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Question

Given the function f(x)=x3−2x2−x+1. Then the value(s) of c satisfying the conditions of the mean value theorem for the function on the interval [−2,2], is

A
±1
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B
23
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C
23
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D
2
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Solution

The correct option is C 23
The given function is a cubic polynomial and, hence, it is continuous and differentiable everywhere. Therefore the LMVT is applicable to this function.
f(2)=(2)32(2)2(2)+1
f(2)=88+2+1=13
f(2)=232222+1
f(2)=882+1=1
f(x)=3x24x1

According to LMVT, we have:
f(c)=f(b)f(a)ba
3c24c1=1(13)2(2),
3c24c1=3,
3c24c4=0
(3c+2)(c2)=0c=23,2
Only one root c1=23 belongs to the open interval (2,2)

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