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Let f(x) be a function such that, $f (0) = f^{'} (0) = 0$ , $f^{''} (x) = \sec^{4} x + 4$ then the function is

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a

\log | (\sin x) | + \frac{1}{3} \tan^{3} x + x

b

\frac{2}{3} \log | (\sec x) | + \frac{1}{6} \tan^{2} x + 2 x^{2}

c

\log | (\cos x) | + \frac{1}{6} \cos^{2} x + \frac{x}{5}

d

None of the above

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detailed solution

Correct option is B

Since, $f^{''} (x) = \sec^{4} x + 4$

$f^{''} (x) = (1 + \tan^{2} x) \sec^{2} x + 4$

$\Rightarrow f^{'} (x) = \tan x + \frac{\tan^{3} x}{3} + 4 x + C \Rightarrow f^{'} (0) = 0 \Rightarrow 0 = C$

$Then, f^{'} (x) = \tan x + \frac{\tan^{3} x}{3} + 4 x$

$\begin{aligned} f^{'} (x) = \tan x + \frac{\tan^{3} x}{3} + 4 x \\ = \tan x + \frac{1}{3} \tan x (\sec^{2} x - 1) + 4 x \\ \Rightarrow f^{'} (x) = \frac{2}{3} \tan x + \frac{1}{3} \tan {xsec}^{2} x + 4 x \\ ∴ f (x) = \frac{2}{3} \log | \sec x | + \frac{\tan^{2} x}{6} + 2 x^{2} + d \\ But f (0) = 0 \Rightarrow d = 0 \end{aligned}$

$Then, f (x) = \frac{2}{3} \log | \sec x | + \frac{1}{6} \tan^{2} x + 2 x^{2}$

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