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

Answer: (B)

a singleton set

Answer 2

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.