AQA C3 2012 January — Question 3 6 marks

Exam BoardAQA
ModuleC3 (Core Mathematics 3)
Year2012
SessionJanuary
Marks6
PaperDownload PDF ↗
TopicStandard Integrals and Reverse Chain Rule
TypeIntegration with substitution given
DifficultyModerate -0.3 Part (a) is trivial differentiation. Part (b) is a standard 'reverse chain rule' integration where the numerator is exactly half the derivative of the denominator, making the substitution straightforward once recognized. This is a textbook exercise testing pattern recognition rather than problem-solving, though the 'hence' connection requires some insight, making it slightly easier than average overall.
Spec1.07i Differentiate x^n: for rational n and sums1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)

3
  1. Given that \(y = 4 x ^ { 3 } - 6 x + 1\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
    (l mark)
  2. Hence find \(\int _ { 2 } ^ { 3 } \frac { 2 x ^ { 2 } - 1 } { 4 x ^ { 3 } - 6 x + 1 } \mathrm {~d} x\), giving your answer in the form \(p \ln q\), where \(p\) and \(q\) are rational numbers.

3
\begin{enumerate}[label=(\alph*)]
\item Given that $y = 4 x ^ { 3 } - 6 x + 1$, find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.\\
(l mark)
\item Hence find $\int _ { 2 } ^ { 3 } \frac { 2 x ^ { 2 } - 1 } { 4 x ^ { 3 } - 6 x + 1 } \mathrm {~d} x$, giving your answer in the form $p \ln q$, where $p$ and $q$ are rational numbers.
\end{enumerate}

\hfill \mbox{\textit{AQA C3 2012 Q3 [6]}}