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Question

f(x)=(x2)(x1)x3,x>3. The minimum value of f(x) is equal to

A
3+22
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B
3+23
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C
32+2
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D
322
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Solution

The correct option is A 3+22
Given, f(x)=(x1)(x2)x3 and x>3

Taking the derivative and equating to zero. We get

f(x)=x1x3+x2x3(x1)(x2)(x3)2

f(x)=1x3(x1+x2(x22xx+2)(x3))

f(x)=1x3(2x3(x23x+2)(x3))

f(x)=1(x3)((2x3)(x3)(x23x+2)(x3))

f(x)=1(x3)2(2x23x6x+9x2+3x2)

f(x)=1(x3)2(x26x+7)

f(x)=01(x3)2(x26x+7)=0

therefore (x26x+7)=0 and (x3)20 since, denominator cannot be equal to zero.

The roots of (x26x+7)=0 are
6+627421 and 6627421

6+82 and 682

6+222 and 6222

3+2 and 32

x32 since x>3

Therefore x=3+2

substituting x=3+2 in f(x) we get

f(x)=(3+21)(3+22)3+23

=(2+2)(1+2)2

=(2+2+22+2)2

=(4+32)2=3+22

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