JEE Advanced 2025 Paper 2 Q08 Mathematics Differentiation & Applications Maxima & Minima Hard

JEE Advanced 2025 Paper 2 · Q08 · Maxima & Minima

Let $\mathbb{R}$ denote the set of all real numbers. Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $$f(x) = \frac{6x + \sin x}{2x + \sin x},\ \text{if } x \ne 0;\ \text{and } f(0) = \frac{7}{3}.$$

Then which of the following statements is (are) TRUE?

  1. A. The point $x = 0$ is a point of local maxima of $f$
  2. B. The point $x = 0$ is a point of local minima of $f$
  3. C. Number of points of local maxima of $f$ in the interval $[\pi, 6\pi]$ is $3$
  4. D. Number of points of local minima of $f$ in the interval $[2\pi, 4\pi]$ is $1$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:B, C, D

Solution

Rewrite $f(x) = 3 - \frac{2 \sin x}{2x + \sin x}$. For small $x$: $\sin x = x - x^3/6 + \ldots$, $2x + \sin x = 3x - x^3/6$, ratio $\sin x / (2x + \sin x) \sim (x - x^3/6) / (3x - x^3/6) = (1/3)(1 - x^2/6)(1 - x^2/18)^{-1} \sim (1/3)(1 - x^2/9)$. So $f(x) \sim 7/3 + (2/27) x^2$ near $0$; $x = 0$ is a local minimum. (A) FALSE, (B) TRUE. Differentiate $f$: $f'(x) = -\frac{4(x \cos x - \sin x)}{(2x + \sin x)^2}$. Critical points satisfy $x \cos x = \sin x$, i.e., $\tan x = x$. The classical equation $\tan x = x$ has exactly one root in each open interval $((2k+1)\pi/2, (2k+3)\pi/2)$, and the sign of $f'(x)$ alternates between consecutive roots. In $[\pi, 6\pi]$ there are $5$ such roots, alternating max, min, max, min, max $\to$ $3$ maxima. (C) TRUE. In $[2\pi, 4\pi]$ there are $2$ roots, one max and one min $\to$ $1$ minimum. (D) TRUE. Correct: (B), (C), (D).

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