Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12


Last updated at Jan. 3, 2020 by Teachoo
Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12
Transcript
Example 42 Verify Rolleβs theorem for the function y = x2 + 2, a = β 2 and b = 2. y = x2 + 2, a = β2 and b = 2 Let π(π₯) = π₯^2+2 Rolleβs theorem is satisfied if Condition 1 π(π₯) is continuous at (β2 , 2) Since π(π₯) is a polynomial . it is continuous 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₯) is differentiable at (β2 , 2) Since π(π₯) is a polynomial . it is differentiable Condition 3 π(β2) = (β2)^2+2 = 4+2 = 6 π(2) = 2^2+2 = 4+2 = 6 Hence, π(2) = π(β2) 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 Now, π(π₯) = π₯^2+2 π^β² (π₯) = 2x So π^β² (π) = 2π Since all 3 conditions are satisfied π^β² (π) = 0 2π = 0 π = 0 Value of c i.e. 0 lies between β2 and 2. Hence c = 0 β (βπ, π) Thus , Rolleβs theorem is satisfied.
About the Author