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

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  1. Chapter 6 Class 12 Application of Derivatives
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Ex 6.2,19 The interval in which 𝑦 = 𝑥2 ﷐𝑒﷮–𝑥﷯ is increasing is (A) (– 𕔴, 𕔴) (B) (– 2, 0) (C) (2, 𕔴) (D) (0, 2) Let f﷐𝑥﷯ = 𝑥2e –𝑥 Step 1: Finding f’﷐𝑥﷯ f’﷐𝑥﷯ = ﷐﷐𝑥﷮2﷯﷐𝑒﷮−𝑥﷯﷯′ f’﷐𝑥﷯ = ﷐𝑥2﷯′ e –𝑥 + ﷐﷐𝑒﷮−𝑥﷯﷯’ ﷐𝑥2﷯ f’﷐𝑥﷯ = ﷐2𝑥﷯ e –𝑥 + ﷐﷐−𝑒﷮−𝑥﷯﷯ ﷐𝑥2﷯ f’﷐𝑥﷯ = 2𝑥 e-𝑥 – e-𝑥 𝑥2 f’﷐𝑥﷯ = 𝑥 e –𝑥 ﷐2−𝑥﷯ Step 2: Putting f’﷐𝑥﷯=0 𝑥 e-𝑥 ﷐2−𝑥﷯ = 0 𝑥 ﷐2−𝑥﷯ = 0 So, 𝑥=0 & 𝑥 = 2 Step 3: Plotting points on real line The points 𝑥 = 0 & 2 divide the real line into 3 disjoint intervals i.e ﷐−𕔴 0﷯ , ﷐0 , 2﷯ & ﷐2 , 𕔴uc1﷯ Step 4: Hence f﷐𝑥﷯ is strictly decreasing on ﷐−𕔴 0﷯ & ﷐2 , 𕔴uc1﷯ & f﷐𝑥﷯ is strictly increasing on ﷐0 , 2﷯ Hence correct answer is ﷐𝐃﷯

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.