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

To find the number of solutions for the given equation \(\tan^{-1}(4x) + \tan^{-1}(6x) = \frac{\pi}{6}\) within the interval \(-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}\), we will use properties of inverse trigonometric functions.

Using the formula for the sum of two inverse tangents:

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

Applying this to the problem, we have:

\(a = 4x\) and \(b = 6x\). Thus,

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

The given equation can now be rewritten as:

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

This implies:

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

Since \(\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}\)

We have:

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

Cross-multiply to find:

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

Simplifying further, we get:

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

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

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 24\), \(b = 10\sqrt{3}\), and \(c = -1\), we solve for \(x\).

First, calculate the discriminant:

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

\(= 300 + 96 = 396\)

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

\(= \frac{-10\sqrt{3} \pm \sqrt{4 \cdot 99}}{48} = \frac{-10\sqrt{3} \pm 2 \cdot \sqrt{99}}{48}\)

\(x = \frac{-5\sqrt{3} \pm \sqrt{99}}{24}\)

Now, simplify \(\sqrt{99}\):

\(\sqrt{99} = 3\sqrt{11}\)

Therefore:

\(x = \frac{-5\sqrt{3} \pm 3\sqrt{11}}{24}\)

Checking these roots within the given interval \(-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}\), we find that only one of these values, \(x = \frac{-5\sqrt{3} + 3\sqrt{11}}{24}\), satisfies the interval, while the other one does not.

Therefore, the number of solutions is 1.