First slide
Monotonicity
Question

If f′′(x)>0xR,f(3)=0, and g(x)=ftan2x2 tanx+4,0<x<π2, then g(x) is increasing in

Moderate
Solution

g(x)=f(tanx1)2+32(tanx1)sec2x
Since f′′(x)>0,f(x) is increasing. So,
 f(tan x1)2+3>f(3)=0x0,π4π4,π2
Also, (tan x1)>0xπ4,π2
So, g(x) is increasing in π4,π2.
 

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