A-Level Mathematics May/June 2025 Q1: Solve the equation ex+2e-x / ex-3 = 4. Give your answer correct to 3 decimal places.

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  The given equation is: $\frac{e^x + 2e^{-x}}{e^x - 3} = 4$

  First, multiply both sides by $(e^x - 3)$ to eliminate the denominator: $e^x + 2e^{-x} = 4(e^x - 3)$ $e^x + 2e^{-x} = 4e^x - 12$

  To form a quadratic equation in terms of $e^x$, we can write $e^{-x}$ as $\frac{1}{e^x}$ and then multiply the entire equation by $e^x$: $e^x \left( e^x + \frac{2}{e^x} \right) = e^x (4e^x - 12)$ $e^{2x} + 2 = 4e^{2x} - 12e^x$

  Now, rearrange the terms to form a standard quadratic equation of the form $ay^2 + by + c = 0$, where $y = e^x$: $4e^{2x} - e^{2x} - 12e^x - 2 = 0$ $3e^{2x} - 12e^x - 2 = 0$

  Let $y = e^x$. The equation becomes: $3y^2 - 12y - 2 = 0$

  This does not factorise easily, so we use the quadratic formula, $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $e^x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(3)(-2)}}{2(3)}$ $e^x = \frac{12 \pm \sqrt{144 + 24}}{6}$ $e^x = \frac{12 \pm \sqrt{168}}{6}$

  This gives two possible values for $e^x$: $e^x = \frac{12 + \sqrt{168}}{6} \quad \text{or} \quad e^x = \frac{12 - \sqrt{168}}{6}$ $e^x \approx \frac{12 + 12.961}{6} \quad \text{or} \quad e^x \approx \frac{12 - 12.961}{6}$ $e^x \approx 4.16025... \quad \text{or} \quad e^x \approx -0.16025...$

  Since $e^x$ must be positive for any real $x$, we discard the negative solution. We proceed with $e^x = 4.16025...$. In exact form, this is $e^x = \frac{6+\sqrt{42}}{3}$.

  To find $x$, we take the natural logarithm of both sides: $x = \ln\left(\frac{12 + \sqrt{168}}{6}\right)$ $x = \ln(4.16025...)$ $x = 1.4255...$

  Rounding the answer to 3 decimal places: $x = 1.426$

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  How the marks are awarded

  • M1 — For the correct method of removing the fraction and using $e^{-x} = 1/e^x$ (often by multiplying the whole equation by $e^x$) to obtain a 3-term quadratic in $e^x$.
  • A1 — For obtaining the correct quadratic equation $3e^{2x} - 12e^x - 2 = 0$ or an equivalent form.
  • DM1 — For correctly applying a valid method, such as the quadratic formula, to solve their 3-term quadratic for $e^x$. This mark is dependent on the first M1 mark being awarded.
  • A1 — For obtaining the correct positive root for $e^x$, which is approximately $4.16$ or exactly $\frac{12+\sqrt{168}}{6}$ or $\frac{6+\sqrt{42}}{3}$.
  • A1 — For obtaining the final answer $x = 1.426$ and no other solutions. This requires the negative root for $e^x$ to be correctly discarded.

  Common mistakes

  • Sign errors when rearranging the equation to form the quadratic. For example, moving terms across the equals sign and not changing their sign, leading to an incorrect quadratic like $3e^{2x} + 12e^x - 2 = 0$.
  • Incorrectly handling the algebraic manipulation. A common error is to multiply by $e^x$ incorrectly, for instance getting $e^{x^2} + 2$ instead of $e^{2x} + 2$.
  • Failing to discard the negative solution for $e^x$. Students may try to calculate $\ln(-0.160...)$ which is undefined, and either stop or incorrectly state 'no solution' for the entire problem.
  • Premature rounding. Calculating a decimal value for $e^x$ and rounding it too early (e.g., to 4.2) before taking the natural logarithm, which can lead to an inaccurate final answer.

  Examiner tip: Recognise that equations involving both $e^x$ and $e^{-x}$ can usually be transformed into a standard quadratic by making the substitution $y = e^x$.

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