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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
Ex 6.2,15
Ex 6.2, 16
Ex 6.2,17 Important
Ex 6.2,18
Ex 6.2,19 (MCQ) Important You are here
Last updated at March 16, 2023 by Teachoo
Ex 6.2, 19 The interval in which π¦ = π₯2 π^(βπ₯) is increasing is (A) (β β, β) (B) (β 2, 0) (C) (2, β) (D) (0, 2)Let f(π₯) = π₯^2 π^(βπ₯) Finding fβ(π) fβ(π₯) = (π₯^2 π^(βπ₯) )β² Using product rule fβ(π₯) = (π₯2)β² π^(βπ₯) + (π^(βπ₯) )β (π₯2) fβ(π₯) = (2π₯) π^(βπ₯) + (γβπγ^(βπ₯) ) (π₯2) fβ(π₯) = 2π₯ π^(βπ₯)βπ^(βπ₯) π₯2 fβ(π) = π e βπ (πβπ) Putting fβ(π)=π π e βπ (πβπ)=π π₯ (2βπ₯) = 0 So, π₯=0 & π₯ = 2 Plotting points on real line (As e βπ₯ is always positive for all π₯ β R) Hence, f(π₯) is strictly increasing on (0 , 2) Therefore, correct answer is (π)