1. Chapter 5 Class 12 Continuity and Differentiability
2. Serial order wise
3. Ex 5.8

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 ā²(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„ šļ·®ā²ļ·Æ š„ļ·Æ = 3š„2 ā10š„ ā 3 š„ā 1, 3ļ·Æ šļ·®ā²ļ·Æ šļ·Æ = 3 šļ·Æļ·®2ļ·Æ ā10 šļ·Æ ā 3 š„ā 1, 3ļ·Æ šā² šļ·Æ = 3 šļ·®2ļ·Æā10šā3 Also, š š„ļ·Æ = š„3 ā 5š„2 ā 3š„ š šļ·Æ = š 1ļ·Æ = 1ļ·Æļ·®3ļ·Æā5 1ļ·Æļ·®2ļ·Æā3 1ļ·Æ = 1ā5ā3 = ā7 š šļ·Æ = š 3ļ·Æ = 3ļ·Æļ·®3ļ·Æā5 3ļ·Æļ·®2ļ·Æā3 3ļ·Æ = 27ā45ā9 = ā27 By Mean Value Theorem šļ·®ā²ļ·Æ šļ·Æ = š šļ·Æ ā š šļ·Æļ·®š ā šļ·Æ 3 šļ·®2ļ·Æā10šā3 = ā27 ā ā7ļ·Æļ·®3 ā 1ļ·Æ 3 šļ·®2ļ·Æā10šā3 = ā27 + 7ļ·®2ļ·Æ 3 šļ·®2ļ·Æā10šā3 = ā20ļ·®2ļ·Æ 3 šļ·®2ļ·Æā10šā3 = ā10 3 šļ·®2ļ·Æā10šā3+10 = 0 3 šļ·®2ļ·Æā10š+7 = 0 3 šļ·®2ļ·Æā3šā7š+7 = 0 3š šā1ļ·Æā7 šā1ļ·Æ = 0 3šā7ļ·Æ šā1ļ·Æ = 0 So, c = 7ļ·®3ļ·Æ & c = 1 Now c = 7ļ·®3ļ·Æ lies between 1 & 3 c = šļ·®šļ·Æ ā 1, 3ļ·Æ Thus, Mean Value Theorem is verified. Now, We need to find cā 1, 3ļ·Æ For which šļ·®ā²ļ·Æ šļ·Æ = 0 šļ·®ā²ļ·Æ šļ·Æ = 0 3 šļ·®2ļ·Æā10šā3 = 0 The above equation is of the form š“ š„ļ·®2ļ·Æ+šµš„+š¶ x = āšµ Ā± ļ·® šµļ·®2ļ·Æ ā4š“š¶ļ·Æ ļ·®2š“ļ·Æ c = ā ā10ļ·Æ Ā± ļ·® ā10ļ·Æļ·®2ļ·Æ ā 4 ā3ļ·Æ 3ļ·Æļ·Æ ļ·®2š“ļ·Æ c = 10 Ā± ļ·®100 + 36ļ·Æ ļ·®2 ā3ļ·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 = ā0.28 Both value does not lie between 1, 3ļ·Æ Hence, there exists no value of šā 1, 3ļ·Æ for which šļ·®ā²ļ·Æ šļ·Æ = 0

Ex 5.8