If x is a real number,then the number of solu | vedclass

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(C) Let $f(x) = \operatorname{Tan}^{-1}(\sqrt{x(x+1)}) + \operatorname{Sin}^{-1}(\sqrt{x^2+x+1})$.For the expression to be defined,the arguments must

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$2$

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(C) Let $f(x) = \operatorname{Tan}^{-1}(\sqrt{x(x+1)}) + \operatorname{Sin}^{-1}(\sqrt{x^2+x+1})$.For the expression to be defined,the arguments must satisfy the domain conditions:$1$. $x(x+1) \ge 0 \implies x \in (-\infty, -1] \cup [0, \infty)$.$2$. $0 \le x^2+x+1 \le 1$.Since $x^2+x+1 = (x+1/2)^2 + 3/4$,the minimum value is $3/4$. Thus,$x^2+x+1 \le 1 \implies x^2+x \le 0 \implies x(x+1) \le 0 \implies x \in [-1, 0]$.Combining the conditions from $(1)$ and $(2)$,we get $x \in \{-1, 0\}$.Case $1$: If $x = 0$,then $\operatorname{Tan}^{-1}(0) + \operatorname{Sin}^{-1}(1) = 0 + \pi/2 = \pi/2$. This is a solution.Case $2$: If $x = -1$,then $\operatorname{Tan}^{-1}(0) + \operatorname{Sin}^{-1}(1) = 0 + \pi/2 = \pi/2$. This is a solution.Thus,there are $2$ solutions.

If $x$ is a real number,then the number of solutions of $\operatorname{Tan}^{-1}(\sqrt{x(x+1)})+\operatorname{Sin}^{-1}(\sqrt{x^2+x+1})=\frac{\pi}{2}$ is:

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