The number of real solutions of tan^{-1}sqrt{ | 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 mus

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$2$

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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 $x(x + 1) \ge 0$.For $\sin^{-1}\sqrt{x^2 + x + 1}$ to be defined,the argument must satisfy $0 \le \sqrt{x^2 + x + 1} \le 1$,which implies $0 \le x^2 + x + 1 \le 1$.Since $x^2 + x + 1 = (x + \frac{1}{2})^2 + \frac{3}{4}$,the minimum value of $x^2 + x + 1$ is $\frac{3}{4}$.Thus,the condition $0 \le x^2 + x + 1 \le 1$ simplifies to $\frac{3}{4} \le x^2 + x + 1 \le 1$,which implies $x^2 + x \le 0$.Combining $x(x + 1) \ge 0$ and $x(x + 1) \le 0$,we must have $x(x + 1) = 0$.This gives $x = 0$ or $x = -1$.Checking $x = 0$: $	an^{-1}(0) + \sin^{-1}(1) = 0 + \frac{\pi}{2} = \frac{\pi}{2}$. This is a solution.Checking $x = -1$: $	an^{-1}(0) + \sin^{-1}(1) = 0 + \frac{\pi}{2} = \frac{\pi}{2}$. This is a solution.Therefore,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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