Examples

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

### Transcript

Example 11 Find all the points of discontinuity of the function f defined by π(π₯)={β(&π₯+2 ,ππ π₯<1@0 , ππ π₯=1@&π₯β2 ,ππ π₯>1)β€ π(π₯)={β(&π₯+2 ,ππ π₯<1@0 , ππ π₯=1@&π₯β2 ,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 = 1 When x < 1 When x > 1Case 1 : 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 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) (1ββ)+2 = limβ¬(hβ0) (3ββ) = 3 β 0 = 3 RHL at x β 1 limβ¬(xβ1^+ ) f(x) = limβ¬(hβ0) f(1 + h) = limβ¬(hβ0) (1+β)β2 = limβ¬(hβ0) (β1+β) = β1 + 0 = β1 Since L.H.L β  R.H.L f(x) is not continuous at x=1 Case 2 : When x < 1 For x < 1, f(x) = x + 2 Since this a polynomial It is continuous β΄ f(x) is continuous for x < 1 Case 3 : When x > 1 For x > 1, f(x) = x β 2 Since this a polynomial It is continuous β΄ f(x) is continuous for x > 1 Hence, only π₯=1 is point is discontinuity. f is continuous at all real numbers except 1 Thus, f is continuous for πβ R β {1}