Question

Let f be differentiable for all x. If f(1) = - 2 and f′(x) ≥ 2 for x ∈ [1, 6] , then

f(6) ≥ 8

f(6) < 8

f(6) < 5

f(6) = 5

Solution

f(6) ≥ 8

As f(1) = - 2 & f′(x) ≥ 2 ∀ x ∈ [1, 6]
Applying Lagrange’s mean value theorem
$fraction numerator straight f left parenthesis 6 right parenthesis space minus straight f left parenthesis 1 right parenthesis over denominator 5 end fraction space equals space straight f apostrophe space left parenthesis straight c right parenthesis space greater or equal than 2 rightwards double arrow space straight f left parenthesis 6 right parenthesis space greater or equal than space 10 plus space straight f left parenthesis 1 right parenthesis rightwards double arrow space straight f left parenthesis 6 right parenthesis space greater or equal than space 10 minus 2 rightwards double arrow space straight f left parenthesis 6 right parenthesis space greater or equal than space 8$

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Continuity And Differentiability

Let f be differentiable for all x. If f(1) = - 2 and f′(x) ≥ 2 for x ∈ [1, 6] , then

f(6) ≥ 8

f(6) < 8

f(6) < 5

f(6) = 5

Some More Questions From Continuity and Differentiability Chapter

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Continuity And Differentiability

Let f be differentiable for all x. If f(1) = - 2 and f′(x) ≥ 2 for x ∈ [1, 6] , then f(6) ≥ 8 f(6) < 8 f(6) < 5 f(6) = 5

Some More Questions From Continuity and Differentiability Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Let f be differentiable for all x. If f(1) = - 2 and f′(x) ≥ 2 for x ∈ [1, 6] , then

f(6) ≥ 8

f(6) < 8

f(6) < 5

f(6) = 5