1(a)

1(b)

30

B1

4

 x   ⇒  y =    z

 1 2 2   11  1    1 1 3   −5  9     3 b 1  4 

3 PQ = k I  1 2 2   −5 a 4   1 0 0       1 1 3   8 −5 c  = k  0 1 0     0 0 1 4 c   3 b 1  c   a12 : a − 10 + 8 = 0 - - - -(1) a32 : 3a − 5b + 4 = 0 - - - -(2) a =2, b=2 a31 : −15 + 8b + c = 0 c = −1 a11 : k = −5 + 16 − 2 = 9  1 2 2 1   Q −1 =  1 1 3  9   3 b 1  1/ 9 2 / 9 2 / 9    =  1/ 9 1/ 9 1/ 3     1/ 3 2 / 9 1/ 9   −5 a 4   x   11       8 − 5 c   y  =  −5  ,  4 c   z   4   c  x   1      y  = 2      z   3 x = 1, y = 2, z = 3

2

z =8

4(a) zz = z = 64 M1

A1

M1

A1

5

D1 (End-point dotted, symmetry in y = x ) 3

D1 ( graph of f −1 )

D1 (graph of f )

A1

M1

A1

M1

2014-1-JOH-SULTAN ISMAIL-MARKING SCHEME f ( f −1 ( x)) = x [ f −1 ( x)]2 + 2 = x

y

20

x

Since f is defined for x ∈ [−5, −1] , f −1 ( x) < 0 f −1 ( x ) = − x − 2 R f = [3, 27]

20 10

10

f −1 : x a − x − 2, x ∈ [3, 27]

−10 −10

2(a) T2 = ar = 24 T5 = ar 4 = −3 1 r = − , a = −48 2 S∞ − Sn < 0.005

−48 −48[1 − (−0.5)n ] − < 0.005 1 − (−0.5) 1 − (−0.5) 32(0.5)n < 0.005 log(0.00015625) n> log(0.5) n > 12.64 the smallest value of n = 13

B1

M1

A1

A1

B1

M1

A1

M1

A1

A1

B1

M1

M1

A1

 4 3  2π arg(−4 + 4 3i ) = π − tan −1   =  4  3

A1

−4 + 4 3i = 42 + (4 3) 2 = 8

2π 2π   z = 8  cos + i sin  3 3    10π  z 5 = 85 cis    3   2π  = 32768cis  −   3  2π 2π   = 32768  cos − i sin  3 3  

4(b)

 2π  w3 = 8cis    3 

4

6

3

 2π / 3 + 2kπ  w = 2cis   , k = 0,1, 2 3    2π   8π   14π  w1 = 2cis   , w2 = 2cis   , w3 = 2cis    9   9   9   2π   8π   4π  w1 = 2cis   , w2 = 2cis   , w3 = 2cis  −   9   9   9  w = 1.5321 + 1.2856i, −1.8794 + 0.6840i, 0.3473 − 1.9696i M1 M1

A1

B1 B1

B1M1

A1

M1

M1

B1

D1

D1

M1

Vertices = (0, −5), (0,5) A1

x

−5 −4 −3 −2 −1 −1 1 2 3 4 5 6 −2 −3 −4 −5 Let (a, b) be a point on the ellipse

4 3 2 1

c 2 = 25 − 9, ⇒ c = ±4 Foci = (0, −4), (0, 4) y

5(a) Centre = (0, 0)

(b) a 2 b2 25 + = 1, ⇒ b 2 = 25 − a 2 - - - (1) 9 25 9 (d1 + d 2 ) 2 = ( a 2 + (b + 4) 2 + a 2 + (b − 4) 2 ) 2 = a 2 + (b + 4) 2 + a 2 + (b − 4) 2 + 2 a 2 + (b − 4) 2 a 2 + (b − 4) 2

2

 −1    2 2 2 2 2 2  2  = 2 + 3 + 4 1 + 2 + 2 cos θ    −2 

1

= 2a 2 + 2b 2 + 32 + 2 (a 2 + b 2 + 16) 2 − (8b) 2 - - - - (2) Substitute (1) into (2), … … … 50 16 = 2a 2 + 50 − a 2 + 32 + 2 (9 + a 2 ) 2 9 9 50 32 = 2a 2 + 50 − a 2 + 32 + 18 + a 2 9 9 = 100 ∴ (d + d ) 2 is constant.

6

2    3   4

5

−2 + 6 − 8 = 3 29 cos θ

4 cos θ = − 29 87 θ = 104.34° The acute angle is θ = 75.66°

2

M1

A1

A1

M1

A1

A1

B1

M1

A1

B1

M1

A1

3

3

4

B1 (or long division)

Section B [15 marks] ( MARK ONLY ONE question in this section – the FIRST ONE AFTER QUESTION 6 )

1

or x = 3 9 = 1.277 x = 9 9 = 1.277 cos( x − 17°13 ') < −0.5 .

3

(a − b − c)2 = (2 b c ) 2 = 4bc log 3 x 2 + log 3 x = log 27 9 log 3 9 log 3 x 3 = log 3 27 1 2 3 log 3 x = log 3 9 or 3 log 3 x = 3 3 1 2 or log x = log 3 9 log 3 x = 9 9

a −b−c = 2 b c

a =b+c+2 b c

x3 + 2 x 2 + 2 x + 3 2 3 ≡ x+ + x2 + 2 x − 3 x −1 x + 3 a= b+ c ( a )2 = ( b + c )2

x3 + 2 x 2 + 2 x + 3 x3 + 2 x 2 − 3x + 5 x + 3 7(a) ≡ x2 + 2 x − 3 x2 + 2 x − 3 5x + 3 ≡ x+ x2 + 2 x − 3 5x + 3 5x + 3 A B ≡ ≡ + x 2 + 2 x − 3 ( x − 1)( x + 3) x − 1 x + 3 5 x + 3 ≡ A( x + 3) + B ( x − 1) x = 1, 8 = 4 A ⇒ A=2 x = −3, − 12 = −4 B ⇒ B=3

7(b)

7(c)

7(d)

M1

−3

−2

−1

2 1

−1 −2

y

1

2

x 3

−3 cos( x − 17°13 ') = −0.5 = − cos 60° x − 17°13 ' = −120°,120° x = −102°47 ',137°13 '

32 + 122 + 62

AB = 4i − 9 j − k − (i + 3 j + 5k ) = 3i − 12 j − 6k 3i − 12 j − 6k

5  2 1

AD = BC  1  2   4        OD −  3  =  −1 −  −9         5   3   −1   −1    OD =  11 = −i + 11j + 9k    9

DABC is a parallelogram,

p=2

21 4 21 2 21 = i− j− k 21 21 21 AC = kAB  p   1  3        −1 −  3  = k  −12         3  5   −6  −4 = −12k or −2 = −6k 1 k= 3 1 p − 1 = (3) 3

The unit vector =

Solution set = { x : −180° ≤ x < −102 °47 ',137 °13 ' < x ≤ 180°}

(iii)

(ii)

8(a) (i)

(b)

 4 2 −3   1 1 −2   2 −1 3

M1 A1 A1A1

B1 M1 A1

M1

M1

A1 B1

M1

A1

B1

3

3

3

 4 2 −3   →  0 2 −5   0 −4 9 4 R2 − R1 → R2 2 R3 − R1 → R3

 4 2 −3  R3 + 2 R2 → R3  →  0 2 −5   0 0 −1 −z = 3 2 y − 5z = 3 4 x + 2 y − 3z = 5 x = 2, y = −6, z = −3

5  3 −3  5  3 3 

M1

M1

A1

M1

A1

6