Question

The number of solutions of \[ \tan^{-1}(4x) + \tan^{-1}(6x) = \frac{\pi}{6}, \] where \[ -\frac{1}{2\sqrt{6}}<x<\frac{1}{2\sqrt{6}}, \] is equal to

• A. \(1\)
• B. \(2\)
• C. \(0\)
• D. \(3\)

Hint:

Always verify the condition \(ab<1\) before applying the inverse tangent addition formula, and check final solutions within the given interval.

Answer 1

The Correct Option is A

We are given \[ \tan^{-1}(4x) + \tan^{-1}(6x) = \frac{\pi}{6}. \]
Step 1: Check applicability of the inverse tangent addition formula.
The identity \[ \tan^{-1} a + \tan^{-1} b = \tan^{-1}\!\left(\frac{a+b}{1-ab}\right) \] is valid when \( ab<1 \). Here, \[ ab = (4x)(6x) = 24x^2. \] Given \[ -\frac{1}{2\sqrt{6}}<x<\frac{1}{2\sqrt{6}}, \] we have \[ 24x^2<1. \] Hence, the formula is applicable.
Step 2: Apply the identity.
Using the formula, \[ \tan^{-1}(4x) + \tan^{-1}(6x) = \tan^{-1}\!\left(\frac{4x+6x}{1-24x^2}\right) = \tan^{-1}\!\left(\frac{10x}{1-24x^2}\right). \] Thus, \[ \tan^{-1}\!\left(\frac{10x}{1-24x^2}\right) = \frac{\pi}{6}. \]
Step 3: Take tangent on both sides.
\[ \frac{10x}{1-24x^2} = \tan\!\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}. \] Multiplying both sides by \( \sqrt{3}(1-24x^2) \), \[ 10\sqrt{3}x = 1 - 24x^2. \] Rearranging, \[ 24x^2 + 10\sqrt{3}x - 1 = 0. \]
Step 4: Solve the quadratic equation.
The discriminant is \[ D = (10\sqrt{3})^2 + 96 = 300 + 96 = 396 = 36 \times 11. \] Hence, \[ x = \frac{-10\sqrt{3} \pm 6\sqrt{11}}{48}. \]
Step 5: Check solutions in the given interval.
On checking both roots against \[ -\frac{1}{2\sqrt{6}}<x<\frac{1}{2\sqrt{6}}, \] we find that **only one root** satisfies the given condition.
Therefore, the number of solutions is \(1\).
Final Answer: \[ \boxed{1} \]