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Interactive practice · Further Maths

Interactive Exam Questions
OCR Further Maths

Work through exam-style OCR Further Maths solutions one line at a time. At each step, choose the line you think comes next, some of the wrong options are exactly what you'd write after a slip earlier on. Press Mark to see which steps you got right and why the others are wrong. This is practice, not a study guide: do it actively.

Subject A-Level Maths Further Maths
Board Edexcel OCR AQA
1 · Read the questionEach card is an OCR Further Maths-style question. Try it on paper first.
2 · Choose each linePick the next correct step. Wrong options mirror real, common mistakes.
3 · Mark & learnSee which steps were right, with feedback on every slip, then reveal the full solution.

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The 10 OCR questions on this page

Every question below is fully interactive above: open it, build the solution line by line, and be marked against an authentic OCR-style mark scheme with per-line feedback. The stems are listed here in full.

  1. Q1. Modulus and argument of a complex number (Pure Core (Y540/Y541) · typically 3 marks, 3 marks · Pure Core · Complex numbers)
    The complex number \(z=-2-2\sqrt{3}\,i\). Find its modulus \(|z|\) and its argument \(\arg z\), where \(-\pi<\arg z\le\pi\).
    Show worked solution & mark scheme

    \(|z|=\sqrt{(-2)^{2}+(-2\sqrt{3})^{2}}=\sqrt{4+12}=4\) (B1).

    The acute angle to the real axis is \(\arctan\sqrt3=\tfrac{\pi}{3}\); both parts are negative so \(z\) is in the third quadrant (M1). Measuring clockwise, \(\arg z=-(\pi-\tfrac{\pi}{3})=-\tfrac{2\pi}{3}\) (A1).

    \(|z|=4,\ \arg z=-\tfrac{2\pi}{3}\). Total: 3 marks.

  2. Q2. Apply De Moivre's theorem (Pure Core (Y540/Y541) · typically 3 marks, 3 marks · Pure Core · Complex numbers)
    Given \(z=3\left(\cos\tfrac{\pi}{4}+i\sin\tfrac{\pi}{4}\right)\), use De Moivre's theorem to find \(z^{4}\) in the form \(a+bi\).
    Show worked solution & mark scheme

    By De Moivre, \(z^{4}=3^{4}\big(\cos(4\cdot\tfrac{\pi}{4})+i\sin(4\cdot\tfrac{\pi}{4})\big)\) (M1) \(=81(\cos\pi+i\sin\pi)\) (A1).

    Since \(\cos\pi=-1\) and \(\sin\pi=0\):

    \(z^{4}=-81\). (A1) Total: 3 marks.

  3. Q3. Multiply two 2×2 matrices (Pure Core (Y540/Y541) · typically 3 marks, 3 marks · Pure Core · Matrices)
    Let \(A=\begin{pmatrix}3&-1\\2&4\end{pmatrix}\) and \(B=\begin{pmatrix}1&2\\5&-3\end{pmatrix}\). Find the product \(AB\).
    Show worked solution & mark scheme

    \(AB=\begin{pmatrix}3&-1\\2&4\end{pmatrix}\begin{pmatrix}1&2\\5&-3\end{pmatrix}=\begin{pmatrix}3(1)+(-1)(5)&3(2)+(-1)(-3)\\2(1)+4(5)&2(2)+4(-3)\end{pmatrix}\) (M1 A1 top row, A1 bottom row).

    \(AB=\begin{pmatrix}-2&9\\22&-8\end{pmatrix}\). Total: 3 marks.

  4. Q4. Determinant and inverse of a 2×2 matrix (Pure Core (Y540/Y541) · typically 3 marks, 3 marks · Pure Core · Matrices)
    The matrix \(M=\begin{pmatrix}4&3\\2&5\end{pmatrix}\) represents a transformation of the plane. Find \(\det M\) and hence the inverse \(M^{-1}\).
    Show worked solution & mark scheme

    \(\det M=(4)(5)-(3)(2)=14\) (B1); since it is non-zero, \(M\) is invertible.

    Adjugate: swap the leading diagonal and negate the rest to get \(\begin{pmatrix}5&-3\\-2&4\end{pmatrix}\) (M1).

    \(M^{-1}=\dfrac{1}{14}\begin{pmatrix}5&-3\\-2&4\end{pmatrix}\). (A1) Total: 3 marks.

  5. Q5. Sum and product of the roots of a cubic (Pure Core (Y540/Y541) · typically 3 marks, 3 marks · Pure Core · Roots of polynomials)
    The cubic equation \(3x^{3}+4x^{2}-6x+8=0\) has roots \(\alpha,\beta,\gamma\). Find \(\alpha+\beta+\gamma\), \(\alpha\beta+\beta\gamma+\gamma\alpha\) and \(\alpha\beta\gamma\).
    Show worked solution & mark scheme

    For \(ax^{3}+bx^{2}+cx+d=0\): \(\alpha+\beta+\gamma=-\tfrac{b}{a}\), \(\sum\alpha\beta=\tfrac{c}{a}\), \(\alpha\beta\gamma=-\tfrac{d}{a}\).

    Here \(a=3,\ b=4,\ c=-6,\ d=8\), so \(\alpha+\beta+\gamma=-\tfrac{4}{3}\) (B1), \(\alpha\beta+\beta\gamma+\gamma\alpha=\tfrac{-6}{3}=-2\) (B1), \(\alpha\beta\gamma=-\tfrac{8}{3}\) (B1).

    \(\alpha+\beta+\gamma=-\tfrac{4}{3},\quad \sum\alpha\beta=-2,\quad \alpha\beta\gamma=-\tfrac{8}{3}\). Total: 3 marks.

  6. Q6. Proof by induction: sum of cubes (Pure Core (Y540/Y541) · typically 6 marks, 6 marks · Pure Core · Proof by induction)
    Prove by induction that \(\displaystyle\sum_{r=1}^{n} r^{3}=\tfrac{1}{4}n^{2}(n+1)^{2}\) for all positive integers \(n\).
    Show worked solution & mark scheme

    Base case \(n=1\): LHS \(=1\), RHS \(=\tfrac14(1)^2(2)^2=1\) (B1).

    Inductive step: assume \(\sum_{r=1}^{k}r^3=\tfrac14 k^2(k+1)^2\). Then \(\sum_{r=1}^{k+1}r^3=\tfrac14 k^2(k+1)^2+(k+1)^3\) (M1) \(=\tfrac14(k+1)^2\big[k^2+4(k+1)\big]\) (dM1).

    \(k^2+4(k+1)=k^2+4k+4=(k+2)^2\) (A1), so \(\sum_{r=1}^{k+1}r^3=\tfrac14(k+1)^2(k+2)^2\), the formula with \(n=k+1\). True for \(n=1\), and \(P(k)\Rightarrow P(k+1)\), so true for all \(n\in\mathbb{Z}^{+}\) (A1).

    \(\sum_{r=1}^{n} r^{3}=\tfrac{1}{4}n^{2}(n+1)^{2}\). Total: 6 marks.

  7. Q7. Method of differences (Pure Core (Y540/Y541) · typically 4 marks, 4 marks · Pure Core · Series)
    Given that \(\dfrac{1}{r(r+2)}=\dfrac{1}{2}\!\left(\dfrac{1}{r}-\dfrac{1}{r+2}\right)\), use the method of differences to find \(\displaystyle\sum_{r=1}^{n}\dfrac{1}{r(r+2)}\).
    Show worked solution & mark scheme

    \(\sum_{r=1}^{n}\tfrac{1}{r(r+2)}=\tfrac12\sum_{r=1}^{n}\big(\tfrac1r-\tfrac1{r+2}\big)=\tfrac12\big[(\tfrac11-\tfrac13)+(\tfrac12-\tfrac14)+\dots+(\tfrac1{n}-\tfrac1{n+2})\big]\) (M1).

    The gap of 2 means survivors are \(+\tfrac11,+\tfrac12,-\tfrac1{n+1},-\tfrac1{n+2}\) (M1), giving \(\tfrac12\big(1+\tfrac12-\tfrac1{n+1}-\tfrac1{n+2}\big)\) (A1).

    \(=\tfrac34-\tfrac{2n+3}{2(n+1)(n+2)}=\dfrac{n(3n+5)}{4(n+1)(n+2)}\) (A1).

    \(\sum_{r=1}^{n}\dfrac{1}{r(r+2)}=\dfrac{n(3n+5)}{4(n+1)(n+2)}\). Total: 4 marks.

  8. Q8. Solve a hyperbolic equation (Pure Core (Y540/Y541) · typically 5 marks, 5 marks · Pure Core · Hyperbolic functions)
    Solve \(\cosh x=\dfrac{5}{4}\), giving your answers as exact logarithms.
    Show worked solution & mark scheme

    \(\cosh x=\tfrac{e^{x}+e^{-x}}{2}=\tfrac54\Rightarrow 2(e^{x}+e^{-x})=5\) (M1). Times \(e^{x}\): \(2e^{2x}-5e^{x}+2=0\) (M1).

    \((2e^{x}-1)(e^{x}-2)=0\Rightarrow e^{x}=\tfrac12\ \text{or}\ 2\) (dM1). Both are positive, so \(x=\ln2\) (A1) and \(x=\ln\tfrac12=-\ln2\) (A1).

    \(x=\pm\ln 2\). Total: 5 marks.

  9. Q9. Volume of revolution about the x-axis (Pure Core (Y540/Y541) · typically 4 marks, 4 marks · Pure Core · Further calculus)
    The region bounded by the curve \(y=\sqrt{x}\), the \(x\)-axis and the line \(x=4\) is rotated through \(2\pi\) about the \(x\)-axis. Find the exact volume of the solid generated.
    Show worked solution & mark scheme

    \(V=\pi\int_{0}^{4} y^{2}\,dx=\pi\int_{0}^{4} x\,dx\) (M1) \(=\pi\left[\tfrac{x^{2}}{2}\right]_{0}^{4}\) (A1) \(=\pi\left(\tfrac{16}{2}-0\right)\) (dM1).

    \(V=8\pi\). (A1) Total: 4 marks.

  10. Q10. Scalar product and the angle between two vectors (Pure Core (Y540/Y541) · typically 4 marks, 4 marks · Pure Core · Vectors)
    The vectors are \(\mathbf{a}=2\mathbf{i}-\mathbf{j}+2\mathbf{k}\) and \(\mathbf{b}=\mathbf{i}+2\mathbf{j}+2\mathbf{k}\). Find the acute angle \(\theta\) between \(\mathbf{a}\) and \(\mathbf{b}\), giving \(\cos\theta\) exactly.
    Show worked solution & mark scheme

    \(\mathbf{a}\cdot\mathbf{b}=(2)(1)+(-1)(2)+(2)(2)=4\) (M1).

    \(|\mathbf{a}|=\sqrt{4+1+4}=3\) and \(|\mathbf{b}|=\sqrt{1+4+4}=3\) (B1).

    \(\cos\theta=\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}=\dfrac{4}{(3)(3)}\) (M1) \(=\dfrac{4}{9}\) (A1), so \(\theta\approx63.6^{\circ}\) (acute).

    \(\cos\theta=\tfrac{4}{9}\). Total: 4 marks.

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