| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | March |
| Marks | 14 |
| Topic | Tangents, normals and gradients |
| Type | Find stationary points |
| Difficulty | Standard +0.8 This is a multi-part question requiring differentiation of a product involving exponential functions to find stationary points, analysis of function behavior for range determination, understanding of one-one functions, and integration by parts (likely twice) for the area calculation. While the techniques are standard M1/C3 content, the combination of multiple sophisticated steps, particularly the integration by parts with exponential and polynomial terms, places this above average difficulty. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Attempt product rule for \(y\) | M1 | Attempt must be of the form \((ax + b)e^{-x} \pm (cx^2 + dx)e^{-x}\) |
| \(y' = (4x - 3)e^{-x} - (2x^2 - 3x)e^{-x}\) | A1 | Correct derivative, in any form |
| \(y' = 0 \Rightarrow (4x - 3) - (2x^2 - 3x) = 0\) | M1 | Set \(y' = 0\) and eliminate exponentials |
| Obtain quadratic in \(x\) and attempt to solve | M1 | Dependent on both previous M marks; \(2x^2 - 7x + 3 = 0\) |
| \(x = \frac{1}{2}\), \(x = 3\) | A1 | Correct values from correct equation |
| \(-e^{-\frac{1}{2}} \leq y \leq 9e^{-3}\) | A1 | Correct range, including correct inequality signs and either y, f or f(x) used for range notation (not x); Allow 'closed interval' notation \([-e^{-\frac{1}{2}}, 9e^{-3}]\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(k = 3\) | B1ft | FT their larger value of \(x\) from (i) |
| Answer | Marks | Guidance |
|---|---|---|
| Use integration by parts with \(u = 2x^2 - 3x\) and \(v' = e^{-x}\) | M1 | Must obtain result \(f(x) \pm \int g(x)\,dx\) |
| \(\int(2x^2 - 3x)e^{-x}\,dx = -(2x^2 - 3x)e^{-x} + \int(4x - 3)e^{-x}\,dx\) | A1 | |
| Attempt parts again with \(u = ax + b\) and \(v' = e^{-x}\) | M1 | Dependent on previous M mark |
| \(\int(2x^2 - 3x)e^{-x}\,dx = -(2x^2 + x + 1)e^{-x} (+c)\) | A1 | oe; accept unsimplified (but all bracketing must be correct) |
| \(2x^2 - 3x = 0 \Rightarrow x = \frac{3}{2}\) (and \(x = 0\)) | B1 | |
| Correct use of correct limits | M1 | Dependent on both previous M marks |
| Integral is \(1 - 7e^{-\frac{3}{2}} < 0\) so area is \(7e^{-\frac{3}{2}} - 1\) | A1 |
## Part (i):
Attempt product rule for $y$ | M1 | Attempt must be of the form $(ax + b)e^{-x} \pm (cx^2 + dx)e^{-x}$
$y' = (4x - 3)e^{-x} - (2x^2 - 3x)e^{-x}$ | A1 | Correct derivative, in any form
$y' = 0 \Rightarrow (4x - 3) - (2x^2 - 3x) = 0$ | M1 | Set $y' = 0$ and eliminate exponentials
Obtain quadratic in $x$ and attempt to solve | M1 | Dependent on both previous M marks; $2x^2 - 7x + 3 = 0$
$x = \frac{1}{2}$, $x = 3$ | A1 | Correct values from correct equation
$-e^{-\frac{1}{2}} \leq y \leq 9e^{-3}$ | A1 | Correct range, including correct inequality signs and either y, f or f(x) used for range notation (not x); Allow 'closed interval' notation $[-e^{-\frac{1}{2}}, 9e^{-3}]$
## Part (ii):
$k = 3$ | B1ft | FT their larger value of $x$ from (i)
## Part (iii):
Use integration by parts with $u = 2x^2 - 3x$ and $v' = e^{-x}$ | M1 | Must obtain result $f(x) \pm \int g(x)\,dx$
$\int(2x^2 - 3x)e^{-x}\,dx = -(2x^2 - 3x)e^{-x} + \int(4x - 3)e^{-x}\,dx$ | A1 |
Attempt parts again with $u = ax + b$ and $v' = e^{-x}$ | M1 | Dependent on previous M mark
$\int(2x^2 - 3x)e^{-x}\,dx = -(2x^2 + x + 1)e^{-x} (+c)$ | A1 | oe; accept unsimplified (but all bracketing must be correct)
$2x^2 - 3x = 0 \Rightarrow x = \frac{3}{2}$ (and $x = 0$) | B1 |
Correct use of correct limits | M1 | Dependent on both previous M marks
Integral is $1 - 7e^{-\frac{3}{2}} < 0$ so area is $7e^{-\frac{3}{2}} - 1$ | A1 |In this question you must show detailed reasoning.
\includegraphics{figure_5}
The function f is defined for the domain $x \geqslant 0$ by
$$\mathrm{f}(x) = (2x^2 - 3x)\mathrm{e}^{-x}.$$
The diagram shows the curve $y = \mathrm{f}(x)$.
\begin{enumerate}[label=(\roman*)]
\item Find the range of f. [6]
\end{enumerate}
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item The function g is defined for the domain $x \geqslant k$ by
$$\mathrm{g}(x) = (2x^2 - 3x)\mathrm{e}^{-x}.$$
Given that g is a one-one function, state the least possible value of $k$. [1]
\end{enumerate}
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Find the exact area of the shaded region enclosed by the curve and the $x$-axis. [7]
\end{enumerate}
\hfill \mbox{\textit{OCR H240/03 2018 Q5 [14]}}