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

The equation to be solved is \( \tan^{-1}(x) + \tan^{-1}(2x) = \frac{\pi}{4} \). We apply the sum of inverse tangents identity:

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

With \( a = x \) and \( b = 2x \), the condition \( ab = 2x^2 < 1 \) must be satisfied.

Substituting \( a \) and \( b \) into the identity yields:

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

Equating this to the given value:

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

This implies:

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

Solving for \( x \):

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

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

Rearranging into a quadratic equation:

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

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 2 \), \( b = 3 \), and \( c = -1 \):

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

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

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

For positive real values, we select:

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

Verification of the condition \( ab < 1 \):

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

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

This value is confirmed to be less than 1, satisfying the condition.

Thus, there is one positive real solution to the equation, resulting in an answer of 1.