Question
The function $f$ is defined by $f(x)=(\arcsin x)^2,-1 \leqslant x \leqslant 1$.
The function $f$ satisfies the equation $\left(1-x^2\right) f^{\prime \prime}(x)-x f^{\prime}(x)-2=0$
a. Show that $f^{\prime}(0)=0$.
b. By differentiating the above equation twice, show that
$[4]$
$$
\left(1-x^2\right) f^{(4)}(x)-5 x f^{(3)}(x)-4 f^{\prime \prime}(x)=0
$$
where $f^{(3)}(x)$ and $f^{(4)}(x)$ denote the 3rd and 4th derivative of $f(x)$ respectively.
c. Hence show that the Maclaurin series for $f(x)$ up to and including the term in $x^4$ is $x^2+\frac{1}{3} x^4$.
d. Use this series approximation for $f(x)$ with $x=\frac{1}{2}$ to find an approximate value for $\pi^2$.
▶️Answer/Explanation
Markscheme
a. ${ }^{\star}$ This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
$$
f^{\prime}(x)=\frac{2 \arcsin (x)}{\sqrt{1-x^2}} \quad \text { M1A1 }
$$
Note: Award $\boldsymbol{M} 1$ for an attempt at chain rule differentiation.
Award $M O A O$ for $f^{\prime}(x)=2 \arcsin (x)$.
$$
f^{\prime}(0)=0 \quad A G
$$
[2 marks]
b. differentiating gives $\left(1-x^2\right) f^{(3)}(x)-2 x f^{\prime \prime}(x)-f^{\prime}(x)-x f^{\prime \prime}(x)(=0) \quad$ M1A1
differentiating again gives $\left(1-x^2\right) f^{(4)}(x)-2 x f^{(3)}(x)-3 f^{\prime \prime}(x)-3 x f^{(3)}(x)-f^{\prime \prime}(x)(=0) \quad$ M1A1
Note: Award $\boldsymbol{M} \boldsymbol{1}$ for an attempt at product rule differentiation of at least one product in each of the above two lines. Do not penalise candidates who use poor notation.
$$
\left(1-x^2\right) f^{(4)}(x)-5 x f^{(3)}(x)-4 f^{\prime \prime}(x)=0 \quad \boldsymbol{A G}
$$
[4 marks]
c. attempting to find one of $f^{\prime \prime}(0), f^{(3)}(0)$ or $f^{(4)}(0)$ by substituting $x=0$ into relevant differential equation(s)
(M1)
Note: Condone $f^{\prime \prime}(0)$ found by calculating $\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{2 \arcsin (x)}{\sqrt{1-x^2}}\right)$ at $x=0$.
$$
\begin{aligned}
& \left(f(0)=0, f^{\prime}(0)=0\right) \\
& f^{\prime \prime}(0)=2 \text { and } f^{(4)}(0)-4 f^{\prime \prime}(0)=0 \Rightarrow f^{(4)}(0)=8 \\
& f^{(3)}(0)=0 \text { and so } \frac{2}{2 !} x^2+\frac{8}{4 !} x^4 \quad \text { A1 }
\end{aligned}
$$
A1
Note: Only award the above $\boldsymbol{A 1}$, for correct first differentiation in part (b) leading to $f^{(3)}(0)=0$ stated or $f^{(3)}(0)=0$ seen from use of the general Maclaurin series.
Special Case: Award (M1)AOA1 if $f^{(4)}(0)=8$ is stated without justification or found by working backwards from the general Maclaurin series. so the Maclaurin series for $f(x)$ up to and including the term in $x^4$ is $x^2+\frac{1}{3} x^4 \quad \boldsymbol{A G}$
[3 marks]
d. substituting $x=\frac{1}{2}$ into $x^2+\frac{1}{3} x^4 \quad \boldsymbol{M 1}$ the series approximation gives a value of $\frac{13}{48}$
$$
\begin{aligned}
& \text { so } \pi^2 \simeq \frac{13}{48} \times 36 \\
& \simeq 9.75\left(\simeq \frac{39}{4}\right) \quad \boldsymbol{A 1}
\end{aligned}
$$
Note: Accept 9.76.
[2 marks]