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  4. If f'( x ) = tan -1(secx + tanx), -(π/2) < x < (π/2), and f(0 ) = 0, then f(1) is equal to :

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

JEE MainJEE Main 2020Differential Equations

Solution:

$f '\left(x\right)=tan^{-1}\left(sec\,x+tan\,x\right)$
$f '\left(x\right)=tan^{\cdot1}\left(\frac{1+sin\,x}{cos\,x}\right)=tan^{-1}\left(\frac{1+tan \frac{x}{2}}{1-tan \frac{x}{2}}\right)$
$=tan^{-1}\left(tan\left(\frac{\pi}{4}+\frac{\pi}{2}\right)\right)
\because-\frac{\pi}{2} < x < \frac{\pi}{2} \Rightarrow 0 < \frac{\pi}{4}+\frac{\pi}{2} < \frac{\pi}{2}$
$\Rightarrow f '\left(x\right)=\frac{\pi}{4}+\frac{x}{2}$
$\therefore f \left(x\right)=\frac{\pi}{4}.x+\frac{x^{2}}{4}+c$
$\because f \left(0\right)=0 \Rightarrow c=0$
$\Rightarrow f \left(x\right)=\frac{\pi}{4}x+\frac{x^{2}}{4}$
$\therefore f \left(1\right)=\frac{\pi+1}{4} $