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JEE Advanced · Mathematics · 27. Definite Integration

Let f :RR  be a function defined by fx= x,x 20,x>2 , where [x] is the greatest integer less than or equal to x . If I= -12xf(x2)2+f(x+1) dx, then the value of (4I-1) is

  1. A \(0\)
  2. B 5
  3. C 9
  4. D 10
Verified Solution

Answer & Solution

Correct Answer

(A) \(0\)

Step-by-step Solution

Detailed explanation

fx+1=x+1,x+120,x+1>2=x+1,x10,x>1
=0,-1x<01,0x<10,1<x<2
fx2=x2.x220,x2>2

=x2.|x|20,|x|>2
= { 0, - 1 x < 0 0, 0 x < 1 1, 1 x < 2 0, 2 < x < 2
l=-100dx+010dx+12xdx2+0+220dx
=14x212=142-1=14
4l=1
4l-1=0
From JEE Advanced
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