Definite Integration
Integral Calculus-2
star_batch_jee_advanced_2025
Grade 12
Question:
Let $f(x)$ be a continuous function with continuous first derivative on $(a, b)$, where $b > a$, and let $\lim_{x \to a^+} f(x) = \infty$, $\lim_{x \to b^-} f(x) = -\infty$ and $f'(x) + f^2(x) \geq -1$, for all $x$ in $(a, b)$, if the minimum value of $(b-a)$ equals to $k$ then $k$ is ____.
Step-by-Step Solution
Key Concept: Transform the inequality involving $f'$ and $f^2$ into an integral inequality using the antiderivative $\tan^{-1}(f(x))$.
Given $f'(x) + f^2(x) \geq -1$, we derive that $f^2(x) + 1 \geq -f'(x)$, which gives $1 \geq \frac{f'(x)}{1+f^2(x)}$ in $(a,b)$. Integrating both sides: $\int_a^b dx \geq \int_a^b \frac{f'(x)}{1+f^2(x)} dx$, which yields $b-a \geq [\tan^{-1}(f(x))]_a^b = \left[-\frac{\pi}{2} - \frac{\pi}{2}\right] = -\pi$. Therefore $(b-a) \geq \pi$.
Correct Answer: 3.14