Question

Consider the function $f^{\prime \prime }\left(x\right)={\sec }^{4}x+4$ with $f\left(0\right)=0$ and $f^{\prime }\left(0\right)=0$. What is $f\left(x\right)$ equal to?

• A. $\frac{4ln\sec x}{3}+\frac{{\tan }^{2}x}{6}+2x^2$
• B. $\frac{3ln\sec x}{2}+\frac{{\cot }^{2}x}{6}+2x^2$
• C. $\frac{4ln\sec x}{3}+\frac{{\sec }^{2}x}{6}+2x^2$
• D. $ln\sec x+\frac{{\tan }^{4}x}{12}+2x^2$

Answer 1

The correct option is A $\frac{4ln\sec x}{3}+\frac{{\tan }^{2}x}{6}+2x^2$

Given that $f^{\prime \prime }\left(x\right)={\sec }^{2}x+4$

Integrating on both sides, we get $f^{\prime }\left(x\right)=\int ({\sec }^{4}x+4)dx$

$=\int {\sec }^{2}x(1+{\tan }^{2}x)dx+4x+c$

$=\int ({\sec }^{2}x+{\sec }^{2}x{\tan }^{2}x)dx+4x+c$

$=\int {\sec }^{2}xdx+\int {\sec }^{2}x{\tan }^{2}xdx+4x+c$

$=\tan x+\frac{{\tan }^{3}x}{3}+4x+c$

substitute $f^{\prime }\left(0\right)=0$ we get $c=0$

$f^{\prime }\left(x\right)=\tan x+\frac{{\tan }^{3}x}{3}+4x$

Integrating on both sides, we get

$f\left(x\right)=\int \tan xdx+\int \frac{{\tan }^{3}x}{3}dx+\int 4xdx$

$f\left(x\right)=ln|secx|+\int \frac{\tan x({\sec }^{2}x-1)}{3}dx+2x^2+c$

$f\left(x\right)=ln|\sec x|+\int \frac{\tan x{\sec }^{2}x}{3}-\int \frac{tanx}{3}+2x^2+c$

$f\left(x\right)=ln|\sec x|+\frac{{\tan }^{2}x}{6}+\frac{\ln |\sec x|}{3}+2x^2+c$

$f\left(x\right)=\frac{{\tan }^{2}x}{6}+\frac{4\ln |\sec x|}{3}+2x^2+c$

Given that $f\left(0\right)=0$ substituting in the above equation , we get $c=0$

$f\left(x\right)=\frac{{\tan }^{2}x}{6}+\frac{4\ln |\sec x|}{3}+2x^2$