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Ex 5.8

Ex 5.8, 1
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Last updated at March 30, 2023 by Teachoo

Ex 5.8, 4 Verify Mean Value Theorem, if π (π₯) = π₯2 β 4π₯ β 3 in the interval [π, π], where π= 1 πππ π= 4 π (π₯) = π₯2 β 4π₯ β 3 π₯β[π, π] where a = 1 & b = 4 Mean Value Theorem satisfied if Condition 1 π(π₯) is continuous π(π₯)=π₯2 β 4π₯ β 3 π(π₯) is a polynomial & Every polynomial function is continuous β π(π₯) is continuous at π₯β[1, 4] Conditions of Mean value theorem π(π₯) is continuous at (π , π) π(π₯) is derivable at (π , π) If both conditions satisfied, then there exist some c in (π , π) such that πβ²(π) = (π(π) β π(π))/(π β π) Conditions of Mean value theorem π(π₯) is continuous at (π , π) π(π₯) is derivable at (π , π) If both conditions satisfied, then there exist some c in (π , π) such that πβ²(π) = (π(π) β π(π))/(π β π) Condition 2 If π(π₯) is differentiable π(π₯) = π₯2 β 4π₯ β 3 π(π₯) is a polynomial & Every polynomial function is differentiable β π(π₯) is differentiable at π₯β[1, 4] Condition 3 π(π₯) = π₯2 β 4π₯ β 3 π^β² (π₯) = 2π₯β4 π^β² (π) = 2πβ4 Conditions of Mean value theorem π(π₯) is continuous at (π , π) π(π₯) is derivable at (π , π) If both conditions satisfied, then there exist some c in (π , π) such that πβ²(π) = (π(π) β π(π))/(π β π) π(π) = π(1) = (1)^2β4(1)β3 = 1 β 4 β 3 = β6 π(π) = π(4) = (4)^2β4(4)β3 = 16 β 16 β 3 = β 3 By Mean Value Theorem π^β² (π) = (π(π) β π(π))/(π β π) π^β² (π) = (β3 β (β6))/(4 β 1) π^β² (π) = (β3 + 6)/3 π^β² (π) = 3/3 π^β² (π) = 1 2c β 4 = 1 2c = 1 + 4 2c = 5 c = 5/2 Value of c = 5/2 which is lies between (1, 4) c = π/πβ(π, π) Hence Mean Value Theorem satisfied