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Question

Value of c of Lagrange's mean value theorem for f(x)=2x−x2 in [0,1]

A
1
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B
0
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C
12
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D
1
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Solution

The correct option is C 12
Lagrange's mean value theorem states that if f(x) be continuous on [a,b] and differentiable on (a,b) then there exists some c between a and b such that f(c)=f(b)f(a)ba

Given f(x)=2xx2 and [a,b]=[0,1]

f(x)=22x

Therefore, f(c)=(2bb2)(2aa2)ba

22c=(21)010

22c=1

2c=1

c=12

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