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 Let f(x) be a function such that,f(0)=f(0)=0 ,f′′(x)=sec4 x+4 then the function is 

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a
log ⁡|(sin ⁡x)|+13tan3 ⁡x+x
b
23log⁡ |(sec⁡ x)|+16tan2⁡ x+2x2
c
log ⁡|(cos ⁡x)|+16cos2 ⁡x+x5
d
None of the above

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detailed solution

Correct option is B

Since,f′′(x)=sec4⁡ x+4f′′(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+4xf′(x)=tan ⁡x+tan3  ⁡x3+4x=tan ⁡x+13tan⁡ xsec2⁡x−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


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