JEE Mathematics 2025 Question with Solution

Given: $2x + 3\tan x = \pi$ with $x \in [-2\pi, 2\pi] - \left\{ \pm \frac{\pi}{2}, \pm \frac{3\pi}{2} \right\}$.

Find: The number of solutions in the given interval.

Rearrange the equation as

Let

So we need the number of intersection points of the graphs of $f(x)$ and $g(x)$ in the given interval.

The function $f(x) = \tan x$ has vertical asymptotes at

and $g(x) = \frac{\pi - 2x}{3}$ is a straight line with slope $-\frac{2}{3}$ and intercept $\frac{\pi}{3}$.

The interval is split into

By sketching the graphs or analyzing their behavior on each interval, we observe a total of $5$ intersection points.

Therefore, the number of solutions is $5$. Hence, the correct option is B.

Given: $2x + 3\tan x = \pi$.

Find: Count the solutions for $x \in [-2\pi, 2\pi] - \left\{ \pm \frac{\pi}{2}, \pm \frac{3\pi}{2} \right\}$.

Since $\tan x$ is undefined at $\pm \frac{\pi}{2}$ and $\pm \frac{3\pi}{2}$, the interval is divided into regions separated by these points.

In each interval of the form

we know that $\tan x$ increases from $-\infty$ to $+\infty$. The equation

can therefore be studied by looking at intersections of the curve with the constant level $\pi$, or equivalently by comparing $\tan x$ with the line $\frac{\pi - 2x}{3}$.

Checking the regions

• $(-2\pi, -\frac{3\pi}{2})$
• $(-\frac{3\pi}{2}, -\frac{\pi}{2})$
• $(-\frac{\pi}{2}, \frac{\pi}{2})$
• $(\frac{\pi}{2}, \frac{3\pi}{2})$
• $(\frac{3\pi}{2}, 2\pi)$ shows that the graphs intersect a total of $5$ times in the required domain.

Thus, the correct answer is $5$, so the correct option is B.