The range of the real valued function f(x) = | vedclass

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(B) Let $g(x) = 9x^2 - 12x + 22$. We can rewrite this as $g(x) = (3x - 2)^2 + 18$. The minimum value of $g(x)$ is $18$ when $3x - 2 = 0$,i.e.,$x = \fr

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$\left[\frac{\pi}{4}, \frac{\pi}{2}\right)$

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(B) Let $g(x) = 9x^2 - 12x + 22$. We can rewrite this as $g(x) = (3x - 2)^2 + 18$. The minimum value of $g(x)$ is $18$ when $3x - 2 = 0$,i.e.,$x = \frac{2}{3}$. As $x 	o \pm \infty$,$g(x) 	o \infty$. Thus,the range of $g(x)$ is $[18, \infty)$. Consequently,the range of $\sqrt{g(x)}$ is $[\sqrt{18}, \infty) = [3\sqrt{2}, \infty)$. Now,consider the argument of $\operatorname{Cos}^{-1}$,which is $u = \frac{3}{\sqrt{g(x)}}$. Since $\sqrt{g(x)} \geq 3\sqrt{2}$,we have $0 So,$u \in (0, \frac{1}{\sqrt{2}}]$. Since $\operatorname{Cos}^{-1}(u)$ is a decreasing function,the range of $f(x) = \operatorname{Cos}^{-1}(u)$ is $[\operatorname{Cos}^{-1}(\frac{1}{\sqrt{2}}), \operatorname{Cos}^{-1}(0))$. This simplifies to $[\frac{\pi}{4}, \frac{\pi}{2})$. Therefore,the correct option is $B$.

The range of the real valued function $f(x) = \operatorname{Cos}^{-1}\left(\frac{3}{\sqrt{9x^2 - 12x + 22}}\right)$ is

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