Ex 5.1

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

### Transcript

Ex 5.1, 15 Discuss the continuity of the function f, where f is defined by π(π₯)={ β(2π₯, ππ π₯<0@ 0, ππ 0β€π₯β€1@ 4π₯, ππ π₯>1 )β€ Since we need to find continuity at of the function We check continuity for different values of x When x < 0 When x = 0 When 0 < x < 1 When x = 1 When x > 1 Case 1 : When x < 0 For x < 0, f(x) = 2x Since this a polynomial It is continuous β΄ f(x) is continuous for x < 0 Case 2 : When x = 0 f(x) is continuous at π₯ =0 if L.H.L = R.H.L = π(0) if limβ¬(xβ0^β ) π(π₯)=limβ¬(xβ0^+ ) " " π(π₯)= π(0) Since there are two different functions on the left & right of 0, we take LHL & RHL . LHL at x β 0 limβ¬(xβ0^β ) f(x) = limβ¬(hβ0) f(0 β h) = limβ¬(hβ0) f(βh) = limβ¬(hβ0) 2(ββ) = 2 Γ 0 = 0 RHL at x β 0 limβ¬(xβ1^+ ) f(x) = limβ¬(hβ0) f(0 + h) = limβ¬(hβ0) f(h) = limβ¬(hβ0) 0 = 0 & π(0) = 0 Hence, L.H.L = R.H.L = π(0) β΄ f is continuous at x = 0 Case 3 : When 0 < x < 1 For 0 < x < 1 f(x) = 0 Since this constant It is continuous β΄ f(x) is continuous for 0 < x < 1 Case 4 : When x = 1 f(x) is continuous at π₯ =1 if L.H.L = R.H.L = π(1) if limβ¬(xβ1^β ) π(π₯)=limβ¬(xβ1^+ ) " " π(π₯)= π(1) Since L.H.L β  R.H.L f(x) is not continuous at x = 1 Since there are two different functions on the left & right of 1, we take LHL & RHL . LHL at x β 1 limβ¬(xβ1^β ) f(x) = limβ¬(hβ0) f(1 β h) = limβ¬(hβ0) 0 = 0 RHL at x β 1 limβ¬(xβ1^+ ) f(x) = limβ¬(hβ0) f(1 + h) = limβ¬(hβ0) 4(1+β) = 4(1 + 0) = 4 Case 5: When x > 1 For x > 1, f(x) = 4x Since this a polynomial It is continuous β΄ f(x) is continuous for x > 1 Hence points of discontinuity are x = 1 Thus, f is continuous for all x β R β {π}