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Ex 6.2,14 - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at April 14, 2021 by

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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