Limits, Continuity & Differentiability
Differential Calculus-1
star_batch_jee_advanced_2025
Grade 12
Question:
Consider a parabola $y = \frac{x^2}{4}$ and the point $F(0,1)$. Let $A_1(x_1, y_1), A_2(x_2, y_2), A_3(x_3, y_3), \ldots, A_N(x_n, y_n)$ are 'n' points on the parabola such that $x_k > 0$ and $\angle OFA_k = \frac{k\pi}{2n}$ $(k = 1,2,\ldots,n)$. If the value of $\lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^{n} FA_k = \frac{m}{\pi}$, then $m$ is ______.
Step-by-Step Solution
Key Concept: Converting the discrete sum to a Riemann integral as $n \to \infty$ allows exact evaluation of the limiting arc length sum.
For points $A_k = (2t_k^2, t_k)$ on the parabola, the slope of $FA_k$ is $\frac{t_k^2 - 1}{2t_k} = \tan(2\theta_k)$, where $t_k = \tan \phi_k$. This gives $\phi_k = \frac{\theta_k}{2} - \frac{k\pi}{4n}$. The tangent at $A_k$ makes angle $\frac{\pi}{4}$ with the $x$-axis. Computing $\lim_{n \to \infty} \sum_{k=1}^n FA_k = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \sec^2\left(\frac{k\pi}{4n}\right) = \int_0^{\pi/4} \sec^2 x \, dx = \frac{4}{\pi}$. Hence $m = 4$.
Correct Answer: 4