MAT-style calculus and graphs questions

Differentiation and integration used to answer questions about graphs: how many solutions, which area, how many tangents. These are original problems in the style of the MAT. The MAT is no longer used for Oxford maths (from 2026 Oxford uses the TMUA), so treat them as problem-solving practice for the TMUA, STEP and interviews.

Practice questions

Try each one before you open a hint. The multiple-choice questions should take a few minutes each; give the longer ones 20 minutes or more.

Question 1: Integral of an absolute value · multiple choice

What is \(\displaystyle\int_0^2\left|x^2-1\right|\,dx\)?

  1. \(\tfrac23\)
  2. \(\tfrac43\)
  3. \(2\)
  4. \(\tfrac83\)
Hint 1

Split the interval where \(x^2-1\) changes sign.

Hint 2

On \([0,1]\) the integrand is \(1-x^2\); on \([1,2]\) it is \(x^2-1\).

Full solution

Answer: C: \(2\)

\(\int_0^1(1-x^2)dx=\tfrac23\) and \(\int_1^2(x^2-1)dx=\left(\tfrac83-2\right)-\left(\tfrac13-1\right)=\tfrac43\). Total \(2\).

Question 2: Tangents through the origin · multiple choice

Lines through the origin touch the curve \(y=x^2+1\). What is the sum of the squares of their gradients?

  1. \(2\)
  2. \(4\)
  3. \(8\)
  4. \(16\)
Hint 1

\(y=mx\) touches the curve when \(x^2-mx+1=0\) has a repeated root.

Hint 2

Repeated root: discriminant zero.

Full solution

Answer: C: \(8\)

\(m^2-4=0\) gives \(m=\pm2\), and \(2^2+(-2)^2=8\).

Question 3: How many tangents from a point? · longer question

Let \(C\) be the curve \(y=x^3-x\).

  1. Show that the tangent to \(C\) at the point with \(x=a\) is \(y=(3a^2-1)x-2a^3\).
  2. Show that this tangent passes through \((2,k)\) if and only if \(2a^3-6a^2+2+k=0\).
  3. Find the set of values of \(k\) for which exactly three tangents to \(C\) pass through \((2,k)\).
Hint 1

Different values of \(a\) give different tangents here, so count real roots \(a\).

Hint 2

Sketch \(h(a)=2a^3-6a^2+2\) and look at where the line \(h=-k\) crosses it three times.

Full solution

(i) \(y'=3a^2-1\), so \(y-(a^3-a)=(3a^2-1)(x-a)\), giving \(y=(3a^2-1)x-2a^3\).

(ii) Put \(x=2\), \(y=k\): \(k=6a^2-2-2a^3\), that is \(2a^3-6a^2+2+k=0\).

(iii) A cubic has different tangents at different points (the gradient \(3a^2-1\) and intercept \(-2a^3\) cannot both agree for \(a\ne b\)), so we need three distinct real roots of \(h(a)=-k\), where \(h(a)=2a^3-6a^2+2\). \(h'(a)=6a(a-2)\): \(h(0)=2\) (maximum), \(h(2)=-6\) (minimum). Three roots iff \(-6<-k<2\), that is \(-2

Results: \(-2

2 more multiple-choice and 1 more longer question

Counting crossings; Area between two parabolas; The best constant. Each with two hints and a full solution.

Included with A Level plans, and with IB, IGCSE or CBSE plans.

Next steps

Other areas: Algebra and functions · Counting, sequences and number · Geometry and logic. Then: TMUA practice · STEP topic guides · Strategies

The MAT was set by the University of Oxford. A Level Math Revision is independent: it is not affiliated with or endorsed by the University of Oxford or any university. Every problem here is our own, written in the style of the MAT; for real papers use Oxford's own past papers page.