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Question: The number of real solutions of \(tan^{- 1}\sqrt{x(x + 1)} + sin^{- 1}\sqrt{x^{2} + x + 1} = \frac{...

The number of real solutions of

tan−1x(x+1)+sin−1x2+x+1=π2tan^{- 1}\sqrt{x(x + 1)} + sin^{- 1}\sqrt{x^{2} + x + 1} = \frac{\pi}{2} is

A

Zero

B

One

C

Two

D

Infinite

Answer

Two

Explanation

Solution

The given equation holds if

tan⁡−1x(x+1)=cos⁡−1x2+x+1\tan^{- 1}\sqrt{x(x + 1)} = \cos^{- 1}\sqrt{x^{2} + x + 1}

⇒cos⁡−11x2+x+1=cos⁡−1x2+x+1\cos^{- 1}\frac{1}{\sqrt{x^{2} + x + 1}} = \cos^{- 1}\sqrt{x^{2} + x + 1}

⇒ x2 + x + 1 = 1 or x(x + 1) = 0 ⇒ x = 0, –1