If the equation sin^{-1} sqrt{x} + cos^{-1} s | vedclass

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(C) For $\sin^{-1} \sqrt{x}$ to be defined,we must have $0 \leq \sqrt{x} \leq 1$,which implies $x \in [0, 1]$.For $\cos^{-1} \sqrt{x^2 - 1}$ to be def

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$3$

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(C) For $\sin^{-1} \sqrt{x}$ to be defined,we must have $0 \leq \sqrt{x} \leq 1$,which implies $x \in [0, 1]$.For $\cos^{-1} \sqrt{x^2 - 1}$ to be defined,we must have $0 \leq \sqrt{x^2 - 1} \leq 1$,which implies $0 \leq x^2 - 1 \leq 1$,so $1 \leq x^2 \leq 2$. Since $x \geq 0$,we have $x \in [1, \sqrt{2}]$.Combining these,the only possible value for $x$ is $x = 1$.Substituting $x = 1$ into the equation,we get $\sin^{-1}(1) + \cos^{-1}(0) + 	an^{-1}(	an y) = a$.This simplifies to $\frac{\pi}{2} + \frac{\pi}{2} + y = a$,where $y$ is any real number such that $	an y$ is defined.Thus,$a = \pi + y$. Since $	an^{-1}(	an y)$ is a periodic function with range $(-\frac{\pi}{2}, \frac{\pi}{2})$,we have $a \in (\pi - \frac{\pi}{2}, \pi + \frac{\pi}{2})$,which is $a \in (\frac{\pi}{2}, \frac{3\pi}{2})$.Using $\pi \approx 3.14$,the interval is $(\frac{3.14}{2}, \frac{3 	imes 3.14}{2}) \approx (1.57, 4.71)$.The integral values of $a$ in this interval are $2, 3, 4$.Therefore,there are $3$ integral values.

If the equation $\sin^{-1} \sqrt{x} + \cos^{-1} \sqrt{x^2 - 1} + \tan^{-1} (\tan y) = a$ has at least one solution,then the number of integral values of $a$ is:

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