Verify Rolles theorem

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

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Question 4 Verify Rolleβs theorem for the function y = x2 + 2, a = β 2 and b = 2. y = x2 + 2, a = β2 and b = 2 Let π(π₯) = π₯^2+2 Rolleβs theorem is satisfied if Condition 1 Since π(π₯) is a polynomial, it is continuous β΄ π(π₯) is continuous at (β2 , 2) Conditions of Rolleβs theorem π(π₯) is continuous at (π , π) π(π₯) is differentiable at (π , π) π(π) = π(π) If all 3 conditions satisfied then there exist some c in (π , π) such that πβ²(π) = 0 Condition 2 Since π(π₯) is a polynomial, it is differentiable β΄ π(π₯) is differentiable at (β2 , 2) Also, π(βπ) = (β2)^2+2 = 4+2 = 6 π(π) = 2^2+2 = 4+2 = 6 Hence, π(π) = π(βπ) Conditions of Rolleβs theorem π(π₯) is continuous at (π , π) π(π₯) is differentiable at (π , π) π(π) = π(π) If all 3 conditions satisfied then there exist some c in (π , π) such that πβ²(π) = 0 Also, π(π₯) = π₯^2+2 π^β² (π₯) = 2x So, π^β² (π) = ππ Since all 3 conditions are satisfied π^β² (π) = π 2π = 0 π = π Value of c i.e. 0 lies between β2 and 2. Hence c = 0 β (βπ, π) Thus, Rolleβs theorem is satisfied.