CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

For f(x)=x+11(xtx2)f(t) dt
(I) limx0f(x)=0(II) f(13)=1
(III) f is continuous and derivable on R
(IV)  maximum value of f(x) does not exist

Which of the following is/are correct?
(correct answer + 2, wrong answer - 0.50)
  1. Only (I)
  2. (I),(II) and (III)
  3. All of them 
  4. Only (II) and (III)


Solution

The correct option is B (I),(II) and (III)
Given : f(x)=x+11(xtx2)f(t) dt
f(x)=x+x11f(t) dtx211tf(t) dtf(x)=axbx2
Where a=1+11f(t) dt,  b=11tf(t) dt

Now, 
a=1+11f(t) dta=1+11atbt2 dta=12b3(1)b=11tf(t) dtb=11t(atbt2) dtb=2a3(2)
From equation (1) and (2), we get
a=14a9a=913b=613f(x)=9x6x213f(x)=912x13f(13)=1313=1
limx0f(x)=f(0)=0

As f(x) is a polynomial of degree 2, so it is continuous and derivable on R
As it is a downward parabola, so it has a finite maximum value.

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
View More


People also searched for
View More



footer-image