Question

Let $\alpha \in R$ be such that the function

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

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