1. Chapter 5 Class 12 Continuity and Differentiability
2. Serial order wise
3. Ex 5.1

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€ We have, π(π₯)={β(2π₯+3, ππ π₯β€2@&2π₯β3, ππ π₯>2)β€ Case 1 At π₯ =2 f is continuous at x = 2 if L.H.L = R.H.L = π(2) i.e. limβ¬(xβ2^β ) π(π₯)=limβ¬(xβ2^+ ) π(π₯)=π(2) L.H.L limβ¬(xβ2^β ) π(π₯) = limβ¬(xβ2^β ) 2π₯+3 Putting π₯ =2 = 2(2) + 3 = 4 + 3 = 7 R.H.L limβ¬(xβ2^+ ) π(π₯) = limβ¬(xβ2^+ ) 2π₯β3 Putting π₯ =2 =2(2) β 3 = 4 β 3 = 1 Since, L.H.L β  R.H.L β΄ f is not continuous at x=2. Case 2 At π₯ =π where c < 2 π(π₯)= 2π₯+3 (As π₯ =π, where c < 2) f is continuous at x=c if limβ¬(xβπ) π(π₯)=π(π) L.H.L limβ¬(xβπ) π(π₯) = limβ¬(xβπ) 2π₯+3 Putting π₯=π = 2π+3 R.H.L π(π)= 2π+3 Hence, limβ¬(xβπ) π(π₯)=π(π) β΄ f is continuous at x=c where c<2 Thus, f is continuous at all real number less than 2. Case 3 At π₯ =π where c > 2 β΄ π(π₯)= 2 π₯β3 (As π₯ =π, c > 2) f is continuous at x=c if limβ¬(xβπ) π(π₯)=π(π) L.H.L limβ¬(xβπ) π(π₯) = limβ¬(xβπ) 2 π₯β3 Putting π₯=π = 2 (π)β3 = 2 πβ3 R.H.L π(π₯)= 2π₯β3 π(π)=2πβ3 Hence, limβ¬(xβπ) π(π₯)=π(π) β f is continuous at x=c where c>2 β f is continuous at all real number greater than 2 Hence, only x=2 is point is discontinuity. β f is continuous at all real numbers except 2. Thus, f is continuous for π± β R β {2}.

Ex 5.1