CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Let f(x) be a polynomial of degree four having extreme values at x=1 and x=2. If limx0(1+f(x)x2)=3, then the maximum value of f(x) in x[0,2] is
  1. 12
  2. 2 
  3. 1


Solution

The correct option is A 12
Let f(x)=ax4+bx3+cx2+dx+e
Since, limx0(1+f(x)x2)=3
limx0(1+ax4+bx3+cx2+dx+ex2)=3

In order to limit exists, d,e must be 0
limx0(1+ax2+bx+c)=3
1+c=3
c=2
f(x)=ax4+bx3+2x2

Now f(1)=0
f(1)=4a+3b+4=0   (1)
f(2)=032a+12b+8=0   (2)
From equation (1) and (2), we have
a=0.5, b=2
f(x)=0.5x42x3+2x2 

For minimum/maximum value
f(x)=02x36x2+4x=0x(x23x+2)=0x(x1)(x2)=0x=0,1,2

Now, f′′(x)=6x212x+4
f′′(0)=f′′(2)=4>0
and f′′(1)=2<0
Hence, the maximum value of f(x) in the given interval is 12.

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
QuestionImage
QuestionImage
View More...



footer-image