| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find gradient at a point - given gradient condition |
| Difficulty | Challenging +1.2 This is a multi-part question requiring product rule differentiation, solving equations with exponentials/logarithms, and integration by parts. While it involves several steps and the 'equally inclined' condition requires geometric insight (equal gradients), each individual technique is standard C3 material. The structured parts guide students through the solution, making it moderately harder than average but not requiring exceptional problem-solving. |
| Spec | 1.07q Product and quotient rules: differentiation1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(xe^{-2x} = mx\) | M1 | May be implied from 2nd line |
| \(\Rightarrow e^{-2x} = m\) | M1 | Dividing by \(x\), or subtracting \(\ln x\) |
| \(\Rightarrow -2x = \ln m\) | ||
| \(\Rightarrow x = -\frac{1}{2}\ln m\) | A1 | NB AG |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| If \(x = -\frac{1}{2}\ln m\), \(y = -\frac{1}{2}\ln m \times e^{\ln m}\) | M1 | Substituting correctly |
| \(= -\frac{1}{2}\ln m \times m\) | A1 | |
| so P lies on \(y = mx\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Let \(u = x\), \(u' = 1\), \(v = e^{-2x}\), \(v' = -2e^{-2x}\) | M1* | Product rule consistent with their derivs |
| \(dy/dx = e^{-2x} - 2xe^{-2x}\) | A1 | o.e. correct expression |
| \(= e^{-2(-\frac{1}{2}\ln m)} - 2\cdot(-\frac{1}{2}\ln m)e^{-2(-\frac{1}{2}\ln m)}\) | M1dep | Subst \(x = -\frac{1}{2}\ln m\) into their deriv, dep M1* |
| \(= e^{\ln m} + e^{\ln m}\ln m\ [= m + m\ln m]\) | A1cao | Condone \(e^{\ln m}\) not simplified |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(m + m\ln m = -m\) | M1 | Their gradient from (ii) \(= -m\) |
| \(\Rightarrow \ln m = -2\) | ||
| \(\Rightarrow m = e^{-2}\) | A1 | NB AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y + \frac{1}{2}m\ln m = m(1 + \ln m)(x + \frac{1}{2}\ln m)\), \(x = -\ln m\), \(y=0 \Rightarrow \frac{1}{2}m\ln m = m(1+\ln m)(-\frac{1}{2}\ln m) \Rightarrow 1 + \ln m = -1\), \(\ln m = -2\), \(m = e^{-2}\) | B2 | For fully correct methods finding \(x\)-intercept of equation of tangent and equating to \(-\ln m\) |
| At P, \(x = 1\) | B1 | |
| \(\Rightarrow y = e^{-2}\) | B1 | isw approximations; not \(e^{-2} \times 1\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area under curve \(= \int_0^1 xe^{-2x}dx\) | ||
| \(u = x\), \(u' = 1\), \(v' = e^{-2x}\), \(v = -\frac{1}{2}e^{-2x}\) | M1 | Parts, condone \(v = ke^{-2x}\), provided used consistently; ignore limits until 3rd A1 |
| \(= \left[-\frac{1}{2}xe^{-2x}\right]_0^1 + \int_0^1 \frac{1}{2}e^{-2x}dx\) | A1ft | ft their \(v\) |
| \(= \left[-\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x}\right]_0^1\) | A1 | \(-\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x}\) o.e. |
| \(= (-\frac{1}{2}e^{-2} - \frac{1}{4}e^{-2}) - (0 - \frac{1}{4}e^0)\) \([= \frac{1}{4} - \frac{3}{4}e^{-2}]\) | A1 | Correct expression; need not be simplified |
| Area of triangle \(= \frac{1}{2}\) base \(\times\) height \(= \frac{1}{2} \times 1 \times e^{-2}\) | M1 | ft their 1, \(e^{-2}\); or \([e^{-2}x^2/2]\); o.e. using isosceles triangle; M1 may be implied from \(0.067\ldots\) |
| A1 | ||
| So area enclosed \(= \frac{1}{4} - \frac{5e^{-2}}{4}\) | A1cao | o.e. must be exact, two terms only |
| [7] |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $xe^{-2x} = mx$ | M1 | May be implied from 2nd line |
| $\Rightarrow e^{-2x} = m$ | M1 | Dividing by $x$, or subtracting $\ln x$ |
| $\Rightarrow -2x = \ln m$ | | |
| $\Rightarrow x = -\frac{1}{2}\ln m$ | A1 | **NB AG** |
| | **[3]** | |
**OR**
| Answer | Marks | Guidance |
|--------|-------|----------|
| If $x = -\frac{1}{2}\ln m$, $y = -\frac{1}{2}\ln m \times e^{\ln m}$ | M1 | Substituting correctly |
| $= -\frac{1}{2}\ln m \times m$ | A1 | |
| so P lies on $y = mx$ | A1 | |
---
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let $u = x$, $u' = 1$, $v = e^{-2x}$, $v' = -2e^{-2x}$ | M1* | Product rule consistent with their derivs |
| $dy/dx = e^{-2x} - 2xe^{-2x}$ | A1 | o.e. correct expression |
| $= e^{-2(-\frac{1}{2}\ln m)} - 2\cdot(-\frac{1}{2}\ln m)e^{-2(-\frac{1}{2}\ln m)}$ | M1dep | Subst $x = -\frac{1}{2}\ln m$ into their deriv, dep M1* |
| $= e^{\ln m} + e^{\ln m}\ln m\ [= m + m\ln m]$ | A1cao | Condone $e^{\ln m}$ not simplified |
| | **[4]** | |
---
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $m + m\ln m = -m$ | M1 | Their gradient from (ii) $= -m$ |
| $\Rightarrow \ln m = -2$ | | |
| $\Rightarrow m = e^{-2}$ | A1 | **NB AG** |
**OR**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y + \frac{1}{2}m\ln m = m(1 + \ln m)(x + \frac{1}{2}\ln m)$, $x = -\ln m$, $y=0 \Rightarrow \frac{1}{2}m\ln m = m(1+\ln m)(-\frac{1}{2}\ln m) \Rightarrow 1 + \ln m = -1$, $\ln m = -2$, $m = e^{-2}$ | B2 | For fully correct methods finding $x$-intercept of equation of tangent and equating to $-\ln m$ |
| At P, $x = 1$ | B1 | |
| $\Rightarrow y = e^{-2}$ | B1 | isw approximations; not $e^{-2} \times 1$ |
| | **[4]** | |
---
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area under curve $= \int_0^1 xe^{-2x}dx$ | | |
| $u = x$, $u' = 1$, $v' = e^{-2x}$, $v = -\frac{1}{2}e^{-2x}$ | M1 | Parts, condone $v = ke^{-2x}$, provided used consistently; ignore limits until 3rd A1 |
| $= \left[-\frac{1}{2}xe^{-2x}\right]_0^1 + \int_0^1 \frac{1}{2}e^{-2x}dx$ | A1ft | ft their $v$ |
| $= \left[-\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x}\right]_0^1$ | A1 | $-\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x}$ o.e. |
| $= (-\frac{1}{2}e^{-2} - \frac{1}{4}e^{-2}) - (0 - \frac{1}{4}e^0)$ $[= \frac{1}{4} - \frac{3}{4}e^{-2}]$ | A1 | Correct expression; need not be simplified |
| Area of triangle $= \frac{1}{2}$ base $\times$ height $= \frac{1}{2} \times 1 \times e^{-2}$ | M1 | ft their 1, $e^{-2}$; or $[e^{-2}x^2/2]$; o.e. using isosceles triangle; M1 may be implied from $0.067\ldots$ |
| | A1 | |
| So area enclosed $= \frac{1}{4} - \frac{5e^{-2}}{4}$ | A1cao | o.e. must be exact, two terms only |
| | **[7]** | |
---2 Fig. 9 shows the curve $y = x \mathrm { e } ^ { - 2 x }$ together with the straight line $y = m x$, where $m$ is a constant, with $0 < m < 1$. The curve and the line meet at O and P . The dashed line is the tangent at P .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d1206ce8-7716-4205-b98e-664e7ead8a25-2_433_979_472_591}
\captionsetup{labelformat=empty}
\caption{Fig. 9}
\end{center}
\end{figure}
(i) Show that the $x$-coordinate of P is $- \frac { 1 } { 2 } \ln m$.\\
(ii) Find, in terms of $m$, the gradient of the tangent to the curve at P .
You are given that OP and this tangent are equally inclined to the $x$-axis.\\
(iii) Show that $m = \mathrm { e } ^ { - 2 }$, and find the exact coordinates of P .\\
(iv) Find the exact area of the shaded region between the line OP and the curve.
END OF QUESTION PAPER
\hfill \mbox{\textit{OCR MEI C3 Q2 [18]}}