If a = lim n → ∞ ∑ k = 1 n 2 n n 2 + k 2 and f x = 1 - cos x 1 + cos x , x ∈ 0 , 1 ,…

If a=limnk=1n2nn2+k2 and fx=1-cosx1+cosx,x0,1, then:
  1. 22fa2=f'a2
  2. fa2f'a2=2
  3. 2fa2=f'a2
  4. fa2=2f'a2

Solution

Given, a=limnk=1n2nn2+k2

a=1nk=1n21+kn2=0121+x2dx

a=2tan-1x01

a=2π4-0=π2

Now, fx=1-cosx1+cosx

fx=tanx2;x0,1

So, fa2=fπ4

fπ4=2-1

And f'π4=12sec2π8=22+1

So, f'π4=2fπ4

Asked in: JEE Main 2022 (26 Jul Shift 1)

Practice more Definite Integration questions on Aicharya