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

Last updated at April 14, 2021 by Teachoo

Ex 6.2

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Ex 6.2, 6 (a)

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Ex 6.2,14 Important You are here

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Ex 6.2,19 (MCQ) Important

Last updated at April 14, 2021 by Teachoo

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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 f(π₯) = π₯2 + aπ₯ + 1 And, fβ(π₯) = 2π₯ + a Given f is strictly increasing on (1 ,2) β΄ fβ(π₯) > 0 on (1 ,2) Putting value of fβ(x) 2π₯ + a > 0 on (1 ,2) Letβs put x = 1 and x = 2 and find value of a So, a > β2 satisfies both equations Thus, we can say that When a > β2 , f(π₯) = π₯2 + aπ₯ + 1 is strictly increasing on (1 , 2) Hence, least value of a is β2 Putting π = 1 2(1) + a > 0 2 + a > 0 a > β2 Putting π = 2 2(2) + a > 0 4 + a > 0 a > β4