AQA FP2 2015 June — Question 6 8 marks

Exam BoardAQA
ModuleFP2 (Further Pure Mathematics 2)
Year2015
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Substitution
TypeAlgebraic manipulation before substitution
DifficultyChallenging +1.2 This is a structured Further Maths question with clear guidance ('show that' and 'hence'). Part (a) requires product rule and chain rule with inverse trig differentiation—standard FP2 techniques. Part (b) is a direct application of the result from (a), requiring only substitution of limits and arithmetic. While it involves multiple techniques, the scaffolding makes it more routine than problem-solving.
Spec1.08h Integration by substitution4.08g Derivatives: inverse trig and hyperbolic functions

6
  1. Given that \(y = ( x - 2 ) \sqrt { 5 + 4 x - x ^ { 2 } } + 9 \sin ^ { - 1 } \left( \frac { x - 2 } { 3 } \right)\), show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = k \sqrt { 5 + 4 x - x ^ { 2 } }$$ where \(k\) is an integer.
  2. Hence show that $$\int _ { 2 } ^ { \frac { 7 } { 2 } } \sqrt { 5 + 4 x - x ^ { 2 } } \mathrm {~d} x = p \sqrt { 3 } + q \pi$$ where \(p\) and \(q\) are rational numbers.
    [0pt] [3 marks]

(a) [5 marks]
Given that \(y = (x-2)\sqrt{5 + 4x - x^2} + 9\sin^{-1}\left(\frac{x-2}{3}\right)\), show that
\[\frac{dy}{dx} = k\sqrt{5 + 4x - x^2}\]
where \(k\) is an integer.
(b) [3 marks]
Hence show that
\[\int_2^7 \sqrt{5 + 4x - x^2}\,dx = p\sqrt{3} + q\pi\]
where \(p\) and \(q\) are rational numbers.
**(a) [5 marks]**
Given that $y = (x-2)\sqrt{5 + 4x - x^2} + 9\sin^{-1}\left(\frac{x-2}{3}\right)$, show that
$$\frac{dy}{dx} = k\sqrt{5 + 4x - x^2}$$
where $k$ is an integer.

**(b) [3 marks]**
Hence show that
$$\int_2^7 \sqrt{5 + 4x - x^2}\,dx = p\sqrt{3} + q\pi$$
where $p$ and $q$ are rational numbers.
6
\begin{enumerate}[label=(\alph*)]
\item Given that $y = ( x - 2 ) \sqrt { 5 + 4 x - x ^ { 2 } } + 9 \sin ^ { - 1 } \left( \frac { x - 2 } { 3 } \right)$, show that

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = k \sqrt { 5 + 4 x - x ^ { 2 } }$$

where $k$ is an integer.
\item Hence show that

$$\int _ { 2 } ^ { \frac { 7 } { 2 } } \sqrt { 5 + 4 x - x ^ { 2 } } \mathrm {~d} x = p \sqrt { 3 } + q \pi$$

where $p$ and $q$ are rational numbers.\\[0pt]
[3 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA FP2 2015 Q6 [8]}}