Question

Considering only the principal value of an inverse function, the set: A= { x ≥ 0, tan^-1x + tan^-16x = \(\frac{\pi}{4}\)}, then A is... 
 

• A. an empty set
• B. a singleton set
• C. consists of two elements
• D. contains more than two elements

Answer 1

The Correct Option is B

We are tasked with finding the set \( A \) that satisfies the equation: \[ \tan^{-1}(x) + \tan^{-1}(6x) = \frac{\pi}{4} \] Let's solve this step by step.

Step 1: Apply the sum identity for inverse tangents:
Using the identity: \[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left( \frac{a + b}{1 - ab} \right) \] we can rewrite the equation as: \[ \tan^{-1}\left( \frac{x + 6x}{1 - x(6x)} \right) = \frac{\pi}{4} \]

Step 2: Simplify the numerator and denominator:
Simplifying the expression inside the inverse tangent: \[ \tan^{-1}\left( \frac{7x}{1 - 6x^2} \right) = \frac{\pi}{4} \]

Step 3: Eliminate the inverse tangent:
We take the tangent of both sides of the equation to eliminate the inverse tangent function: \[ \tan\left( \tan^{-1}\left( \frac{7x}{1 - 6x^2} \right) \right) = \tan\left( \frac{\pi}{4} \right) \] Since \( \tan\left( \frac{\pi}{4} \right) = 1 \), we get: \[ \frac{7x}{1 - 6x^2} = 1 \]

Step 4: Solve for \( x \):
Multiply both sides of the equation by \( 1 - 6x^2 \): \[ 7x = 1 - 6x^2 \] Rearranging the equation: \[ 6x^2 + 7x - 1 = 0 \] This is a quadratic equation in terms of \( x \).

Step 5: Solve the quadratic equation:
We can solve the quadratic equation \( 6x^2 + 7x - 1 = 0 \) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 6 \), \( b = 7 \), and \( c = -1 \). Substituting these values into the formula: \[ x = \frac{-7 \pm \sqrt{7^2 - 4(6)(-1)}}{2(6)} \] Simplifying: \[ x = \frac{-7 \pm \sqrt{49 + 24}}{12} = \frac{-7 \pm \sqrt{73}}{12} \]

Step 6: Determine the valid solution:
Since \( x \) must be greater than or equal to 0 according to the set \( A \), we discard the negative solution. Thus, we take the positive root: \[ x = \frac{-7 + \sqrt{73}}{12} \] Therefore, the set \( A \) consists of the single element: \[ A = \left\{ \frac{-7 + \sqrt{73}}{12} \right\} \]

Final Answer:
The correct answer is (B) a singleton set.