| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find gradient at point |
| Difficulty | Standard +0.3 Part (a) is a straightforward numerical integration using mid-ordinate rule with given strip width. Part (b) requires product rule application to differentiate e^(2-x)ln(3x-2) and chain rule for both factors, then substitution - standard C3 technique with no novel insight, but slightly above average due to the combination of exponential and logarithmic functions requiring careful execution. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Strip width \(h = \frac{5.5-1.5}{4} = 1\) | B1 | Correct strip width stated or implied |
| Mid-ordinates at \(x = 1.75, 2.75, 3.75, 4.75\) | M1 | At least 3 correct mid-x values |
| \(f(1.75) = e^{0.25}\ln(3.25) \approx 1.4994\) | ||
| \(f(2.75) = e^{-0.75}\ln(6.25) \approx 0.9290\) | ||
| \(f(3.75) = e^{-1.75}\ln(9.25) \approx 0.3779\) | ||
| \(f(4.75) = e^{-2.75}\ln(12.25) \approx 0.1344\) | A1 | All four y-values correct |
| Estimate \(= 1 \times (1.4994 + 0.9290 + 0.3779 + 0.1344) \approx 2.941\) | A1 | Correct answer to 3 d.p. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = e^{2-x} \cdot \frac{3}{3x-2} + \ln(3x-2) \cdot (-e^{2-x})\) | M1 | Attempt product rule |
| \(= e^{2-x}\left(\frac{3}{3x-2} - \ln(3x-2)\right)\) | A1 | Correct derivative |
| At \(x=2\): \(e^{0}\left(\frac{3}{4} - \ln 4\right)\) | M1 | Substituting \(x=2\) |
| \(= \frac{3}{4} - \ln 4\) | A1 | Exact answer |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Strip width $h = \frac{5.5-1.5}{4} = 1$ | B1 | Correct strip width stated or implied |
| Mid-ordinates at $x = 1.75, 2.75, 3.75, 4.75$ | M1 | At least 3 correct mid-x values |
| $f(1.75) = e^{0.25}\ln(3.25) \approx 1.4994$ | | |
| $f(2.75) = e^{-0.75}\ln(6.25) \approx 0.9290$ | | |
| $f(3.75) = e^{-1.75}\ln(9.25) \approx 0.3779$ | | |
| $f(4.75) = e^{-2.75}\ln(12.25) \approx 0.1344$ | A1 | All four y-values correct |
| Estimate $= 1 \times (1.4994 + 0.9290 + 0.3779 + 0.1344) \approx 2.941$ | A1 | Correct answer to 3 d.p. |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = e^{2-x} \cdot \frac{3}{3x-2} + \ln(3x-2) \cdot (-e^{2-x})$ | M1 | Attempt product rule |
| $= e^{2-x}\left(\frac{3}{3x-2} - \ln(3x-2)\right)$ | A1 | Correct derivative |
| At $x=2$: $e^{0}\left(\frac{3}{4} - \ln 4\right)$ | M1 | Substituting $x=2$ |
| $= \frac{3}{4} - \ln 4$ | A1 | Exact answer |
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\begin{enumerate}[label=(\alph*)]
\item Use the mid-ordinate rule with four strips to find an estimate for $\int _ { 1.5 } ^ { 5.5 } \mathrm { e } ^ { 2 - x } \ln ( 3 x - 2 ) \mathrm { d } x$, giving your answer to three decimal places.\\[0pt]
[4 marks]
\item Find the exact value of the gradient of the curve $y = \mathrm { e } ^ { 2 - x } \ln ( 3 x - 2 )$ at the point on the curve where $x = 2$.\\[0pt]
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2015 Q1 [8]}}