JEE Mathematics 2026 Question with Solution

Given: $\tan^{-1}(4x) + \tan^{-1}(6x) = \frac{\pi}{6}$ with $-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$.

Find: The number of solutions in the given interval.

First verify the condition for applying the inverse tangent addition formula.

$$\tan^{-1} a + \tan^{-1} b = \tan^{-1}\!\left(\frac{a+b}{1-ab}\right)$$

This is valid here because

$$ab = (4x)(6x) = 24x^2$$

and from $-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$, we get

$$24x^2 < 1$$

So the formula is applicable.

Now apply the identity:

$$\tan^{-1}(4x) + \tan^{-1}(6x) = \tan^{-1}\!\left(\frac{4x+6x}{1-24x^2}\right) = \tan^{-1}\!\left(\frac{10x}{1-24x^2}\right)$$

Hence,

$$\tan^{-1}\!\left(\frac{10x}{1-24x^2}\right) = \frac{\pi}{6}$$

Taking tangent on both sides,

$$\frac{10x}{1-24x^2} = \tan\!\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$$

Multiplying through,

$$10\sqrt{3}x = 1 - 24x^2$$

So,

$$24x^2 + 10\sqrt{3}x - 1 = 0$$

Now solve the quadratic. Its discriminant is

$$D = (10\sqrt{3})^2 + 96 = 300 + 96 = 396 = 36 \times 11$$

Therefore,

$$x = \frac{-10\sqrt{3} \pm 6\sqrt{11}}{48}$$

On checking both roots in the interval $-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$, only one root satisfies the condition.

Therefore, the number of solutions is $1$. The correct option is A.

Given: The roots are

$$x = \frac{-10\sqrt{3} \pm 6\sqrt{11}}{48}$$

with interval

$$-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$$

Find: How many of these roots lie inside the interval.

After forming the quadratic

$$24x^2 + 10\sqrt{3}x - 1 = 0$$

we obtain two roots. The solution working states that after checking both against the interval, only one of them lies in the admissible range.

Hence exactly one value of $x$ satisfies both the equation and the given interval restriction.

Therefore, the number of solutions is $1$.