SPS SPS FM Pure 2021 May — Question 7 9 marks

Exam BoardSPS
ModuleSPS FM Pure (SPS FM Pure)
Year2021
SessionMay
Marks9
TopicDifferentiating Transcendental Functions
TypeDifferentiate inverse trigonometric functions
DifficultyStandard +0.3 Part (a) is a standard textbook derivation of the derivative of arcsin using implicit differentiation (3 marks of routine work). Part (b) requires computing a mean value integral involving splitting the integrand, recognizing one part as arcsin derivative and the other requiring substitution, then simplifying—this is methodical A-level integration practice with multiple steps but no novel insight. The 9 total marks reflect length rather than conceptual difficulty, making this slightly easier than average overall.
Spec1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs4.08e Mean value of function: using integral4.08g Derivatives: inverse trig and hyperbolic functions

Given that \(y = \arcsin x\), \(-1 \leqslant x < 1\),
  1. show that \(\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}\). [3]
Given that \(f(x) = \frac{3x + 2}{\sqrt{4 - x^2}}\),
  1. show that the mean value of \(f(x)\) over the interval \([0, \sqrt{2}]\), is $$\frac{\pi\sqrt{2}}{4} + A\sqrt{2} - A,$$ where \(A\) is a constant to be determined. [6]

Given that $y = \arcsin x$, $-1 \leqslant x < 1$,

\begin{enumerate}[label=(\alph*)]
\item show that $\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}$. [3]
\end{enumerate}

Given that $f(x) = \frac{3x + 2}{\sqrt{4 - x^2}}$,

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item show that the mean value of $f(x)$ over the interval $[0, \sqrt{2}]$, is

$$\frac{\pi\sqrt{2}}{4} + A\sqrt{2} - A,$$

where $A$ is a constant to be determined. [6]
\end{enumerate}

\hfill \mbox{\textit{SPS SPS FM Pure 2021 Q7 [9]}}