Ex 5.1

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

### Transcript

Ex 5.1, 6 Find all points of discontinuity of f, where f is defined by π(π₯)={β(2π₯+3, ππ π₯β€2@&2π₯β3, ππ π₯>2)β€ π(π₯)={β(2π₯+3, ππ π₯β€2@&2π₯β3, ππ π₯>2)β€ Since we need to find continuity at of the function We check continuity for different values of x When x = 2 When x < 2 When x > 2 Case 1 : When x = 2 f(x) is continuous at π₯ =2 if L.H.L = R.H.L = π(2) if limβ¬(xβ2^β ) π(π₯)=limβ¬(xβ2^+ ) " " π(π₯)= π(2) Since there are two different functions on the left & right of 2, we take LHL & RHL . LHL at x β 2 limβ¬(xβ2^β ) f(x) = limβ¬(hβ0) f(2 β h) = limβ¬(hβ0) (2(2ββ)+3) = limβ¬(hβ0) (4β2β+3) = limβ¬(hβ0) (7β2β) = 7 β 0 = 7 RHL at x β 2 limβ¬(xβ2^+ ) f(x) = limβ¬(hβ0) f(2 + h) = limβ¬(hβ0) (2(2+β)β3) = limβ¬(hβ0) (4β2ββ3) = limβ¬(hβ0) (1β2β) = 1 β 0 = 1 Since L.H.L β  R.H.L f(x) is not continuous at x = 2 Case 2 : When x < 2 For x < 2, f(x) = 2x + 3 Since this a polynomial It is continuous β΄ f(x) is continuous for x < 2 Case 3 : When x > 2 For x > 2, f(x) = 2x β 3 Since this a polynomial It is continuous β΄ f(x) is continuous for x > 2 Hence, only x = 2 is point is discontinuity. f is continuous at all real numbers except 2. Thus, f is continuous for π± β R β {2}