Range of f(x) = sin^{-1} (sqrt{x^2 + x + 1}) | vedclass

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(D) Let $g(x) = x^2 + x + 1$. Since $x^2 + x + 1 = (x + 1/2)^2 + 3/4$,the minimum value of $g(x)$ is $3/4$ at $x = -1/2$. As $x 	o \pm \infty$,$g(x)

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$\left[ \frac{\pi}{3}, \frac{\pi}{2} \right]$

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(D) Let $g(x) = x^2 + x + 1$. Since $x^2 + x + 1 = (x + 1/2)^2 + 3/4$,the minimum value of $g(x)$ is $3/4$ at $x = -1/2$. As $x 	o \pm \infty$,$g(x) 	o \infty$. However,for $\sin^{-1}(\sqrt{g(x)})$ to be defined,we must have $0 \le \sqrt{g(x)} \le 1$,which implies $0 \le g(x) \le 1$. Thus,$3/4 \le x^2 + x + 1 \le 1$. Taking the square root,we get $\sqrt{3}/2 \le \sqrt{x^2 + x + 1} \le 1$. Applying the $\sin^{-1}$ function,we get $\sin^{-1}(\sqrt{3}/2) \le \sin^{-1}(\sqrt{x^2 + x + 1}) \le \sin^{-1}(1)$. This results in $\pi/3 \le f(x) \le \pi/2$. Therefore,the range is $\left[ \frac{\pi}{3}, \frac{\pi}{2} ight]$.

Range of $f(x) = \sin^{-1} (\sqrt{x^2 + x + 1})$ is -

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