Ex 6.2,14 - Chapter 6 Class 12 Application of Derivatives
Last updated at May 29, 2018 by Teachoo
Transcript
Ex 6.2,14 Find the least value of a such that the function f given by π (π₯) = π₯2 + ππ₯ + 1 is strictly increasing on (1, 2). We have fο·π₯ο·― = π₯2 + aπ₯ + 1 Given f is strictly increasing on ο·1 ,2ο·― β fβο·π₯ο·― > 0 on ο·1 ,2ο·― fβο·π₯ο·― = π₯2 + aπ₯ + 1 fβο·π₯ο·― = 2π₯ + a. Thus, 2π₯ + a > 0 on ο·1 ,2ο·― β΄ When a > β2 , fο·π₯ο·― = π₯2 + aπ₯ + 1 is strictly increasing on ο·π , πο·― Hence, least value of a is β2
Chapter 6 Class 12 Application of Derivatives
Serial order wise
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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.
here there is open interval 1,2 (1,2) that means we shouldn't include 1&2 right but u hv taken as x=1 & x=2 why..?