$2 {tan}^{- 1} (x) + {sin}^{- 1} (\frac{2 x}{1 + x^{2}})$ $2tan^-1(x)+sin^-1((2x)/(1+x^2))$ is independent of $x$ $x$ , then?

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$2 {tan}^{- 1} (x) + {sin}^{- 1} (\frac{2 x}{1 + x^{2}})$ $2tan^-1(x)+sin^-1((2x)/(1+x^2))$ is independent of $x$ $x$ , then?

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

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Answer 1 · 2018-05-27T03:45:39

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

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

So we have two inequalities we must solve, which are

$\frac{2 x}{1 + x^{2}} \geq - 1$ $(2x)/(1 + x^2) ≥ -1$

AND

$\frac{2 x}{1 + x^{2}} \leq 1$ $(2x)/(1 + x^2) ≤ 1$

Let's solve!

$2 x \geq - x^{2} - 1$ $2x ≥ -x^2 - 1$

$x^{2} + 2 x + 1 \geq 0$ $x^2 + 2x + 1 ≥ 0$

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

As for the second, we have:

$2 x \leq x^{2} + 1$ $2x ≤ x^2 + 1$

$0 \leq x^{2} - 2 x + 1$ $0 ≤ x^2 - 2x + 1$

$0 \leq {(x - 1)}^{2}$ $0 ≤ (x -1)^2$

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 $d$ $d$ , or none of these. We can confirm graphically that the domain is all real numbers.

enter image source here

Hopefully this helps!

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$2 {tan}^{- 1} (x) + {sin}^{- 1} (\frac{2 x}{1 + x^{2}})$ $2tan^-1(x)+sin^-1((2x)/(1+x^2))$ is independent of $x$ $x$ , then?

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

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2tan^-1(x)+sin^-1((2x)/(1+x^2))2tan−1(x)+sin−1(2x1+x2)2tan^-1(x)+sin^-1((2x)/(1+x^2)) is independent of xxx, then?

a) x in [1, oo)x∈[1,∞)x in [1, oo) b) x in (-oo, -1]x∈(−∞,−1]x in (-oo, -1] c) x in [-1,1]x∈[−1,1]x in [-1,1] d) None of these

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Impact of this question

$2 {tan}^{- 1} (x) + {sin}^{- 1} (\frac{2 x}{1 + x^{2}})$ $2tan^-1(x)+sin^-1((2x)/(1+x^2))$ is independent of $x$ $x$ , then?

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