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Question

Let f be differential for all x. If f(1)=2 and f(x)2 for xϵ[1,6], then ?

A
f(6)=5
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B
f(6)<5
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C
f(6)<8
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D
f(6)5
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Solution

The correct option is A f(6)=5
As f(1)=2f(1)=−2 and f(x)2x[1,6]f′(x)≥2∀x∈[1,6]
Applying Langrange's mean value theorem
f(6)f(1)5f(6)−f(1)5=f(c)2=f′(c)≥2
f(6)10+f(1)⇒f(6)≥10+f(1)
f(6)102⇒f(6)≥10−2
f(6)8

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