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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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Correct option is D

g′(x)=f′(tan⁡x−1)2+32(tan⁡x−1)sec2⁡xSince f′′(x)>0,f′(x) is increasing. So,∴ f′(tan ⁡x−1)2+3>f′(3)=0∀x∈0,π4∪π4,π2Also, (tan x−1)>0x∈π4,π2So, g(x) is increasing in π4,π2.

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If f′′(x)>0∀x∈R,f′(3)=0, and g(x)=ftan2⁡x−2 tan⁡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