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

To find the set A satisfying the equation tan^-1(x) + tan^-1(6x) = \(\frac{\pi}{4}\), let's work on solving the equation step by step.
Using the identity tan^-1(a) + tan^-1(b) = tan^-1\(\frac{(a+b)}{(1-ab)}\), we can rewrite the equation as:
tan^-1\(\frac{(x + 6x)}{(1 - x(6x)}\) = \(\frac{\pi}{4}\)
Simplifying the numerator and denominator:
tan^-1\(\frac{7x}{ (1 - 6x^2)}\) = \(\frac{\pi}{4}\)
Next, we can take the tangent of both sides to eliminate the inverse tangent function:
tan(tan^-1\(\frac{7x}{ (1 - 6x^2)}\)) = tan(\(\frac{\pi}{4}\))
Simplifying further:
\(\frac{7x}{ (1 - 6x^2)}\)= 1
Multiplying both sides by (1 - 6x²):
7x = 1 - 6x²
Rearranging the equation:
6x² + 7x - 1 = 0
Now, we can solve this quadratic equation for x. Using the quadratic formula:
x = \(-b\pm\frac{\sqrt{b^2-4ac}}{2a}\)
where a = 6, b = 7, and c = -1, we can substitute these values in:|
x = \(-7\pm\frac{\sqrt{7^2-4\times6\times(-1)}}{2\times6}\)
x = -7 ± \(\frac{\sqrt{73}}{12}\)
Since x must be greater than or equal to 0 according to the set A, we discard the negative solution. Therefore, the set A consists of a single element:
A = -7 ± \(\frac{\sqrt{73}}{12}\)
So, the correct answer is (B) a singleton set.