Question

Let $\alpha \in R$ 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 $X$. Then:

• A. $\alpha=\frac{\pi}{\sqrt{2}}$
• B. $\alpha=0$
• C. no such $\alpha$ exists
• D. $\alpha=\frac{\pi}{1}$

Answer 1

Answer: (C)

$\displaystyle\lim_{x \rightarrow 0^{+}} f(x)=f(0)=\displaystyle\lim_{x \rightarrow 0^{-}}(x)$
$\displaystyle\lim_{x \rightarrow 0^{+}} \frac{\cos ^{-1}\left(1-x^{2}\right) \cdot \sin ^{-1}(1-x)}{x(1-x)(1+x)}$
$\displaystyle\lim_{x \rightarrow 0^{+}} \frac{\cos ^{-1}\left(1-x^{2}\right)}{x \cdot 1 \cdot 1} \cdot \frac{\pi}{2}$
Let $1 - x^2 = \cos\,\theta$
$\frac{\pi}{2} \displaystyle\lim_{\pi \rightarrow 0^+} \frac{\theta}{\sqrt{1 - \cos\,\theta}}$
$\frac{\pi}{2} \displaystyle\lim_{\theta \rightarrow 0^+} \frac{\theta}{\sqrt{1 - \sin\,\frac{\theta}{2}}} = \frac{\pi}{\sqrt{2}}$
Now, $\displaystyle\lim_{x \rightarrow 0^{-}} \frac{\cos ^{-1}\left(1-(1+x)^{2}\right) \sin ^{-1}(-x)}{(1+x)-(1+x)^{3}}$
$\displaystyle\lim_{x \rightarrow 0^{-}} \frac{\frac{\pi}{2}\left(-\sin ^{-1} x\right)}{(1+x)(2+x)(-x)}$
$\displaystyle\lim_{x \rightarrow 0^{-}} \frac{\frac{\pi}{2}}{1 \cdot 2} \cdot \frac{\sin ^{-1} x}{x}=\frac{\pi}{4}$
$\Rightarrow RHL \neq LHL$
Function can't be continuous
$\Rightarrow $ No value of $\alpha$ exist