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 ⇒ − <θ < − [all correct] 2 6 6 sin θ < −
−π
[can be implied]
π
5π π π π <θ < − , − <θ < − } 6 2 2 6
5
1 2 3 3 2 3 8 4 3 2 17 1 1 2 3 3 R2 − 2 R1 → R2 → 0 −1 2 − 2 3 2 17 1 1 2 3 3 R3 −3 R1 → R3 → 0 −1 2 − 2 0 − 4 8 − 8 1 2 3 3 R3 − 4 R2 → R3 → 0 −1 2 − 2 0 0 0 0 ∴ x = −1 − 7t, y = 2t + 2, z = t where t ∈R
∴ the solution set = {θ : −
6
(a) f(y) = y3 − y3 = 0 x − y is a factor of x3 − y3 [factor theorem] x3 − y3 = (x − y)(x2 + xy + y2) Putting y = −y
−2−
A1A1
A1
M1 M1
M1
M1
A1
B1
M1
M1
M1
A1
M1 A1 B1
6
5
5
x3 + y3 = (x + y)(x2 − xy + y2) x6 − y6 = (x3 − y3)(x3 + y3) x6 − y6 = (x − y)(x + y) (x2 + xy + y2)(x2 − xy + y2) (b) f(x) = (x − 2)(x2 + kx + 4) f(x) = 0 (x − 2)(x2 + kx + 4) = 0 x = 2 or x2 + kx + 4 = 0, let q(x) = x2 + kx + 4 f(x) has at least 2 real distinct roots, x2 + kx + 4 = 0 has two real roots and q(2) ≠ 0 b2 − 4ac ≥ 0 and q(2) ≠ 0 k2 − 4(1)(4) ≥ 0 and 4 + 2k + 4 ≠ 0 k2 ≥ 16 and 2k + 8 ≠ 0 k ≥ 4 and 2k ≠ −8 k ≥ 4, k ≤ −4 and k ≠ −4 ∴ the set of values of k = {k | k∈R, k ≥ 4 or k < −4}
−3−
B1 B1
B1
M1
A1A1 A1
5
7
6 − 8 − 3 0 0 − 1 2 4 13 AB = 3 1 − 2 − 17 − 9 10 = 0 − 3 0 6 − 7 0 0 − 3 1 4 5 11 1 0 0 AB = − 3 0 1 0 = −3I 0 0 1 A−1AB = A−1(−3I) IB = −3A−1 −3A−1 = B 1 A−1 = − B 3 − 1 2 4 x 8 3 1 − 2 y = − 1 1 4 5 z 13 AX = C A−1AX = A−1C IX = A−1C X = A−1C 6 − 8 8 x 13 1 y = − − 17 − 9 10 − 1 3 6 − 7 13 z 11 x 2 y = − 1 z 3 ∴ x = 2, y = −1, z = 3 (b) AB = −3I ABB−1 = −3IB−1 AI = −3B−1 −3B−1 = A 1 B−1 = − A 3 6 − 8 p − 13 13 − 17 − 9 10 q = 20 6 − 7 r − 14 11 BY = D B−1BY = B−1D IY = B−1D Y = B−1D p − 1 2 4 − 13 1 q = − 3 1 − 2 20 3 r 1 4 5 − 14
−4−
B1
B1
B1
M1
M1
A1
A2, 1, 0
B1
B1
M1
M1
8
8
p 1 q = − 3 r 1 ∴ p = 1, q = −3, r = 1
(a) (3 x )(3)(4 x )(4 2 ) = (52 ) x (5−1 )
−5−
A1
A2,1,0
M1
4
3 } 2
25 ( ) x = 240 12 25 x lg( ) = lg 240 12 x = 7.467 (4sf) (b) f(1) = 0 6 − 7 + a + b − 12 = 0 a + b = 13 … (1) f(−1) = −50 6 − 7(−1) + a + b(−1) − 12 = −50 a − b = −51 … (2) a = −19, b = 32 (a) f(−2) = 0 ⇒ x + 2 is a factor of f(x) (b) f(x) = (x − 1)(x + 2)(6x2 + Ax + 6) = (x2 + x − 2)(6x2 + Ax + 6) By long division or comparing coefficients or any valid methods, A = −13 f(x) = (x − 1)(x + 2)(2x − 3)(3x − 2) (c) (x − 1)(x + 2)(2x − 3)(3x − 2) > 0 By graphical or number line method or any valid method, 2 Set of values of x = {x | x∈R, x < −2 or < x < 1 or 3 x>
−6−
M1
M1
A1
M1
M1
A1(both)
M1A1
M1
M1A1
A1
M1A1
11