| 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 has several steps and involves transcendental functions, each individual technique is standard C3 material. The 'equally inclined' condition in part (iii) requires understanding that gradients sum to zero for perpendicular lines through the origin, which adds modest problem-solving demand beyond routine application. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07q Product and quotient rules: differentiation1.08e Area between curve and x-axis: using definite integrals1.08h Integration by substitution |
| 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 \(\ln x\); o.e. e.g. \([\ln x] - 2x = \ln m + [\ln x]\) or factorising: \(x(e^{-2x} - m) = 0\) |
| \(\Rightarrow -2x = \ln m\) | ||
| \(\Rightarrow x = -\tfrac{1}{2}\ln m\) * | A1 [3] | NB AG |
| or If \(x = -\tfrac{1}{2}\ln m\), \(y = -\tfrac{1}{2}\ln m \times e^{\ln m}\) | M1 | Substituting correctly |
| \(= -\tfrac{1}{2}\ln m \times m\) | 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 |
| \(= e^{-2(-\frac{1}{2}\ln m)} - 2(-\tfrac{1}{2}\ln m)e^{-2(-\frac{1}{2}\ln m)}\) | M1dep | Subst \(x = -\tfrac{1}{2}\ln m\) into their deriv dep M1* |
| \(= e^{\ln m} + e^{\ln m}\ln m\ [= m + m\ln m]\) | A1cao [4] | Condone \(e^{\ln m}\) not simplified; but not \(-2(-\tfrac{1}{2}\ln m)\); mark final answer |
| 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] | |
| or \(y + \tfrac{1}{2}m\ln m = m(1 + \ln m)(x + \tfrac{1}{2}\ln m)\), \(x = -\ln m\) | B2 | For fully correct methods finding \(x\)-intercept of equation of tangent and equating to \(-\ln m\) |
| \(y = 0 \Rightarrow \tfrac{1}{2}m\ln m = m(1+\ln m)(-\tfrac{1}{2}\ln m)\) | ||
| \(\Rightarrow 1 + \ln m = -1\), \(\ln m = -2\), \(m = e^{-2}\) | ||
| At P, \(x = 1\) | B1 | |
| \(\Rightarrow y = e^{-2}\) | B1 [4] | isw approximations; not \(e^{-2} \times 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area under curve \(= \int_0^1 xe^{-2x}\,dx\) | ||
| \(u = x\), \(u' = 1\), \(v' = e^{-2x}\), \(v = -\tfrac{1}{2}e^{-2x}\) | M1 | Parts, condone \(v = ke^{-2x}\), provided used consistently in parts formula; ignore limits until 3rd A1 |
| \(= \left[-\tfrac{1}{2}xe^{-2x}\right]_0^1 + \int_0^1 \tfrac{1}{2}e^{-2x}\,dx\) | A1ft | ft their \(v\) |
| \(= \left[-\tfrac{1}{2}xe^{-2x} - \tfrac{1}{4}e^{-2x}\right]_0^1\) | A1 | \(-\tfrac{1}{2}xe^{-2x} - \tfrac{1}{4}e^{-2x}\) o.e. |
| \(= (-\tfrac{1}{2}e^{-2} - \tfrac{1}{4}e^{-2}) - (0 - \tfrac{1}{4}e^0)\) | A1 | Correct expression; need not be simplified |
| \([= \tfrac{1}{4} - \tfrac{3}{4}e^{-2}]\) | ||
| Area of triangle \(= \tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} \times 1 \times e^{-2}\) | 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 \(= \tfrac{1}{4} - \tfrac{5e^{-2}}{4}\) | A1cao [7] | o.e. must be exact, two terms only; isw |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $xe^{-2x} = mx$ | M1 | May be implied from 2nd line |
| $\Rightarrow e^{-2x} = m$ | M1 | Dividing by $x$, or subtracting $\ln x$; o.e. e.g. $[\ln x] - 2x = \ln m + [\ln x]$ or factorising: $x(e^{-2x} - m) = 0$ |
| $\Rightarrow -2x = \ln m$ | | |
| $\Rightarrow x = -\tfrac{1}{2}\ln m$ * | A1 **[3]** | **NB AG** |
| **or** If $x = -\tfrac{1}{2}\ln m$, $y = -\tfrac{1}{2}\ln m \times e^{\ln m}$ | M1 | Substituting correctly |
| $= -\tfrac{1}{2}\ln m \times m$ | A1 | |
| So P lies on $y = mx$ | A1 | |
---
## Question 4(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 |
| $= e^{-2(-\frac{1}{2}\ln m)} - 2(-\tfrac{1}{2}\ln m)e^{-2(-\frac{1}{2}\ln m)}$ | M1dep | Subst $x = -\tfrac{1}{2}\ln m$ into their deriv dep M1* |
| $= e^{\ln m} + e^{\ln m}\ln m\ [= m + m\ln m]$ | A1cao **[4]** | Condone $e^{\ln m}$ not simplified; but not $-2(-\tfrac{1}{2}\ln m)$; mark final answer |
---
## Question 4(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** $y + \tfrac{1}{2}m\ln m = m(1 + \ln m)(x + \tfrac{1}{2}\ln m)$, $x = -\ln m$ | B2 | For fully correct methods finding $x$-intercept of equation of tangent and equating to $-\ln m$ |
| $y = 0 \Rightarrow \tfrac{1}{2}m\ln m = m(1+\ln m)(-\tfrac{1}{2}\ln m)$ | | |
| $\Rightarrow 1 + \ln m = -1$, $\ln m = -2$, $m = e^{-2}$ | | |
| At P, $x = 1$ | B1 | |
| $\Rightarrow y = e^{-2}$ | B1 **[4]** | isw approximations; not $e^{-2} \times 1$ |
---
## Question 4(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area under curve $= \int_0^1 xe^{-2x}\,dx$ | | |
| $u = x$, $u' = 1$, $v' = e^{-2x}$, $v = -\tfrac{1}{2}e^{-2x}$ | M1 | Parts, condone $v = ke^{-2x}$, provided used consistently in parts formula; ignore limits until 3rd A1 |
| $= \left[-\tfrac{1}{2}xe^{-2x}\right]_0^1 + \int_0^1 \tfrac{1}{2}e^{-2x}\,dx$ | A1ft | ft their $v$ |
| $= \left[-\tfrac{1}{2}xe^{-2x} - \tfrac{1}{4}e^{-2x}\right]_0^1$ | A1 | $-\tfrac{1}{2}xe^{-2x} - \tfrac{1}{4}e^{-2x}$ o.e. |
| $= (-\tfrac{1}{2}e^{-2} - \tfrac{1}{4}e^{-2}) - (0 - \tfrac{1}{4}e^0)$ | A1 | Correct expression; need not be simplified |
| $[= \tfrac{1}{4} - \tfrac{3}{4}e^{-2}]$ | | |
| Area of triangle $= \tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} \times 1 \times e^{-2}$ | 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 $= \tfrac{1}{4} - \tfrac{5e^{-2}}{4}$ | A1cao **[7]** | o.e. must be exact, two terms only; isw |4 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]{6ea594c5-52ba-4467-a098-cb66004b5a38-2_431_977_728_602}
\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.
\hfill \mbox{\textit{OCR MEI C3 Q4 [18]}}