If f′′(x)>0∀x∈R,f′(3)=0, and g(x)=ftan2x−2 tanx+4,0<x<π2, then g(x) is increasing in
0,π4
π6,π3
0,π3
π4,π2
g′(x)=f′(tanx−1)2+32(tanx−1)sec2xSince 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.