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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

A
12
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B
2
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C
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D
1
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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.

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