lf f x = sin lim t→0 2x/(π) cot^(−1) x/(t^2) , then ∫-(π)/2(π)/2fx dx is equal to (where, x≠0)

Question

MARKS Content Team · Accepted Answer

Correct answer: (B) -1. Let y=limt→0⁡2x/(π)cot^(-1)⁡x/(t^2) Case-I : when x>0 then y=2x/(π)limt→0⁡cot^(-1)⁡x/(t^2)=2x/(π)×0=0 Case-II : when x<0 then y=2x/(π) limt→0⁡cot^(-1)⁡x/(t^2)=2x/(π)×π=2x f(x)= sin0 x>0 sin2x x<0 Now, ∫ (−π)/2 (π)/2 f( x )dx= ∫ −(π)/2 0 sin2x dx+ ∫ 0 (π)/2 0 dx=− ( cos2x/2 ) −(π)/2 0 =−1/2 ( 1−( −1 ) )=−1

lf $f (x) = sin (\lim_{t \to 0} \frac{2 x}{π} \cot^{- 1} \frac{x}{t^{2}}),$ then $\int_{- \frac{π}{2}}^{\frac{π}{2}} f (x) d x$ is equal to (where, $x \neq 0$ )

Solution

More Definite Integration Questions

lf f x = sin lim t→0 2x π cot −1 x t 2 , then ∫-π2π2fx dx is equal to (where, x≠0)

Solution

More Definite Integration Questions

lf $f (x) = sin (\lim_{t \to 0} \frac{2 x}{π} \cot^{- 1} \frac{x}{t^{2}}),$ then $\int_{- \frac{π}{2}}^{\frac{π}{2}} f (x) d x$ is equal to (where, $x \neq 0$ )