Question

Let $f\left(x\right)=\frac{{\cos }^{-1}(1-\{x{\}}^2\left){\sin }^{-1}\right(1-\left\{x\right\})}{\left\{x\right\}-\{x{\}}^3},x\ne 0$, where $\left\{x\right\}$ denotes fractional part of $x$. Then
(correct answer + 1, wrong answer - 0.25)

• A. If $f\left(0\right)=\frac{\pi }{4},$ then $f\left(x\right)$ is continuous at $x=0$
• B. If $f\left(0\right)=\frac{\pi }{\sqrt{2}},$ then $f\left(x\right)$ is continuous at $x=0$
• C. If $f\left(0\right)=\frac{\pi }{2\sqrt{2}},$ then $f\left(x\right)$ is continuous at $x=0$
• D. $f\left(x\right)$ is a discontinuous function.

Answer 1

The correct option is D $f\left(x\right)$ is a discontinuous function.

Given : $f\left(x\right)=\frac{{\cos }^{-1}(1-\{x{\}}^2\left){\sin }^{-1}\right(1-\left\{x\right\})}{\left\{x\right\}-\{x{\}}^3},x\ne 0$
Finding $R.H.L.$
$=\underset{x\to 0^{+}}{lim}\frac{{\cos }^{-1}(1-x^2){\sin }^{-1}(1-x)}{x-x^3}=\underset{x\to 0^{+}}{lim}\frac{{\cos }^{-1}(1-x^2)}{x}\times \frac{{\sin }^{-1}(1-x)}{(1-x^2)}=\underset{x\to 0^{+}}{lim}\frac{{\cos }^{-1}(1-x^2)}{x}\times \frac{\pi }{2}$
Using L'Hospital's Rule,
$=\frac{\pi }{2}\times \underset{x\to 0^{+}}{lim}\frac{-\frac{-2x}{\sqrt{1-(1-x^2{)}^2}}}{1}=\frac{\pi }{2}\times \underset{x\to 0^{+}}{lim}\frac{2}{\sqrt{-x^2+2}}=\frac{\pi }{\sqrt{2}}$

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

As $\text{L.H.L}\ne \text{R.H.L}$, therefore no value of $f\left(0\right)$ can make $f\left(x\right)$ continuous at $x=0$.