The number of real solutions of the equation | vedclass

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(C) Let $u = \sqrt{x^2+x}$. The domain requires $x^2+x \ge 0$ and $0 \le x^2+x+1 \le 1$. Since $x^2+x+1 \le 1 \Rightarrow x^2+x \le 0$. Combining $x^2

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$2$

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(C) Let $u = \sqrt{x^2+x}$. The domain requires $x^2+x \ge 0$ and $0 \le x^2+x+1 \le 1$. Since $x^2+x+1 \le 1 \Rightarrow x^2+x \le 0$. Combining $x^2+x \ge 0$ and $x^2+x \le 0$,we must have $x^2+x = 0$. If $x^2+x = 0$,then $u = 0$. The equation becomes $	an^{-1}(0) + \sin^{-1}(1) = 0 + \pi/2 = \pi/2$. This satisfies the equation. $x^2+x = 0 \Rightarrow x(x+1) = 0$,which gives $x = 0$ or $x = -1$. Both values are valid. Thus,there are $2$ real solutions.

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

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