Answer Maths T STPM 2014 Sem 1 Trial SMJK Jit Sin

2014-1-PEN-JIT SIN Marking scheme :

1

No

Working/Answer

f −1 ( x ) =

D f −1 = [0, ∞) , R f −1

r cos α = 3 , r sin α = 3 r2 = 32 + ( 3 ) 2

3 tan α = 3

π π

) ≤1

3 cos θ + 3 sin θ = 2 3 cos(θ − (a) −1 ≤ cos(θ −

6

π , 2π −

log 2 y =4 log 2 4

=±

, 2π

6

π

3

π

(b) θ −

θ = 0,

(a) log 2 x −

6

π

6

)

(changing base)

− 2 3 ≤ 2 3 cos(θ − ) ≤ 2 3 6 − 2 3 ≤ 3 cos θ + 3 sin θ ≤ 2 3 Maximum value is 2 3 , Minimum value is − 2 3

6

π

3 cos θ + 3 sin θ = r cos α cos θ + r sin α sin θ

3 = [− , ∞) 4

2 + 9x − x 2 A B C (a) ≡ + + (1 + x )(1 − x )2 1 + x 1 − x (1 − x )2 Try to solve for A(−2), B(−1) and C(5) : 2 + 9 −1 = C(1+1) or 2 − 9 − 1 = A(1 + 1)2 or −1 = A − B or equivalent 2 + 9x − x 2 2 1 5 ≡− − + (1 + x )(1 − x )2 1 + x 1 − x (1 − x ) 2 (b) f is one to one function Try to find inverse x2 − 3 4

2

3

x2 = 28 (eliminate log) y (2 y + 32) 2 = 256 y (eliminate one of the variables) y = 16, x = 64 (b) 4x − 1 > 3 − x or 4x − 1 < −(3 − x) (definition)

−1−

Partial marks B1 M1

A1 B1 M1 A1 B1B1

B1 M1 M1

M1 A1(both) M1 A1

M1 M1 M1 A1A1 M1

Total marks 8

7

5

4

4

x>

4 2 or x < − 5 3

∴ set of values of x = {x | x ∈R, x < −

π

2 4 or x > } 3 5

1 − 2sin2θ > 3sin θ + 2 (2sin θ + 1)(sin θ + 1) < 0 1 1 OR sin θ > −1 , sin θ < − [allow − 1 < sin θ < − 2 2 separate inequalities without “and”.]

−

5π 6

−

2

π

−

6

π

0

sin θ > −1 ⇒ − π < θ < π , θ ≠ − 2 1 5π π ⇒ −