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

Question

a) #x in [1, oo)#
b) #x in (-oo, -1]#
c) #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 #arctanx# has a domain of #(-oo, oo)#. Therefore, all we have to look at #arcsin((2x)/(1 + x^2))#

Recall that #arcsinx# 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 + 2x + 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.

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Hopefully this helps!