Differential Calculus
Differential Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
If $f(x+h) - f(x) + h f'(x+\theta h), 0 < \theta < 1$, the value of $40$, when $f(x) = Ax^2 + Bx + C$ is_____
Step-by-Step Solution
Key Concept: Telescoping series on the cotangent difference reduces the problem to a geometric series that converges to a fixed value.
From the given condition $\sin(x_{n+1} - x_n) + 2^{-(n+1)} \sin x_n \sin x_{n+1} = 0$, we derive $\cot x_{n+1} - \cot x_n = 2^{-(n+1)}$. Summing telescopically: $\cot x_{n+1} - \cot x_1 = \frac{1}{2^2} + \frac{1}{2^3} + \ldots + \frac{1}{2^{n+1}}$. As $n \to \infty$, the right side converges to $\frac{1}{4}$, so $\lim_{n \to \infty} x_{n+1} = \frac{\pi}{4}$ and $l = \frac{\pi}{4}$, giving $4l = \pi$.
Correct Answer: 2