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: Understanding the Concept

This equation involves the inverse tangent function, \( \tan^{-1}x \). The goal is to simplify both sides using addition and subtraction formulas for inverse tangent and then compare their arguments to determine the possible values of \(x\).

Step 2: Key Formula or Approach

The given equation is:

\[ \tan^{-1}(2x - 1) + \tan^{-1}(2x + 1) = \tan^{-1}(4x) - \tan^{-1}(2x) \]

We use the formulas:

\[ \tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A + B}{1 - AB}\right), \quad \tan^{-1}A - \tan^{-1}B = \tan^{-1}\left(\frac{A - B}{1 + AB}\right) \]

These are valid when the corresponding products of \(A\) and \(B\) satisfy \(AB < 1\).

Step 3: Detailed Explanation

Applying the addition formula to the left-hand side (LHS):

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

Applying the subtraction formula to the right-hand side (RHS):

\[ \tan^{-1}(4x) - \tan^{-1}(2x) = \tan^{-1}\left(\frac{2x}{1 + 8x^2}\right) \]

Equating the arguments:

\[ \frac{4x}{2 - 4x^2} = \frac{2x}{1 + 8x^2} \]

Simplify:

\[ 4x(1 + 8x^2) = 2x(2 - 4x^2) \] \[ 4x + 32x^3 = 4x - 8x^3 \] \[ 40x^3 = 0 \quad \Rightarrow \quad x = 0 \]

Checking \(x = 0\) in the original equation:

\[ \tan^{-1}(-1) + \tan^{-1}(1) = \tan^{-1}(0) - \tan^{-1}(0) \] \[ -\frac{\pi}{4} + \frac{\pi}{4} = 0 \]

Hence, \(x = 0\) satisfies the equation. Testing other values such as \(x = \tfrac{1}{2}\) and \(x = \tfrac{3}{4}\) shows that they do not satisfy the equation, confirming this is the only valid solution.

Step 4: Final Answer

\[ \boxed{x = 0} \]