| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2020 |
| Session | November |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Trigonometric substitution: direct evaluation |
| Difficulty | Standard +0.3 Part (a)(i) is routine differentiation using chain rule twice. Part (a)(ii) requires solving a quadratic inequality from the second derivative. Part (b) is a standard trigonometric substitution with x=tan(θ) leading to a straightforward integral of cos²(θ), which is textbook material for this topic. The question tests standard techniques without requiring novel insight, making it slightly easier than average. |
| Spec | 1.07f Convexity/concavity: points of inflection1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{-4x}{(x^2+1)^3}\) | M1 | Attempt to differentiate; chain or quotient or product rule |
| A1 | Correct first derivative; don't have to simplify | |
| \(\frac{d^2y}{dx^2} = \frac{-4(x^2+1)^3 + 4x \cdot 2x \cdot 3(x^2+1)^2}{(x^2+1)^6}\) | M1 | Attempt to use quotient rule to find second derivative; for each M1 allow one error |
| A1 | Any correct expression for second derivative | |
| \(\frac{d^2y}{dx^2} = \frac{-4(x^2+1)+24x^2}{(x^2+1)^4}\) | A1 | Simplifying by cancelling \((x^2+1)\) and correct completion (AG) |
| \(\Rightarrow \frac{d^2y}{dx^2} = \frac{20x^2-4}{(x^2+1)^4}\) | ||
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For concave downwards, \(\frac{20x^2-4}{(x^2+1)^4} < 0\) so \(20x^2 - 4 < 0\) | M1 | 2.2a |
| \(5x^2 < 1\) so \(x^2 < \frac{1}{5}\) | M1 | Or \(x = \pm\frac{1}{\sqrt{5}}\); condone \(x^2 > \frac{1}{5}\) following \(y'' > 0\) for M1 |
| \(-\frac{1}{\sqrt{5}} < x < \frac{1}{\sqrt{5}}\) | A1 | Correct solution correctly expressed; allow decimals (\(\pm 0.447\)) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dx}{d\theta} = \sec^2\theta\) | M1 | o.e.; condone \(\sec^2 x\) |
| \(\int_{x=-1}^{x=1} \frac{\sec^2\theta}{(1+\tan^2\theta)^2}d\theta\) | M1 | Substitution for either \(x\) or \(dx\) (limits may be missing) |
| \(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\sec^2\theta}{(1+\tan^2\theta)^2}d\theta\) | M1 | Limits (may be done at any point) |
| \(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{\sec^2\theta}\, d\theta\) | M1 | Use of \(1+\tan^2\theta = \sec^2\theta\) |
| \(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2\theta\, d\theta\) | M1 | Use of \(\sec\theta = \frac{1}{\cos\theta}\) |
| \(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2}(\cos 2\theta + 1)d\theta\) | M1 | Use of \(\cos 2\theta = 2\cos^2\theta - 1\) |
| \(\frac{1}{2}\left[\frac{1}{2}\sin 2\theta + \theta\right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\) | M1 | Integration; must include at least one trig term; limits may be wrong or missing |
| \(\frac{1}{2}\left(1+\frac{\pi}{2}\right) = \frac{1}{2}+\frac{\pi}{4}\) | A1 | Exact form |
| [8] |
## Question 8(a)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{-4x}{(x^2+1)^3}$ | M1 | Attempt to differentiate; chain or quotient or product rule |
| | A1 | Correct first derivative; don't have to simplify |
| $\frac{d^2y}{dx^2} = \frac{-4(x^2+1)^3 + 4x \cdot 2x \cdot 3(x^2+1)^2}{(x^2+1)^6}$ | M1 | Attempt to use quotient rule to find second derivative; for each M1 allow one error |
| | A1 | Any correct expression for second derivative |
| $\frac{d^2y}{dx^2} = \frac{-4(x^2+1)+24x^2}{(x^2+1)^4}$ | A1 | Simplifying by cancelling $(x^2+1)$ and correct completion (AG) |
| $\Rightarrow \frac{d^2y}{dx^2} = \frac{20x^2-4}{(x^2+1)^4}$ | | |
| **[5]** | | |
---
## Question 8(a)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| For concave downwards, $\frac{20x^2-4}{(x^2+1)^4} < 0$ so $20x^2 - 4 < 0$ | M1 | 2.2a |
| $5x^2 < 1$ so $x^2 < \frac{1}{5}$ | M1 | Or $x = \pm\frac{1}{\sqrt{5}}$; condone $x^2 > \frac{1}{5}$ following $y'' > 0$ for M1 |
| $-\frac{1}{\sqrt{5}} < x < \frac{1}{\sqrt{5}}$ | A1 | Correct solution correctly expressed; allow decimals ($\pm 0.447$) |
| **[3]** | | |
---
## Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dx}{d\theta} = \sec^2\theta$ | M1 | o.e.; condone $\sec^2 x$ |
| $\int_{x=-1}^{x=1} \frac{\sec^2\theta}{(1+\tan^2\theta)^2}d\theta$ | M1 | Substitution for either $x$ or $dx$ (limits may be missing) |
| $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\sec^2\theta}{(1+\tan^2\theta)^2}d\theta$ | M1 | Limits (may be done at any point) |
| $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{\sec^2\theta}\, d\theta$ | M1 | Use of $1+\tan^2\theta = \sec^2\theta$ |
| $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2\theta\, d\theta$ | M1 | Use of $\sec\theta = \frac{1}{\cos\theta}$ |
| $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2}(\cos 2\theta + 1)d\theta$ | M1 | Use of $\cos 2\theta = 2\cos^2\theta - 1$ |
| $\frac{1}{2}\left[\frac{1}{2}\sin 2\theta + \theta\right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}$ | M1 | Integration; must include at least one trig term; limits may be wrong or missing |
| $\frac{1}{2}\left(1+\frac{\pi}{2}\right) = \frac{1}{2}+\frac{\pi}{4}$ | A1 | Exact form |
| **[8]** | | |
---8
\begin{enumerate}[label=(\alph*)]
\item The curve $y = \frac { 1 } { \left( 1 + x ^ { 2 } \right) ^ { 2 } }$ is shown in Fig. 8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a13f7a05-e2d3-4354-a0c7-ef7283eff514-08_495_1058_1105_315}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Show that $\frac { d ^ { 2 } y } { d x ^ { 2 } } = \frac { 20 x ^ { 2 } - 4 } { \left( 1 + x ^ { 2 } \right) ^ { 4 } }$.
\item In this question you must show detailed reasoning.
Find the set of values of $x$ for which the curve is concave downwards.
\end{enumerate}\item Use the substitution $x = \tan \theta$ to find the exact value of $\int _ { - 1 } ^ { 1 } \frac { 1 } { \left( 1 + x ^ { 2 } \right) ^ { 2 } } d x$.
Answer all the questions.\\
Section B (15 marks)
The questions in this section refer to the article on the Insert. You should read the article before attempting the questions.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 2020 Q8 [16]}}