If a function is everywhere continuous and differentiable such that f′(x)≥6 for all x∈[2,4] and f(2)=−4, then
A
f(4)<8
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B
f(4)≥8
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C
f(4)≥2
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D
None of these
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Solution
The correct option is Bf(4)≥8 Since, f(x) is everywhere continuous and differentiable. Therefore, by Lagrange's meanvalue theorem there exists C∈(2,4) such that f′(c)=f(4)−f(2)4−2 ⇒f(4)+42≥6 [f′(x)≥6 for all x∈[2,4]] ⇒f(4)≥8