Question

The number of values of \( x \) satisfying \( \tan^{-1}(4x) + \tan^{-1}(6x) = \frac{\pi}{6} \) and \( x<\left[ \frac{-1}{2\sqrt{6}} , \frac{1}{2\sqrt{6}} \right] \) is:

• A. 1
• B. 0
• C. 2
• D. 3

Hint:

When solving trigonometric equations involving arctangents, use the addition formula and carefully solve the resulting equation.

Answer 1

The Correct Option is A

To solve the equation \(\tan^{-1}(4x) + \tan^{-1}(6x) = \frac{\pi}{6}\), we will make use of the identity for the sum of two inverse tangents:

\(\tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1} \left( \frac{a+b}{1-ab} \right)\), provided \(ab < 1\). 

Applying this identity to the given equation where \(a = 4x\) and \(b = 6x\), we have:

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

The equation now becomes:

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

Which implies:

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

Since \(\tan \left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}\), the equation becomes:

\(\frac{10x}{1 - 24x^2} = \frac{1}{\sqrt{3}}\)

Cross-multiplying gives us:

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

Rearranging terms, we get a quadratic equation:

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

Let’s solve this quadratic equation using the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a=24\), \(b=10\sqrt{3}\), and \(c=-1\).

First, calculate the discriminant:

\(b^2 - 4ac = (10\sqrt{3})^2 - 4 \cdot 24 \cdot (-1)\)

\(= 300 + 96 = 396\)

Since the discriminant is positive, there are two real solutions:

\(x = \frac{-10\sqrt{3} \pm \sqrt{396}}{48}\)

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

Considering that \(x\) is bounded by \(\left[ \frac{-1}{2\sqrt{6}}, \frac{1}{2\sqrt{6}} \right]\). Let's check if any solutions lie within this interval.

The value \(\frac{-1}{2\sqrt{6}} \approx -0.204\) and \(\frac{1}{2\sqrt{6}} \approx 0.204\).

Evaluate the solutions for \(x\):

• \(x_1 = \frac{-10\sqrt{3} + 6\sqrt{11}}{48}\)
• \(x_2 = \frac{-10\sqrt{3} - 6\sqrt{11}}{48}\)

After calculations, it is found that only one solution \(x_1\) lies within the interval \(\left[ \frac{-1}{2\sqrt{6}}, \frac{1}{2\sqrt{6}} \right]\).

Therefore, the number of values of \(x\) that satisfy the given condition is 1.