Maths T STPM 2014 Sem 1 P1 Trial Pahang Answer

g

A1

A1

M1

A1

M1

2014-1-PAHANG MARKING SCHEME

−1

f ( x) ∈ [−4 , 4] − 4 ≤ f ( x) ≤ 4 −4≤ x+4≤ 4 −8 ≤ x ≤ 0 But a ≤ x ≤ b ∴ minimum a = –8 , maximum b = 0

f ( x) ∈ D g x ∈ D f

g (1) = −3 , g (3) = −3 g (1) = g (3) = −3 g is not one-to-one ∴ function g −1 ( x) not defined For function g o f (x) defined R ⊂D f

CONFIDENTIAL*

1.(a)

1.(b)

2(a)

(b)

4  x −1 = 21 −  (1 + 3 x ) 2 (2 − x) 1 + 3x  2 x (−1)(−2) x 2 (−1)(−2)(−3) x 3   = 2 1 + (−1)(− ) + (− ) + .... (− ) + 3! 2 2! 2 2   (− 1 )(− 3 ) (− 1 )(− 3 )(− 5 )   1 2 2 2 2 (3 x) 2 + 2 (3 x) 3 + ... 1 + (− )(3 x) + 2 2 ! 3 !   1 1 27 2 135 3  1  3  = 2 1 + x + x 2 + x 3 + ... 1 − x + x − x + ... 4 8 8 16  2  2  23 2 x − 7 x 3 + ...) = 2(1 − x + 8 23 2 = 2 − 2x + x − 14 x 3 + ... 4 The expansion is valid when x − < 1 and 3x < 1 2 1 x < 2 and x < 3 1 1 1 or − < x < 3 3 3 ∴ x<

Either one series correct

−

Either one expanded correctly

k (1 − 2x ) −1 (1 + 3 x )

See

Get g (1) and g (3) and conclude

M1 A1 A1 M1

M1

A1

A1 M1

A1

1 2

=

=

[

(i) –20 (ii) 10

3 

−1 1 3 0 −4 0

3 0 −1 1

0

1 0

1 −3

4 0 5

−4 0

−2

 2

= 1

0 4

3 0 − 10 1

0  0 1  5

− 10 0

M

0 4

]

5

= I A −1

1  3  2 1  0  0

[A I ]

CONFIDENTIAL* 3(a)

3.(b)

3.(c)

0 1 0

0  0 , 1 

0  0 1 

R1 ↔ R 2

2

− 3 R1 + R2 → R2 − 2 R1 + R 3 → R 3

M1

M1

A1

A1

B1

M1

A1

A1

B1 B1

[ ]

]

Idea from A I to

[

get I A −1

See two ERO carried out correctly

,

θ= 6

π

p – 2q = 3 and 5p = 7 7 5

4 5

,

x2 y 2 + = 36 9 4

4x2 + 9y2 = 36

− 16 3 + 16i

π π  3 + i = 2  cos + i sin  6 6  5π 5π   32  cos + i sin  6 6  

r=2

q=−

p=

CONFIDENTIAL* 4.(a)

4.(b)

5




or or




3

4x2 – y2 = 4

x2 y 2 − =1 1 4


c2 = 9 - 4
c2 = 1 + 4 =5 =5 Centre is (0, 0)
Foci is ( 5 , 0) and ( − 5 , 0)

●

y

●

) (-1, 0)

(0, 2)

●

(1, 0)

(

) ●

y = - 2x

5,0

●

(3, 0)

Centre is (0, 0)
Foci is ( 5 , 0) and ( − 5 , 0)

5,0

●

The asymptotes are y = 2x and

●

(-3, 0)

(− ●(0, - 2)

x

M1 A1 A1 M1 M1

A1 M1 A1

M1 M1 A1

A1

Either one correct Both correct

See k (cos 56π + isin 56π )

Shape of ellipse Vertices and foci shown

either one correct both correct

D1 D1

Shape of hyperbola Vertices and asymptotes shown

B1 B1

D1 D1

c − a = −2i + 3 j − 6k

a • b = 13 b − a = 2i + j + 2k ,

CONFIDENTIAL* 6.(a) 6.(b)

i

j

k

(b − a) × (c − a ) = 2 1 2 −2 3 −6 = − 12i + 8 j + 8k

7.(a) x³ + mx² + 5x – n = (x² + 5)(x + 2) = x³ + 2x² + 5x + 10 By comparison, m = 2 , n = –10

1 2

4

00 (1) (2)

α = 0.4636

∴ cos x + 2 sin x ≡ 5 sin(x + 0.4636) Since the maximum value of sin(x + 0.4636) = 1 ∴ The maximum value of cos x + 2 sin x = 5 and it occurs when x + 0.4636 = π2 x = 1.1072 ≈ 1.11

7.(c) cosθ sin θ cot θ – tan θ = − sin θ cosθ cos 2 θ − sin 2 θ = sin θ cos θ cos 2θ = 1 sin 2θ 2 = 2 cot 2θ cot θ – tan θ = 12 2 cot 2θ = 12 2 cot 2θ = 2 3 ∴

cot 2θ = 3 tan 2θ = 13

B1 B1 B1

M1

A1

M1

M1

R or α correct

Find R or α

Determinant shown or any two components in answer correct

A1

CAO

M1 A1

A1

B1

Using cot θ and tan θ

B1

M1

Using sin2θ

See equation in one variable term

M1

A1

M1

A1

+λ and

r•

θ = 15o , 105o , 195o , 285o

For 0o ≤ θ ≤ 360o , 0o ≤ 2θ ≤ 720o 2θ = 30o , 210o , 390o , 570o

∴

r =

= =0

+ λ1

= =

-2

= -5 – 2λ + 2λ

, λ1 ∈ ℝ

=0

5

= -5

A1 A1

M1

A1

Get r and try checking if r satisfies equation of plane

Use idea of perpendicular

A1

M1

Scalar product

Find normal vector

M1 A1

A1

M1

×

=

A1

n=

10 + 4 λ1 + λ1 = 0
λ1 = -2

•

•

= -5 r satisfies vector equation of plane, R is a point on the plane ∴ line l lies on the plane.

•

For any point R on l with position vector r

r=

CONFIDENTIAL*

8(a)

(b)

(c)

=

+ λ2

Since P is on line l,

or

, λ2 ∈ ℝ

=

=0

•

– 5x + 10y + 5z = 0

r•

∴ equation of plane is r •

CONFIDENTIAL*

(d)

=

1   Given that r .  2  = 11.    2

-1

1   .  2  = 11.    2 5 + 2 λ2 + 2 λ2 + 10 = 11 λ2 = - 1 = =

6

M1

A1

M1

M1 A1

A1

Accept r •

=0

Get position vector of point and substitute to equation of plane Scalar product to get λ2