| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find derivative of quotient |
| Difficulty | Standard +0.3 Part (a) is a standard bookwork proof of tan x derivative using quotient rule. Part (b) is routine integration by parts with sec²x. Part (c) requires volume of revolution but is straightforward application of the formula once parts (a) and (b) are established. The question guides students through connected steps with no novel insight required, making it slightly easier than average. |
| Spec | 1.07q Product and quotient rules: differentiation1.08i Integration by parts4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d}{dx}(\tan x) = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x}\) | M1 | Applying quotient rule correctly |
| \(= \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x\) | A1 | Using \(\sin^2 x + \cos^2 x = 1\) to reach \(\sec^2 x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Let \(u = x\), \(\frac{dv}{dx} = \sec^2 x\), so \(\frac{du}{dx} = 1\), \(v = \tan x\) | M1 | Correct choice of parts |
| \(\int x\sec^2 x\, dx = x\tan x - \int \tan x\, dx\) | A1 | Correct expression after applying parts |
| \(= x\tan x - \int \frac{\sin x}{\cos x}\, dx\) | M1 | Attempting to integrate \(\tan x\) |
| \(= x\tan x + \ln | \cos x | + c\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(V = \pi \int_0^1 y^2\, dx = \pi \int_0^1 (5\sqrt{x})^2 \sec^2 x\, dx = 25\pi \int_0^1 x\sec^2 x\, dx\) | M1 | Correct volume formula with \(y^2\) |
| \(= 25\pi \left[x\tan x + \ln | \cos x | \right]_0^1\) |
| \(= 25\pi(\tan 1 + \ln(\cos 1)) = 25\pi(1.5574... - 0.6137...) \approx 74\) | A1 | Correct answer to 2 s.f. |
# Question 5:
## Part (a):
| $\frac{d}{dx}(\tan x) = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x}$ | M1 | Applying quotient rule correctly |
|---|---|---|
| $= \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x$ | A1 | Using $\sin^2 x + \cos^2 x = 1$ to reach $\sec^2 x$ |
## Part (b):
| Let $u = x$, $\frac{dv}{dx} = \sec^2 x$, so $\frac{du}{dx} = 1$, $v = \tan x$ | M1 | Correct choice of parts |
|---|---|---|
| $\int x\sec^2 x\, dx = x\tan x - \int \tan x\, dx$ | A1 | Correct expression after applying parts |
| $= x\tan x - \int \frac{\sin x}{\cos x}\, dx$ | M1 | Attempting to integrate $\tan x$ |
| $= x\tan x + \ln|\cos x| + c$ | A1 | Correct final answer |
## Part (c):
| $V = \pi \int_0^1 y^2\, dx = \pi \int_0^1 (5\sqrt{x})^2 \sec^2 x\, dx = 25\pi \int_0^1 x\sec^2 x\, dx$ | M1 | Correct volume formula with $y^2$ |
|---|---|---|
| $= 25\pi \left[x\tan x + \ln|\cos x|\right]_0^1$ | A1 | Using result from (b) |
| $= 25\pi(\tan 1 + \ln(\cos 1)) = 25\pi(1.5574... - 0.6137...) \approx 74$ | A1 | Correct answer to 2 s.f. |
---5
\begin{enumerate}[label=(\alph*)]
\item By writing $\tan x$ as $\frac { \sin x } { \cos x }$, use the quotient rule to show that $\frac { \mathrm { d } } { \mathrm { d } x } ( \tan x ) = \sec ^ { 2 } x$.\\[0pt]
[2 marks]
\item Use integration by parts to find $\int x \sec ^ { 2 } x \mathrm {~d} x$.\\[0pt]
[4 marks]
\item The region bounded by the curve $y = ( 5 \sqrt { x } ) \sec x$, the $x$-axis from 0 to 1 and the line $x = 1$ is rotated through $2 \pi$ radians about the $x$-axis to form a solid.
Find the value of the volume of the solid generated, giving your answer to two significant figures.\\[0pt]
[3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2015 Q5 [9]}}