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.

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Pure Mathematics

Algebra and Functions

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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

Details

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

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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

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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

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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

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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

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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

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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

Details

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

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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

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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

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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

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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

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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

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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

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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

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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

Details

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

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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

Details

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

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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

Details

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

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Statistics

Data Representation

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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

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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

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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 $x$.

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

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

If you get this wrong, revise: Correlation and Regression

Probability

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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

Details

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

Details

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

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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

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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

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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

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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

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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

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Mechanics

Kinematics

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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

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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

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

If you get this wrong, revise: Kinematics

Forces and Newton's Laws

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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.