OCR Further Pure Core 1 2021 June — Question 4 6 marks

Exam BoardOCR
ModuleFurther Pure Core 1 (Further Pure Core 1)
Year2021
SessionJune
Marks6
TopicIntegration using inverse trig and hyperbolic functions
TypeIntegration by parts with inverse trig
DifficultyChallenging +1.2 Part (a) requires chain rule differentiation of an inverse trig function with a square root composition—straightforward for Further Maths students. Part (b) is a multi-step integration problem requiring recognition that the integrand relates to the derivative from part (a), then applying integration by parts or substitution, followed by careful evaluation at limits. While requiring several techniques and insight to connect parts (a) and (b), this is a standard Further Pure question type with clear signposting.
Spec4.08g Derivatives: inverse trig and hyperbolic functions4.08h Integration: inverse trig/hyperbolic substitutions

4 You are given that \(y = \tan ^ { - 1 } \sqrt { 2 x }\).
  1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
  2. Show that \(\int _ { \frac { 1 } { 6 } } ^ { \frac { 1 } { 2 } } \frac { \sqrt { x } } { \left( x + 2 x ^ { 2 } \right) } \mathrm { d } x = k \pi\) where \(k\) is a number to be determined in exact form.

4 You are given that $y = \tan ^ { - 1 } \sqrt { 2 x }$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Show that $\int _ { \frac { 1 } { 6 } } ^ { \frac { 1 } { 2 } } \frac { \sqrt { x } } { \left( x + 2 x ^ { 2 } \right) } \mathrm { d } x = k \pi$ where $k$ is a number to be determined in exact form.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 1 2021 Q4 [6]}}