A Level H1 Mathematics Algebra Functions Quiz

A-Level Maths H1 Quiz - Algebra Functions (Answer Key)

Topic: Algebra & Functions
Total Marks: 40
Duration: 60 minutes

Section A: Functions and Equations (Q1–10)

Q1 [2 marks]
Solve $2^x = 8$.
Since $8 = 2^3$, we have $2^x = 2^3 \Rightarrow x = 3$.
Teaching note: Express both sides with same base; equate exponents.
Common mistake: Writing $x = \log_2 8$ without simplifying to 3 loses no mark but full credit for 3.

Q2 [2 marks]
$\ln y = 3 \Rightarrow y = e^3$.
Teaching note: Definition of natural log: if $\ln y = a$ then $y = e^a$.

Q3 [2 marks]
$e^x = 5 \Rightarrow x = \ln 5$.
Teaching note: Inverse relationship between $e^x$ and $\ln x$.

Q4 [2 marks]
For two equal real roots, discriminant $= 0$: $k^2 - 4(1)(4) = 0 \Rightarrow k^2 = 16 \Rightarrow k = \pm 4$.
Marking: 1 mark for condition $b^2 - 4ac = 0$, 1 mark for values.

Q5 [2 marks]
$x^2 - 6x + 10$: discriminant $= 36 - 40 = -4 < 0$, coefficient of $x^2 > 0$, so always positive.
Teaching note: No real roots and upward parabola → always positive.

Q6 [2 marks]
Substitute: $2x+1 = x^2+3x-2 \Rightarrow x^2 + x - 3 = 0$.
$x = \frac{-1 \pm \sqrt{1+12}}{2} = \frac{-1 \pm \sqrt{13}}{2}$.
Then $y = 2x+1$ gives corresponding $y$.
Marking: 1 mark substitution, 1 mark solutions.

Q7 [2 marks]
$(x-2)(x-3) < 0 \Rightarrow 2 < x < 3$.
Teaching note: Quadratic inequality sign chart.

Q8 [2 marks]
Using GC: intersect of $y=e^x$ and $y=x^2+2$ at $x \approx 1.319$ (positive).
Marking: 2 marks for correct approx.

Q9 [2 marks]
$f(g(7)) = e^{\ln 7} = 7$.
Teaching note: $e^{\ln x} = x$ for $x>0$.

Q10 [2 marks]
$\ln(2x-1)=0 \Rightarrow 2x-1 = e^0 = 1 \Rightarrow x = 1$.
Check: $2(1)-1=1>0$ valid.

Section B: Inequalities and Graphical Methods (Q11–15)

Q11 [2 marks]
Always positive $\Rightarrow$ discriminant $< 0$: $(2k)^2 - 36 < 0 \Rightarrow 4k^2 < 36 \Rightarrow k^2 < 9 \Rightarrow -3 < k < 3$.

Q12 [2 marks]
$3^x > 9 = 3^2 \Rightarrow x > 2$ (base >1, inequality unchanged).

Q13 [2 marks]
GC intersection at $x \approx 1.073$ (solve $e^x = 4-x$).
Marking: 1 mark sketch, 1 mark x-value.

Q14 [2 marks]
Substitute $y=x+3$: $x^2+(x+3)^2=25 \Rightarrow 2x^2+6x-16=0 \Rightarrow x^2+3x-8=0$.
$x = \frac{-3 \pm \sqrt{41}}{2}$, $y = x+3$.
Marking: 1 mark equation, 1 mark roots.

Q15 [2 marks]
As $x \to -\infty$, $e^x \to 0$, so $y \to -2$. Horizontal asymptote: $y = -2$.

Section C: Applied and Structured (Q16–20)

Q16 [2 marks]
$200 = 100e^{0.05t} \Rightarrow 2 = e^{0.05t} \Rightarrow \ln 2 = 0.05t \Rightarrow t = \frac{\ln 2}{0.05} \approx 13.86$.

Q17 [2 marks]
$\frac{1}{2}A_0 = A_0 e^{-0.1t} \Rightarrow \frac{1}{2} = e^{-0.1t} \Rightarrow \ln(1/2) = -0.1t \Rightarrow t = \frac{\ln 2}{0.1} = 10\ln 2$.

Q18 [2 marks]
Let integers be $n, n+1$: $n(n+1)=72 \Rightarrow n^2+n-72=0$.
Marking: 1 mark setup, 1 mark equation.

Q19 [2 marks]
$2^{x+1}=16=2^4 \Rightarrow x+1=4 \Rightarrow x=3$.

Q20 [2 marks]
$\ln 3 + \ln b = \ln 12 \Rightarrow \ln(3b)=\ln 12 \Rightarrow 3b=12 \Rightarrow b=4$.