jee-main 2023 Q79

jee-main · India · session2_13apr_shift1 Chain Rule Chain Rule Combined with Fundamental Theorem of Calculus
Let $f: \mathbb{R} \to \mathbb{R}$ be a differentiable function such that $f\left(\frac{\pi}{4}\right) = \sqrt{2}$, $f\left(\frac{\pi}{2}\right) = 0$ and $f'\left(\frac{\pi}{2}\right) = 1$ and let $g(x) = \int_x^{\pi/4} (f'(t)\sec t + \tan t \cdot f(t)\sec t)\,dt$. Then $\lim_{x \to \pi/2} \frac{g(x)}{(x - \pi/2)^2}$ is equal to $\_\_\_\_$.
Let $f: \mathbb{R} \to \mathbb{R}$ be a differentiable function such that $f\left(\frac{\pi}{4}\right) = \sqrt{2}$, $f\left(\frac{\pi}{2}\right) = 0$ and $f'\left(\frac{\pi}{2}\right) = 1$ and let $g(x) = \int_x^{\pi/4} (f'(t)\sec t + \tan t \cdot f(t)\sec t)\,dt$. Then $\lim_{x \to \pi/2} \frac{g(x)}{(x - \pi/2)^2}$ is equal to $\_\_\_\_$.