| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2020 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Integration by parts with inverse trig |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring proof of inverse trig derivative using implicit differentiation, integration by parts with inverse trig (non-standard due to the 4x), and mean value calculation. While the techniques are standard for FM students, the algebraic manipulation and the specific form of the answer in part (b) require careful work beyond routine application. |
| Spec | 4.08e Mean value of function: using integral4.08g Derivatives: inverse trig and hyperbolic functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \tan^{-1}x \Rightarrow \tan y = x \Rightarrow \frac{\text{d}x}{\text{d}y} = \sec^2 y\) or \(\frac{\text{d}y}{\text{d}x}\sec^2 y = 1\) | M1 | Makes progress establishing derivative by taking tan of both sides and differentiating w.r.t. \(y\) or implicitly w.r.t. \(x\) |
| \(\frac{\text{d}x}{\text{d}y} = 1 + \tan^2 y\) or \(\frac{\text{d}y}{\text{d}x}(1 + \tan^2 y) = 1\) | M1 | Use of correct identity |
| \(\frac{\text{d}y}{\text{d}x} = \frac{1}{1 + \tan^2 y} = \frac{1}{1 + x^2}\) | A1* | Fully correct proof |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{\text{d}}{\text{d}x}(\tan^{-1}4x) = \frac{4}{1 + 16x^2}\) | B1 | Correct differentiation stated |
| \(\int x\tan^{-1}4x\, \text{d}x = \alpha x^2 \tan^{-1}4x - \int \alpha x^2 \times \frac{4}{1+16x^2}\,\text{d}x\) | M1 | Integration by parts in correct direction |
| \(\int x\tan^{-1}4x\, \text{d}x = \frac{x^2}{2}\tan^{-1}4x - \int \frac{x^2}{2} \times \frac{4}{1+16x^2}\,\text{d}x\) | A1 | Correct expression |
| \(= \ldots - \frac{1}{8}\int \frac{16x^2 + 1 - 1}{1+16x^2}\,\text{d}x = \ldots - \frac{1}{8}\int\left(1 - \frac{1}{1+16x^2}\right)\text{d}x\) | M1 | Correct approach to remaining integral |
| \(= \frac{x^2}{2}\tan^{-1}4x - \frac{1}{8}x + \frac{1}{32}\tan^{-1}4x + k\) | A1 | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mean value \(= \left(\frac{1}{\frac{\sqrt{3}}{4} - 0}\right)\left[\frac{x^2}{2}\tan^{-1}4x - \frac{1}{8}x + \frac{1}{32}\tan^{-1}4x\right]_0^{\frac{\sqrt{3}}{4}}\) \(= \frac{4}{\sqrt{3}}\left(\left(\frac{3}{32} \times \frac{\pi}{3} - \frac{1}{8} \times \frac{\sqrt{3}}{4} + \frac{1}{32} \times \frac{\pi}{3}\right) - 0\right)\) | M1 | Correct application of mean value formula |
| \(= \frac{\sqrt{3}}{72}(4\pi - 3\sqrt{3})\) or \(\frac{\sqrt{3}}{18}\pi - \frac{1}{8}\) | A1 | Correct final answer |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \tan^{-1}x \Rightarrow \tan y = x \Rightarrow \frac{\text{d}x}{\text{d}y} = \sec^2 y$ or $\frac{\text{d}y}{\text{d}x}\sec^2 y = 1$ | M1 | Makes progress establishing derivative by taking tan of both sides and differentiating w.r.t. $y$ or implicitly w.r.t. $x$ |
| $\frac{\text{d}x}{\text{d}y} = 1 + \tan^2 y$ or $\frac{\text{d}y}{\text{d}x}(1 + \tan^2 y) = 1$ | M1 | Use of correct identity |
| $\frac{\text{d}y}{\text{d}x} = \frac{1}{1 + \tan^2 y} = \frac{1}{1 + x^2}$ | A1* | Fully correct proof |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\text{d}}{\text{d}x}(\tan^{-1}4x) = \frac{4}{1 + 16x^2}$ | B1 | Correct differentiation stated |
| $\int x\tan^{-1}4x\, \text{d}x = \alpha x^2 \tan^{-1}4x - \int \alpha x^2 \times \frac{4}{1+16x^2}\,\text{d}x$ | M1 | Integration by parts in correct direction |
| $\int x\tan^{-1}4x\, \text{d}x = \frac{x^2}{2}\tan^{-1}4x - \int \frac{x^2}{2} \times \frac{4}{1+16x^2}\,\text{d}x$ | A1 | Correct expression |
| $= \ldots - \frac{1}{8}\int \frac{16x^2 + 1 - 1}{1+16x^2}\,\text{d}x = \ldots - \frac{1}{8}\int\left(1 - \frac{1}{1+16x^2}\right)\text{d}x$ | M1 | Correct approach to remaining integral |
| $= \frac{x^2}{2}\tan^{-1}4x - \frac{1}{8}x + \frac{1}{32}\tan^{-1}4x + k$ | A1 | Correct final answer |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mean value $= \left(\frac{1}{\frac{\sqrt{3}}{4} - 0}\right)\left[\frac{x^2}{2}\tan^{-1}4x - \frac{1}{8}x + \frac{1}{32}\tan^{-1}4x\right]_0^{\frac{\sqrt{3}}{4}}$ $= \frac{4}{\sqrt{3}}\left(\left(\frac{3}{32} \times \frac{\pi}{3} - \frac{1}{8} \times \frac{\sqrt{3}}{4} + \frac{1}{32} \times \frac{\pi}{3}\right) - 0\right)$ | M1 | Correct application of mean value formula |
| $= \frac{\sqrt{3}}{72}(4\pi - 3\sqrt{3})$ or $\frac{\sqrt{3}}{18}\pi - \frac{1}{8}$ | A1 | Correct final answer |\begin{enumerate}
\item (a)
\end{enumerate}
$$y = \tan ^ { - 1 } x$$
Assuming the derivative of $\tan x$, prove that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 1 + x ^ { 2 } }$$
$$\mathrm { f } ( x ) = x \tan ^ { - 1 } 4 x$$
(b) Show that
$$\int \mathrm { f } ( x ) \mathrm { d } x = A x ^ { 2 } \tan ^ { - 1 } 4 x + B x + C \tan ^ { - 1 } 4 x + k$$
where $k$ is an arbitrary constant and $A , B$ and $C$ are constants to be determined.\\
(c) Hence find, in exact form, the mean value of $\mathrm { f } ( x )$ over the interval $\left[ 0 , \frac { \sqrt { 3 } } { 4 } \right]$
\hfill \mbox{\textit{Edexcel CP2 2020 Q5 [10]}}