Question

If $f^{\prime }\left(x\right)=ta{n}^{-1}\left(secx+tanx\right),-\frac{\pi }{2}<x<\frac{\pi }{2}$, 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

Answer: (A)

$f^{\prime }\left(x\right)=ta{n}^{-1}\left(secx+tanx\right)$
$f^{\prime }\left(x\right)=ta{n}^{\cdot 1}\left(\frac{1+sinx}{cosx}\right)=ta{n}^{-1}\left(\frac{1+tan\frac{x}{2}}{1-tan\frac{x}{2}}\right)$
$=ta{n}^{-1}\left(tan\left(\frac{\pi }{4}+\frac{\pi }{2}\right)\right)<br/>\because -\frac{\pi }{2}<x<\frac{\pi }{2}\Rightarrow 0<\frac{\pi }{4}+\frac{\pi }{2}<\frac{\pi }{2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{\pi }{4}+\frac{x}{2}$
$\therefore f\left(x\right)=\frac{\pi }{4}.x+\frac{x^2}{4}+c$
$\because f\left(0\right)=0\Rightarrow c=0$
$\Rightarrow f\left(x\right)=\frac{\pi }{4}x+\frac{x^2}{4}$
$\therefore f\left(1\right)=\frac{\pi +1}{4}$