Ex 5.8, 5 - Verify Mean Value Theorem f(x) = x3 - 5x2 - 3x - Ex 5.8

Slide18.JPG
Slide19.JPG Slide20.JPG Slide21.JPG Slide22.JPG Slide23.JPG

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

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‘ āˆˆ (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‘„āˆˆ 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‘„āˆˆ 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‘„āˆˆ 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‘„āˆˆ 1, 3ļ·Æ š‘“ā€² š‘ļ·Æ = 3 š‘ļ·®2ļ·Æāˆ’10š‘āˆ’3 Also, š‘“ š‘„ļ·Æ = š‘„3 ā€“ 5š‘„2 ā€“ 3š‘„ š‘“ š‘Žļ·Æ = š‘“ 1ļ·Æ = 1ļ·Æļ·®3ļ·Æāˆ’5 1ļ·Æļ·®2ļ·Æāˆ’3 1ļ·Æ = 1āˆ’5āˆ’3 = āˆ’7 š‘“ š‘ļ·Æ = š‘“ 3ļ·Æ = 3ļ·Æļ·®3ļ·Æāˆ’5 3ļ·Æļ·®2ļ·Æāˆ’3 3ļ·Æ = 27āˆ’45āˆ’9 = āˆ’27 By Mean Value Theorem š‘“ļ·®ā€²ļ·Æ š‘ļ·Æ = š‘“ š‘ļ·Æ āˆ’ š‘“ š‘Žļ·Æļ·®š‘ āˆ’ š‘Žļ·Æ 3 š‘ļ·®2ļ·Æāˆ’10š‘āˆ’3 = āˆ’27 āˆ’ āˆ’7ļ·Æļ·®3 āˆ’ 1ļ·Æ 3 š‘ļ·®2ļ·Æāˆ’10š‘āˆ’3 = āˆ’27 + 7ļ·®2ļ·Æ 3 š‘ļ·®2ļ·Æāˆ’10š‘āˆ’3 = āˆ’20ļ·®2ļ·Æ 3 š‘ļ·®2ļ·Æāˆ’10š‘āˆ’3 = āˆ’10 3 š‘ļ·®2ļ·Æāˆ’10š‘āˆ’3+10 = 0 3 š‘ļ·®2ļ·Æāˆ’10š‘+7 = 0 3 š‘ļ·®2ļ·Æāˆ’3š‘āˆ’7š‘+7 = 0 3š‘ š‘āˆ’1ļ·Æāˆ’7 š‘āˆ’1ļ·Æ = 0 3š‘āˆ’7ļ·Æ š‘āˆ’1ļ·Æ = 0 So, c = 7ļ·®3ļ·Æ & c = 1 Now c = 7ļ·®3ļ·Æ lies between 1 & 3 c = šŸ•ļ·®šŸ‘ļ·Æ āˆˆ 1, 3ļ·Æ Thus, Mean Value Theorem is verified. Now, We need to find cāˆˆ 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·Æāˆ’10š‘āˆ’3 = 0 The above equation is of the form š“ š‘„ļ·®2ļ·Æ+šµš‘„+š¶ x = āˆ’šµ Ā± ļ·® šµļ·®2ļ·Æ āˆ’4š“š¶ļ·Æ ļ·®2š“ļ·Æ c = āˆ’ āˆ’10ļ·Æ Ā± ļ·® āˆ’10ļ·Æļ·®2ļ·Æ āˆ’ 4 āˆ’3ļ·Æ 3ļ·Æļ·Æ ļ·®2š“ļ·Æ c = 10 Ā± ļ·®100 + 36ļ·Æ ļ·®2 āˆ’3ļ·Æļ·Æ c = 10 Ā± ļ·®136ļ·Æ ļ·®6ļ·Æ c = 10 Ā± ļ·®2 Ɨ 2 Ɨ 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œāˆˆ 1, 3ļ·Æ for which š‘“ļ·®ā€²ļ·Æ š‘ļ·Æ = 0

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.