A Level H1 Mathematics Algebra Functions Quiz

A-Level Maths H1 Quiz - Algebra Functions

Answer Key and Marking Scheme

Section A: Functions and Their Properties

1. f(x) = e^(2x) − 3

(a) f(0) = e^(0) − 3 = 1 − 3 = −2 [1 mark]

(b) e^(2x) − 3 = 5
e^(2x) = 8
2x = ln 8
x = (ln 8)/2 ≈ 1.04 [2 marks: 1 for correct equation, 1 for correct solution]

2. g(x) = ln(3x − 1)

(a) g(x) is not defined when 3x − 1 ≤ 0, i.e., x ≤ 1/3 [1 mark]

(b) g(2) = ln(3(2) − 1) = ln 5 ≈ 1.61 [1 mark]

(c) ln(3x − 1) = 0
3x − 1 = e^0 = 1
3x = 2
x = 2/3 [2 marks: 1 for setting argument to 1, 1 for solving]

3. P = 2e^(0.4t)

(a) Initial population: t = 0, P = 2e^0 = 2 thousand [1 mark]

(b) Double to 4 thousand: 4 = 2e^(0.4t)
e^(0.4t) = 2
0.4t = ln 2
t = (ln 2)/0.4 ≈ 1.73 hours [2 marks: 1 for equation, 1 for solution]

(c) dP/dt = 2(0.4)e^(0.4t) = 0.8e^(0.4t)
At t = 5: dP/dt = 0.8e^(2) ≈ 5.91 thousand per hour [2 marks: 1 for derivative, 1 for evaluation]

4. y = e^(−x)

(a) Horizontal asymptote: y = 0 [1 mark]

(b) Decreasing (as x increases, e^(−x) decreases) [1 mark]

(c) y-intercept: when x = 0, y = e^0 = 1; coordinates (0, 1) [1 mark]

5. e^(2x) − 4e^x + 3 = 0
Let u = e^x, then u² − 4u + 3 = 0
(u − 1)(u − 3) = 0
u = 1 or u = 3
e^x = 1 → x = 0
e^x = 3 → x = ln 3 ≈ 1.10
Solutions: x = 0, x = ln 3 [4 marks: 1 for substitution, 1 for factorising, 1 for each solution]

Section B: Equations and Inequalities

6. 2x² − 5x − 3 = 0
(2x + 1)(x − 3) = 0
x = −1/2 or x = 3 [2 marks: 1 for factorisation, 1 for both solutions]

7. x² + kx + 4 = 0 has no real roots when discriminant < 0.
Δ = k² − 4(1)(4) = k² − 16 < 0
k² < 16
−4 < k < 4 [3 marks: 1 for discriminant, 1 for inequality, 1 for range]

8. x² − 6x + 5 ≤ 0
(x − 1)(x − 5) ≤ 0
Critical values: x = 1, x = 5
Since coefficient of x² is positive, the parabola opens upward.
Solution: 1 ≤ x ≤ 5 [3 marks: 1 for factorisation, 1 for critical values, 1 for correct interval]

9. 2x² + 3x − 2 > 0
(2x − 1)(x + 2) > 0
Critical values: x = 1/2, x = −2
Since coefficient of x² is positive, the parabola opens upward.
Solution: x < −2 or x > 1/2 [3 marks: 1 for factorisation, 1 for critical values, 1 for correct intervals]

10. x² + px + 9 is always positive when discriminant < 0 and coefficient of x² > 0 (which it is).
Δ = p² − 4(1)(9) = p² − 36 < 0
p² < 36
−6 < p < 6 [2 marks: 1 for discriminant condition, 1 for range]

Section C: Simultaneous Equations and Applications

11. y = 3x − 1 and y = x² + x − 4
x² + x − 4 = 3x − 1
x² − 2x − 3 = 0
(x − 3)(x + 1) = 0
x = 3 or x = −1
When x = 3: y = 3(3) − 1 = 8
When x = −1: y = 3(−1) − 1 = −4
Solutions: (3, 8) and (−1, −4) [4 marks: 1 for equating, 1 for solving quadratic, 1 for each y-value]

12. Intersection: x² − x + 1 = 2x + k
x² − 3x + (1 − k) = 0
For two distinct points, discriminant > 0:
(−3)² − 4(1)(1 − k) > 0
9 − 4 + 4k > 0
4k > −5
k > −5/4 [4 marks: 1 for equation, 1 for discriminant, 1 for inequality, 1 for range]

13. Tangent condition: x² + 3x + 1 = mx + 2 has exactly one solution.
x² + (3 − m)x − 1 = 0
Discriminant = 0: (3 − m)² − 4(1)(−1) = 0
(3 − m)² + 4 = 0
(3 − m)² = −4
No real solutions for m.
Wait — check: x² + 3x + 1 = mx + 2 → x² + (3 − m)x − 1 = 0.
Δ = (3 − m)² − 4(1)(−1) = (3 − m)² + 4.
Since (3 − m)² ≥ 0, Δ ≥ 4 > 0 always.
Therefore, the line always cuts the curve at two distinct points. There are no values of m for which the line is a tangent.
[Alternative interpretation: the question may have a typo; if the curve were y = x² + 3x − 1, then Δ = (3 − m)² − 4(1)(−1) = (3 − m)² + 4, still no tangent. Accept "no possible values" with reasoning.]
[3 marks: 1 for setting up equation, 1 for discriminant, 1 for conclusion with reasoning]

14. 2x + y = 5 → y = 5 − 2x
Substitute into x² + y² = 13:
x² + (5 − 2x)² = 13
x² + 25 − 20x + 4x² = 13
5x² − 20x + 12 = 0
x = [20 ± √(400 − 240)] / 10 = [20 ± √160] / 10 = [20 ± 4√10] / 10 = 2 ± (2√10)/5
x = 2 + (2√10)/5 or x = 2 − (2√10)/5
When x = 2 + (2√10)/5: y = 5 − 2(2 + (2√10)/5) = 1 − (4√10)/5
When x = 2 − (2√10)/5: y = 5 − 2(2 − (2√10)/5) = 1 + (4√10)/5
[4 marks: 1 for substitution, 1 for quadratic, 1 for solving, 1 for y-values]

15. P > 0: −2x² + 20x − 32 > 0
Divide by −2 (reverse inequality): x² − 10x + 16 < 0
(x − 2)(x − 8) < 0
Solution: 2 < x < 8
The company makes a profit when production is between 200 and 800 units. [3 marks: 1 for inequality, 1 for factorisation, 1 for range with interpretation]

Section D: Exponential and Logarithmic Applications

16. V = 5000e^(0.06t)

(a) t = 10: V = 5000e^(0.6) ≈ 5000 × 1.82212 ≈ $9110.60 ≈$9110 (3 s.f.) [1 mark]

(b) 10 000 = 5000e^(0.06t)
e^(0.06t) = 2
0.06t = ln 2
t = (ln 2)/0.06 ≈ 11.55 years ≈ 11.6 years (3 s.f.) [2 marks: 1 for equation, 1 for solution]

17. T = 25 + 75e^(−0.1t)

(a) As t → ∞, e^(−0.1t) → 0, so T → 25°C. Room temperature is 25°C. [1 mark]

(b) t = 15: T = 25 + 75e^(−1.5) ≈ 25 + 75(0.22313) ≈ 25 + 16.73 ≈ 41.7°C (3 s.f.) [1 mark]

(c) 40 = 25 + 75e^(−0.1t)
15 = 75e^(−0.1t)
e^(−0.1t) = 0.2
−0.1t = ln 0.2
t = −(ln 0.2)/0.1 ≈ 16.09 ≈ 16.1 minutes (3 s.f.) [2 marks: 1 for equation, 1 for solution]

18. ln(x + 2) + ln(x − 1) = ln 10
ln[(x + 2)(x − 1)] = ln 10
(x + 2)(x − 1) = 10
x² + x − 2 = 10
x² + x − 12 = 0
(x + 4)(x − 3) = 0
x = −4 or x = 3
Check domain: x + 2 > 0 and x − 1 > 0 → x > 1.
Therefore x = 3 is the only valid solution. [3 marks: 1 for combining logs, 1 for solving quadratic, 1 for checking domain and selecting valid solution]

19. m = m₀e^(−kt), half-life = 8 days.

(a) When t = 8, m = m₀/2:
m₀/2 = m₀e^(−8k)
1/2 = e^(−8k)
−8k = ln(1/2) = −ln 2
k = (ln 2)/8 [2 marks: 1 for setting up half-life equation, 1 for deriving k]

(b) After 20 days: m = m₀e^(−20k) = m₀e^(−20(ln 2)/8) = m₀e^(−2.5 ln 2) = m₀(2^(−2.5))
Percentage remaining = 2^(−2.5) × 100% ≈ 0.17678 × 100% ≈ 17.7% (3 s.f.) [2 marks: 1 for substitution, 1 for percentage]

20. S = 50/(1 + 4e^(−0.3t))

(a) t = 0: S = 50/(1 + 4e^0) = 50/(1 + 4) = 50/5 = 10 thousand [1 mark]

(b) As t → ∞, e^(−0.3t) → 0, so S → 50/1 = 50 thousand [1 mark]

(c) 40 = 50/(1 + 4e^(−0.3t))
1 + 4e^(−0.3t) = 50/40 = 1.25
4e^(−0.3t) = 0.25
e^(−0.3t) = 0.0625
−0.3t = ln 0.0625
t = −(ln 0.0625)/0.3 ≈ 9.24 months (3 s.f.) [2 marks: 1 for equation, 1 for solution]

END OF ANSWER KEY