# Ex 5.8, 2 (iii) - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Last updated at Aug. 19, 2021 by Teachoo

Ex 5.8

Ex 5.8, 1
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Last updated at Aug. 19, 2021 by Teachoo

Ex 5.8, 2 Examine if Rolleβs theorem is applicable to the functions. Can you say some thing about the converse of Rolleβs theorem from this function? (πππ) π (π₯) = π₯2 β 1 πππ π₯ β [1, 2]π (π₯) = π₯2 β 1 πππ π₯ β [1 , 2] Condition 1 π(π₯) = π₯2 β 1 π(π₯) is a polynomial & Every polynomial function is continuous β π(π₯) is continuous at π₯β[1, 2] Conditions of Rolleβs theorem π(π₯) is continuous at (π , π) π(π₯) is derivable at (π , π) π(π) = π(π) If all 3 conditions satisfied then there exist some c in (π , π) such that πβ²(π) = 0 Condition 2 π(π₯)=π₯2 β 1 π(π₯) is a polynomial & Every polynomial function is differentiable β π(π₯) is differentiable at π₯β[1, 2] Condition 3 π(π₯) = π₯2 β 1 π(1) = (1)^2+(1) = 1 β 1 = 0 & π(2) = (2)^2β1= 4β1 = 3 Since π(π) β π(π) Thus third condition of Rolleβs Theorem is not satisfied. Therefore Rolleβs theorem is not applicable Conditions of Rolleβs theorem π(π₯) is continuous at (π , π) π(π₯) is derivable at (π , π) π(π) = π(π) If all 3 conditions satisfied then there exist some c in (π , π) such that πβ²(π) = 0 Converse of Rolleβs Theorem If π [π, π]βπ for some πβ[π, π] for which π^β² (π)=0 then (i) π(π) = π(π) (ii) π is continuous at [π, π] (iii) & Differentiable at [π, π] Now, π(π₯)=π₯^2β1 π^β² (π₯)=2π₯ π^β² (π)=2π Conditions of Rolleβs theorem π(π₯) is continuous at (π , π) π(π₯) is derivable at (π , π) π(π) = π(π) If all 3 conditions satisfied then there exist some c in (π , π) such that πβ²(π) = 0 If π^β² (π)=0 2π=0 π=0 Since π=0 does not belong in (1, 2) i.e. c = 0 β (1 , 2) β There is no value of c for which π^β² (π)=0 β΄ Converse of Rolleβs Theorem is also not applicable.