Let f(x) and f′(x) be differentiable for 0≤x≤1, such that f(0)=2,g(0)=0,f(1)=6. Let there exist a real number c in [0,1] such that f′(c)=2g′(c), then the value of g(1) must be
A
1
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B
2
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C
−2
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D
−1
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Solution
The correct option is D2
Given that f(0)=2,f(1)=6,g(0)=0
Given f(x),f′(x) are differentiable on (0,1).
Therefore, Lagrange's mean value theorem can be applied.
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)b−a