# Ex 5.8, 5

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Ex 5.8, 5 Verify Mean Value Theorem, if š (š„) = š„3 ā 5š„2 ā 3š„ in the interval [a, b], where a = 1 and b = 3. Find all š ā (1, 3) for which š ā²(š) = 0. š (š„) = š„3 ā 5š„2 ā 3š„ in [a, b], where a = 1 and b = 3 Condition 1 š (š„) = š„3 ā 5š„2 ā 3š„ š š„ļ·Æ is a polynomial & Every polynomial function is continuous ā š š„ļ·Æ is continuous at š„ā 1, 3ļ·Æ Condition 2 š š„ļ·Æ = š„3 ā 5š„2 ā 3š„ š š„ļ·Æ is a polynomial & Every polynomial function is differentiable ā š š„ļ·Æ is differentiable at š„ā 1, 3ļ·Æ Now š š„ļ·Æ = š„3 ā 5š„2 ā 3š„ šļ·®ā²ļ·Æ š„ļ·Æ = 3š„2 ā10š„ ā 3 š„ā 1, 3ļ·Æ šļ·®ā²ļ·Æ šļ·Æ = 3 šļ·Æļ·®2ļ·Æ ā10 šļ·Æ ā 3 š„ā 1, 3ļ·Æ šā² šļ·Æ = 3 šļ·®2ļ·Æā10šā3 Also, š š„ļ·Æ = š„3 ā 5š„2 ā 3š„ š šļ·Æ = š 1ļ·Æ = 1ļ·Æļ·®3ļ·Æā5 1ļ·Æļ·®2ļ·Æā3 1ļ·Æ = 1ā5ā3 = ā7 š šļ·Æ = š 3ļ·Æ = 3ļ·Æļ·®3ļ·Æā5 3ļ·Æļ·®2ļ·Æā3 3ļ·Æ = 27ā45ā9 = ā27 By Mean Value Theorem šļ·®ā²ļ·Æ šļ·Æ = š šļ·Æ ā š šļ·Æļ·®š ā šļ·Æ 3 šļ·®2ļ·Æā10šā3 = ā27 ā ā7ļ·Æļ·®3 ā 1ļ·Æ 3 šļ·®2ļ·Æā10šā3 = ā27 + 7ļ·®2ļ·Æ 3 šļ·®2ļ·Æā10šā3 = ā20ļ·®2ļ·Æ 3 šļ·®2ļ·Æā10šā3 = ā10 3 šļ·®2ļ·Æā10šā3+10 = 0 3 šļ·®2ļ·Æā10š+7 = 0 3 šļ·®2ļ·Æā3šā7š+7 = 0 3š šā1ļ·Æā7 šā1ļ·Æ = 0 3šā7ļ·Æ šā1ļ·Æ = 0 So, c = 7ļ·®3ļ·Æ & c = 1 Now c = 7ļ·®3ļ·Æ lies between 1 & 3 c = šļ·®šļ·Æ ā 1, 3ļ·Æ Thus, Mean Value Theorem is verified. Now, We need to find cā 1, 3ļ·Æ For which šļ·®ā²ļ·Æ šļ·Æ = 0 šļ·®ā²ļ·Æ šļ·Æ = 0 3 šļ·®2ļ·Æā10šā3 = 0 The above equation is of the form š“ š„ļ·®2ļ·Æ+šµš„+š¶ x = āšµ Ā± ļ·® šµļ·®2ļ·Æ ā4š“š¶ļ·Æ ļ·®2š“ļ·Æ c = ā ā10ļ·Æ Ā± ļ·® ā10ļ·Æļ·®2ļ·Æ ā 4 ā3ļ·Æ 3ļ·Æļ·Æ ļ·®2š“ļ·Æ c = 10 Ā± ļ·®100 + 36ļ·Æ ļ·®2 ā3ļ·Æļ·Æ c = 10 Ā± ļ·®136ļ·Æ ļ·®6ļ·Æ c = 10 Ā± ļ·®2 Ć 2 Ć 34ļ·Æļ·®6ļ·Æ c = 10 Ā± 2 ļ·®34ļ·Æļ·®6ļ·Æ c = 2 5 Ā± ļ·®34ļ·Æ ļ·Æļ·®6ļ·Æ c = 5 Ā± ļ·®34ļ·Æļ·®3ļ·Æ So, Thus, c = 3.61 & c = ā0.28 Both value does not lie between 1, 3ļ·Æ Hence, there exists no value of šā 1, 3ļ·Æ for which šļ·®ā²ļ·Æ šļ·Æ = 0

Chapter 5 Class 12 Continuity and Differentiability

Ex 5.1, 9
Important

Ex 5.1, 13 Important

Ex 5.1, 16 Important

Ex 5.1, 18 Important

Ex 5.1, 28 Important

Ex 5.1, 30 Important

Ex 5.1, 34 Important

Ex 5.2, 5 Important

Ex 5.2, 9 Important

Ex 5.2, 10 Important

Ex 5.3, 10 Important

Ex 5.3, 14 Important

Example 32 Important

Example 33 Important

Ex 5.5,6 Important

Ex 5.5, 7 Important

Ex 5.5, 11 Important

Ex 5.5, 16 Important

Ex 5.6, 7 Important

Ex 5.6, 11 Important

Example 41 Important

Ex 5.7, 14 Important

Example 42 Important

Ex 5.8, 5 Important You are here

Example 44 Important

Example 45 Important

Example 47 Important

Misc 6 Important

Misc 15 Important

Misc 16 Important

Misc 23 Important

Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.