Differential Calculus
Differential Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
The figure shows two regions in the first quadrant. $A(t)$ is the area under the curve $y = \sin x^2$ from $0$ to $t$ and $B(t)$ is the area of the triangle with vertices $O$, $P$ and $M(t, 0)$. If $\lim_{t \to 0} \frac{A(t)}{B(t)} = \frac{1}{k}$, then $k$ is______.
Step-by-Step Solution
Key Concept: L'Hôpital's rule applied systematically to indeterminate forms $\frac{0}{0}$ reduces the integral ratio to elementary limits.
Given $A(t) = \int_0^t \sin^2 x \, dx$ and $B(t) = \frac{t \sin^2 t}{2}$, we compute $\lim_{t \to 0} \frac{A(t)}{B(t)} = \lim_{t \to 0} \frac{2\int_0^t \sin^2 x \, dx}{t \sin^2 t}$. Applying L'Hôpital's rule twice: $\lim_{t \to 0} \frac{2\sin^2 t}{3t^2}$, then $\lim_{t \to 0} \frac{2\sin(t^2)}{3t^2} = \frac{2}{3}$. Therefore $m + n = 2 + 3 = 5$.
Correct Answer: 1