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Question: \(2\tan^{- 1}x + \sin^{- 1}\frac{2x}{1 + x^{2}}\) is independent of x , then...

2tan⁡−1x+sin⁡−12x1+x22\tan^{- 1}x + \sin^{- 1}\frac{2x}{1 + x^{2}} is independent of x , then

A

x∈[1,+∞)x \in \lbrack 1, + \infty)

B

x∈[−1,1]x \in \lbrack - 1,1\rbrack

C

x∈(−∞,−1x \in ( - \infty, - 1]

D

None of these

Answer

x∈[1,+∞)x \in \lbrack 1, + \infty)

Explanation

Solution

Let x=tan⁡θx = \tan\theta. Then

sin⁡−12x1+x2=sin⁡−12tan⁡θ1+tan⁡2θ=sin⁡−1(sin⁡2θ)\sin^{- 1}\frac{2x}{1 + x^{2}} = \sin^{- 1}\frac{2\tan\theta}{1 + \tan^{2}\theta} = \sin^{- 1}(\sin 2\theta)

∴2tan⁡−1x+sin⁡−12x1+x2=2θ+sin⁡−1(sin⁡2θ)\therefore 2\tan^{- 1}x + \sin^{- 1}\frac{2x}{1 + x^{2}} = 2\theta + \sin^{- 1}(\sin 2\theta) If −π2≤2θ≤π2,2tan⁡−1x+sin⁡−12x1+x2- \frac{\pi}{2} \leq 2\theta \leq \frac{\pi}{2},2\tan^{- 1}x + \sin^{- 1}\frac{2x}{1 + x^{2}}

= 2θ+2θ=4tan⁡−1x≠2\theta + 2\theta = 4\tan^{- 1}x \neqindependent of x.

If −π2≤π−2θ≤π2,2tan⁡−1x+sin⁡−12x1+x2- \frac{\pi}{2} \leq \pi - 2\theta \leq \frac{\pi}{2},2\tan^{- 1}x + \sin^{- 1}\frac{2x}{1 + x^{2}}

= 2θ+sin⁡−1[sin⁡(π−2θ)]=2θ+π−2θ2\theta + \sin^{- 1}\lbrack\sin(\pi - 2\theta)\rbrack = 2\theta + \pi - 2\theta= π\pi = independent of x.

∴\therefore θ∉[−π4,π4]\theta \notin \left\lbrack - \frac{\pi}{4},\frac{\pi}{4} \right\rbrack butθ∈[π4,3π4]\theta \in \left\lbrack \frac{\pi}{4},\frac{3\pi}{4} \right\rbrack

and from the principal value of tan⁡−1x\tan^{- 1}x.

θ∈(−π2,π2)\theta \in \left( - \frac{\pi}{2},\frac{\pi}{2} \right). Hence, θ∈(π4,π2)\theta \in \left( \frac{\pi}{4},\frac{\pi}{2} \right)

∴\therefore θ∈(π4,π2)\theta \in \left( \frac{\pi}{4},\frac{\pi}{2} \right) ⇒\Rightarrow tan⁡−1x+sin⁡−12x1+x2=π\tan^{- 1}x + \sin^{- 1}\frac{2x}{1 + x^{2}} = \pi.

Also at

θ=π4,2tan⁡−1x+sin⁡−12x1+x2=2.π4+sin⁡−11=π2+π2=π\theta = \frac{\pi}{4},2\tan^{- 1}x + \sin^{- 1}\frac{2x}{1 + x^{2}} = 2.\frac{\pi}{4} + \sin^{- 1}1 = \frac{\pi}{2} + \frac{\pi}{2} = \pi.

∴\therefore The given function = π\pi = constant if θ∈[π4,π2)\theta \in \left\lbrack \frac{\pi}{4},\frac{\pi}{2} \right). i.e., x∈[1,+∞)x \in \lbrack 1, + \infty)