| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Arc length of parametric curve |
| Difficulty | Challenging +1.3 Part (a) is a standard arc length formula application requiring differentiation of ln(1-x²) and algebraic manipulation—routine FP2 work. Part (b) requires recognizing a partial fractions decomposition combined with a substitution (likely x = tan θ or similar), which is more challenging but follows established FP2 integration techniques. The 10-mark total and multi-step nature elevate this above average, but it remains a textbook-style question testing standard Further Pure methods without requiring novel insight. |
| Spec | 1.07s Parametric and implicit differentiation1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{dy}{dx} = \frac{2x}{(1-x^2)}\) | B1 | |
| \(1 + \left(\frac{dy}{dx}\right)^2 = 1 + \frac{(2x)^2}{(1-x^2)^2}\) | M1 | FT their \(\frac{dy}{dx}\) |
| \(\frac{1 - 2x^2 + x^4 + 4x^2}{(1-x^2)^2} = \frac{(1+x^2)^2}{(1-x^2)^2}\) | m1 | Allow m1 if sign error in \(\frac{dy}{dx}\) |
| \(s = \int_0^3 \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx\) | ||
| \(s = \int_0^3 \frac{1+x^2}{(1-x^2)} dx\) | A1 cso | 4 marks |
| (b) \(\frac{1+x^2}{1-x^2} = \frac{A}{1-x^2} + B\) | M1 | and attempt to find constants \(B \neq 0\) |
| \(\frac{1+x^2}{1-x^2} = \frac{2}{1-x^2} - 1\) | A1 | |
| \(\left(\frac{A}{2}\ln\left | \frac{1+x}{1-x}\right | \right)\) or \(A\tanh^{-1}x + Bx\) |
| \(\ln\left(\frac{1+x}{1-x}\right) - x\) | A1 | or \(2\tanh^{-1}x - x\) |
| \(\ln\left(\frac{1+\frac{3}{4}}{1-\frac{3}{4}}\right) - \frac{3}{4}\) OE | A1 | PI by next A1 or \(\left(s = \right)2\tanh^{-1}\left(\frac{3}{4}\right) - \frac{3}{4}\) |
| \(-\frac{3}{4} + \ln 7\) | A1 | 6 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1+x^2}{1-x^2} = \frac{C}{1+x} + \frac{D}{1-x} + E\) | (M1) | and attempt to find constants \(E \neq 0\) |
| \(\frac{1+x^2}{1-x^2} = \frac{1}{1+x} + \frac{1}{1-x} - 1\) | (A1) | |
| \(C\ln(1+x) - D\ln(1-x) + Ex\) | (m1) | FT integral of their \(\frac{C}{1+x} + \frac{D}{1-x} + E\) |
| \(= \ln(1+x) - \ln(1-x) - x\) | (A1) | correct |
| \((s) = \ln\left(\frac{7}{4}\right) - \ln\left(\frac{1}{4}\right) - \frac{3}{4}\) OE | (A1) | correct unsimplified |
| \((s) = \ln 7 - \frac{3}{4}\) | (A1) | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| - (b) If M1 is not earned, award SC B1 for sight of \(\int\frac{1}{1-x^2}dx = \frac{1}{2}\ln\left | \frac{1+x}{1-x}\right | \) or \(\tanh^{-1}x\) or SC B1 for sight of \(\int\frac{p}{1+x} + \frac{q}{1-x}dx = p\ln(1+x) - q\ln(1-x)\) |
**(a)** $\frac{dy}{dx} = \frac{2x}{(1-x^2)}$ | B1 |
$1 + \left(\frac{dy}{dx}\right)^2 = 1 + \frac{(2x)^2}{(1-x^2)^2}$ | M1 | FT their $\frac{dy}{dx}$
$\frac{1 - 2x^2 + x^4 + 4x^2}{(1-x^2)^2} = \frac{(1+x^2)^2}{(1-x^2)^2}$ | m1 | Allow m1 if sign error in $\frac{dy}{dx}$
$s = \int_0^3 \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$ | |
$s = \int_0^3 \frac{1+x^2}{(1-x^2)} dx$ | A1 cso | **4 marks** | AG must have $dx$ and limits on final line
**(b)** $\frac{1+x^2}{1-x^2} = \frac{A}{1-x^2} + B$ | M1 | and attempt to find constants $B \neq 0$
$\frac{1+x^2}{1-x^2} = \frac{2}{1-x^2} - 1$ | A1 |
$\left(\frac{A}{2}\ln\left|\frac{1+x}{1-x}\right|\right)$ or $A\tanh^{-1}x + Bx$ | m1 | FT integral of their $\frac{A}{1-x^2} + B$
$\ln\left(\frac{1+x}{1-x}\right) - x$ | A1 | or $2\tanh^{-1}x - x$ | correct
$\ln\left(\frac{1+\frac{3}{4}}{1-\frac{3}{4}}\right) - \frac{3}{4}$ OE | A1 | PI by next A1 or $\left(s = \right)2\tanh^{-1}\left(\frac{3}{4}\right) - \frac{3}{4}$
$-\frac{3}{4} + \ln 7$ | A1 | **6 marks** | or $(s) = \ln 7 - \frac{3}{4}$
**Alternative:**
$\frac{1+x^2}{1-x^2} = \frac{C}{1+x} + \frac{D}{1-x} + E$ | (M1) | and attempt to find constants $E \neq 0$
$\frac{1+x^2}{1-x^2} = \frac{1}{1+x} + \frac{1}{1-x} - 1$ | (A1) |
$C\ln(1+x) - D\ln(1-x) + Ex$ | (m1) | FT integral of their $\frac{C}{1+x} + \frac{D}{1-x} + E$
$= \ln(1+x) - \ln(1-x) - x$ | (A1) | correct
$(s) = \ln\left(\frac{7}{4}\right) - \ln\left(\frac{1}{4}\right) - \frac{3}{4}$ OE | (A1) | correct unsimplified
$(s) = \ln 7 - \frac{3}{4}$ | (A1) | (6)
**Total: 10 marks**
**Notes:**
- **(a)** Condone omission of brackets in final line or poor use of brackets if recovered for A1cso
- **(b)** If M1 is not earned, award SC B1 for sight of $\int\frac{1}{1-x^2}dx = \frac{1}{2}\ln\left|\frac{1+x}{1-x}\right|$ or $\tanh^{-1}x$ or SC B1 for sight of $\int\frac{p}{1+x} + \frac{q}{1-x}dx = p\ln(1+x) - q\ln(1-x)$
---The arc of the curve with equation $y = 4 - \ln(1-x^2)$ from $x = 0$ to $x = \frac{3}{4}$ has length $s$.
\begin{enumerate}[label=(\alph*)]
\item Show that $s = \int_0^{\frac{3}{4}} \frac{\sqrt{1+x^2}}{1-x^2} \, dx$. [4 marks]
\item Find the value of $s$, giving your answer in the form $p + \ln N$, where $p$ is a rational number and $N$ is an integer. [6 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2016 Q3 [10]}}