JEE Advanced 2025 Paper 2, Mathematics Section 2 Q4: Maxima & Minima

Step 1: Behaviour at $x=0$.

${\lim }_{x\to 0^{+}}f(x)={\lim }_{x\to 0^{-}}f(x)=3$

Since $3>\frac{7}{3}$, therefore

$3>f(0)$

$\Rightarrow x=0$ is a local minima.

Therefore, Option (B) is correct.

Step 2: Derivative for $x\ne 0$.

$f(x)=1+\frac{4x}{2x+\sin x}\Rightarrow f^{\prime }(x)=\frac{4(\sin x-x\cos x)}{(2x+\sin x{)}^2}=\frac{4\cos x(\tan x-x)}{(2x+\sin x{)}^2}$.

The sign of $f^{\prime }(x)$ is governed by $\cos x(\tan x-x)$.

Step 3: Critical points beyond $x=0$.

$\tan x=x$ has, in each interval $\left(n\pi -\frac{\pi }{2},n\pi +\frac{\pi }{2}\right)$ for $n\ge 1$, exactly one solution $x_n$ close to (but less than) $n\pi +\frac{\pi }{2}$. Also $\cos x=0$ at the half-odd multiples of $\pi$.

A sign chart on $[\pi ,6\pi ]$ gives the alternating pattern: maxima at $x_1$ (near $3\pi /2$), $x_3$ (near $7\pi /2$), $x_5$ (near $11\pi /2$). That is 3 maxima in $[\pi ,6\pi ]$. Option (C) TRUE.

Step 4: Minima on $[2\pi ,4\pi ]$.

Within $[2\pi ,4\pi ]$ the relevant critical points are $x_2$ (near $5\pi /2$, a minimum) and $x_3$ (near $7\pi /2$, a maximum). Exactly one local minimum lies in this interval. Option (D) TRUE.

$\therefore$ The correct options are (B), (C) and (D).