Question

Considering only the principal values of inverse trigonometric functions, the number of positive real values of \( x \) satisfying \[ \tan^{-1}(x) + \tan^{-1}(2x) = \frac{\pi}{4} \] is:

• A. More than 2
• B. 1
• C. 2
• D. 0

Answer 1

The Correct Option is B

To solve the equation \( \tan^{-1}(x) + \tan^{-1}(2x) = \frac{\pi}{4} \), we will use the identity for the sum of two inverse tangents:

\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right), \quad \text{if } ab < 1 \]

Given: \( a = x \) and \( b = 2x \), thus the condition \( ab = 2x^2 < 1 \) must hold true.

Using the identity, substitute the values of \( a \) and \( b \): 

\[ \tan^{-1}(x) + \tan^{-1}(2x) = \tan^{-1}\left(\frac{x + 2x}{1 - 2x^2}\right) = \tan^{-1}\left(\frac{3x}{1 - 2x^2}\right) \]

Given that:

\[ \tan^{-1}\left(\frac{3x}{1 - 2x^2}\right) = \frac{\pi}{4} \]

This implies that:

\[ \frac{3x}{1 - 2x^2} = 1 \]

Solving the equation \( \frac{3x}{1 - 2x^2} = 1 \):

Multiply both sides by \( 1 - 2x^2 \):

\[ 3x = 1 - 2x^2 \]

Rearrange it as a quadratic equation:

\[ 2x^2 + 3x - 1 = 0 \]

To find the roots, use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = 3 \), and \( c = -1 \):

\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]

\[ x = \frac{-3 \pm \sqrt{9 + 8}}{4} \]

\[ x = \frac{-3 \pm \sqrt{17}}{4} \]

Since we are looking for positive real values, consider:

\[ x = \frac{-3 + \sqrt{17}}{4} \]

Checking whether \( ab < 1 \):

Calculate \( 2x^2 \) for \( x = \frac{-3 + \sqrt{17}}{4} \):

\[ 2x^2 = 2 \left(\frac{-3 + \sqrt{17}}{4}\right)^2 \]

After simplification, it can be verified that \( 2x^2 < 1 \). Thus, the condition holds.

Therefore, there is exactly 1 positive real value satisfying the given equation, which makes the answer 1.

Answer 2

Given: \(\tan^{-1} x + \tan^{-1} 2x = \frac{\pi}{4}\), where \(x > 0\).

\(\implies \tan^{-1} 2x = \frac{\pi}{4} - \tan^{-1} x\)

Taking tangent on both sides:

\(\implies 2x = \frac{1 - x}{1 + x}\)

\(\implies 2x(1 + x) = 1 - x\)

\(\implies 2x^2 + 3x - 1 = 0\)

Solving the quadratic equation:

\(x = \frac{-3 \pm \sqrt{9 + 8}}{4}\)

\(x = \frac{-3 \pm \sqrt{17}}{4}\)

Since \(x > 0\), the only possible solution is:

\(x = \frac{-3 + \sqrt{17}}{4}\)

Thus, the number of positive real values of \(x\) is \(1\).

The Correct answer is: 1