Let F(x) = \int^{x}_{0} \frac{cost}{(1+t^2)} dt, 0 ≤ x ≤ 2π. Then

Correct Answer is : F is increasing in (0,π2) and (3π2,2π) and decreasing in (π2,3π2)

F is increasing in (\pi/2,{3\pi}/2) and decreasing in (0,\pi/2) and ({3\pi}/2,2\pi)

F is increasing in (0,π) and decreasing in (π,2π)

F is increasing in (π,2π) and decreasing in (0,π)

F is increasing in (0,\pi/2) and (\frac{3\pi}{2},2\pi) and decreasing in (\pi/2,\frac{3\pi}{2})

WBJEE 2013 Math Solved Paper

Question : 69 of 80

Marks: +1, -0

Let F(x) =

∫

cost

(1+t2)

dt, 0 ≤ x ≤ 2π. Then

Go to Question:

Prev Question

Next Question