JEE Advanced 2025 — Paper 2 — Question 24

For $x \neq 0$, rewrite $f(x) = \dfrac{6x+\sin x}{2x+\sin x} = 1 + \dfrac{4x}{2x+\sin x}$. Compute limit as $x \to 0$: $\lim_{x \to 0} f(x) = \lim_{x \to 0} \dfrac{6x+\sin x}{2x+\sin x} = \dfrac{6+1}{2+1} = 3$, using $\sin x \sim x$. Since $f(0) = \dfrac{7}{3} < 3$, $x = 0$ is a point of local minima.

Hence option B is correct and A is false. Differentiate: $f'(x) = \dfrac{d}{dx}\left(1 + \dfrac{4x}{2x+\sin x}\right) = \dfrac{4[(2x+\sin x)\cdot 1 - x(2+\cos x)]}{(2x+\sin x)^2} = \dfrac{4(\sin x - x \cos x)}{(2x+\sin x)^2}$. Denominator $(2x+\sin x)^2 > 0$ for all $x \neq 0$, so sign of $f'$ = sign of $\sin x - x \cos x = \cos x (\tan x - x)$. Critical points: $f'(x) = 0 \Rightarrow \tan x = x$.

For $x > 0$, this equation has infinitely many solutions, one in each interval $(\frac{\pi}{2} + k\pi, \frac{3\pi}{2} + k\pi)$ for integer $k \geq 0$.

The first few positive solutions are approximately $4.49, 7.73, 10.90, 14.07, \dots$.

Sign analysis: In $(0, \frac{\pi}{2})$: $\cos > 0$, $\tan x > x \Rightarrow f' > 0$.

In $(\frac{\pi}{2}, \pi)$: $\cos < 0$, $\tan x - x < 0 \Rightarrow f' > 0$.

In $(\pi, \frac{3\pi}{2})$: $\cos < 0$, $\tan x - x$ changes from negative to positive $\Rightarrow f'$ changes from positive to negative, so first critical point ($4.49$) is a local maximum.

Next critical point ($7.73$) is a local minimum, next ($10.90$) is a local maximum, next ($14.07$) is a local minimum, etc. In the interval $[\pi, 6\pi] \approx [3.14, 18.85]$, the local maxima occur at $x \approx 4.49, 10.90,$ and $17.00$ (the third maximum in this interval), giving $3$ local maxima. Hence option C is correct. In the interval $[2\pi, 4\pi] \approx [6.28, 12.57]$, the only local minimum is at $x \approx 7.73$ (since $14.07 > 12.57$).

Hence number of local minima is $1$.

Therefore option D is correct. Thus the correct options are B, C, D.