Ex 5.1

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

### Transcript

Ex 5.1, 3 Examine the following functions for continuity. (d) f (x) = |x β 5| f(x) = |π₯β5| = {β((π₯β5), π₯β5β₯0@β(π₯β5), π₯β5<0)β€ = {β((π₯β5), π₯β₯5@β(π₯β5), π₯<5)β€ Since we need to find continuity at of the function We check continuity for different values of x When x = 5 When x < 5 When x > 5 Case 1 : When x = 5 f(x) is continuous at π₯ = 5 if L.H.L = R.H.L = π(5) if limβ¬(xβ5^β ) π(π₯)=limβ¬(xβ5^+ ) " " π(π₯)= π(5) Since there are two different functions on the left & right of 5, we take LHL & RHL . LHL at x β 5 limβ¬(xβ5^β ) f(x) = limβ¬(hβ0) f(5 β h) = limβ¬(hβ0) |(5ββ)β5| = limβ¬(hβ0) |ββ| = limβ¬(hβ0) β = 0 RHL at x β 5 limβ¬(xβ5^+ ) f(x) = limβ¬(hβ0) f(5 + h) = limβ¬(hβ0) |(5+β)β5| = limβ¬(hβ0) |β| = limβ¬(hβ0) β = 0 & π(π) = |π₯β5| = |5β5| = 0 Hence, L.H.L = R.H.L = π(5) β΄ f is continuous at x = 5 Case 2 : When x < 5 For x < 5, f(x) = β (x β 5) Since this a polynomial It is continuous β΄ f(x) is continuous for x < 5 Case 3 : When x > 5 For x > 5, f(x) = (x β 5) Since this a polynomial It is continuous β΄ f(x) is continuous for x > 5 Hence, π(π₯)= |π₯β5| is continuous at all points. i.e. f is continuous at π β R.