1. Class 12
2. Important Question for exams Class 12
3. Chapter 5 Class 12 Continuity and Differentiability

Transcript

Ex 5.2, 9 Prove that the function f given by π (π₯) = | π₯ β 1|, π₯ β π is not differentiable at x = 1. f(x) = |π₯β1| f(x) is a differentiable at x = 1 if LHD = RHD L H D (πππ)β¬(hβ0) (π(π₯) β π(π₯ β β))/β = (πππ)β¬(hβ0) (π(1) β π(1 β β))/β = (πππ)β¬(hβ0) (|1 β 1| β |(1 β β) β 1|)/β = (πππ)β¬(hβ0) (|0| β |ββ|)/β = (πππ)β¬(hβ0) (0 β β)/β = (πππ)β¬(hβ0) (ββ)/β = (πππ)β¬(hβ0) (β1) = β1 R H D (πππ)β¬(hβ0) (π(π₯ + β) β π(π₯))/β = (πππ)β¬(hβ0) (π(1 + β) β π(1))/β = (πππ)β¬(hβ0) (|(1 + β) β 1| β |1 β 1|)/β = (πππ)β¬(hβ0) (|β| β |0|)/β = (πππ)β¬(hβ0) (β β 0)/β = (πππ)β¬(hβ0) β/β = (πππ)β¬(hβ0) (1) = 1 Since LHD β  RHD β΄ f(x) is not differentiable at x = 1 Hence proved

Chapter 5 Class 12 Continuity and Differentiability

Class 12
Important Question for exams Class 12