| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find derivative of quotient |
| Difficulty | Standard +0.3 Part (a) is a guided proof using quotient rule on sin x/cos x, which is straightforward application of a standard technique. Part (b) requires integrating tan²x using the identity tan²x = sec²x - 1, then evaluating definite integrals—this involves recognizing a trigonometric identity and careful integration, but follows a well-established method. Overall slightly easier than average due to the scaffolding in part (a) and standard nature of the integration technique. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07q Product and quotient rules: differentiation1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Recalls \(\tan x = \frac{\sin x}{\cos x}\) | B1 (AO 1.2) | |
| \(\frac{d}{dx}(\tan x) = \frac{\cos x \cos x - (-\sin x)\sin x}{\cos^2 x}\) uses correct quotient rule | M1 (AO 1.1a) | Condone sign error in differentiation of sin or cosine |
| \(= \frac{\sin^2 x + \cos^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x\) | R1 (AO 2.1) | Must use \(\sin^2 x + \cos^2 x = 1\) or \(\tan^2 x + 1 = \sec^2 x\) explicitly. Must include \(\frac{d}{dx}(\tan x) = \ldots\) or \(\frac{dy}{dx} = \ldots\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Writes down integral of the form \(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan^2 x\, dx\) or \(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (1 - \tan^2 x)\, dx\) | M1 (AO 3.1a) | Condone missing or incorrect limits and missing \(dx\) |
| Uses \(\tan^2 x + 1 = \sec^2 x\) to write integrand in integrable form: \(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan^2 x\, dx = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec^2 x - 1\, dx\) | M1 (AO 3.1a) | Condone sign error |
| Integrates to form \(A\sec^2 x + B\): \(= [\tan x - x]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\) | A1F (AO 1.1b) | |
| \(= \left(\tan\frac{\pi}{4} - \frac{\pi}{4}\right) - \left(\tan\left(-\frac{\pi}{4}\right) - \left(-\frac{\pi}{4}\right)\right) = 1 - \frac{\pi}{4} + 1 - \frac{\pi}{4} = 2 - \frac{\pi}{2}\) | B1 (AO 1.1b) | Forms expression for rectangle: \(2\frac{\pi}{4}\tan^2\frac{\pi}{4}\) or \(\frac{\pi}{4}\tan^2\frac{\pi}{4}\). Could be implicit within integral |
| Area of shaded region \(= \frac{\pi}{2} - \left(2 - \frac{\pi}{2}\right) = \pi - 2\) | R1 (AO 2.1) | Substitution of consistent limits must be explicit, no slips in algebra, use of \(dx\) required. AG |
## Question 10(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Recalls $\tan x = \frac{\sin x}{\cos x}$ | B1 (AO 1.2) | |
| $\frac{d}{dx}(\tan x) = \frac{\cos x \cos x - (-\sin x)\sin x}{\cos^2 x}$ uses correct quotient rule | M1 (AO 1.1a) | Condone sign error in differentiation of sin or cosine |
| $= \frac{\sin^2 x + \cos^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x$ | R1 (AO 2.1) | Must use $\sin^2 x + \cos^2 x = 1$ or $\tan^2 x + 1 = \sec^2 x$ explicitly. Must include $\frac{d}{dx}(\tan x) = \ldots$ or $\frac{dy}{dx} = \ldots$ |
**Subtotal: 3 marks**
---
## Question 10(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Writes down integral of the form $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan^2 x\, dx$ or $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (1 - \tan^2 x)\, dx$ | M1 (AO 3.1a) | Condone missing or incorrect limits and missing $dx$ |
| Uses $\tan^2 x + 1 = \sec^2 x$ to write integrand in integrable form: $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan^2 x\, dx = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec^2 x - 1\, dx$ | M1 (AO 3.1a) | Condone sign error |
| Integrates to form $A\sec^2 x + B$: $= [\tan x - x]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}$ | A1F (AO 1.1b) | |
| $= \left(\tan\frac{\pi}{4} - \frac{\pi}{4}\right) - \left(\tan\left(-\frac{\pi}{4}\right) - \left(-\frac{\pi}{4}\right)\right) = 1 - \frac{\pi}{4} + 1 - \frac{\pi}{4} = 2 - \frac{\pi}{2}$ | B1 (AO 1.1b) | Forms expression for rectangle: $2\frac{\pi}{4}\tan^2\frac{\pi}{4}$ or $\frac{\pi}{4}\tan^2\frac{\pi}{4}$. Could be implicit within integral |
| Area of shaded region $= \frac{\pi}{2} - \left(2 - \frac{\pi}{2}\right) = \pi - 2$ | R1 (AO 2.1) | Substitution of consistent limits must be explicit, no slips in algebra, use of $dx$ required. AG |
**Subtotal: 5 marks**
---10
\begin{enumerate}[label=(\alph*)]
\item Given that
$$y = \tan x$$
use the quotient rule to show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \sec ^ { 2 } x$$
10
\item The region enclosed by the curve $y = \tan ^ { 2 } x$ and the horizontal line, which intersects the curve at $x = - \frac { \pi } { 4 }$ and $x = \frac { \pi } { 4 }$, is shaded in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{042e248a-9efa-4844-957d-f05715900ffc-17_1059_967_461_539}
Show that the area of the shaded region is
$$\pi - 2$$
Fully justify your answer.\\
Do not write outside the box\\
\includegraphics[max width=\textwidth, alt={}, center]{042e248a-9efa-4844-957d-f05715900ffc-19_2488_1716_219_153}
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2021 Q10 [8]}}