A-Level Mathematics May/June 2024 Q5(a): A curve has equation y = (1+e2x)/(1+3x). The curve has exactly one stationary point P.…

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  The curve has the equation $y = \frac{1+e^{2x}}{1+3x}$.

  To find the stationary point, we first need to find the derivative, $\frac{dy}{dx}$, using the quotient rule. Let $u = 1+e^{2x}$ and $v = 1+3x$. Then $\frac{du}{dx} = 2e^{2x}$ and $\frac{dv}{dx} = 3$.

  Using the quotient rule, $\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$: $\frac{dy}{dx} = \frac{(1+3x)(2e^{2x}) - (1+e^{2x})(3)}{(1+3x)^2}$

  At a stationary point, $\frac{dy}{dx} = 0$. This occurs when the numerator is zero. $(1+3x)(2e^{2x}) - 3(1+e^{2x}) = 0$

  Expand the brackets: $2e^{2x} + 6xe^{2x} - 3 - 3e^{2x} = 0$

  Collect terms: $6xe^{2x} - e^{2x} - 3 = 0$

  Now, we rearrange the equation to make x the subject. $6xe^{2x} = 3 + e^{2x}$

  Divide both sides by $e^{2x}$: $6x = \frac{3}{e^{2x}} + \frac{e^{2x}}{e^{2x}}$ $6x = 3e^{-2x} + 1$

  Finally, divide by 6: $x = \frac{1}{6} + \frac{3}{6}e^{-2x}$ $x = \frac{1}{6} + \frac{1}{2}e^{-2x}$

  This shows that the x-coordinate of the stationary point P satisfies the given equation.

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  How the marks are awarded

  • M1 — The M1 mark is awarded for correctly applying the quotient rule. This involves differentiating the numerator and denominator and substituting them into the formula $\frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$.
  • A1 — The A1 mark is for obtaining the correct, unsimplified expression for the derivative: $\frac{(1+3x)(2e^{2x}) - (1+e^{2x})(3)}{(1+3x)^2}$.
  • DM1 — This dependent method mark is awarded for setting the derivative to zero and starting the algebraic rearrangement to make x the subject. This includes clearing the denominator and collecting terms to reach a form like $6xe^{2x} = 3 + e^{2x}$.
  • A1 — The final A1 mark is for correctly completing the rearrangement with clear steps to show the required result, $x = \frac{1}{6} + \frac{1}{2}e^{-2x}$, with no errors.

  Common mistakes

  • Incorrectly differentiating $e^{2x}$ as $e^{2x}$ instead of $2e^{2x}$ by forgetting the chain rule.
  • Mixing up the terms in the quotient rule numerator, for example calculating $u \frac{dv}{dx} - v \frac{du}{dx}$, which results in sign errors.
  • Making an algebraic error when expanding the brackets in the numerator, such as getting $2e^{2x} + 3xe^{2x}$ instead of $2e^{2x} + 6xe^{2x}$.
  • Incorrectly simplifying the equation after setting the numerator to zero, for instance cancelling $e^{2x}$ terms improperly before rearranging.

  Examiner tip: Master the quotient rule and be meticulous with algebraic rearrangement, especially when dealing with exponential terms, as showing a given result requires every step to be accurate.

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