Question

If $f^{\prime }\left(x\right)={\tan }^{-1}(\sec x+\tan x),-\frac{\pi }{2}<x<\frac{\pi }{2},$ and $f\left(0\right)=0$, then $f\left(1\right)$ is equal to:

• A. $\frac{\pi +1}{4}$
• B. $\frac{\pi +2}{4}$
• C. $\frac{1}{4}$
• D. $\frac{\pi -1}{4}$

Answer 1

The correct option is A $\frac{\pi +1}{4}$

$f^{\prime }\left(x\right)={\tan }^{-1}(\sec x+\tan x)\Rightarrow f^{\prime }\left(x\right)={\tan }^{-1}\left(\frac{1+\sin x}{\cos x}\right)\Rightarrow f^{\prime }\left(x\right)={\tan }^{-1}\left(\frac{{(\cos \frac{x}{2}+\sin \frac{x}{2})}^2}{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}\right)\Rightarrow f^{\prime }\left(x\right)={\tan }^{-1}\left[\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right]\Rightarrow f^{\prime }\left(x\right)={\tan }^{-1}\left[\tan (\frac{\pi }{4}+\frac{x}{2})\right]\Rightarrow f^{\prime }\left(x\right)=\frac{\pi }{4}+\frac{x}{2}\Rightarrow f\left(x\right)=\frac{\pi }{4}x+\frac{x^2}{4}+c\Rightarrow f\left(0\right)=0\Rightarrow c=0f\left(1\right)=\frac{\pi }{4}+\frac{1}{4}=\frac{\pi +1}{4}$