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

(D) $\alpha=\frac{\pi}{\sqrt{2}}$

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Continuity and Differentiability

Let

α \in R

be such that the function

f (x) = {\frac{c o s ^{- 1} ( 1 - { x } ^{2} ) s i n ^{- 1} ( 1 - { x })}{{ x } - { x } ^{3}} α, x \neq = 0, x = 0

is continuous at

x = 0

, where

{x} = x - [x], [x]

is the greatest integer less than or equal to

x

. Then :

Answer: (C)

For

x \to 0^{+}

the fractional part is

{x} = x

. Since

cos^{- 1} (1 - h) \sim 2 h

for small

h > 0

, here

cos^{- 1} (1 - x^{2}) \sim 2 x

, while

sin^{- 1} (1 - x) \to \frac{π}{2}

, so

x \to 0^{+} lim \frac{cos ^{- 1} ( 1 - x ^{2} ) sin ^{- 1} ( 1 - x )}{x - x ^{3}} = x \to 0^{+} lim \frac{2 x \cdot \frac{π}{2}}{x ( 1 - x ^{2} )} = \frac{π}{2}

For

x \to 0^{-}

the fractional part is

{x} = x + 1

, so with

t = {x} \to 1^{-}

the quotient is

\frac{cos ^{- 1} ( 1 - t ^{2} ) sin ^{- 1} ( 1 - t )}{t ( 1 - t ) ( 1 + t )}

. Here

cos^{- 1} (1 - t^{2}) \to cos^{- 1} 0 = \frac{π}{2}

and

sin^{- 1} (1 - t) \sim 1 - t

, so

x \to 0^{-} lim f (x) = \frac{π}{2} \cdot \frac{1}{1 \cdot 2} = \frac{π}{4}

The two one-sided limits

\frac{π}{2}

and

\frac{π}{4}

differ, so

lim_{x \to 0} f (x)

does not exist and no such

α

exists.

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Single correct option