$f(x)=\left\{\begin{array}{ll} \frac{\cos ^{-1}\left(1-\{x\}^{2}\right) \sin ^{-1}(1-\{x\})}{\{x\}-\{x\}^{3}}, & x \neq 0 \\ \alpha, & x=0 \end{array}\right.$
is continuous at $x=0,$ where $\{x\}=x-[x],[x]$ is the greatest integer less than or equal to $X$.
Then :
$\operatorname{Lim}_{x \rightarrow 0^{+}} \frac{\cos ^{-1}\left(1-x^{2}\right) \cdot \sin ^{-1}(1-x)}{x(1-x)(1+x)}$
$\operatorname{Lim}_{x \rightarrow 0^{+}} \frac{\cos ^{-1}\left(1-x^{2}\right)}{x \cdot 1 \cdot 1} \cdot \frac{\pi}{2}$
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, $\operatorname{Lim}_{x \rightarrow 0^{-}} \frac{\cos ^{-1}\left(1-(1+x)^{2}\right) \sin ^{-1}(-x)}{(1+x)-(1+x)^{3}}$
$\operatorname{Lim}_{x \rightarrow 0^{-}} \frac{\frac{\pi}{2}\left(-\sin ^{-1} x\right)}{(1+x)(2+x)(-x)}$
$\operatorname{Lim}_{x \rightarrow 0^{-}} \frac{\frac{\pi}{2}}{1 \cdot 2} \cdot \frac{\sin ^{-1} x}{x}=\frac{\pi}{4}$
$\Rightarrow RHL \neq LHL$
Function can't be continuous
$\Rightarrow$ No value of $\alpha$ exist
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
${I_1} = \int_{1 - k}^k {x\,f\left\{ {x(1 - x)} \right\}} \,dx$, ${I_2} = \int_{1 - k}^k {\,f\left\{ {x(1 - x)} \right\}} \,dx$
when $2k - 1 > 0.$ Then ${I_1}/{I_2}$ is