The number of real solutions of tan ^{-1} sqr | vedclass

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(C) The given equation is $	an ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2}$.For $	an ^{-1} \sqrt{x(x+1)}$ to be defined,we must have

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two.

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(C) The given equation is $	an ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2}$.For $	an ^{-1} \sqrt{x(x+1)}$ to be defined,we must have $x(x+1) \geq 0$.For $\sin ^{-1} \sqrt{x^2+x+1}$ to be defined,the argument must satisfy $0 \leq \sqrt{x^2+x+1} \leq 1$,which implies $0 \leq x^2+x+1 \leq 1$.The inequality $x^2+x+1 \leq 1$ simplifies to $x^2+x \leq 0$,or $x(x+1) \leq 0$.Combining $x(x+1) \geq 0$ and $x(x+1) \leq 0$,we must have $x(x+1) = 0$.This gives $x = 0$ or $x = -1$.If $x = 0$,the equation becomes $	an ^{-1} \sqrt{0} + \sin ^{-1} \sqrt{1} = 0 + \frac{\pi}{2} = \frac{\pi}{2}$.If $x = -1$,the equation becomes $	an ^{-1} \sqrt{0} + \sin ^{-1} \sqrt{1} = 0 + \frac{\pi}{2} = \frac{\pi}{2}$.Both values satisfy the equation,so there are $2$ real solutions.

The number of real solutions of $\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2}$ is

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