Question

If $0 \le x \le \frac{3}{4}$, then the number of values of $x$ satisfying the equation $\text{Tan}^{-1}(2x-1) + \text{Tan}^{-1}2x = \text{Tan}^{-1}4x - \text{Tan}^{-1}(2x+1)$ is

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

When solving equations with inverse trig functions, always be mindful of the domains of the functions and the conditions under which the sum/difference formulas are valid (e.g., $AB<1$). If an algebraic solution seems straightforward but leads to a different number of solutions than expected, carefully re-examine these conditions for different intervals of the variable.

Answer 1

The Correct Option is C

Step 1: Rearrange the Equation: \[ \tan^{-1}(2x-1) + \tan^{-1}(2x+1) = \tan^{-1}(4x) - \tan^{-1}(2x) \]
Step 2: Apply Inverse Tangent Formula: LHS: \( \tan^{-1} A + \tan^{-1} B = \tan^{-1} \frac{A+B}{1-AB} \) (Check \( AB<1 \)) \[ \text{LHS} = \tan^{-1} \frac{(2x-1)+(2x+1)}{1 - (2x-1)(2x+1)} = \tan^{-1} \frac{4x}{1 - (4x^2-1)} = \tan^{-1} \frac{4x}{2-4x^2} = \tan^{-1} \frac{2x}{1-2x^2} \] RHS: \( \tan^{-1} A - \tan^{-1} B = \tan^{-1} \frac{A-B}{1+AB} \) \[ \text{RHS} = \tan^{-1} \frac{4x-2x}{1 + (4x)(2x)} = \tan^{-1} \frac{2x}{1+8x^2} \]
Step 3: Equate Arguments: \[ \frac{2x}{1-2x^2} = \frac{2x}{1+8x^2} \] \[ 2x \left( \frac{1}{1-2x^2} - \frac{1}{1+8x^2} \right) = 0 \] This gives two possibilities: 1. \( 2x = 0 \implies x = 0 \). 2. \( 1-2x^2 = 1+8x^2 \implies 10x^2 = 0 \implies x = 0 \).
Step 4: Check Domain: The given domain is \( 0 \le x<3/4 \). \( x=0 \) lies in this interval. Also check if denominator \( 1-2x^2 \) is zero for LHS validity. At \( x=0 \), denom is 1. Valid. Thus, there is exactly 1 solution: \( x=0 \).