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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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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.