JEE Main 2025 (Online) 3rd April Morning ShiftQuestion 1
The number of solutions of the equation $2x+3\tan x=\pi$, $x\in[-2\pi,2\pi]-\left\{\pm\frac{\pi}{2},\pm\frac{3\pi}{2}\right\}$ is:
Trigonometric Ratio & Identities
Solution
Concept. $\tan x=\frac{\pi-2x}{3}$: the line meets each branch of $y=\tan x$ exactly once, because $f(x)=2x+3\tan x-\pi$ is increasing on every branch. Formulas used. (1) $\dfrac{d}{dx}\tan x=\sec^2x>0$Step 1 - Monotonic. $f'(x)=2+3\sec^2x>0$ by (1), so $f$ has at most one root on each branch. Step 2 - The branches. In $[-2\pi,2\pi]$ the branches are $$\left[-2\pi,-\tfrac{3\pi}2\right),\ \left(-\tfrac{3\pi}2,-\tfrac\pi2\right),\ \left(-\tfrac\pi2,\tfrac\pi2\right),\ \left(\tfrac\pi2,\tfrac{3\pi}2\right),\ \left(\tfrac{3\pi}2,2\pi\right].$$ Step 3 - Middle three branches. On each, $f$ runs from $-\infty$ to $+\infty$: one root each. Step 4 - End branches. $f(-2\pi)=-5\pi<0$ and $f\to+\infty$ as $x\to-\frac{3\pi}{2}^-$: one root. $f\to-\infty$ as $x\to\frac{3\pi}{2}^+$ and $f(2\pi)=3\pi>0$: one root. In the figure the straight line is $y=\frac{\pi-2x}{3}$ and the dashed lines are the asymptotes of $\tan x$:Step 5 - Total. $3+2=5$. Answer: (C)