
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Ex 6.2
Ex 6.2,2
Ex 6.2,3 Important
Ex 6.2,4
Ex 6.2, 5 Important
Ex 6.2, 6 (a)
Ex 6.2, 6 (b) Important
Ex 6.2, 6 (c) Important
Ex 6.2, 6 (d)
Ex 6.2, 6 (e) Important
Ex 6.2, 7
Ex 6.2,8 Important
Ex 6.2,9 Important
Ex 6.2,10
Ex 6.2,11
Ex 6.2, 12 (A)
Ex 6.2, 12 (B) Important
Ex 6.2, 12 (C) Important
Ex 6.2, 12 (D)
Ex 6.2, 13 (MCQ) Important
Ex 6.2,14 Important You are here
Ex 6.2,15
Ex 6.2, 16
Ex 6.2,17 Important
Ex 6.2,18
Ex 6.2,19 (MCQ) Important
Last updated at May 29, 2023 by Teachoo
Ex 6.2, 14 Find the least value of a such that the function f given by 𝑓 (𝑥) = 𝑥2 + 𝑎𝑥 + 1 is strictly increasing on (1, 2).We have f(𝑥) = 𝑥2 + a𝑥 + 1 And, f’(𝑥) = 2𝑥 + a Given f is strictly increasing on (1 ,2) ∴ f’(𝑥) > 0 on (1 ,2) Putting value of f’(x) 2𝑥 + a > 0 on (1 ,2) Let’s put x = 1 and x = 2 and find value of a So, a > −2 satisfies both equations Thus, we can say that When a > –2 , f(𝑥) = 𝑥2 + a𝑥 + 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