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Ex 6.2,19 (MCQ) - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at March 16, 2023 by

Ex 6.2, 19 - The interval in which y = x2 e-x is increasing

Ex 6.2,19 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.2,19 - Chapter 6 Class 12 Application of Derivatives - Part 3

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Ex 6.2, 19 The interval in which 𝑦 = π‘₯2 𝑒^(–π‘₯) is increasing is (A) (– ∞, ∞) (B) (– 2, 0) (C) (2, ∞) (D) (0, 2)Let f(π‘₯) = π‘₯^2 𝑒^(βˆ’π‘₯) Finding f’(𝒙) f’(π‘₯) = (π‘₯^2 𝑒^(βˆ’π‘₯) )β€² Using product rule f’(π‘₯) = (π‘₯2)β€² 𝑒^(βˆ’π‘₯) + (𝑒^(βˆ’π‘₯) )’ (π‘₯2) f’(π‘₯) = (2π‘₯) 𝑒^(βˆ’π‘₯) + (γ€–βˆ’π‘’γ€—^(βˆ’π‘₯) ) (π‘₯2) f’(π‘₯) = 2π‘₯ 𝑒^(βˆ’π‘₯)βˆ’π‘’^(βˆ’π‘₯) π‘₯2 f’(𝒙) = 𝒙 e –𝒙 (πŸβˆ’π’™) Putting f’(𝒙)=𝟎 𝒙 e –𝒙 (πŸβˆ’π’™)=𝟎 π‘₯ (2βˆ’π‘₯) = 0 So, π‘₯=0 & π‘₯ = 2 Plotting points on real line (As e –π‘₯ is always positive for all π‘₯ ∈ R) Hence, f(π‘₯) is strictly increasing on (0 , 2) Therefore, correct answer is (𝐃)

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