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Question

# Let f(x) satisfy the requirements of Lagrange's mean value theorem in [0,2]. If f(0)=0 and |f′(x)|≤12 for all x∈[0,2], then

A
f(x)2
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B
|f(x)|2x
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C
|f(x)|1
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D
f(x)=3, for at least one x[0,2]
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Solution

## The correct option is C |f(x)|≤1By Lagrange's mean value theorem in [0,2], there exists c∈(0,2) such thatf′(c)=f(2)−f(0)2⇒2f′(c)=f(2)−f(0)⇒2f′(c)=f(2) , since f(0)=0Now, since |f′(x)|≤12 for all x∈[0,2],⇒|f′(c)|≤12⇒2|f′(c)|≤1⇒|f(2)|≤1Also, |f′(k)|=∣∣∣f(k)−f(0)k∣∣∣≤12∴|f(k)|≤k2 for k∈[0,2]∴|f(x)|≤1

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