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UPSC NDA II 2014 Mathematics Paper

For the next two (02) items that follow:

Consider the function

f''(x)=sec^4x+4

with f(0)=0 nd

f'(0)=0

[NDA II 2014]

Question : 105 of 120

Marks: +1, -0

What is f(x) equal to ?

Solution:

f''(x)=\sec^4 x+4

\Rightarrow f''(x)=\sec^2 x(1+\tan^2 x)+4

\Rightarrow f'(x)=\tan x+\frac{\tan^3 x}{3}+4x+C_1

f'(0)=0\Rightarrow C_1=0

\Rightarrow f'(x)=\tan x+\frac{\tan^3 x}{3}+4x

\Rightarrow f(x)=\ln \sec x+\frac{1}{3}\left(\frac{\tan^2 x}{2}-\ln \sec x\right)+2x^2+C_2

\Rightarrow f(x)=\frac{2\ln \sec x}{3}+\frac{\tan^2 x}{6}+2x^2+C_2

f(0)=0\Rightarrow C_2=0

\Rightarrow f(x)=\frac{2\ln \sec x}{3}+\frac{\tan^2 x}{6}+2x^2

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