Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

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

1) α = π/4

2) No such α exists

3) α = 0

4) α = π/√2

Solution:

RHL =

$$\begin{array}{l}\lim_{x\to 0^{+}}\frac{cos^{-1}(1-x^{2})sin^{-1}(1-x)}{x(1-x^{2})}\end{array}$$

=

$$\begin{array}{l}\frac{\pi }{2}\lim_{x\to 0^{+}}\frac{cos^{-1}(1-x^{2})}{x}\end{array}$$

=

$$\begin{array}{l}\frac{\pi }{2}\lim_{x\to 0^{+}}\frac{2x}{\sqrt{1-(1-x^{2})^{2}}}\end{array}$$

(L’Hospital’s rule)

=

$$\begin{array}{l}\pi\lim_{x\to 0^{+}}\frac{x}{\sqrt{2x^{2}-x^{4}}} = \pi\lim_{x\to 0^{+}}\frac{1}{\sqrt{2-x^{2}}}=\frac{\pi }{\sqrt{2}}\end{array}$$

LHL =

$$\begin{array}{l}\lim_{x\to 0^{-}}\frac{cos^{-1}(1-(1+x)^{2})sin^{-1}(-x)}{(1+x)-(1+x)^{3}}\end{array}$$

=

$$\begin{array}{l}\frac{\pi }{2}\lim_{x\to 0^{-}}\frac{sin^{-1}(x)}{(1+x)((1+x)^{2}-1)}\end{array}$$

=

$$\begin{array}{l}\frac{\pi }{2}\lim_{x\to 0^{-}}\frac{sin^{-1}(x)}{(x^{2}+2x)}\end{array}$$

=

$$\begin{array}{l}\frac{\pi }{2}\times \frac{1}{2}=\frac{\pi }{4}\end{array}$$

As LHL not equal to RHL, f(x) is not continuous at x = 0.

Hence option b is the answer.