Let x denote the fractional part of x and f(x)= ^-1(1-x^2) ^-1(1-x)x-x^3,x 0.

Question

Let ${x}$ denote the fractional part of $x$ and $f(x)=\frac{{\cos }^{-1}(1-{x{}}^{2}){\sin }^{-1}(1-{x})}{{x}-{x{}}^{3}},x
e 0$. If $L$ and $R$ respectively denotes the left hand limit and the right hand limit of $f(x)$ at $x=0$, then $\frac{32}{{\pi }^{2}}({L}^{2}+{R}^{2})$ is equal to \_\_\_\_\_\_\_\_\_\_\_.

Answer: 18

Accepted Answer

Finding right hand limit

R = $\begin{gathered} \underset{x	o {0}^{+}}{lim}f(x)=\underset{h	o 0}{lim}f(0+h)=\underset{h	o 0}{lim}f(h) \end{gathered}$

$\begin{gathered} =\underset{h	o 0}{lim}\frac{{\cos }^{-1}(1-{h}^{2}){\sin }^{-1}(1-h)}{h(1-{h}^{2})} \ =\underset{h	o 0}{lim}\frac{{\cos }^{-1}(1-{h}^{2})}{h}(\frac{{\sin }^{-1}1}{1}) \end{gathered}$

Let ${\cos }^{-1}(1-{h}^{2})=	heta \Rightarrow \cos 	heta =1-{h}^{2}$

$\begin{gathered} =\frac{\pi }{2}\underset{	heta 	o 0}{lim}\frac{	heta }{\sqrt{1-\cos 	heta }} \ =\frac{\pi }{2}\underset{	heta 	o 0}{lim}\frac{1}{\sqrt{\frac{1-\cos 	heta }{{	heta }^{2}}}} \ =\frac{\pi }{2}\frac{1}{\sqrt{1/2}} \ 	herefore R=\frac{\pi }{\sqrt{2}} \end{gathered}$

Now finding left hand limit

$\begin{gathered} L=\underset{x	o {0}^{-}}{lim}f(x)=\underset{h	o 0}{lim}f(-h) \end{gathered}$

$\begin{gathered} =\underset{h	o 0}{lim}\frac{{\cos }^{-1}(1-{-h{}}^{2}){\sin }^{-1}(1-{-h})}{{-h}-{-h{}}^{3}} \ =\underset{h	o 0}{lim}\frac{{\cos }^{-1}(1-(-h+1{)}^{2}){\sin }^{-1}(1-(-h+1))}{(-h+1)-(-h+1{)}^{3}} \ =\underset{h	o 0}{lim}\frac{{\cos }^{-1}(-{h}^{2}+2h){\sin }^{-1}h}{(1-h)(1-(1-h{)}^{2})} \end{gathered}$

$\begin{gathered} =\underset{h	o 0}{lim}(\frac{\pi }{2})\frac{{\sin }^{-1}h}{(1-(1-h{)}^{2})} \ =\frac{\pi }{2}\underset{h	o 0}{lim}(\frac{{\sin }^{-1}h}{-{h}^{2}+2h}) \ =\frac{\pi }{2}\underset{h	o 0}{lim}(\frac{{\sin }^{-1}h}{h})(\frac{1}{-h+2}) \ \, L=\frac{\pi }{4} \end{gathered}$

$\begin{gathered} \frac{32}{{\pi }^{2}}({L}^{2}+{R}^{2})=\frac{32}{{\pi }^{2}}(\frac{{\pi }^{2}}{2}+\frac{{\pi }^{2}}{16}) \ =16+2 \ =18 \end{gathered}$

Solutions · 1