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JEE MAIN 2021
16-03-2021 S2
Question
Let $\alpha \in \mathrm{R}$ be such that the function $$ f(x)= \begin{cases}\frac{\cos ^{-1}\left(1-\{x\}^2\right) \sin ^{-1}(1-\{x\})}{\{x\}-\{x\}^3}, & x \neq 0 \\ \alpha, & x=0\end{cases} $$ is continuous at $\mathrm{x}=0$, where $\{\mathrm{x}\}=\mathrm{x}-[\mathrm{x}]$, [x] is the greatest integer less than or equal to x . Then :
Select the correct option:
A
$\alpha=\frac{\pi}{\sqrt{2}}$
B
$\alpha=0$
C
no such $\alpha$ exists
D
$\alpha=\frac{\pi}{4}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Question Tags
JEE Main
Mathematics
Medium
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