If f''(x) is a positive function for all x in | vedclass

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(D) Given that $f''(x) > 0$ for all $x \in R$,it implies that $f'(x)$ is a strictly increasing function for all $x \in R$.Since $f'(3) = 0$,we have $f

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$(\frac{\pi}{4}, \frac{\pi}{2})$

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(D) Given that $f''(x) > 0$ for all $x \in R$,it implies that $f'(x)$ is a strictly increasing function for all $x \in R$.Since $f'(3) = 0$,we have $f'(x)  0$ for $x > 3$.Now,consider $g(x) = f(	an^2 x - 2 	an x + 4)$.Differentiating with respect to $x$,we get $g'(x) = f'(	an^2 x - 2 	an x + 4) \cdot \frac{d}{dx}(	an^2 x - 2 	an x + 4)$.$g'(x) = f'(	an^2 x - 2 	an x + 4) \cdot (2 	an x \sec^2 x - 2 \sec^2 x) = f'(	an^2 x - 2 	an x + 4) \cdot 2 \sec^2 x (	an x - 1)$.Let $u = 	an^2 x - 2 	an x + 4 = (	an x - 1)^2 + 3$. Since $(	an x - 1)^2 \ge 0$,we have $u \ge 3$.For $u > 3$,we know $f'(u) > 0$ because $f'(x)$ is increasing and $f'(3) = 0$.For $g(x)$ to be increasing,we require $g'(x) > 0$.Since $f'(u) > 0$ and $2 \sec^2 x > 0$ for $x \in (0, \frac{\pi}{2})$,the sign of $g'(x)$ depends on $(	an x - 1)$.Thus,$g'(x) > 0$ when $	an x - 1 > 0$,which means $	an x > 1$.For $x \in (0, \frac{\pi}{2})$,$	an x > 1$ implies $x \in (\frac{\pi}{4}, \frac{\pi}{2})$.

If $f''(x)$ is a positive function for all $x \in R$,$f'(3) = 0$ and $g(x) = f(\tan^2 x - 2 \tan x + 4)$ for $0 < x < \frac{\pi}{2}$,then the interval in which $g(x)$ is increasing is

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