Question

The value of ${\int }_0^{100}\left[ta{n}^{-1}x\right]dx$, is?

• A. $100$
• B. $100-ta{n}^{-1}$ 1
• C. $100-tan1$
• D. none of these

Answer 1

The correct option is C $100-tan1$

${\int }_0^{100}\left[{\tan }^{-1}x\right]dx$

${\tan }^{-1}x$ is strictly an increasing function

$\Rightarrow f\left(x\right)={\tan }^{-1}x\Rightarrow f^{\prime }\left(x\right)=\frac{1}{1+x^2}>0$

now in $1<x>\tan 1$

$\Rightarrow \left[{\tan }^{-1}\right(0\left)\right]=0$

And in $\tan 1<x<100$

$\Rightarrow \left[{\tan }^{-1}\right(0\left)\right]=1\Rightarrow {\int }_0^{100}\left[{\tan }^{-1}x\right]dx\Rightarrow {\int }_1^{{\tan }^{-1}}0dx+{\int }_{\tan {}^{-1}}^{100}1dx\Rightarrow 0+(100-\tan 1)\Rightarrow 100-\tan 1$

Hence the correct value of above equation is $100-\tan 1$