# Rvhs h2 Math p1 Solutions

July 29, 2017 | Author: jimmytanlimlong | Category: Plane (Geometry), Equations, Space, Elementary Mathematics, Analysis

#### Short Description

Rvhs h2 Math p1 Solutions...

#### Description

RVHS Yr 6 H2 Maths 2011 Preliminary Exam Paper 1 (Solutions) Question 1 [4 marks] Given that y  ax 

b c . x 1

Differentiating with respect to x:

dy b a 2 dx  x  1 Since it is given that the curve passes through the

 13   1   and   , 2  and that the curve has a  2  2  turning point at x  3 .

points 1,

We can set up the following equations.

13 1 13 1 a bc   a bc 2 11 2 2 1 1 1 2 a b  c  2   a  2b  c 1 2 2  1 2 1 1 0a b 0a b 2 4  3  1 Solving the system of linear equations gives

a  1, b  4, c  192 So, the equation of the curve is y   x 

(i)

(ii)

4 19   x  1 2

Question 2 [6 marks] n r Sn   r  2 ( r  1)! 1 11 59 , S4  . Using GC, S 2  , S3  3 24 120 1 1 1 S2    3 2 3! 11 1 1 S3    24 2 4! 59 1 1 S4    120 2 5! n r 1 1 Conjecture: S n     2 (n  1)! r  2 ( r  1)! n r 1 1 Let P (n) be the statement  for   2 (n  1)! r  2 ( r  1)! Pg 1

n  2,3, 4, ... . When n  2 , 2 r 2 1 LHS     3! 3 r  2 ( r  1)! 1 1 2 1 RHS     2 3! 6 3 Since LHS = RHS, therefore P (2) is true. Assume that P( k ) is true for some k  k

i.e.

r

1

, k  2.

1

 (r  1)!  2  (k  1)! r 2

Need to prove that P ( k  1) is true k 1

i.e.

r

1

1

 (r  1)!  2  (k  2)! r 2

LHS 

k 1

r

 (r  1)! r 2

k

r  (k  1)th term r  2 ( r  1)! k r k 1   (k  2)! r  2 ( r  1)! 1 1 k 1    2 (k  1)! (k  2)! 1  1 k 1     2  (k  1)! (k  2)! 

1  k 2 k 1    2  (k  2)! (k  2)!

1  k  2  (k  1)   2  (k  2)!  1 1 = RHS   2 (k  2)! Thus P(k ) is true implies P(k  1) is true. 

Since ( P (2) is true) and ( P(k ) is true implies P (k  1) is true), by mathematical induction, P (n) is true for n  2,3, ... .

Question 3 [7 marks] (i)

Locus of P is a half-line from and excluding the point representing the complex number 1  i 2 that makes an Pg 2

angle of  with the positive real axis.

(ii)

(iii)

EF  5 and EG  radius  2 2 sin   5    0.287 Locus of P meets the locus of Q more than once when 0.287    0.287 . (iv)

E  6  i 2 , F  1  i 2 and EF  5 Minimum value of z  w

l r

 5sin

 4

 2

5  2 2

3 3 2 or 2 2

Question 4 [9 marks]

(i)

(2r  3)

= 2(r  1)  Ar  B(r  1)  (2  A  B )r  (2  B ) Pg 3

Solve simultaneously, A  1 , B  1 .

 (2r  3)  2(r  1)  r  (r  1) . (ii)

n

 (2r  3)2 r = r 1

n

 (2(r  1)  r  (r  1))2 r 1

n

=

r

 (2(r  1)2

r

 r  2 r  (r  1)2 r )

r 1

=

(2)(2)(2)  (2)

 (0)

 (2)(3)(4)  (2)(4)  (1)(4)  (2)(4)(8)  (3)(8)  (2)(8)  (2)(5)(16)  (4)(16)  (3)(16)  ......

 (2)(n)(2 n 1 )  (n  1)(2 n 1 )  (n  2)(2 n 1 )  (2)(n  1)(2 n )  (n)(2 n )

 (n  1)(2 n )

=  2  n  2 n  ( n  1)  2 n 1  n  2 n =  2  2  n  2 n  ( n  1)  2 n 1 =  2  n  2 n 1  ( n  1)  2 n 1 =  2  ( 2n  1)  2 n 1

(iii)

n

 (2r  5)2

r

r 1

n

=

 ( 2 r  3  2) 2

r

r 1

=

n

n

r 1

r 1

 (2r  3)2 r   2 r 1

 

=  2  ( 2n  1)  2 n 1  2 2  2 3  ...  2 n 1

 

 4 2n  1  2 1

=  2  (2n  1)  2 n 1  

 

=  2  (2n  1)  2 n 1  2  21 n  4 = (2n  3)  2 n 1  6

Alternative Method Replace r by k –1: n

 (2r  5)2

r

r 1

Pg 4

n 1

=

 (2k  3)2

k 1

k 2

=

1 n 1 (2k  3)2k  2 k 2

1  n 1  (2k  3)2k  (2  3)21    2  k 1  1 1 =  2  (2n  3)2n  2  10  =  12  (2n  3)2n  2  2 2

=

Question 5 [9 marks] a 1  r10   100 ----- (1) 1 r

a 1  (r 2 )5 

ar 1  (r 2 )5 

 10 1 r2 1 r2 a 1  r10  r (1  r10 )   10 1 r2  a 1  r10  r  r11   10 ----- (2) (1  r )(1  r ) 

(1)  (2): (1  r10 )(1  r )  10 1  r10  r  r11 1  r  r10  r11  10  10r10  10r  10r11 11r11  9r10  11r  9  0 (shown) Solve using GC, r  0.81818 or r  1 (rej.)  a  65.992 --- from (1)

 Area of the smallest sector = ar 9 = 10.8 cm2 a  100 1  0.7 a  30 Hence, the maximum area = 30 cm2 (or 94.2 cm2)

Question 6 [9 marks] We have that

y  tan ln 1  x  

(1)

Differentiating the equation gives, Pg 5

1 dy  sec 2 ln 1  x   (1  x) dx dy (1  x)  1  tan 2  ln(1  x)  dx dy 1  x   1  y 2 dx

(2)

Differentiate (2) w.r.t x

d 2 y dy dy   2y 2 dx dx dx 2 d y dy 1  x  2  1  2 y   0 dx dx

1  x 

(3)

Differentiate (3) w.r.t x 2

d3y d2y d2y dy 1  x  3  2  1  2 y  2  2    0 dx dx dx  dx  2

d3y d2y dy 1  x  3  2 1  y  2  2    0 (4) dx dx  dx  Substitute x  0 into (1), (2), (3), and (4):

y x 0

dy  0, dx

x 0

d2y  1, dx 2

x 0

d3y  1, dx3

 4. x 0

Therefore, by Maclaurin’s Theorem,

1 1  x 2    4 x3   2! 3! 1 2 2 3 y  x x  x 2 3

y  0 x

1 2 tan ln 1  x    x  x 2  x3 2 3 Differentiating with respect to x

sec2 ln 1  x  

1 x 2 sec ln 1  x  

(1  x)(1  x) sec2 ln 1  x  

(1  x ) sec ln 1  x   2

 1  x  2 x2

 (1  x  2 x 2 )(1  x) 1

 (1  x  2 x 2 )(1  x  x 2 )

2

(1  x ) 2

 1  2 x2

Question 7 [9 marks] Pg 6

 7  3  2       Given OA   0  , line l: r   15      7  ,   R   5  4    1       The position vector of any point lying on l is   7  3     15  7  , for some   R   5  4    

(7  3  2) 2  (15  7 ) 2  (5  4  1) 2  10

2  4  3  0   1 or 3   7  3(1)    4      So, OB   15  7(1)    8    5  4(1)    1      

  7  3(3)   2      OC   15  7(3)     6    5  4(3)   7      

 (4  2)    1    1 P is midpoint of BC  OP   (8  6)    1  2     1 7   3    3      AP  OP  OA   1   4    

(i)

 3   1  3  Equation of  1 : r •  1    1   1   16 4 3 4       So, r • (–3i + j + 4k) = 16. Express equations of  1 ,  2 and  3 as system of equations:  3 1 4 16    2 3 5   1  1 2 1 1   From GC, x  15, y  13, z  12  15  So, position vector of D, OD   13  . 12    Find base area BCD: 

(ii)

Pg 7

 6    BC  OC  OB    14   8    

19    BD  OD  OB   5  13    

Area of BCD 

 1  BC  BD = 2

  222   1 1  142376  74   2 2    296    3   Perpendicular height, AP   1   26  4    

1 142376  26 6 = 320.67 = 321 units3. (to 3 s.f.)

So, volume =

Question 8 [10 marks] (i)

Using dot product theorem,

 6 5    2 2 2 2 2 2   5     1  6  5  4 5  1  3 cos    4  3     cos  

30  5  12 77 35

  63.70163401 = 63.7

(ii)

 m  6       n     5   32  6    4    

 6m  5n  8 --- (1)

 m  5       n     1  24 6  3     

 5m  n  6 --- (2)

Solve simultaneously, m  2 , n  4 .

(iii)

direction vector of line l

 6   5    19  1         =   5     1    38   19  2    4   3   19    1         Pg 8

 2   position vector of A =  4  6    2   Vector equation of line l: r =  4  +  6    Cartesian equation of l : x  2  Alternative Method Solve 6 x  5 y  4 z  32 & 5 x  y  3 z  24

y4  6 z 2

by G.C.

 2   Vector equation of line l: r =  4  +  6    Cartesian equation of l : x  2 

(iv)

1    2 ,  R  1  

1    2 ,  R  1  

y4  6 z 2

 1   2  1        3: r   2    4 2   2  8  6  4  1  6   1       So, Cartesian equation of  3 : x  2 y  z  4 . Cartesian equation of  4 is 9 x  2 y  5 z  40 . Since the system of equations has an infinite number of solutions, it means that the planes  1 ,  2 and

 4 intersect at a common line.

(a)

Question 9 [11 marks] a 1 0 (a 2  u 2 )2 du = =

4 0

4 0

1 a3 1 = 3 a =

1 a sec 2 x dx 2 2 2 2 (a  a tan x)

u  a tan x 

du  a sec2 x dx

1 dx a sec 2 x 3

4 0

4 0

cos 2 x dx 1 (cos2x +1) dx 2 

1  sin 2 x 4 =  x  2a 3  2 0 Pg 9

1 1    2a 3  2 4  2 = 8a 3

=

(b) (i)

(ii)

Area of R 1 x 1  1  =    dx 0 1  x2 4   = 0.410 units2 Required volume 1 x 1  1 dx      dx 2 2 0 (1  x ) 0  4   2   1 2 =     ( x  2 x  1) dx  8  16 0

= 

2

1

1

3   2    x =    x2  x     8  16  3 0  2     7  =      8  16  3   (6  5) = units3 48

Question 10 [13 marks]

(i)

vertical asymptote at x  1  x  c  0 when x  1 So, c  1

ax 2  bx  5 x 1  x  1 2ax  b    ax 2  bx  5 1 dy  2 dx  x  1 y

ax 2  2ax   5  b 

 x  1

2

Since there is a turning point on the y-axis, so

dy  0 when x  0 dx  ax 2  2ax   5  b   0 when x  0 Pg 10

 0  0  5  b  0 (ii)

b5 ax 2  5 x  5 y x 1 C has no x-intercept 

(iii)

ax 2  5 x  5  0 has no real roots x 1

 ax 2  5 x  5  0 has no real roots  52  4a  5   0 25  20a  0 5  a   (shown) 4 ax 2  5 x  5 y  x 1 ax  x  1   5  a  x  5  x 1  5  a  x  1  a  ax  5  a  a  ax  x 1 x 1 2 ax  2ax   5  5  dy  2 dx  x  1

dy  0  ax 2  2ax  0  x  0 or x  2 dx y x 1

 0,5 

y  ax  5  a 

a x 1

5a 5 a a O

x

 2, 4a  5

y  ax  5  a (iv)

Add the line y  ax  1 . It has the same gradient as the oblique asymptote of C, but with a smaller y-intercept. Pg 11

Solving for intersection between C and y  ax  1 :

a  ax  1 x 1 a  4a 0 x 1   4  a  x  1  a ax  5  a 

 4 x  4  ax  a  a  4  a x  4 x

4 4a  

Hence set of values of x is  x : x  R ,x  1 or x 

(v)

4  . 4 a

y

x 1

O

2

x

y  f  x ya

Question 11 [13 marks] (a) (i)

f : x  ln  x 2  2 x  3 , for x  2 Rf   ln 2,     0.693,  

(ii)

Since there exists a line y  k for ln 2  k  ln 3 such that it cuts the graph at more than 1 point.  f is not 1 – 1 function.  f 1 does not exist Pg 12

(iii)

For gf to exist, Rg  Df ,

 Rg  (2, )

 Rg  (2, 6) 1  2 x2  x   18 or 18 Thus, 6 

Least value of b =

1 8

g f ( 18 , )  (2, 6)  [ln 2, ln 51)

 Rfg  [ln 2, ln 51) (b) (i)

f : x  ln  x 2  2 x  3 , x  a f : x  ln ( x  1) 2  2  ,

xa

Thus least value of a  1 . Let y  ln ( x  1) 2  2  , x  1

e y  ( x  1) 2  2 x  1  e y  2 (reject x  1  e y  2 as x  1 )

 f 1  x   1  e x  2 , Df 1  Rf  (ln 2, ) (ii)

Pg 13