Let f(x) be a function such that,f(0)=f′(0)=0 ,f′′(x)=sec4 x+4 then the function is
log |(sin x)|+13tan3 x+x
23log |(sec x)|+16tan2 x+2x2
log |(cos x)|+16cos2 x+x5
None of the above
Since,f′′(x)=sec4 x+4
f′′(x)=1+tan2 xsec2 x+4
⇒ f′(x)=tan x+tan3 x3+4x+C⇒f′(0)=0⇒0=C
Then, f′(x)=tan x+tan3 x3+4x
f′(x)=tan x+tan3 x3+4x=tan x+13tan xsec2x−1+4x⇒ f′(x)=23tan x+13tan xsec2 x+4x∴ f(x)=23log |sec x|+tan2 x6+2x2+d But f(0)=0⇒d=0
Then, f(x)=23log |sec x|+16tan2 x+2x2