Ex 6.2, 14 - Find least value of a such that f(x) = x2+ax+1 - To show increasing/decreasing in intervals

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
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Ex 6.2,14 Find the least value of a such that the function f given by 𝑓 (π‘₯) = π‘₯2 + π‘Žπ‘₯ + 1 is strictly increasing on (1, 2). We have f﷐π‘₯ο·― = π‘₯2 + aπ‘₯ + 1 Given f is strictly increasing on ﷐1 ,2ο·― β‡’ f’﷐π‘₯ο·― > 0 on ﷐1 ,2ο·― f’﷐π‘₯ο·― = π‘₯2 + aπ‘₯ + 1 f’﷐π‘₯ο·― = 2π‘₯ + a. Thus, 2π‘₯ + a > 0 on ﷐1 ,2ο·― ∴ When a > –2 , f﷐π‘₯ο·― = π‘₯2 + aπ‘₯ + 1 is strictly increasing on ﷐𝟏 , 𝟐﷯ Hence, least value of a is –2

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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.