A-Level Calculus exam-style questions
Use Clevolab for A-Level exam-style practice in calculus. This page shows what the topic covers, what skills the current set targets, and a few real examples from the reviewed question bank.
About this topic
Clevolab treats calculus as repeated practice with explanation, not just answer checking. This page is designed to make the topic legible before you open the app.
The current A-Level set covers differentiation, integration, rates of change, areas, and applications. 34 reviewed questions currently published for this page.
What you can practise
- Gradients and rates of change
- Differentiation rules
- Stationary points
- Integration basics
- Applied calculus questions
Real sample questions from the current set
These examples come from the reviewed questions currently stored for this topic. They are here so the page shows the actual flavour of Clevolab, not just a summary.
Sample question
A rectangle has its base on the $x$-axis and top corners on $y=9-x^2$. What dimensions maximise its area?
- Awidth $2\sqrt{3}$, height $6$Correct answer
- Bwidth $2\sqrt{6}$, height $3$Answer option
- Cwidth $2\sqrt{3}$, height $3$Answer option
- Dwidth $2$, height $8$Answer option
Why this answer is right
Let top corners be $(\pm a,9-a^2)$. Area $A=2a(9-a^2)$. Maximise: $A'=18-6a^2=0\Rightarrow a=\sqrt{3}$. Height $=9-3=6$.
By symmetry, take top corners $(\pm a,9-a^2)$ with $a>0$. Then width $=2a$, height $=9-a^2$. Area $$A(a)=2a(9-a^2).$$ Differentiate: $$A'(a)=18-6a^2=0\;\Rightarrow\;a^2=3\;\Rightarrow\;a=\sqrt{3}.$$ Height is $9-a^2=6$. Thus width $2\sqrt{3}$, height $6$. A second derivative check $A''(a)=-12<0$ confirms a maximum.
Sample question
Given $g'(x)=3-x^2$ and $g(0)=2$, at which $x$ does $g$ attain a local maximum?
- A$x=-\sqrt{3}$Answer option
- B$x=\sqrt{3}$Correct answer
- C$x=0$Answer option
- DNo local extremumAnswer option
Why this answer is right
Stationary points at $x=\pm\sqrt{3}$. With $g''(x)=-2x$, $g''(\sqrt{3})<0$ so $x=\sqrt{3}$ is a local maximum; $x=-\sqrt{3}$ is a local minimum.
Solve $g'(x)=0$: $$3-x^2=0\;\Rightarrow\;x=\pm\sqrt{3}.$$ Second derivative is $$g''(x)=-2x.$$ At $x=\sqrt{3}$, $g''<0$, so $g$ is concave down and has a local maximum. At $x=-\sqrt{3}$, $g''>0$, so there is a local minimum.
Sample question
A function has derivative $f'(x)=x(2-x)$ for $0\le x\le 3$ and $f(0)=0$. What is the maximum value of $f$ on $[0,3]$?
- A$0$Answer option
- B$1$Answer option
- C$\dfrac{4}{3}$Correct answer
- D$2$Answer option
Why this answer is right
Integrate to get $f(x)=x^2-\tfrac{x^3}{3}$. Stationary points at $x=0,2$. Compare $f(0)=0$, $f(2)=\tfrac{4}{3}$, $f(3)=0$. Max is $\tfrac{4}{3}$.
Given $$f'(x)=x(2-x)=2x-x^2,$$ integrate with $f(0)=0$: $$f(x)=\int_0^x(2t-t^2)\,dt=x^2-\frac{x^3}{3}.$$ Critical points satisfy $f'(x)=0$ at $x=0,2$. Evaluate: $$f(0)=0,\quad f(2)=4-\frac{8}{3}=\frac{4}{3},\quad f(3)=9-9=0.$$ Thus the maximum on $[0,3]$ is $\frac{4}{3}$ at $x=2$.
How this page fits into Clevolab
Clevolab is broader than any one exam mode. GCSE and A-level pages are useful entry points, while the wider project is about sharpening understanding through repeated topic practice.