If f′′(x)>0∀x∈R,f′(3)=0 and g(x)=ftan2⁡x−2tan⁡x+4,0<x<π2, Then g(x) is increasing in

gI(x)=fITan2⁡x−2Tan⁡x+4(2Tan⁡x−2)Sec2⁡x>0gI(x)>0Tan⁡x−1>0x∈π4,π2

If f′′(x)>0∀x∈R,f′(3)=0 and g(x)=ftan2⁡x−2tan⁡x+4,0<x<π2, Then g(x) is increasing in

gI(x)=fITan2⁡x−2Tan⁡x+4(2Tan⁡x−2)Sec2⁡x>0gI(x)>0Tan⁡x−1>0x∈π4,π2

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If $f^{''} (x) > 0 \forall x \in R, f^{'} (3) = 0 and g (x) = f (\tan^{2} x - 2 \tan x + 4), 0 < x < \frac{π}{2}, Then g (x)$ is increasing in

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$(0, \frac{π}{4})$

$(\frac{π}{6}, \frac{π}{3})$

$(0, \frac{π}{3})$

$(\frac{π}{4}, \frac{π}{2})$

answer is D.

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$\begin{array}{l} g^{I} (x) = f^{I} ({Tan}^{2} x - 2 Tan x + 4) (2 Tan x - 2) {Sec}^{2} x > 0 \\ g^{I} (x) > 0 \\ Tan x - 1 > 0 \\ x \in (\frac{π}{4}, \frac{π}{2}) \end{array}$

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If f′′(x)>0∀x∈R,f′(3)=0 and g(x)=ftan2⁡x−2tan⁡x+4,0<x<π2, Then g(x) is increasing in

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If $f^{''} (x) > 0 \forall x \in R, f^{'} (3) = 0 and g (x) = f (\tan^{2} x - 2 \tan x + 4), 0 < x < \frac{π}{2}, Then g (x)$ is increasing in