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Ex 6.2, 14 - Find least value of a such that f(x) = x2+ax+1

Ex 6.2,14 - Chapter 6 Class 12 Application of Derivatives - Part 2


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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‘₯) = π‘₯2 + aπ‘₯ + 1 And, f’(π‘₯) = 2π‘₯ + a Given f is strictly increasing on (1 ,2) ∴ f’(π‘₯) > 0 on (1 ,2) Putting value of f’(x) 2π‘₯ + a > 0 on (1 ,2) Let’s put x = 1 and x = 2 and find value of a So, a > βˆ’2 satisfies both equations Thus, we can say that When a > –2 , f(π‘₯) = π‘₯2 + aπ‘₯ + 1 is strictly increasing on (1 , 2) Hence, least value of a is –2 Putting 𝒙 = 1 2(1) + a > 0 2 + a > 0 a > –2 Putting 𝒙 = 2 2(2) + a > 0 4 + a > 0 a > –4

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.