# Ex 5.1 ,6 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at Nov. 12, 2018 by Teachoo

Last updated at Nov. 12, 2018 by Teachoo

Transcript

Ex 5.1, 6 Find all points of discontinuity of f, where f is defined by π(π₯)={β(2π₯+3, ππ π₯β€2@&2π₯β3, ππ π₯>2)β€ We have, π(π₯)={β(2π₯+3, ππ π₯β€2@&2π₯β3, ππ 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¬(xβ2^β ) π(π₯)=limβ¬(xβ2^+ ) π(π₯)=π(2) L.H.L limβ¬(xβ2^β ) π(π₯) = limβ¬(xβ2^β ) 2π₯+3 Putting π₯ =2 = 2(2) + 3 = 4 + 3 = 7 R.H.L limβ¬(xβ2^+ ) π(π₯) = limβ¬(xβ2^+ ) 2π₯β3 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¬(xβπ) π(π₯)=π(π) L.H.L limβ¬(xβπ) π(π₯) = limβ¬(xβπ) 2π₯+3 Putting π₯=π = 2π+3 R.H.L π(π)= 2π+3 Hence, limβ¬(xβπ) 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₯β3 (As π₯ =π, c > 2) f is continuous at x=c if limβ¬(xβπ) π(π₯)=π(π) L.H.L limβ¬(xβπ) π(π₯) = limβ¬(xβπ) 2 π₯β3 Putting π₯=π = 2 (π)β3 = 2 πβ3 R.H.L π(π₯)= 2π₯β3 π(π)=2πβ3 Hence, limβ¬(xβπ) 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}.

Checking continuity using LHL and RHL

Example 12

Example 10

Example 13

Ex 5.1, 10

Ex 5.1, 11

Ex 5.1 ,6 You are here

Ex 5.1, 13

Ex 5.1, 12

Example 11

Example 7

Ex 5.1 ,3 Important

Ex 5.1, 14

Ex 5.1, 16

Ex 5.1, 15 Important

Ex 5.1 ,7 Important

Ex 5.1, 25

Ex 5.1, 23

Ex 5.1, 24 Important

Ex 5.1 ,8

Ex 5.1, 9 Important

Ex 5.1, 29

Ex 5.1, 27

Ex 5.1, 28 Important

Ex 5.1, 17 Important

Ex 5.1, 18 Important

Ex 5.1, 26 Important

Ex 5.1, 30 Important

Example 15

Chapter 5 Class 12 Continuity and Differentiability

Concept wise

- Checking continuity at a given point
- Checking continuity at any point
- Checking continuity using LHL and RHL
- Algebra of continous functions
- Continuity of composite functions
- Checking if funciton is differentiable
- Finding derivative of a function by chain rule
- Finding derivative of Implicit functions
- Finding derivative of Inverse trigonometric functions
- Finding derivative of Exponential & logarithm functions
- Logarithmic Differentiation - Type 1
- Logarithmic Differentiation - Type 2
- Derivatives in parametric form
- Finding second order derivatives - Normal form
- Finding second order derivatives- Implicit form
- Proofs
- Verify Rolles theorem
- Verify Mean Value Theorem

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.