Answer:
Option D
Explanation:
Here , f(x)={[x],x≤20,x>2
I=∫2−1xf(x2)2+f(x+1)dx
I=∫0−1xf(x2)2+f(x+1)dx + ∫10xf(x2)2+f(x+1)dx+ ∫√21xf(x2)2+f(x+1)dx
+ ∫√3√2xf(x2)2+f(x+1)dx + ∫2√3xf(x2)2+f(x+1)dx
= ∫0−1odx+∫100dx+∫√21x.12+0dx+∫√3√2odx+∫2√3o.dx
{ ∵−1<x<0⇒0<x2<1⇒[x2]=0,
o<x<1⇒0<x2<1⇒[x2]=0,
1<x<√2⇒ {1<x2<2⇒[x2]=12<x+1<1+√2⇒f(x+1)=0
⇒ f(x2)= 0,
and √3<x<2⇒3<x2<4
⇒ f(x2 )=0 }
⇒ I=∫√21x2dx=[x24]√21
=14(2−1)=14
∴ 4l =1 ⇒ 4l-1=0