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Question

Let f(x) be a function such that f(0)=f0 and f′′(x)=sec4x+4 then f(x)=

A
log(sinx)+13tan3x+6x
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B
23log(secx)+16tan2x+2x2
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C
log(cosx)+16cos2x+x25
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D
log(secx)+16sin3x+x2
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Solution

The correct option is B 23log(secx)+16tan2x+2x2
f′′(x)=sec4(x)+4
We write above euation as-
f(x)=sec4(x)dx+4dx ......(1)
Let solve I=sec4 x dx
sec4x dx=sec2x.sec2x dx
using Integration by part
udv=uvvdu dx
u=sec2x and dv=sec2x
sec2x.sec2x dx=tanx.sec2x2tan2x.sec2x dx
sec4x dx=tanx.sec2x2(sec2x1).sec2 dx {sec2x1=tan2x}
sec4x dx=tanx.sec2x2(sec4xsec2x) dx
sec4x dx=tanx.sec2x2sec4x+2sec2x dx
3sec4x dx=tanx.sec2x+2sec2x dx
sec4x dx=13tanx.sec2x+23tanx
Lets put this result into equation (1)
f(x)=13tanx.sec2x+23tanx+4x+c where c is integration constant
Now let's put boundary condition f(0)=0
f(0)=0=13tan(0).sec(0)+23tan(0)+4.0+c
hence c=0
f(x)=13tanx.sec2x+23tanx+4x
f(x)=13tanx.sec2x dx+23tanx dx+4x dx
let tanx=t, sec2x dx=dt
f(x)=13t dt+23log(|sec x|)+2x2+c1 {tanx dx=log(|sec x|)}
f(x)=13t22+23log(|sec x|)+2x2+c1
Now let's again put boundary condition f(0)=0
f(0)=0+23log(sec(0))+0+c1=0
f(0)=0+23log(1)+0+c1=0
f(0)=0+23.0+0+c1=0
c1=0
f(x)=t26+23log(|sec x|)+2x2
Now back substituting t value-
f(x)=tan2x6+23log(|sec x|)+2x2
Option b is correct answer.


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