Question

Let $\alpha \in R$ be such that the function $f\left(x\right)=\{\begin{array}{cc}\frac{{\cos }^{-1}(1-\{x{\}}^2){\sin }^{-1}(1-\{x\left\}\right)}{\left\{x\right\}-\{x{\}}^3},& x\ne 0\\ \alpha ,& x=0\end{array}$ is continuous at $x=0,$ where $\left\{x\right\}=x-\left[x\right],$ $\left[x\right]$ is the greatest integer less than or equal to $x.$ Then :

• A. $\alpha =\frac{\pi }{4}$
• B. no such $\alpha$ exists
• C. $\alpha =0$
• D. $\alpha =\frac{\pi }{\sqrt{2}}$

Answer 1

The correct option is B no such $\alpha$ exists

RHL $=\underset{x\to 0^{+}}{lim}\frac{{\cos }^{-1}(1-x^2){\sin }^{-1}(1-x)}{x(1-x^2)}$
$=\frac{\pi }{2}\underset{x\to 0^{+}}{lim}\frac{{\cos }^{-1}(1-x^2)}{x}$
$=\frac{\pi }{2}\underset{x\to 0^{+}}{lim}\frac{-1}{\sqrt{1-(1-x^2{)}^2}}(-2x)$ (L'Hospital Rule)
$=\pi \underset{x\to 0^{+}}{lim}\frac{x}{\sqrt{2x^2-x^4}}$
$=\pi \underset{x\to 0^{+}}{lim}\frac{1}{\sqrt{2-x^2}}=\frac{\pi }{\sqrt{2}}$

LHL $=\underset{x\to 0^{-}}{lim}\frac{{\cos }^{-1}(1-(1+x{)}^2){\sin }^{-1}(-x)}{(1+x)-(1+x{)}^3}$
$=\frac{\pi }{2}\underset{x\to 0^{-}}{lim}\frac{{\sin }^{-1}x}{(1+x)\left[\right(1+x{)}^2-1]}$
$=\frac{\pi }{2}\underset{x\to 0^{-}}{lim}\frac{{\sin }^{-1}x}{x^2+2x}$
$=\frac{\pi }{2}\left(\frac{1}{2}\right)=\frac{\pi }{4}$

As LHL $\ne$ RHL, so $f\left(x\right)$ is not continuous at $x=0.$