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

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.

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

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

$g^{'} (x) = (f^{'} ((\tan x - 1)^{2} + 3)) 2 (\tan x - 1) \sec^{2} x$
Since $f^{''} (x) > 0, f^{'} (x)$ is increasing. So,
$∴ f^{'} ((\tan x - 1)^{2} + 3) > f^{'} (3) = 0 \forall x \in (0, \frac{π}{4}) \cup (\frac{π}{4}, \frac{π}{2})$
Also, $(\tan x - 1) > 0 x \in (\frac{π}{4}, \frac{π}{2})$
So, $g (x) is increasing in (\frac{π}{4}, \frac{π}{2})$ .