Differential Calculus
Differential Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
If $f(x) = \begin{cases} \frac{\left(\frac{\pi}{2} - \sin^{-1}\left|1-\{x\}^2\right|\right) \sin^{-1}(1-\{x\})}{\sqrt{2}\left(\{x\} - \{x\}^3\right)} & x > 0 \\ k & x = 0 \\ \frac{A\sin^{-1}(1-\{x\})\cos^{-1}(1-\{x\})}{\sqrt{2}\{x\}(1-\{x\})} & x < 0 \end{cases}$ is continuous at $x = 0$, then the value of $A$ is______. (where $\{.\}$ denotes fractional part of $x$).
Step-by-Step Solution
Key Concept: Taylor expansions of inverse trigonometric functions near 0 are essential for evaluating the limit; matching LHS and RHS determines the constant $A$.
Computing the RHS limit by careful manipulation of inverse trigonometric functions and simplification yields $\frac{\pi}{2}$ as found in equation (1). For the LHS, using the expansion $\sin^{-1}(h) \approx h$ and $\cos^{-1}(h) \approx \frac{\pi}{2} - h$ as $h \to 0$, the limit evaluates to $\frac{A\pi}{2\sqrt{2}}$. Equating with RHS and solving gives $A = \sqrt{2}$.
Correct Answer: 5