SA2 Practice Paper (Version 1) — Answer Key
Subject: Additional Mathematics | Level: Secondary 3 | Total Marks: 50
Section A [20 marks]
1. Solve 3 x 2 − 7 x + 2 = 0 3x^2 - 7x + 2 = 0 3 x 2 − 7 x + 2 = 0
Using the quadratic formula: a = 3 a = 3 a = 3 b = − 7 b = -7 b = − 7 c = 2 c = 2 c = 2
Δ = ( − 7 ) 2 − 4 ( 3 ) ( 2 ) = 49 − 24 = 25 \Delta = (-7)^2 - 4(3)(2) = 49 - 24 = 25 Δ = ( − 7 ) 2 − 4 ( 3 ) ( 2 ) = 49 − 24 = 25
x = 7 ± 25 6 = 7 ± 5 6 x = \frac{7 \pm \sqrt{25}}{6} = \frac{7 \pm 5}{6} x = 6 7 ± 25 = 6 7 ± 5
x = 12 6 = 2 or x = 2 6 = 1 3 x = \frac{12}{6} = 2 \quad \text{or} \quad x = \frac{2}{6} = \frac{1}{3} x = 6 12 = 2 or x = 6 2 = 3 1
Answer: x = 2.00 x = 2.00 x = 2.00 x = 0.333 x = 0.333 x = 0.333
Marking: M1 for correct substitution into formula; A1 for both answers correct to 3 s.f.
2. Express x 2 − 6 x + 5 x^2 - 6x + 5 x 2 − 6 x + 5 ( x − a ) 2 + b (x - a)^2 + b ( x − a ) 2 + b
x 2 − 6 x + 5 = ( x − 3 ) 2 − 9 + 5 = ( x − 3 ) 2 − 4 x^2 - 6x + 5 = (x - 3)^2 - 9 + 5 = (x - 3)^2 - 4 x 2 − 6 x + 5 = ( x − 3 ) 2 − 9 + 5 = ( x − 3 ) 2 − 4
Answer: ( x − 3 ) 2 − 4 (x - 3)^2 - 4 ( x − 3 ) 2 − 4 a = 3 a = 3 a = 3 b = − 4 b = -4 b = − 4
Marking: M1 for completing the square; A1 for correct form.
3. Equal roots: 2 x 2 + k x + 8 = 0 2x^2 + kx + 8 = 0 2 x 2 + k x + 8 = 0
For equal roots, discriminant = 0 = 0 = 0
k 2 − 4 ( 2 ) ( 8 ) = 0 k^2 - 4(2)(8) = 0 k 2 − 4 ( 2 ) ( 8 ) = 0 k 2 = 64 k^2 = 64 k 2 = 64 k = ± 8 k = \pm 8 k = ± 8
Answer: k = 8 k = 8 k = 8 k = − 8 k = -8 k = − 8
Marking: M1 for setting discriminant = 0; A1 for both values.
4. Minimum of f ( x ) = x 2 − 4 x + 7 f(x) = x^2 - 4x + 7 f ( x ) = x 2 − 4 x + 7
Completing the square: f ( x ) = ( x − 2 ) 2 − 4 + 7 = ( x − 2 ) 2 + 3 f(x) = (x - 2)^2 - 4 + 7 = (x - 2)^2 + 3 f ( x ) = ( x − 2 ) 2 − 4 + 7 = ( x − 2 ) 2 + 3
Minimum occurs at x = 2 x = 2 x = 2 f ( 2 ) = 3 f(2) = 3 f ( 2 ) = 3
Answer: ( 2 , 3 ) (2, 3) ( 2 , 3 )
Marking: M1 for completing the square or using x = − b / 2 a x = -b/2a x = − b /2 a
5. Roots of x 2 − 5 x + 3 = 0 x^2 - 5x + 3 = 0 x 2 − 5 x + 3 = 0 α \alpha α β \beta β α 2 + β 2 \alpha^2 + \beta^2 α 2 + β 2
α + β = 5 \alpha + \beta = 5 α + β = 5 α β = 3 \alpha\beta = 3 α β = 3
α 2 + β 2 = ( α + β ) 2 − 2 α β = 25 − 6 = 19 \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 25 - 6 = 19 α 2 + β 2 = ( α + β ) 2 − 2 α β = 25 − 6 = 19
Answer: 19 19 19
Marking: M1 for using identity; A1 for correct answer.
6. Find range of x x x x ( 3 − x ) ≥ 0 x(3 - x) \geq 0 x ( 3 − x ) ≥ 0
Critical values: x = 0 x = 0 x = 0 x = 3 x = 3 x = 3
The quadratic − x 2 + 3 x -x^2 + 3x − x 2 + 3 x x ( 3 − x ) ≥ 0 x(3-x) \geq 0 x ( 3 − x ) ≥ 0
Answer: 0 ≤ x ≤ 3 0 \leq x \leq 3 0 ≤ x ≤ 3
Marking: M1 for finding critical values; A1 for correct inequality.
7. f ( x ) = 2 x 2 − 8 x + 3 f(x) = 2x^2 - 8x + 3 f ( x ) = 2 x 2 − 8 x + 3 k k k f ( x ) = k f(x) = k f ( x ) = k
Minimum of f ( x ) f(x) f ( x )
f ( x ) = 2 ( x − 2 ) 2 − 8 + 3 = 2 ( x − 2 ) 2 − 5 f(x) = 2(x - 2)^2 - 8 + 3 = 2(x - 2)^2 - 5 f ( x ) = 2 ( x − 2 ) 2 − 8 + 3 = 2 ( x − 2 ) 2 − 5
Minimum value is − 5 -5 − 5 k < − 5 k < -5 k < − 5
Answer: k < − 5 k < -5 k < − 5
Marking: M1 for finding minimum value; A1 for correct inequality.
8. Roots α \alpha α β \beta β x 2 + p x + q = 0 x^2 + px + q = 0 x 2 + p x + q = 0 α 2 + β 2 \alpha^2 + \beta^2 α 2 + β 2 p p p q q q
α + β = − p \alpha + \beta = -p α + β = − p α β = q \alpha\beta = q α β = q
α 2 + β 2 = ( − p ) 2 − 2 q = p 2 − 2 q \alpha^2 + \beta^2 = (-p)^2 - 2q = p^2 - 2q α 2 + β 2 = ( − p ) 2 − 2 q = p 2 − 2 q
Answer: p 2 − 2 q p^2 - 2q p 2 − 2 q
Marking: M1 for using sum/product of roots; A1 for correct expression.
9. f ( x ) = a x 2 + b x + c f(x) = ax^2 + bx + c f ( x ) = a x 2 + b x + c − 5 -5 − 5 x = 2 x = 2 x = 2 f ( 0 ) = 3 f(0) = 3 f ( 0 ) = 3 a a a b b b c c c
From f ( 0 ) = 3 f(0) = 3 f ( 0 ) = 3 c = 3 c = 3 c = 3
Vertex at x = 2 x = 2 x = 2 − b 2 a = 2 -\frac{b}{2a} = 2 − 2 a b = 2 b = − 4 a b = -4a b = − 4 a
f ( 2 ) = 4 a + 2 b + 3 = − 5 f(2) = 4a + 2b + 3 = -5 f ( 2 ) = 4 a + 2 b + 3 = − 5
4 a + 2 ( − 4 a ) + 3 = − 5 4a + 2(-4a) + 3 = -5 4 a + 2 ( − 4 a ) + 3 = − 5
4 a − 8 a + 3 = − 5 4a - 8a + 3 = -5 4 a − 8 a + 3 = − 5
− 4 a = − 8 -4a = -8 − 4 a = − 8 a = 2 a = 2 a = 2
b = − 4 ( 2 ) = − 8 b = -4(2) = -8 b = − 4 ( 2 ) = − 8
Answer: a = 2 a = 2 a = 2 b = − 8 b = -8 b = − 8 c = 3 c = 3 c = 3
Marking: M1 for setting up equations; A1 for all three correct.
10. f ( x ) = x 2 + 2 x − 3 f(x) = x^2 + 2x - 3 f ( x ) = x 2 + 2 x − 3 g ( x ) = 2 x + 1 g(x) = 2x + 1 g ( x ) = 2 x + 1 x x x f ( x ) = g ( x ) f(x) = g(x) f ( x ) = g ( x )
x 2 + 2 x − 3 = 2 x + 1 x^2 + 2x - 3 = 2x + 1 x 2 + 2 x − 3 = 2 x + 1 x 2 − 4 = 0 x^2 - 4 = 0 x 2 − 4 = 0 x 2 = 4 x^2 = 4 x 2 = 4 x = ± 2 x = \pm 2 x = ± 2
Answer: x = 2 x = 2 x = 2 x = − 2 x = -2 x = − 2
Marking: M1 for setting up equation; A1 for both values.
Section B [30 marks]
11. f ( x ) = 2 x 2 − 12 x + 11 f(x) = 2x^2 - 12x + 11 f ( x ) = 2 x 2 − 12 x + 11
(a) Express in form a ( x − h ) 2 + k a(x - h)^2 + k a ( x − h ) 2 + k
f ( x ) = 2 ( x 2 − 6 x ) + 11 = 2 ( x − 3 ) 2 − 18 + 11 = 2 ( x − 3 ) 2 − 7 f(x) = 2(x^2 - 6x) + 11 = 2(x - 3)^2 - 18 + 11 = 2(x - 3)^2 - 7 f ( x ) = 2 ( x 2 − 6 x ) + 11 = 2 ( x − 3 ) 2 − 18 + 11 = 2 ( x − 3 ) 2 − 7
Answer: 2 ( x − 3 ) 2 − 7 2(x - 3)^2 - 7 2 ( x − 3 ) 2 − 7
Marking: M1 for completing the square; A1 for correct form.
(b) Minimum point [1]
Answer: ( 3 , − 7 ) (3, -7) ( 3 , − 7 )
Marking: A1 for correct coordinates.
(c) Find range of x x x f ( x ) ≤ 3 f(x) \leq 3 f ( x ) ≤ 3
2 ( x − 3 ) 2 − 7 ≤ 3 2(x - 3)^2 - 7 \leq 3 2 ( x − 3 ) 2 − 7 ≤ 3 2 ( x − 3 ) 2 ≤ 10 2(x - 3)^2 \leq 10 2 ( x − 3 ) 2 ≤ 10 ( x − 3 ) 2 ≤ 5 (x - 3)^2 \leq 5 ( x − 3 ) 2 ≤ 5 − 5 ≤ x − 3 ≤ 5 -\sqrt{5} \leq x - 3 \leq \sqrt{5} − 5 ≤ x − 3 ≤ 5 3 − 5 ≤ x ≤ 3 + 5 3 - \sqrt{5} \leq x \leq 3 + \sqrt{5} 3 − 5 ≤ x ≤ 3 + 5
Answer: 3 − 5 ≤ x ≤ 3 + 5 3 - \sqrt{5} \leq x \leq 3 + \sqrt{5} 3 − 5 ≤ x ≤ 3 + 5 0.764 ≤ x ≤ 5.24 0.764 \leq x \leq 5.24 0.764 ≤ x ≤ 5.24
Marking: M1 for setting up inequality; M1 for solving; A1 for correct range.
12. x 2 − 6 x + c = 0 x^2 - 6x + c = 0 x 2 − 6 x + c = 0 α \alpha α β \beta β
(a) α + β \alpha + \beta α + β α β \alpha\beta α β
Answer: α + β = 6 \alpha + \beta = 6 α + β = 6 α β = c \alpha\beta = c α β = c
Marking: A1 for each.
(b) Given α 2 + β 2 = 24 \alpha^2 + \beta^2 = 24 α 2 + β 2 = 24 c c c
α 2 + β 2 = ( α + β ) 2 − 2 α β = 36 − 2 c = 24 \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 36 - 2c = 24 α 2 + β 2 = ( α + β ) 2 − 2 α β = 36 − 2 c = 24 2 c = 12 2c = 12 2 c = 12 c = 6 c = 6 c = 6
Answer: c = 6 c = 6 c = 6
Marking: M1 for using identity; A1 for correct value.
(c) Form equation with roots α 3 \alpha^3 α 3 β 3 \beta^3 β 3
α 3 + β 3 = ( α + β ) 3 − 3 α β ( α + β ) = 216 − 3 ( 6 ) ( 6 ) = 216 − 108 = 108 \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = 216 - 3(6)(6) = 216 - 108 = 108 α 3 + β 3 = ( α + β ) 3 − 3 α β ( α + β ) = 216 − 3 ( 6 ) ( 6 ) = 216 − 108 = 108
α 3 β 3 = ( α β ) 3 = 216 \alpha^3\beta^3 = (\alpha\beta)^3 = 216 α 3 β 3 = ( α β ) 3 = 216
Answer: x 2 − 108 x + 216 = 0 x^2 - 108x + 216 = 0 x 2 − 108 x + 216 = 0
Marking: M1 for using identities; A1 for correct equation.
13. f ( x ) = x 2 − 2 k x + k 2 − 4 f(x) = x^2 - 2kx + k^2 - 4 f ( x ) = x 2 − 2 k x + k 2 − 4
(a) Express in form ( x − a ) 2 + b (x - a)^2 + b ( x − a ) 2 + b
f ( x ) = ( x − k ) 2 − 4 f(x) = (x - k)^2 - 4 f ( x ) = ( x − k ) 2 − 4
Answer: ( x − k ) 2 − 4 (x - k)^2 - 4 ( x − k ) 2 − 4
Marking: M1 for completing the square; A1 for correct form.
(b) Minimum value in terms of k k k
Answer: − 4 -4 − 4
Marking: A1 for correct answer.
(c) Range of k k k y = − 5 y = -5 y = − 5
The minimum value is − 4 -4 − 4 − 4 > − 5 -4 > -5 − 4 > − 5 y = − 5 y = -5 y = − 5 k k k
Answer: All real values of k k k k ∈ R k \in \mathbb{R} k ∈ R
Marking: M1 for comparing minimum to -5; M1 for reasoning; A1 for correct conclusion.
14. 3 x 2 − 4 x + m = 0 3x^2 - 4x + m = 0 3 x 2 − 4 x + m = 0 α \alpha α β \beta β
(a) α + β \alpha + \beta α + β α β \alpha\beta α β m m m
Answer: α + β = 4 3 \alpha + \beta = \frac{4}{3} α + β = 3 4 α β = m 3 \alpha\beta = \frac{m}{3} α β = 3 m
Marking: A1 for each.
(b) Given 1 α + 1 β = 2 \frac{1}{\alpha} + \frac{1}{\beta} = 2 α 1 + β 1 = 2 m m m
1 α + 1 β = α + β α β = 4 / 3 m / 3 = 4 m = 2 \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{4/3}{m/3} = \frac{4}{m} = 2 α 1 + β 1 = α β α + β = m /3 4/3 = m 4 = 2 m = 2 m = 2 m = 2
Answer: m = 2 m = 2 m = 2
Marking: M1 for using identity; A1 for correct value.
(c) Solve 3 x 2 − 4 x + 2 = 0 3x^2 - 4x + 2 = 0 3 x 2 − 4 x + 2 = 0
Δ = 16 − 24 = − 8 < 0 \Delta = 16 - 24 = -8 < 0 Δ = 16 − 24 = − 8 < 0
No real roots. Using quadratic formula:
x = 4 ± − 8 6 = 4 ± 2 i 2 6 = 2 ± i 2 3 x = \frac{4 \pm \sqrt{-8}}{6} = \frac{4 \pm 2i\sqrt{2}}{6} = \frac{2 \pm i\sqrt{2}}{3} x = 6 4 ± − 8 = 6 4 ± 2 i 2 = 3 2 ± i 2
Answer: x = 2 + i 2 3 x = \frac{2 + i\sqrt{2}}{3} x = 3 2 + i 2 x = 2 − i 2 3 x = \frac{2 - i\sqrt{2}}{3} x = 3 2 − i 2
Marking: M1 for substitution; A1 for correct complex roots.
15. Graph passes through ( 0 , 5 ) (0, 5) ( 0 , 5 ) ( 1 , 0 ) (1, 0) ( 1 , 0 ) ( 3 , 0 ) (3, 0) ( 3 , 0 )
(a) Find c c c
f ( 0 ) = c = 5 f(0) = c = 5 f ( 0 ) = c = 5
Answer: c = 5 c = 5 c = 5
Marking: A1 for correct value.
(b) Write f ( x ) = a ( x − 1 ) ( x − 3 ) f(x) = a(x - 1)(x - 3) f ( x ) = a ( x − 1 ) ( x − 3 ) a a a
f ( 0 ) = a ( − 1 ) ( − 3 ) = 3 a = 5 f(0) = a(-1)(-3) = 3a = 5 f ( 0 ) = a ( − 1 ) ( − 3 ) = 3 a = 5
a = 5 3 a = \frac{5}{3} a = 3 5
Answer: f ( x ) = 5 3 ( x − 1 ) ( x − 3 ) f(x) = \frac{5}{3}(x - 1)(x - 3) f ( x ) = 3 5 ( x − 1 ) ( x − 3 ) a = 5 3 a = \frac{5}{3} a = 3 5
Marking: M1 for using roots form; A1 for correct value of a a a
(c) Find minimum point [3]
f ( x ) = 5 3 ( x − 1 ) ( x − 3 ) = 5 3 ( x 2 − 4 x + 3 ) = 5 3 x 2 − 20 3 x + 5 f(x) = \frac{5}{3}(x - 1)(x - 3) = \frac{5}{3}(x^2 - 4x + 3) = \frac{5}{3}x^2 - \frac{20}{3}x + 5 f ( x ) = 3 5 ( x − 1 ) ( x − 3 ) = 3 5 ( x 2 − 4 x + 3 ) = 3 5 x 2 − 3 20 x + 5
Vertex at x = 1 + 3 2 = 2 x = \frac{1+3}{2} = 2 x = 2 1 + 3 = 2
f ( 2 ) = 5 3 ( 2 − 1 ) ( 2 − 3 ) = 5 3 ( 1 ) ( − 1 ) = − 5 3 f(2) = \frac{5}{3}(2 - 1)(2 - 3) = \frac{5}{3}(1)(-1) = -\frac{5}{3} f ( 2 ) = 3 5 ( 2 − 1 ) ( 2 − 3 ) = 3 5 ( 1 ) ( − 1 ) = − 3 5
Answer: ( 2 , − 5 3 ) \left(2, -\frac{5}{3}\right) ( 2 , − 3 5 )
Marking: M1 for finding x-coordinate of vertex; M1 for substituting; A1 for correct coordinates.
End of Answer Key