The integral I= displaystyle ∫ 0100 π [t a n- 1 x]dx , (where, [.] represents the greatest integer function) has the value Kπ +tan p, then the value of K+p is equal to

Question

The integral I= displaystyle ∫ 0100 π [t a n- 1 x]dx , (where, [.] represents the greatest integer function) has the value Kπ +tan p, then the value of K+p is equal to (A) 101 (B) 99 (C) 100π  (D) 99π

Tardigrade Admin · Accepted Answer

The given integral is I=∫0 tan 1 0 d x+∫ tan 1100 π 1 d x (as . tan -1 x ∈(1, (π/2)), ∀ x> tan 1) ⇒ I=100 π- tan 1 =100 π+ tan (-1) ∴ k=100, p=-1 ∴ k+p=99

Q. The integral $I = \int_{0}^{100 π} [t a n^{- 1} x] d x$ , (where, $[.]$ represents the greatest integer function) has the value $K π + t an p,$ then the value of $K + p$ is equal to

$101$

$99$

$100 π$

$99 π$

The given integral is $I = \int_{0}^{t a n 1} 0 d x + \int_{t a n 1}^{100 π} 1 d x$
(as $tan^{- 1} x \in (1, \frac{π}{2}), \forall x > tan 1)$
$\Rightarrow I = 100 π - tan 1$
$= 100 π + tan (- 1)$
$∴ k = 100, p = - 1$
$∴ k + p = 99$

Q. The integral I=∫0100π​[tan−1x]dx , (where, [.] represents the greatest integer function) has the value Kπ+tanp, then the value of K+p is equal to

101

99

100π

99π