)21()1ln()

21( −+−++ nnnn

= 11

ln)2

1( −

++

n

nn

= 11

1ln2

1−

+

+

nn

(iii) Note that

+

n

11ln = ...

1

3

11

2

1132

+

+

−

nnn

= ...3

1

2

1132++−

nnn

Hence, 111ln21 −

+

+

nn = 1...

31

211

21

32−

++−

+

nnnn

= 1...4

1

3

1

2

1

2

11

22−

+−++−

nnnn

≈ 212

1

n(for

3

1

nand higher powers of

n

1to be neglected)





7. (a) The first four terms 1u , 2u , 3u , 4u of an arithmetic progression are such that

1524 =− uu and 13 94 uu = .

Find the value of 1u . [3]

(b) The positive numbers n x satisfy the relation

2

1

1 )5( +=+ nn x x , for n = 1, 2, 3, … .

As ∞→n , n x → .

(i) Find (in either order) the value of to 3 decimal places and the exact value

of . [3]

(ii) Prove that ( ) −=−+ nn x x 221 . [2]

(iii) Hence show that, if n x > , then n x > 1+n x > . [3]

- - - - - - (a) Let a be the first term and d be the common difference of the A.P.

Given 1524 =− uu ⇒ 15)()3( =+−+ d ad a ⇒ 2

15=d

Given 13 94 uu = ⇒ ad a 9)2(4 =+ ⇒ d a 85 = ⇒ 1u = 122

15

5

8=

=a .

(b)(i) Given : When ∞→n , n x → . Then 1+n x → .

Since 21

1 )5( +=+ nn x x , therefore 21

)5( += .

⇒ 52

+= …. (*)

⇒ 052

=−−

⇒ 2

211±=

Since n x > 0, 0> . Therefore 791.22

211=

+= (to 3 d.p.)

(b)(ii) Given 21

1 )5( +=+ nn x x

⇒ ( ) 52

1 +=+ nn x x

⇒ ( ) 2221 5 −+=−+ nn x x

⇒ ( ) )5(5221 +−+=−+ nn x x using (*)

⇒ ( ) −=−+ nn x x22

1





(b)(iii) From (b)(ii) : ( ) −=−+ nn x x 221

⇒ ( )( ) −=+− ++ nnn x x x 11 … (1)

Given n x > ⇒ 0>− n x .

Hence (1) becomes ( )( ) 011 >+− ++ nn x x

⇒ −<+1n x or >+1n x …. (2)

(rejected)

From (1) : ( )( ) −=+− ++ nnn x x x 11

⇒ ( )( )

( )

−+

=−

+

+ nn

n x x

x1

1

1

Note that 791.21 ≈>++ n x .

Hence, ( ) 11

01

<+<+ n x .

⇒ ( ) ( ) −<−+ nn x x 1

⇒ nn x x <+1 …. (3)

Combining results (2) and (3), we have n x > 1+n x > .





8. (a) Find ⌡⌠

+

+ x

x

xd

9

122

. [3]

(b) Using the substitution θ tan4 += x , find ⌡⌠

+−

x x x

d178

12

. [5]

(c) Find the volume of revolution when the region bounded by the part of the curve

x y 2cos= between 0= x and π 21= x and by the x- and y-axes is rotated

completely about the x-axis. Give your answer correct to 3 significant figures. [3]

- - - - -

(a) ⌡⌠

+

+ x

x

xd

9

122

= ⌡⌠

+

+⌡⌠

+

x x

x x

xd

3

1d

9

2222

= ( ) C x

x +++−

3tan

3

19ln

12

(b) θ tan4 += x ⇒ θ

θ

2sec=

d

dx.

⌡⌠

+−

x x x

d178

12

= ⌡

⌠

+−

x x

d1)4(

12

= ⌡

⌠

+

θ θ

θ

dsec1)(tan

1 2

2

= ⌡

⌠ θ d1

= C +θ = C x +−−

)4(tan1

(c) Volume required

= ( ) x x dcos2

0

22⌡

⌠ π

π

= x x dcos2

0

4⌡

⌠ π

π

= 1.85 unit3.





9. Consider the curve)6(

63

+

−=

x x

x y .

(i) State the coordinates of any points of intersection with the axes. [1]

(ii) State the equations of the asymptotes. [3]

(iii) Prove, using an algebraic method, that

)6(

63

+

−

x x

xcannot lie between two certain

values (to be determined). [5]

(iv) Draw a sketch of the curve

(a) )6(

63

+

−=

x x

x y ,

(b) )6(

63

+

−=

x x

x y ,

making clear the main relevant features of each curve. [4]

- - - - - -

(i) )6(

63

+

−

= x x

x y .

When y = 0, 063 =− x ⇒ 2= x

(2,0) is the only point of intersection with the x-axis.

(ii) The asymptotes are 0= x , 6−= x and 0= y .

(iii) )6(

63

+

−=

x x

x y ⇒ 636

2−=+ x yx yx

⇒ 06)36(2=+−+ x y yx … (1)

For any valid values of y substituted into (1), there must correspond some values of x.Hence, equation (1) has solution for x.

Hence, we consider discriminant 0≥ .

That is, 0)6(4)36( 2≥−− y y

⇒ 096036 2≥+− y y

⇒ 0320122

≥+− y y

⇒ 0)16)(32( ≥−− y y

⇒ 6

1≤ y or

2

3≥ y

Hence,)6(

63

+

−

x x

x cannot lie between6

1 and2

3 .





(iv)(a)

(iv)(b)

(6,1/6)−2 1/2

2•

•

x = −6

x

y

(6,1/6)−2 1/2

2

•

•

x = −6

y

x





10. (i) Expand 31

)8( x+ in ascending powers of x up to and including the term in2

x , and

state the values of x for which this expansion is valid. [5]

(ii) The function f( x) = 31

)8(3 x+− is defined for 190 ≤≤ x . Find the inverse

function )(f 1 x− . [3]

(iii) Sketch the graphs of y = f( x) and y = )(f 1 x−

on the same axes, and show the

geometrical relation between the two graphs and the line y = x. [3]

(iv) Show that the x-coordinate of the point of intersection of the graphs of y = f( x) and

y = )(f 1 x−

is given by

31

)8( x+ = x−3 . [2]

(v) Use the first two terms of expansion of 31

)8( x+ to obtain an approximation to the

x-coordinate of the point of intersection. Give your answer as a simple fraction. [1]

- - - - - -

(i) 31

)8( x+ =31

31

818

+

x=

( )( )

+

−+

+ ...

8!283

112

232

31

x x

=

+−+ ...

576

1

24

112

2 x x

= ...288

1

12

12

2+−+ x x

Expansion is valid when 18<

x ⇒ 8< x .

(ii) Let y = f( x) = 31

)8(3 x+−

⇒ y x −=+ 3)8( 31

⇒ ( )338 y x −=+

⇒ ( ) 833−−= y x

⇒ ( ) 83)(f 31−−=

− y y

⇒ ( ) 83)(f 31−−=

− x x

Hence, ( ) 83:f 31−−→

− x x , ]1,0[∈ x

y = f( x)

1

19

R f = [0,1]

x





(iii)

The graphs of y = f( x) and y = )(f 1 x− are reflection of each other about the line

y = x.

(iv) The point of intersection of the graphs of y = f( x) and y = )(f 1 x−

is the same as the point

of intersection between the graphs of y = f( x) and y = x .

Hence, x = 31

)8(3 x+−

⇒ 31

)8( x+ = x−3 … (1)

(v) From (i) 31

)8( x+ = ...288

1

12

12

2+−+ x x . … (2)

Substitute the first two terms of (2) into (1) :

x x −≈+ 312

12

⇒ 13

12≈ x

x

y

y = x

y = f( x)

y = f

−1

( x)

1

1

19

19

2011 set b mock paper 1 solutions

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