A Level Mathematics Diagnostic Test

Instructions

This diagnostic test contains 45 questions spanning all A Level Mathematics topics. Each Question tests a specific concept and requires 2-5 steps. Attempt all questions before checking Solutions.

Time: Allow approximately 90 minutes.
Equipment: Calculator permitted where indicated.
Scoring: Each question is worth 1 mark. Use your results to identify weak areas for revision.

Pure Mathematics

Algebra and Functions

Q1. Simplify $\dfrac{x^2 - 9}{x^2 - x - 6}$.

$\dfrac{x^2-9}{x^2-x-6} = \dfrac{(x-3)(x+3)}{(x-3)(x+2)} = \dfrac{x+3}{x+2}$, $x \neq 3, -2$.

If you get this wrong, revise: Algebraic Expressions

Q2. Solve $x^2 - 5x + 6 \geq 0$.

$(x-2)(x-3) \geq 0$. The quadratic opens upward, so $x \leq 2$ or $x \geq 3$.

If you get this wrong, revise: Equations and Inequalities

Q3. Find the inverse of $f(x) = \dfrac{2x+1}{x-3}$, $x \neq 3$.

$y = \dfrac{2x+1}{x-3} \implies y(x-3) = 2x+1 \implies yx - 3y = 2x + 1 \implies x(y-2) = 3y+1$.

$f^{-1}(x) = \dfrac{3x+1}{x-2}$ , $x \neq 2$ .

If you get this wrong, revise: Functions

Q4. Express $\dfrac{3x+5}{(x-1)(x+2)}$ in partial fractions.

$\dfrac{3x+5}{(x-1)(x+2)} = \dfrac{A}{x-1} + \dfrac{B}{x+2}$.

$3x+5 = A(x+2) + B(x-1)$ . $x=1$ : $8 = 3A \implies A = 8/3$ . $x=-2$ : $-1 = -3B \implies B = 1/3$ .

$= \dfrac{8/3}{x-1} + \dfrac{1/3}{x+2} = \dfrac{8}{3(x-1)} + \dfrac{1}{3(x+2)}$ .

If you get this wrong, revise: Algebraic Expressions

Sequences, Series, and Binomial Expansion

Q5. Find the sum of the first 50 terms of the arithmetic series $3 + 7 + 11 + \cdots$.

$a = 3$, $d = 4$. $S_{50} = \dfrac{50}{2}[2(3) + 49(4)] = 25(6 + 196) = 25 \times 202 = 5050$.

If you get this wrong, revise: Sequences and Series

Q6. Find the coefficient of $x^3$ in the expansion of $(2-3x)^5$.

$\binom{5}{3}(2)^2(-3x)^3 = 10 \times 4 \times (-27x^3) = -1080x^3$. Coefficient $= -1080$.

If you get this wrong, revise: Binomial Expansion

Q7. Find the sum to infinity of $0.5 + 0.1 + 0.02 + 0.004 + \cdots$.

$a = 0.5$, $r = 0.2$. $|r| \lt 1$. $S_\infty = \dfrac{0.5}{1-0.2} = \dfrac{0.5}{0.8} = 0.625$.

If you get this wrong, revise: Sequences and Series

Trigonometry

Q8. Solve $\sin 2x = \cos x$ for $0 \leq x \leq 2\pi$.

$2\sin x\cos x = \cos x \implies \cos x(2\sin x - 1) = 0$.

$\cos x = 0 \implies x = \pi/2, 3\pi/2$ . $2\sin x - 1 = 0 \implies \sin x = 1/2 \implies x = \pi/6, 5\pi/6$ .

$x = \pi/6, \pi/2, 5\pi/6, 3\pi/2$ .

If you get this wrong, revise: Trigonometry

Q9. Prove that $\dfrac◆LB◆1-\cos 2x◆RB◆◆LB◆1+\cos 2x◆RB◆ = \tan^2 x$.

$\dfrac◆LB◆1-\cos 2x◆RB◆◆LB◆1+\cos 2x◆RB◆ = \dfrac◆LB◆2\sin^2 x◆RB◆◆LB◆2\cos^2 x◆RB◆ = \tan^2 x$. $\blacksquare$

If you get this wrong, revise: Trigonometry

Exponentials and Logarithms

Q10. Solve $3^{2x-1} = 7$.

$(2x-1)\ln 3 = \ln 7 \implies x = \dfrac◆LB◆\ln 7 + \ln 3◆RB◆◆LB◆2\ln 3◆RB◆ = \dfrac◆LB◆\ln 21◆RB◆◆LB◆2\ln 3◆RB◆ \approx 1.771$.

If you get this wrong, revise: Exponentials and Logarithms

Q11. A population grows from 500 to 2000 in 6 hours. Find the doubling time (assume exponential growth).

$2000 = 500e^{6k} \implies e^{6k} = 4 \implies k = \dfrac◆LB◆\ln 4◆RB◆◆LB◆6◆RB◆ = \dfrac◆LB◆\ln 2◆RB◆◆LB◆3◆RB◆$.

$T_d = \dfrac◆LB◆\ln 2◆RB◆◆LB◆k◆RB◆ = 3$ hours.

If you get this wrong, revise: Exponentials and Logarithms

Differentiation

Q12. Find $\dfrac{dy}{dx}$ where $y = \dfrac◆LB◆x^2 e^x◆RB◆◆LB◆\sin x◆RB◆$.

$u = x^2e^x$$v = \sin x$. $u' = e^x(x^2+2x)$$v' = \cos x$.

$\dfrac{dy}{dx} = \dfrac◆LB◆e^x(x^2+2x)\sin x - x^2e^x\cos x◆RB◆◆LB◆\sin^2 x◆RB◆$ .

If you get this wrong, revise: Differentiation

Q13. Find the stationary points of $y = x^3 - 3x + 2$ and classify them.

$y' = 3x^2 - 3 = 0 \implies x = \pm 1$. $y'' = 6x$.

$x=1$ : $y'' = 6 \gt 0$ Minimum at $(1, 0)$ . $x=-1$ : $y'' = -6 \lt 0$ Maximum at $(-1, 4)$ .

If you get this wrong, revise: Differentiation

Q14. A sphere's radius increases at $3\,\mathrm{cm/s}$. Find $dV/dt$ when $r = 5\,\mathrm{cm}$.

$V = \dfrac{4}{3}\pi r^3$. $\dfrac{dV}{dt} = 4\pi r^2\dfrac{dr}{dt} = 4\pi(25)(3) = 300\pi\,\mathrm{cm}^3/\mathrm{s}$.

If you get this wrong, revise: Differentiation

Integration

Q15. Evaluate $\displaystyle\int_0^1 x e^x\,dx$.

By parts: $u=x$$dv=e^x\,dx$. $I = [xe^x]_0^1 - \int_0^1 e^x\,dx = e - (e-1) = 1$.

If you get this wrong, revise: Integration

Q16. Find the area enclosed between $y = x^2$ and $y = x$.

Intersection: $x^2 = x \implies x = 0, 1$.

$A = \int_0^1 (x - x^2)\,dx = \left[\dfrac{x^2}{2} - \dfrac{x^3}{3}\right]_0^1 = \dfrac{1}{2} - \dfrac{1}{3} = \dfrac{1}{6}$ .

If you get this wrong, revise: Integration

Vectors

Q17. Find the angle between $\mathbf{a} = \begin{pmatrix}1\\2\\-1\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}3\\-1\\2\end{pmatrix}$.

$\mathbf{a}\cdot\mathbf{b} = 3-2-2 = -1$. $|\mathbf{a}| = \sqrt{6}$$|\mathbf{b}| = \sqrt{14}$.

$\cos\theta = \dfrac◆LB◆-1◆RB◆◆LB◆\sqrt{84}◆RB◆ \implies \theta \approx 96.3^\circ$ .

If you get this wrong, revise: Vectors

Q18. Write the equation of the line through $(1,2,-1)$ in direction $\begin{pmatrix}2\\-1\\3\end{pmatrix}$.

$\mathbf{r} = \begin{pmatrix}1\\2\\-1\end{pmatrix} + t\begin{pmatrix}2\\-1\\3\end{pmatrix}$.

If you get this wrong, revise: Vectors

Proof

Q19. Prove by contradiction that $\sqrt{5}$ is irrational.

Suppose $\sqrt{5} = a/b$ in lowest terms. $5b^2 = a^2$So $5 \mid a^2 \implies 5 \mid a$. Write $a = 5k$: $5b^2 = 25k^2 \implies b^2 = 5k^2$So $5 \mid b$. Contradicts $\gcd(a,b)=1$. $\blacksquare$

If you get this wrong, revise: Proof

Q20. Prove by induction that $\displaystyle\sum_{r=1}^{n} r^2 = \dfrac{n(n+1)(2n+1)}{6}$.

*Base ($n=1$):* $1 = 1(2)(3)/6 = 1$. ✓ *Step:* $\dfrac{k(k+1)(2k+1)}{6} + (k+1)^2 = \dfrac{(k+1)[k(2k+1)+6(k+1)]}{6} = \dfrac{(k+1)(2k^2+7k+6)}{6} = \dfrac{(k+1)(k+2)(2k+3)}{6}$. ✓

If you get this wrong, revise: Proof

Numerical Methods

Q21. Show $x^3 - x - 2 = 0$ has a root in $[1, 2]$.

$f(1) = -2 \lt 0$$f(2) = 4 \gt 0$. Sign change, continuous function $\implies$ root in $(1,2)$.

If you get this wrong, revise: Numerical Methods

Q22. Use Newton-Raphson with $x_0=1.5$ to find $x_1$ for $f(x)=x^3-x-2=0$.

$f'(x)=3x^2-1$. $f(1.5)=3.375-1.5-2=-0.125$. $f'(1.5)=6.75-1=5.75$.

$x_1 = 1.5-(-0.125/5.75) = 1.5+0.0217 = 1.5217$ .

If you get this wrong, revise: Numerical Methods

Statistics

Data Representation

Q23. Find the mean and standard deviation of $\{4, 8, 6, 2, 10\}$.

$\bar{x} = 30/5 = 6$. $\sum x^2 = 16+64+36+4+100 = 220$. $\sigma^2 = 220/5 - 36 = 44-36 = 8$. $\sigma = 2\sqrt{2} \approx 2.83$.

If you get this wrong, revise: Data Representation

Q24. Data coded as $y = (x-50)/10$ has $\bar{y}=3$ and $\sigma_y=2$. Find the original mean and SD.

$\bar{x} = 10(3)+50 = 80$. $\sigma_x = 10(2) = 20$.

If you get this wrong, revise: Data Representation

Correlation and Regression

Q25. Given $S_{xx}=40$$S_{xy}=24$$S_{yy}=25$$\bar{x}=5$$\bar{y}=7$Find $r$ and the regression line of $y$ on $x$.

$r = \dfrac◆LB◆24◆RB◆◆LB◆\sqrt{40 \times 25}◆RB◆ = \dfrac◆LB◆24◆RB◆◆LB◆\sqrt{1000}◆RB◆ = \dfrac{24}{31.62} \approx 0.759$.

$b = 24/40 = 0.6$$a = 7 - 0.6(5) = 4$ . Line: $y = 4 + 0.6x$ .

If you get this wrong, revise: Correlation and Regression

Probability

Q26. $P(A)=0.6$$P(B)=0.5$$P(A \cap B)=0.3$. Find $P(A|B)$ and $P(A \cup B)$.

$P(A|B) = 0.3/0.5 = 0.6$. $P(A \cup B) = 0.6+0.5-0.3 = 0.8$.

If you get this wrong, revise: Probability

Q27. A bag has 5 red and 3 blue balls. Two are drawn without replacement. Find $P(\mathrm{both red})$.

$P = \dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}$.

If you get this wrong, revise: Probability

Q28. A disease affects 2% of the population. A test is 95% accurate. Find $P(\mathrm{disease} \mid \mathrm{positive})$.

$P(T^+|D) = 0.95$$P(T^+|D') = 0.05$. $P(T^+) = 0.95(0.02) + 0.05(0.98) = 0.019 + 0.049 = 0.068$.

$P(D|T^+) = 0.019/0.068 = 19/68 \approx 0.279$ .

If you get this wrong, revise: Probability

Statistical Distributions

Q29. $X \sim B(12, 0.3)$. Find $P(X = 4)$.

$P(X=4) = \binom{12}{4}(0.3)^4(0.7)^8 = 495 \times 0.0081 \times 0.0576 \approx 0.2311$.

If you get this wrong, revise: Statistical Distributions

Q30. $X \sim N(100, 64)$. Find $P(X \gt 108)$.

$P(X \gt 108) = P(Z \gt 8/8) = P(Z \gt 1) = 1 - 0.8413 = 0.1587$.

If you get this wrong, revise: Statistical Distributions

Q31. $X \sim \mathrm{Po}(5)$. Find $P(X \leq 3)$.

$P(X \leq 3) = e^{-5}\left(1+5+\dfrac{25}{2}+\dfrac{125}{6}\right) = e^{-5}(1+5+12.5+20.833) = 39.333 \times 0.00674 \approx 0.2650$.

If you get this wrong, revise: Statistical Distributions

Hypothesis Testing

Q32. A coin is tossed 20 times, landing heads 15 times. Test at 5% if biased towards heads.

$H_0: p=0.5$$H_1: p>0.5$. Under $H_0$: $X \sim B(20,0.5)$.

$P(X \geq 15) = 1-P(X \leq 14) \approx 0.0207 \lt 0.05$ . Reject $H_0$ : evidence of bias.

If you get this wrong, revise: Hypothesis Testing

Q33. Define Type I and Type II errors.

Type I: Rejecting $H_0$ when $H_0$ is true (false positive). Type II: Failing to reject $H_0$ when $H_0$ is false (false negative).

If you get this wrong, revise: Hypothesis Testing

Mechanics

Kinematics

Q34. A car accelerates from $15\,\mathrm{m/s}$ to $35\,\mathrm{m/s}$ over $200\,\mathrm{m}$. Find the acceleration.

$v^2 = u^2 + 2as \implies 1225 = 225 + 400a \implies a = 1000/400 = 2.5\,\mathrm{m/s}^2$.

If you get this wrong, revise: Kinematics

Q35. A projectile is launched at $25\,\mathrm{m/s}$ at $50^\circ$ above horizontal. Find the maximum height.

$H = \dfrac◆LB◆(25\sin 50°)^2◆RB◆◆LB◆2(9.8)◆RB◆ = \dfrac{(19.15)^2}{19.6} = \dfrac{366.7}{19.6} \approx 18.71\,\mathrm{m}$.

If you get this wrong, revise: Kinematics

Q36. A particle has velocity $v = 4t - t^2$ m/s. Find the total distance travelled from $t=0$ to $t=4$.

$v=0$ at $t=0,4$. For $00$. $s = \int_0^4(4t-t^2)\,dt = [2t^2-t^3/3]_0^4 = 32-64/3 = 32/3 \approx 10.67\,\mathrm{m}$.

If you get this wrong, revise: Kinematics

Forces and Newton’s Laws

Q37. A $5\,\mathrm{kg}$ block on a rough surface ($\mu=0.4$) is pushed by $30\,\mathrm{N}$ horizontally. Find the acceleration.

$R = 49\,\mathrm{N}$. $F_{\max} = 19.6\,\mathrm{N}$. $a = (30-19.6)/5 = 10.4/5 = 2.08\,\mathrm{m/s}^2$.

If you get this wrong, revise: Forces and Newton’s Laws

Q38. Masses $8\,\mathrm{kg}$ and $5\,\mathrm{kg}$ hang over a smooth pulley. Find the acceleration and tension.

$8g-T=8a$$T-5g=5a$. Adding: $3g=13a \implies a = 3g/13 \approx 2.26\,\mathrm{m/s}^2$.

$T = 5(g+a) = 5(9.8+2.26) = 60.3\,\mathrm{N}$ .

If you get this wrong, revise: Forces and Newton’s Laws

Moments

Q39. A uniform beam of length $6\,\mathrm{m}$ and weight $300\,\mathrm{N}$ is supported at both ends. A $200\,\mathrm{N}$ load is $2\,\mathrm{m}$ from the left end. Find the reactions.

Moments about left end: $R_R \times 6 = 300 \times 3 + 200 \times 2 = 1300 \implies R_R = 216.7\,\mathrm{N}$.

$R_L = 500 - 216.7 = 283.3\,\mathrm{N}$ .

If you get this wrong, revise: Moments

Q40. Find the centre of mass of masses $3\,\mathrm{kg}$$4\,\mathrm{kg}$$5\,\mathrm{kg}$ at $(0,0)$$(6,0)$$(3,4)$.

$\bar{x} = \dfrac{0+24+15}{12} = 39/12 = 3.25$. $\bar{y} = \dfrac{0+0+20}{12} = 5/3 \approx 1.67$.

If you get this wrong, revise: Moments

Energy and Work

Q41. A car of mass $1000\,\mathrm{kg}$ has engine power $40\,\mathrm{kW}$. Find the maximum speed against a resistance of $500\,\mathrm{N}$.

$P = Fv \implies 40000 = 500v \implies v = 80\,\mathrm{m/s}$.

If you get this wrong, revise: Energy and Work

Q42. A $2\,\mathrm{kg}$ ball is dropped from $15\,\mathrm{m}$. Find its speed just before impact using energy conservation.

$mgh = \tfrac{1}{2}mv^2 \implies v = \sqrt{2(9.8)(15)} = \sqrt{294} \approx 17.1\,\mathrm{m/s}$.

If you get this wrong, revise: Energy and Work

Momentum

Q43. A $4\,\mathrm{kg}$ body moving at $6\,\mathrm{m/s}$ collides with a $2\,\mathrm{kg}$ body at rest. They stick together. Find the common velocity.

$4(6) + 2(0) = 6v \implies v = 4\,\mathrm{m/s}$.

If you get this wrong, revise: Momentum

Q44. A ball hits a wall at $10\,\mathrm{m/s}$ and rebounds at $7\,\mathrm{m/s}$. If its mass is $0.15\,\mathrm{kg}$Find the impulse.

$J = m(v-u) = 0.15(-7-10) = 0.15(-17) = -2.55\,\mathrm{Ns}$. Magnitude: $2.55\,\mathrm{Ns}$.

If you get this wrong, revise: Momentum

Q45. Two bodies ($3\,\mathrm{kg}$ at $5\,\mathrm{m/s}$$2\,\mathrm{kg}$ at $-3\,\mathrm{m/s}$) collide with $e=0.5$. Find the velocities after collision.

Momentum: $15-6 = 3v_1+2v_2 \implies 3v_1+2v_2 = 9$. Restitution: $v_2-v_1 = 0.5(5-(-3)) = 4 \implies v_2 = v_1+4$.

$3v_1+2(v_1+4) = 9 \implies 5v_1 = 1 \implies v_1 = 0.2\,\mathrm{m/s}$$v_2 = 4.2\,\mathrm{m/s}$ .

If you get this wrong, revise: Momentum

Scoring and Revision Guide

Score	Action
40–45	Excellent — focus on exam technique
30–39	Good — revise weak topics
20–29	Fair — systematic revision needed
Below 20	Significant revision required

Use the revision links under each question to jump directly to the relevant topic notes.

Common Pitfalls

Forgetting to check that solutions satisfy the original equation (especially with squaring both sides or dividing by variables).
Losing marks by not showing sufficient working — always write out each step, especially in proof questions.
Forgetting the $+c$ constant of integration in indefinite integrals, or misusing boundary conditions in definite integrals.
Dropping negative signs during algebraic manipulation — substitute back to verify your answer.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Mark this page as reviewed