# 2014 H2 Maths Prelim Papers - RJC P1 solution.pdf

August 6, 2017 | Author: cherylhzy | Category: Trigonometric Functions, Perpendicular, Triangle, Algebra, Mathematical Analysis

#### Description

RAFFLES INSTITUTION 2014 YEAR 6 PRELIMINARY EXAMINATION Qn. [Marks] 1

Solution f ( x) 

[4]

x 2  ax  b b  xa x x

f '( x)  1 

b 0 x b x2

x-intercepts

x

b ,0 and  b ,0 y



b, 0

0

b, 0

x=0 2(i) [2]

Using G.C. n6

u6  1.0817  1

n7

u7  0.71334  1

Least n  7 (ii)

Since un  20 for all values of n , then

[2]

1 5 20    20  r(20) 1   r 20  4   100  1    100 

(iii)

4  l  As n  , un  l and un 1  l , l  l  1   3  100 

[2]

l  0 (N.A.) or

4 l  1   1 3  100  l  25

1

MATHEMATICS PAPER 1 (9740 / 1) Higher 2

Qn. [Marks] 3(i) [3]

Solution

f ( x)  1  x  1  x 2 

2

  2  2  2  3 4  1  x  1  x  x  1! 2!   1  x  1  2 x 2  3 x 4 

  

 1  x  2 x 2  2 x3  3x 4  (ii)

Observe that

[3]

1  x 

2 2

 1

 2  x 2   2  3 x 4  1!

2!

 2  3 r!

r

2

,

the coefficient of x 2 r 1 is   1 (r  1)   1 r

4

(r  1)

  1 (r  1) x 2 r 

 1  2 x 2  3x 4  Since f ( x)  1  x  1  x 2

r 1

(r  1).

Let Pn be the statement

[4]

n

 sin(2rx)sin x  sin nx sin(n  1) x for n 

.

r 1

1

L.H.S. =  sin(2rx)sin x  sin 2 x sin x

When n = 1,

r 1

R.H.S. = sin x sin 2 x  L.H.S. = R.H.S.

Hence P1

is

true.

Assume Pk is true for some k 

,

k

i.e.

 sin(2rx)sin x  sin kx sin(k  1) x r 1

To prove that Pk 1 is true, k 1

i.e.

 sin(2rx)sin x  sin  k  1 x sin  k  2 x r 1

k 1

L.H.S.

=  sin(2rx)sin x r 1 k

  sin(2rx )sin x  sin 2(k  1) x sin x r 1

 sin kx sin(k  1) x  sin 2(k  1) x sin x

 sin kx sin(k  1) x  2sin(k  1) x cos(k  1) x  sin x 2

x2r 

Qn. [Marks]

Solution

 sin(k  1) x sin kx  2cos(k  1) x sin x   sin(k  1) x sin kx  sin(k  2) x  sin kx 

 sin  k  2 x sin  k  1 x R.H.S. OR k 1

L.H.S.   sin(2rx )sin x r 1 k

  sin(2rx)sin x  sin 2(k  1) x sin x r 1

 sin kx sin(k  1) x  sin 2(k  1) x sin x 

1 1 cos(2k  1) x  cos x   cos(2k  3) x  cos(2k  1) x  2 2



1 cos(2k  3) x  cos x  2

1  2k  3  1   2k  3  1      2sin   x sin   x 2 2 2     

 sin  k  2 x sin  k  1 x = R.H.S. Hence Pk is true  Pk 1 is true, and since P1 is true, by mathematical induction , Pn is true for all n 

[3]

.

sin 3x sin 4 x dx  sin x 2

4

=

 sin(2rx) dx. 

4

=

3

2

r 1

 2

 sin 2 x  sin 4 x  sin 6 x  dx.

4 

1 1  1 2 =   cos 2 x  cos 4 x  cos6 x  4 6  2  4

= 

1 1 1 1  1  0   1  1   1  0  = 6 2 4 6

3

Qn. [Marks] 5

Solution Equation of the new curve is y  3 

[4]

1 . 2x 1

(0, 4)

x

[4] y

y (0, 4)

(0, 4)

y=3

y=3

x

From the graph, for f ( x)  f  x  , x   6(i) [1] (ii) [2]

a

x

1 1 2 or x . 2 2 3

b represents the length of projection of OA onto OB. b 2

3a  b  102  100 2 2 9 a  b  6a b  100 6a b  9(3)2  (5) 2  100  6

Therefore a b  1. (iii) [2]

Let N be the foot of the perpendicular from A to the line OB. ON  a

b 1  . b 5

Using Pythagoras Theorem,

4

Qn. [Marks]

Solution 2

 1  224 AN  OA  ON  3     . 25 5 2

AN 

2

2

2

224 4  14  2.9933  2.99 (3sf). 25 5

1 Area of triangle OAB  OB  AN  2 14  7.48 (3sf). 2

(iv)

(a  2b)  a  k (2a  3b)  a for some constant k.

[4]

Now, a  0, b  0 and a b  1  a b , so a and b are non-zero and non-parallel vectors.

(   1)a  2b  ka  3kb (   1  k )a  (3k  2)b Hence 3k  2  0  k  7(a)

2 5 and   k  1  . 3 3

C1 is a circle centred at (0,0) with radius 5.

[2]

C2 :

x2 y2   1 is an ellipse centred at (0,0) with length of the 100 100 a b 10 10 ) and vertical axis 2( ) . horizontal axis 2( a b

Note : a  b  length of the horizontal axis > length of vertical axis To get 4 points of intersection, we need :

10 10  5  0  a  4 and 5b  4 a b OR

x2 y 2 x2 y2  1.   1 with C2 : Compare C1 : x  y  25  100 100 25 25 a b 2

2

For them to intersects at 4 points, 100 100  25 and  25 b a

b  4 and 0  a  4 since a  0 is given.

5

Qn. [Marks] (b) [3]

Solution

x 2  y 2  25

C1 :

x2 y2 C2 : 2  10 10 3

 

2

1 

y   25  x 2 100  x 2 y 3

C1 and C2 intersect at x  3.9528 (5 s.f.) (from GC) Thus area of the required region

 3.9528 100  x 2  3.9528  2  dx   25  x 2 dx  10 5 3    22.3 (3 s.f.) OR x 2  y 2  25

C1 :

C2 : x 2  9 y 2  100

x   25  y 2

x   100  9 y 2

C1 and C2 intersect at y  3.0619 (5 s.f.) (from GC) Thus area of the required region  2   0

3.0619

100  9 y 2  25  y 2 dy  

 22.3 (3 s.f.)

(c) [4]

x 2  y 2  25

C1 : C2 :

x2 y2  102 10 2

 

2

1 

Required Volume 5 4 3    x 2 dy    5 5 3 

5

5

 100  4 y  dy  500 3 2

x 2  25  y 2 x 2  100  4 y 2

Note :

4 3   5 is the volume of sphere 3 formed when rotating the circle

5

4  500    100 y  y 3    3  5 3 

500   500   500     500      500    3   3   3   500

6

Qn. [Marks] 8(a)

Solution

z4  8  3  i  0

[6]

z4  8  3  i 5

 16 e 6

i

5   2k  i 6    16 e

 5 1    k  i  24 2 

z  2e

Im

, k  0 , 1 , 2

2

–2

2

–2

2 e 24  i 7

w2  aw *  b  0

(b) [6]

Re

 w  aw *  b  *  0*  w  *   aw * *  b*  0 2

2

 w *2  a  w * *  b  0 , a*  a and b*  b since a real. 2 Hence, w* is a root of z  az * b  0 .

z 2  6 z * 9  0

 x  iy 2  6  x  iy   9  0 x 2  y 2  2ixy  6 x  6iy  9  0 x 2  y 2  6 x  9  2 y  x  3 i  0 Compare imaginary parts, y  0 or x  3 .

Consider real parts: 2 When y  0 , x  6 x  9  0 which gives x  3

When x  3 , 32  y 2  18  9  0 giving y  6 Hence z  3 , 3  6i , 3  6i

7

and b are

Qn. [Marks] 9 (i) [3]

Solution

5 0 5       OA   0  , OC   3  , OR   3  , so 0 0  2       0 5 6        n  AR  CR   3    0   10  .  2   2   15         6  5  6       Therefore  1 : r  10    0   10   30.  15   0   15      

(ii) [2]

The angle between

1 is cos

1

0   and the horizontal base, which has normal  0  , 1  

 6   0     10  .  0   15  1   15      cos 1    142.136 . 36  100  225 0  0  1  19 

The acute angle is 180  142.136  37.9 . (1 dec. pl.) (iii)

By symmetry, the acute angle between  2 and the horizontal base is

[1]

also 37.864 Hence the angle between  1 and  2 is  2(37.864 ) =

75.7 (1 dec. pl.). (iv) [3]

0 0 0     OP   0  , OS   3  , SP   3  , therefore  2  2 0       0 0   0  OX  OS  SX   3     3    3(1   )  .  2 0   2       

l XY :

l XY :

5   r  OY   XY ,   , and with OY   2  , 1    5  5  r   2     3  1 ,   . 1  1     

8

Qn. [Marks]

Solution 5 r   3 ,   .  2  

Clearly lOR : (v) [4]

When the lines XY and OR intersect, 5 5

 2 1    

 1   

5

  3    2     3  1 .

Therefore,

  1  3  2   (3  1) 2  1   Solving gives  

1 2 1 ,    and   . 3 3 3

OW  OR, so OW : OR  2 : 3. 10(i) [3]

Since x  tan  , y  sec and tan2   1  sec2 

x 2  1  y 2 for x  0 and y  1 y C

x (ii)

y

[7]

Q   cot  ,0 

x  tan  , y  sec  ;

x

dy sec tan  tan     sin  dx sec2  sec

Equation of tangent at P is y  sec  sin   x  tan  

9

Qn. [Marks]

Solution At Q , 0  sec  sin   x  tan   .

sin  1 sin 2   1 cos 2  x    cos  sin  cos  sin  cos  sin  cos  x   cot  Tangent at P intersects x-axis at Q   cot  ,0 

Equation of normal at P is y  sec   At R , 0  sec 

x

1  x  tan   sin 

1  x  tan   sin 

sin   tan   2 tan  cos 

Normal at P intersects x-axis at R  2 tan  ,0  Triangle PQR is a right-angled triangle in circle, so QR is a diameter.  2 tan     cot     QR  A      2  2    2

1/ tan       tan    2  

2

2

2

1   A    tan    [shown] 2 tan    (iii) [2]

2

1 1 1  2 2  t    2  t 1 2  2  t 1 2 4t 4t  2t  2

1   t    0  2t 

[1]

For 0   

 2

, t  tan  > 0

 1   So minimizing A    tan    over 0    is equivalent to 2 2 tan    2

2

1  minimizing A    t   for t  0  2t  2

1   A    tan     2 from above result. 2 tan    Hence the minimum value of A = 2

10

Qn. [Marks]

Solution

11(i) When x = 0, [1] (ii)

d2 y d2 y   1   y and y  1 when x  0 ) . (since dx 2 dx 2

Differentiating (1) with respect to x twice gives

[2]

d3 y dy  3 dx dx

and

d4 y d2 y   . dx 4 dx 2

d2 y dy  1 [from (i)] When x  0,  1 (given) and dx dx 2 Hence, when x  0,

d3 y d4 y  1 and 4  1. dx3 dx

Maclaurin’s theorem gives

1 1 1 2 1 2 1 4 x  x  x  x  1! 2! 3! 4! 1 2 1 3 1 4 y  1 x  x  x  x  2 6 24 y  1

(iii)

Differentiating

[4]

dy  u implicitly with respect to x gives dx

d 2 y 1 1 du  dx 2 2 u dx

y 

1 dx du 2 dy dx

Hence,

(since

1 dx dy d2 y = )   y and  u  2 dx dx u dy

du  2 y. (shown) dy

Alternative Method 2

dy  dy   u  u   . dx  dx  2

 dy  Differentiating u    implicitly with respect to x gives  dx  2 du du dx du dy d y  2 y. (shown)   2( y )  2 2 dy dx dy dx dx dx

Integrating

du  2 y with respect to y gives dy

u   2 y dy   y 2  A, where A is an arbitrary constant.

11

Qn. [Marks]

Solution Using u  1 when y  1, we have A  2, Hence, u  2  y 2 .

(iv)

Substituting u  2  y 2 into

[4]

dy  u gives dx

dy  2  y2 , dx

which we can integrate via

1

dy 1 2  y 2 dx 1 dy   1dx  2  y2 y sin 1  xB 2 y  2 sin  x  B  , where B is an arbitrary constant Using y  1 when x  0, we have B  sin 1

1   . 2 4

   Hence y  2 sin  x   , i.e. P  2 and Q  . 4 4  (v) [2]

  y  2 sin  x   4      2 sin x cos  cos x sin  4 4   sin x  cos x (since cos

= sin

4    x2 x4 x3   x     1    3! 2! 4!    2 3 4 x x x  1 x     2 6 24

=

1 ) 2

4   (from MF15) 

12