Question

The real values of \( x \) that satisfy the equation \[ \tan^{-1}x + \tan^{-1}2x = \frac{\pi}{4} \] is:

• A. \( \frac{-3 \pm \sqrt{17}}{4} \)
• B. \( \frac{-1 \pm \sqrt{3}}{2} \)
• C. \( \sqrt{3} - 1 \)
• D. \( \frac{\sqrt{17} - 3}{4} \)

Hint:

For equations involving the inverse trigonometric functions, use the sum identity for arctangents and solve the resulting algebraic equation.

Answer 1

The Correct Option is D

We are given the equation: \[ \tan^{-1}x + \tan^{-1}2x = \frac{\pi}{4}. \] Step 1: Use the identity for the sum of arctangents The identity for the sum of arctangents is: \[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right), \quad \text{for} \quad ab < 1. \] Using this identity, we can write: \[ \tan^{-1} x + \tan^{-1} 2x = \tan^{-1} \left( \frac{x + 2x}{1 - x \cdot 2x} \right) = \tan^{-1} \left( \frac{3x}{1 - 2x^2} \right). \] Step 2: Set the equation equal to \( \frac{\pi}{4} \) We know that \( \tan \frac{\pi}{4} = 1 \), so we set the argument of the arctangent equal to 1: \[ \frac{3x}{1 - 2x^2} = 1. \] Multiply both sides by \( 1 - 2x^2 \): \[ 3x = 1 - 2x^2. \] Step 3: Solve the quadratic equation Rearrange the terms: \[ 2x^2 + 3x - 1 = 0. \] Solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \] where \( a = 2, b = 3, c = -1 \). Substituting these values: \[ x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-1)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 8}}{4} = \frac{-3 \pm \sqrt{17}}{4}. \] Thus, the real values of \( x \) are \( \frac{\sqrt{17} - 3}{4} \).