Q. Let $f(x)$ be a function such that, $f(0) = f^{'}(0) = 0,f^{''}(x) = \sec^{4}{x + 4}$, then the function is
A) $\log{|(}\sin{x)| + \frac{1}{3}\tan^{3}{x + x}}$
B) $\frac{2}{3}\log{|(\sec{x)| + \frac{1}{6}\tan^{2}{x + 2x^{2}}}}$
C) $\log{\left| {(cos}{x)} \right| + \frac{1}{6}\cos^{2}{x + \frac{x^{2}}{5}}}$
D) None of the above
Explanation
(b)
Since, $f^{''}(x) = \sec^{4}{x + 4}$
$f^{''}(x) = (1 + \tan^{2}{x)\sec^{2}{x + 4}}$
$\Rightarrow f^{'}(x) = \tan{x + \frac{\tan^{3}x}{3} + 4x} + c$
$\Rightarrow f^{'}(0) = 0$
$\Rightarrow 0 = c$
Then, $f^{'}(x) = \tan{x + \frac{\tan^{3}x}{3} + 4x}$
$= \tan{x + \frac{1}{3}\tan{x(\sec^{2}{x - 1) + 4x}}}$
$\Rightarrow f^{'}(x) = \frac{2}{3}\tan{x + \frac{1}{3}\tan{x\sec^{2}{x + 4x}}}$
$\therefore f(x) = \frac{2}{3}\log{|\sec{x| + \frac{\tan^{2}x}{6}2x^{2} + d}}$
But $f(0) = 0$
$\Rightarrow d = 0$
Then, $f(x) = \frac{2}{3}\log{|\sec{x| + \frac{1}{6}\tan^{2}{x + 2x^{2}}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.