Question

If $h\left(x\right)=\left(fog\right)\left(x\right)+k$ where $k$ is any constant. If $\frac{d}{dx}\left(h\right(x\left)\right)=-\frac{\sin x}{{\cos }^2\left(\cos x\right)},$ then compute the value of $j\left(0\right)$,
where $j\left(x\right)=\underset{g\left(x\right)}{\overset{f\left(x\right)}{\int }}\frac{f\left(t\right)}{g\left(t\right)}dt,$ $f$ and $g$ are trigonometric functions

• A. $\sec 1$
• B. $1-\sec 1$
• C. $\sec 1-1$
• D. $-\sec 1$

Answer 1

The correct option is B $1-\sec 1$

$h\left(x\right)=\left(fog\right)\left(x\right)+k\frac{dh\left(x\right)}{dx}=f^{\prime }\left(g\right(x\left)\right)g^{\prime }\left(x\right)=\frac{-\sin x}{{\cos }^2\left(\cos x\right)}j\left(x\right)=\underset{g\left(x\right)}{\overset{f\left(x\right)}{\int }}\frac{f\left(t\right)}{g\left(t\right)}dth\left(x\right)=-\int \frac{\sin x}{{\cos }^2\left(\cos x\right)}dx=\int \frac{dt}{{\cos }^{2}t}=\tan th\left(x\right)=\tan \left(\cos x\right)+c\therefore f\left(x\right)=\tan x,g\left(x\right)=\cos x$
$j\left(x\right)=\underset{\cos x}{\overset{\tan x}{\int }}\frac{\tan t}{\cos t}dtj\left(0\right)={\int }_1^0\sec t\tan tdt=1-\sec 1$