Ex 5.1

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

Get live Maths 1-on-1 Classs - Class 6 to 12

### Transcript

Ex 5.1, 23 Find all points of discontinuity of f, where π(π₯)={β(sinβ‘π₯/π₯, ππ π₯<[email protected]&π₯+1, ππ π₯β₯0)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 x > 0 Case 1 : When x < 0 For x < 0, f(x) = sinβ‘π₯/π₯ Since sin x and x are continuous So, sinβ‘π₯/π₯ 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) sinβ‘γ (ββ)γ/((ββ)) = 1 RHL at x β 0 limβ¬(xβ0^+ ) f(x) = limβ¬(hβ0) f(0 + h) = limβ¬(hβ0) f(h) = limβ¬(hβ0) β+1 = 0 + 1 = 1 & π(0) = π₯+1 = 0+1 = 1 Hence, L.H.L = R.H.L = π(0) β΄ f is continuous at x = 0 Case 3 : When x > 0 For x > 0, f(x) = x + 1 Since this a polynomial It is continuous β΄ f(x) is continuous for x > 1 Hence, there is no point of discontinuity