The number of solutions of the equation 2x + 3tan x = π, x ∈ [-2π, 2π] - {-π/2, π/2, 3π/2} is:

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q10 – JEE-MAIN 2025

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Home Maths Trigonometric FunctionsQuestion #492

The number of solutions of the equation

2 x + 3 \tan x = π

x \in [- 2 π, 2 π] - {- \frac{π}{2}, \frac{π}{2}, \frac{3 π}{2}}

is:

Clue 1 of 3: Graphical Analysis

To find the number of solutions, it is often helpful to analyze the intersection points of two functions graphically.

🥷

Ninja Strategy•Count Branches of Tangent

Recognize that a continuous, monotonic function like a straight line will intersect each branch of $\tan x$ exactly once. Count the number of continuous branches of $\tan x$ in the specified domain to find the total number of solutions.

🧠 Full Solution Path

Step 1: Rewrite the Equation and Identify Functions✦ Active

The given equation is $2 x + 3 \tan x = π$ . We can rewrite this as $3 \tan x = π - 2 x$ , or $\tan x = \frac{π - 2 x}{3}$ . Let $f (x) = \tan x$ and $g (x) = \frac{π - 2 x}{3}$ . The number of solutions to the equation is the number of intersection points of the graphs $y = f (x)$ and $y = g (x)$ within the specified domain.

Step 2: Analyze the Linear Function

g (x)

○ Expand

The function $g (x) = - \frac{2}{3} x + \frac{π}{3}$ is a straight line with a negative slope ( $- \frac{2}{3}$ ) and a y-intercept of $\frac{π}{3}$ . We evaluate $g (x)$ at the boundaries of the domain $x \in [- 2 π, 2 π]$ and at the points where $\tan x$ is undefined ( $x = \pm \frac{π}{2}, \pm \frac{3 π}{2}$ ):

g (- 2 π) = - \frac{2}{3} (- 2 π) + \frac{π}{3} = \frac{4 π}{3} + \frac{π}{3} = \frac{5 π}{3} \approx 5.23

g (- \frac{3 π}{2}) = - \frac{2}{3} (- \frac{3 π}{2}) + \frac{π}{3} = π + \frac{π}{3} = \frac{4 π}{3} \approx 4.18

g (- \frac{π}{2}) = - \frac{2}{3} (- \frac{π}{2}) + \frac{π}{3} = \frac{π}{3} + \frac{π}{3} = \frac{2 π}{3} \approx 2.09

g (\frac{π}{2}) = - \frac{2}{3} (\frac{π}{2}) + \frac{π}{3} = - \frac{π}{3} + \frac{π}{3} = 0

g (\frac{3 π}{2}) = - \frac{2}{3} (\frac{3 π}{2}) + \frac{π}{3} = - π + \frac{π}{3} = - \frac{2 π}{3} \approx - 2.09

g (2 π) = - \frac{2}{3} (2 π) + \frac{π}{3} = - \frac{4 π}{3} + \frac{π}{3} = - π \approx - 3.14

Step 3: Count Intersections Graphically○ Expand

The function $y = \tan x$ has vertical asymptotes at $x = \pm \frac{π}{2}, \pm \frac{3 π}{2}$ . The domain excludes these points. We analyze the number of intersections in each continuous interval for $\tan x$ within $[- 2 π, 2 π]$ :

1. **Interval $[- 2 π, - \frac{3 π}{2})$ :** $f (x) = \tan x$ increases from $0$ to $\infty$ . $g (x)$ decreases from $\frac{5 π}{3}$ to $\frac{4 π}{3}$ (both positive). There is **1 solution**.

2. **Interval $(- \frac{3 π}{2}, - \frac{π}{2})$ :** $f (x) = \tan x$ increases from $- \infty$ to $\infty$ . $g (x)$ decreases from $\frac{4 π}{3}$ to $\frac{2 π}{3}$ (both positive). There is **1 solution**.

3. **Interval $(- \frac{π}{2}, \frac{π}{2})$ :** $f (x) = \tan x$ increases from $- \infty$ to $\infty$ . $g (x)$ decreases from $\frac{2 π}{3}$ to $0$ . There is **1 solution**.

4. **Interval $(\frac{π}{2}, \frac{3 π}{2})$ :** $f (x) = \tan x$ increases from $- \infty$ to $\infty$ . $g (x)$ decreases from $0$ to $- \frac{2 π}{3}$ . There is **1 solution**.

5. **Interval $(\frac{3 π}{2}, 2 π]$ :** $f (x) = \tan x$ increases from $- \infty$ to $0$ . $g (x)$ decreases from $- \frac{2 π}{3}$ to $- π$ (both negative). There is **1 solution**.

Summing the solutions from each interval, the total number of solutions is $1 + 1 + 1 + 1 + 1 = 5$ .

💡 Teacher's Secret Hint

Carefully observe the range of both functions in each interval to confirm an intersection.

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Maths Question 10 – JEE-MAIN 2025

Clue 1 of 3: Graphical Analysis

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