A-Level Mathematics October/November 2024 Q11(a): Let f(x) = 2e^(2x) / (e^(2x)-3eˣ+2). Find f'(x) and hence find the exact coordinates of…

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  To find the derivative of $f(x) = \frac{2e^{2x}}{e^{2x}-3e^x+2}$, we use the quotient rule, $\frac{d}{dx}(\frac{u}{v}) = \frac{u'v - uv'}{v^2}$.

  Let $u = 2e^{2x}$ and $v = e^{2x}-3e^x+2$. Then $u' = 4e^{2x}$ and $v' = 2e^{2x}-3e^x$.

  Applying the quotient rule: $f'(x) = \frac{(4e^{2x})(e^{2x}-3e^x+2) - (2e^{2x})(2e^{2x}-3e^x)}{(e^{2x}-3e^x+2)^2}$

  To find the stationary point, we set $f'(x) = 0$. This occurs when the numerator is zero. $(4e^{2x})(e^{2x}-3e^x+2) - (2e^{2x})(2e^{2x}-3e^x) = 0$

  Factor out the common term $2e^{2x}$ (which is never zero): $2e^{2x} [2(e^{2x}-3e^x+2) - (2e^{2x}-3e^x)] = 0$ $2(e^{2x}-3e^x+2) - (2e^{2x}-3e^x) = 0$ $2e^{2x} - 6e^x + 4 - 2e^{2x} + 3e^x = 0$ $-3e^x + 4 = 0$ $3e^x = 4$ $e^x = \frac{4}{3}$ $x = \ln(\frac{4}{3})$

  Now, we find the corresponding $y$-coordinate by substituting $e^x = \frac{4}{3}$ and $e^{2x} = (e^x)^2 = (\frac{4}{3})^2 = \frac{16}{9}$ into the original function $f(x)$.

  $y = \frac{2e^{2x}}{e^{2x}-3e^x+2} = \frac{2(\frac{16}{9})}{(\frac{16}{9}) - 3(\frac{4}{3}) + 2}$ $y = \frac{\frac{32}{9}}{\frac{16}{9} - \frac{12}{3} + 2}$ $y = \frac{\frac{32}{9}}{\frac{16}{9} - \frac{36}{9} + \frac{18}{9}}$ $y = \frac{\frac{32}{9}}{\frac{16 - 36 + 18}{9}}$ $y = \frac{\frac{32}{9}}{\frac{-2}{9}}$ $y = -16$

  The exact coordinates of the stationary point are $(\ln(\frac{4}{3}), -16)$.

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  How the marks are awarded

  • M1 — Correctly applying the quotient rule structure, with an attempt to differentiate the numerator and denominator. This is shown in the line: $f'(x) = \frac{(4e^{2x})(e^{2x}-3e^x+2) - (2e^{2x})(2e^{2x}-3e^x)}{(e^{2x}-3e^x+2)^2}$.
  • A1 — Obtaining the correct derivative. The expression for $f'(x)$ must be fully correct, including the derivatives of $u$ and $v$, even if unsimplified.
  • M1 — Setting the derivative equal to zero. This is demonstrated by setting the numerator of the fraction to zero: $(4e^{2x})(e^{2x}-3e^x+2) - (2e^{2x})(2e^{2x}-3e^x) = 0$.
  • DM1 — Correctly solving the equation from the numerator to find an expression for $x$. This involves simplifying the equation to the form $ae^x = b$ and then taking natural logarithms to get $x = \ln(b/a)$. This mark is dependent on the previous M1.
  • A1 — Obtaining the exact coordinates. This requires the correct $x = \ln(4/3)$ and then substituting this back into the original function $f(x)$ to correctly calculate the exact value $y = -16$. The final coordinate pair must be stated.

  Common mistakes

  • Incorrectly differentiating $e^{2x}$ as $e^{2x}$ instead of $2e^{2x}$ due to forgetting the chain rule.
  • Making algebraic errors when expanding and simplifying the numerator of $f'(x)$, such as sign errors or incorrectly combining terms.
  • Errors when solving the equation for the stationary point, for example, trying to solve a 'hidden' quadratic in $e^x$ without first simplifying the expression by factoring.
  • Mistakes in substituting back to find the y-coordinate, particularly in evaluating $e^{2\ln(a)}$ as $2a$ instead of $a^2$.

  Examiner tip: When a derivative is a fraction, remember that stationary points occur when the numerator is zero, which often simplifies the subsequent algebra significantly.

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