Differential Calculus
Differential Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
Let $f(x) = \tan^{-1} x, |x| \leq 1 = \frac{\pi}{4} \operatorname{sgn} x + \frac{x-1}{2}, |x| > 1$, (where $\operatorname{sgn}$ denotes signum function). Then the value of $4f\left(1^{+}\right)$ equals.
Step-by-Step Solution
Key Concept: The derivative at a point is the limit of the difference quotient as the increment approaches zero.
Use the definition of derivative: $f'(1) = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0} \frac{\frac{π}{4} + \frac{1+h-1}{2} - \frac{π}{4}}{h} = \lim_{h \to 0} \frac{h/2}{h} = \frac{1}{2}$.
Correct Answer: 9