Question

The integral $I=\displaystyle \int _{0}^{100 \pi } \left[t a n^{- 1} x\right]dx$ , (where, $\left[.\right]$ represents the greatest integer function) has the value $K\pi +tan p,$ then the value of $K+p$ is equal to

• A. $101$
• B. $99$
• C. $100\pi $
• D. $99\pi $

Answer 1

Answer: (B)

The given integral is $I=\int_{0}^{\tan 1} 0 d x+\int_{\tan 1}^{100 \pi} 1 d x$
(as $\left.\tan ^{-1} x \in\left(1, \frac{\pi}{2}\right), \forall x>\tan 1\right)$
$\Rightarrow I=100 \pi-\tan 1$
$=100 \pi+\tan (-1)$
$\therefore k=100, p=-1$
$\therefore k+p=99$