# 2tan^-1(x)+sin^-1((2x)/(1+x^2)) is independent of x, then?

## a) $x \in \left[1 , \infty\right)$ b) $x \in \left(- \infty , - 1\right]$ c) $x \in \left[- 1 , 1\right]$ d) None of these

May 27, 2018

We start by noticing that the function $\arctan x$ has a domain of $\left(- \infty , \infty\right)$. Therefore, all we have to look at $\arcsin \left(\frac{2 x}{1 + {x}^{2}}\right)$

Recall that $\arcsin x$ is only defined on -1 ≤ x ≤ 1

So we have two inequalities we must solve, which are

(2x)/(1 + x^2) ≥ -1

AND

(2x)/(1 + x^2) ≤ 1

Let's solve!

2x ≥ -x^2 - 1

x^2 + 2x + 1 ≥ 0

This is true on all real numbers as ${x}^{2} + 2 x + 1$ is a parabola which opens upwards and whose minimum occurs at $y = 0$.

As for the second, we have:

2x ≤ x^2 + 1

0 ≤ x^2 - 2x + 1

0 ≤ (x -1)^2

This also has a solution of all real numbers since it's a parabola which also opens upwards and has it's minimum on the x-axis.

Therefore the answer is $d$, or none of these. We can confirm graphically that the domain is all real numbers. Hopefully this helps!