Let alpha in R be such that the function

Question:

Let $\alpha \in \mathrm{R}$ be such that the function

$f(x)= \begin{cases}\frac{\cos ^{-1}\left(1-\{x\}^{2}\right) \sin ^{-1}(1-\{x\})}{\{x\}-\{x\}^{3}}, & x \neq 0 \\ \alpha, & x=0\end{cases}$

is continuous at $x=0$, where $\{x\}=x-[x],[x]$ is the greatest integer less than or equal to $\mathrm{x}$.

Then:

1. (1) $\alpha=\frac{\pi}{\sqrt{2}}$

2. (2) $\alpha=0$

3. (3) no such $\alpha$ exists

4. (4) $\alpha=\frac{\pi}{4}$

Correct Option: , 3

Solution:

$\operatorname{Lim}_{\mathrm{x} \rightarrow 0^{+}} \mathrm{f}(\mathrm{x})=\mathrm{f}(0)=\operatorname{Lim}_{\mathrm{x} \rightarrow 0^{-}}(\mathrm{x})$

$\operatorname{Lim}_{x \rightarrow 0^{+}} \frac{\cos ^{-1}\left(1-x^{2}\right) \cdot \sin ^{-1}(1-x)}{x(1-x)(1+x)}$

$\operatorname{Lim}_{x \rightarrow 0^{+}} \frac{\cos ^{-1}\left(1-x^{2}\right)}{x \cdot 1 \cdot 1} \cdot \frac{\pi}{2}$

Let $1-x^{2}=\cos \theta$

$\frac{\pi}{2} \operatorname{Lim}_{x \rightarrow 0^{+}} \frac{\theta}{\sqrt{1-\cos \theta}}$

$\frac{\pi}{2} \operatorname{Lim}_{\theta \rightarrow 0^{+}} \frac{\theta}{\sqrt{2} \sin \frac{\theta}{2}}=\frac{\pi}{\sqrt{2}}$

Now, $\operatorname{Lim}_{x \rightarrow 0^{-}} \frac{\cos ^{-1}\left(1-(1+x)^{2}\right) \sin ^{-1}(-x)}{(1+x)-(1+x)^{3}}$

$\operatorname{Lim}_{x \rightarrow 0^{-}} \frac{\frac{\pi}{2}\left(-\sin ^{-1} x\right)}{(1+x)(2+x)(-x)}$

$\operatorname{Lim}_{x \rightarrow 0^{-}} \frac{\frac{\pi}{2}}{1 \cdot 2} \cdot \frac{\sin ^{-1} x}{x}=\frac{\pi}{4}$

$\Rightarrow \mathrm{RHL} \neq \mathrm{LHL}$

Function can't be continuous

$\Rightarrow$ No value of $\alpha$ exist