Standard +0.3 This is a straightforward area between curves problem requiring students to find intersection points, set up the integral ∫(top - bottom)dx, and evaluate using standard exponential integration. The 'show that' format provides the target answer, making it slightly easier than an open-ended question. While it requires multiple steps (finding intersection, setting up integral, integrating e^x and e^{2x}, simplifying to the given form), all techniques are standard A-level content with no novel insight required.
15 The region enclosed between the curves \(y = \mathrm { e } ^ { x } , y = 6 - \mathrm { e } ^ { \overline { 2 } }\) and the line \(x = 0\) is shown shaded in the diagram below.
\includegraphics[max width=\textwidth, alt={}, center]{08e1f291-7052-40a5-b7b2-13fd1d0137c2-28_1155_1009_424_513}
Show that the exact area of the shaded region is
$$6 \ln 4 - 5$$
Fully justify your answer.
\includegraphics[max width=\textwidth, alt={}, center]{08e1f291-7052-40a5-b7b2-13fd1d0137c2-30_2491_1736_219_139}
Forms a single equation eliminating \(x\) or \(y\)
\(e^x + e^{\frac{x}{2}} - 6 = 0\) or \(\left(e^{\frac{x}{2}}+3\right)\left(e^{\frac{x}{2}}-2\right)=0\) or \(e^x + e^{\frac{x}{2}} + \frac{1}{4} = \frac{25}{4}\) OE
A1 (1.1b)
Obtains correct rearranged quadratic equation
\(e^{\frac{x}{2}} = -3\) or \(2\)
M1 (1.1a)
Solves 'their' quadratic; must be a quadratic in \(e^{\frac{x}{2}}\), or if squaring is used must be a quadratic in \(e^x\)
\(e^{\frac{x}{2}} > 0\) so \(-3\) is not a valid solution
E1F (2.4)
Explains that \(e^{\frac{x}{2}}=-3\) is not valid as \(e^{\frac{x}{2}}>0\); or if squaring is used must clearly check both solutions and conclude \(\ln 9\) is not valid. OE
Integrates 'their' expressions involving exponentials fully correctly; must have integrated both \(e^x\) and \(e^{\frac{x}{2}}\) terms; condone missing/incorrect limits
\(= 6\ln 4 - 4 - 4 + 3\)
M1 (1.1a)
Substitutes 0 and 'their' upper limits into 'their' integrated expression; must correctly use \(F(\text{upper limit}) - F(0)\) for each integral
\(= 6\ln 4 - 5\)
R1 (2.1)
Completes rigorous argument by showing explicit evaluation of exponential terms before obtaining final answer. AG; this mark can be achieved without achieving the E1 mark
## Question 15:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $6 - e^{\frac{x}{2}} = e^x$ | M1 (3.1a) | Forms a single equation eliminating $x$ or $y$ |
| $e^x + e^{\frac{x}{2}} - 6 = 0$ or $\left(e^{\frac{x}{2}}+3\right)\left(e^{\frac{x}{2}}-2\right)=0$ or $e^x + e^{\frac{x}{2}} + \frac{1}{4} = \frac{25}{4}$ OE | A1 (1.1b) | Obtains correct rearranged quadratic equation |
| $e^{\frac{x}{2}} = -3$ or $2$ | M1 (1.1a) | Solves 'their' quadratic; must be a quadratic in $e^{\frac{x}{2}}$, or if squaring is used must be a quadratic in $e^x$ |
| $e^{\frac{x}{2}} > 0$ so $-3$ is not a valid solution | E1F (2.4) | Explains that $e^{\frac{x}{2}}=-3$ is not valid as $e^{\frac{x}{2}}>0$; or if squaring is used must clearly check both solutions and conclude $\ln 9$ is not valid. OE |
| $x = 2\ln 2 = \ln 4$ | A1 (1.1b) | Obtains $x=2\ln 2$ or $x=\ln 4$ |
| $\int_0^{\ln 4}\left(6-e^{\frac{x}{2}}-e^x\right)dx$ | M1 (1.1a) | Forms any definite integral which would contribute to finding the required area |
| $= \left[6x - 2e^{\frac{x}{2}} - e^x\right]_0^{\ln 4}$ | A1F (3.1a) | Forms fully correct definite integral(s) which would lead to evaluating the correct area; follow through 'their' incorrect upper limit |
| $= \left(6\ln 4 - 2e^{\frac{\ln 4}{2}} - e^{\ln 4}\right) - (-2-1)$ | B1F (1.1b) | Integrates 'their' expressions involving exponentials fully correctly; must have integrated both $e^x$ and $e^{\frac{x}{2}}$ terms; condone missing/incorrect limits |
| $= 6\ln 4 - 4 - 4 + 3$ | M1 (1.1a) | Substitutes 0 and 'their' upper limits into 'their' integrated expression; must correctly use $F(\text{upper limit}) - F(0)$ for each integral |
| $= 6\ln 4 - 5$ | R1 (2.1) | Completes rigorous argument by showing explicit evaluation of exponential terms before obtaining final answer. AG; this mark can be achieved without achieving the E1 mark |
15 The region enclosed between the curves $y = \mathrm { e } ^ { x } , y = 6 - \mathrm { e } ^ { \overline { 2 } }$ and the line $x = 0$ is shown shaded in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{08e1f291-7052-40a5-b7b2-13fd1d0137c2-28_1155_1009_424_513}
Show that the exact area of the shaded region is
$$6 \ln 4 - 5$$
Fully justify your answer.\\
\includegraphics[max width=\textwidth, alt={}, center]{08e1f291-7052-40a5-b7b2-13fd1d0137c2-30_2491_1736_219_139}
\hfill \mbox{\textit{AQA Paper 1 2020 Q15 [10]}}