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Interactive practice · Further Maths

Interactive Exam Questions
Edexcel Further Maths

Ten Core Pure exam-style questions, worked one line at a time. Open a question, and each line is revealed only after you commit to the previous one, just like building a solution under exam conditions. You're then marked the way Edexcel marks it: method (M) and accuracy (A) marks adding up to the question's real total, with feedback on every slip.

Subject A-Level Maths Further Maths
Board Edexcel OCR AQA
1 · Open a questionEach is an Edexcel-style Core Pure question with its real mark total. Try it on paper first.
2 · Build the solutionPick each line; the next appears once you commit. Wrong options mirror real, common mistakes.
3 · Get your marksPress Mark to see your score out of the total, line-by-line M1/A1 awards and feedback.

Marks follow Edexcel conventions: M = method mark (correct approach), A = accuracy mark (correct result; needs the method), B = an independent accuracy mark, dM = a method mark that depends on the previous one. Each question shows its total, exactly as on the paper.

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The 10 Edexcel 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 Edexcel-style mark scheme with per-line feedback. The stems are listed here in full.

  1. Q1. Modulus and argument of a complex number (Core Pure 1 (9FM0/01) · typically 3 marks, 3 marks · Core Pure · Complex numbers)
    Find the modulus and argument of \(z=-1+\sqrt{3}\,i\), giving the argument in radians in the interval \((-\pi,\pi]\).
    Show worked solution & mark scheme

    \(|z|=\sqrt{(-1)^{2}+(\sqrt{3})^{2}}=\sqrt{1+3}=2\) (B1).

    \((-1,\sqrt3)\) is in the second quadrant, reference angle \(\arctan\frac{\sqrt3}{1}=\frac{\pi}{3}\), so \(\arg z=\pi-\frac{\pi}{3}\) (M1) \(=\frac{2\pi}{3}\) (A1).

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

  2. Q2. Use De Moivre's theorem (Core Pure 1 (9FM0/01) · typically 4 marks, 4 marks · Core Pure · Complex numbers)
    Use De Moivre's theorem to evaluate \((1+i)^{8}\).
    Show worked solution & mark scheme

    \(1+i=\sqrt2\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)\) (M1 A1).

    \((1+i)^{8}=(\sqrt2)^{8}\left(\cos\frac{8\pi}{4}+i\sin\frac{8\pi}{4}\right)\) (M1) \(=16(\cos2\pi+i\sin2\pi)\).

    \(=16\) (A1). Total: 4 marks.

  3. Q3. Divide two complex numbers (Core Pure 1 (9FM0/01) · typically 3 marks, 3 marks · Core Pure · Complex numbers)
    Express \(\dfrac{3+2i}{1-i}\) in the form \(a+bi\), where \(a,b\in\mathbb{R}\).
    Show worked solution & mark scheme

    \(\dfrac{3+2i}{1-i}\times\dfrac{1+i}{1+i}\) (M1) \(=\dfrac{3+3i+2i+2i^{2}}{1-i^{2}}=\dfrac{1+5i}{2}\) (M1).

    \(=\dfrac{1}{2}+\dfrac{5}{2}i\) (A1). Total: 3 marks.

  4. Q4. Determinant and inverse of a 2×2 matrix (Core Pure 1 (9FM0/01) · typically 3 marks, 3 marks · Core Pure · Matrices)
    Given \(\mathbf{A}=\begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\), find \(\det\mathbf{A}\) and hence \(\mathbf{A}^{-1}\).
    Show worked solution & mark scheme

    \(\det\mathbf{A}=(3)(2)-(1)(5)=1\) (B1).

    \(\mathbf{A}^{-1}=\dfrac{1}{\det\mathbf{A}}\begin{pmatrix}2&-1\\-5&3\end{pmatrix}\) (M1) \(=\dfrac{1}{1}\begin{pmatrix}2&-1\\-5&3\end{pmatrix}\).

    \(\mathbf{A}^{-1}=\begin{pmatrix}2&-1\\-5&3\end{pmatrix}\) (A1). Total: 3 marks.

  5. Q5. Multiply two 2×2 matrices (Core Pure 1 (9FM0/01) · typically 3 marks, 3 marks · Core Pure · Matrices)
    Given \(\mathbf{A}=\begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}\) and \(\mathbf{B}=\begin{pmatrix} 1 & 4 \\ 2 & 1 \end{pmatrix}\), find the product \(\mathbf{AB}\).
    Show worked solution & mark scheme

    Top row: \((2)(1)+(1)(2)=4\), \((2)(4)+(1)(1)=9\) (M1).

    Bottom row: \((0)(1)+(3)(2)=6\), \((0)(4)+(3)(1)=3\) (A1).

    \(\mathbf{AB}=\begin{pmatrix}4&9\\6&3\end{pmatrix}\) (A1). Total: 3 marks.

  6. Q6. Symmetric functions of the roots (Core Pure 1 (9FM0/01) · typically 3 marks, 3 marks · Core Pure · Roots of polynomials)
    The quadratic \(2x^{2}+3x-5=0\) has roots \(\alpha\) and \(\beta\). Find \(\alpha^{2}+\beta^{2}\).
    Show worked solution & mark scheme

    \(\alpha+\beta=-\frac{b}{a}=-\frac{3}{2}\), \(\alpha\beta=\frac{c}{a}=-\frac{5}{2}\) (B1).

    \(\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2\alpha\beta\) (M1) \(=\left(-\frac{3}{2}\right)^{2}-2\left(-\frac{5}{2}\right)=\frac{9}{4}+5\).

    \(=\dfrac{29}{4}\) (A1). Total: 3 marks.

  7. Q7. Proof by induction (sum of integers) (Core Pure 1 (9FM0/01) · induction step, typically 4 marks, 4 marks · Core Pure · Proof by induction)
    Prove by induction that \(\displaystyle\sum_{r=1}^{n} r=\dfrac{n(n+1)}{2}\) for all integers \(n\ge 1\). Here we mark the inductive step: assume the result holds for \(n=k\).
    Show worked solution & mark scheme

    Assume true for \(n=k\): \(\sum_{r=1}^{k}r=\frac{k(k+1)}{2}\).

    Then \(\sum_{r=1}^{k+1}r=\frac{k(k+1)}{2}+(k+1)\) (M1) \(=(k+1)\left(\frac{k}{2}+1\right)\) (dM1) \(=\frac{(k+1)(k+2)}{2}\) (A1).

    This is the formula with \(n=k+1\); as it is true for \(n=1\) and true for \(n=k\Rightarrow\) true for \(n=k+1\), it holds for all \(n\ge1\) (A1). Total: 4 marks.

  8. Q8. Method of differences (Core Pure 2 (9FM0/02) · typically 4 marks, 4 marks · Core Pure · Series)
    Given that \(\dfrac{1}{r(r+1)}=\dfrac{1}{r}-\dfrac{1}{r+1}\), use the method of differences to find \(\displaystyle\sum_{r=1}^{n}\dfrac{1}{r(r+1)}\).
    Show worked solution & mark scheme

    \(\sum_{r=1}^{n}\left(\frac{1}{r}-\frac{1}{r+1}\right)=\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\dots+\left(\frac1n-\frac1{n+1}\right)\) (M1 A1).

    All interior terms cancel, leaving \(1-\frac{1}{n+1}\) (dM1).

    \(=\dfrac{n}{n+1}\) (A1). Total: 4 marks.

  9. Q9. Maclaurin series of ln(1+x) (Core Pure 2 (9FM0/02) · typically 3 marks, 3 marks · Core Pure · Maclaurin series)
    Find the first three non-zero terms of the Maclaurin series for \(f(x)=\ln(1+x)\).
    Show worked solution & mark scheme

    \(f'(x)=\frac{1}{1+x},\ f''(x)=-\frac{1}{(1+x)^{2}},\ f'''(x)=\frac{2}{(1+x)^{3}}\); so \(f(0)=0,\,f'(0)=1,\,f''(0)=-1,\,f'''(0)=2\) (M1).

    \(\ln(1+x)=0+(1)x+\frac{-1}{2!}x^{2}+\frac{2}{3!}x^{3}+\dots\) (M1).

    \(=x-\dfrac{x^{2}}{2}+\dfrac{x^{3}}{3}\) (A1). Total: 3 marks.

  10. Q10. Hyperbolic functions (final answer) (Core Pure 2 (9FM0/02) · typically 2 marks, 2 marks · Core Pure · Hyperbolic functions)
    Given that \(\cosh x=\dfrac{5}{4}\) and \(x>0\), use the identity \(\cosh^{2}x-\sinh^{2}x=1\) to find the exact value of \(\sinh x\).
    Show worked solution & mark scheme

    \(\sinh^{2}x=\cosh^{2}x-1=\left(\frac{5}{4}\right)^{2}-1=\frac{25}{16}-\frac{16}{16}=\frac{9}{16}\) (M1).

    Since \(x>0\), \(\sinh x>0\).

    \(\sinh x=\dfrac{3}{4}\) (A1). Total: 2 marks.

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