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Misc 1 (b) Important Deleted for CBSE Board 2022 Exams
Misc 2 Important You are here
Misc 3 Important
Misc 4
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Misc 7
Misc 8 Important
Misc 9 Important
Misc 10
Misc 11 Important
Misc 12 Important
Misc 13 Important
Misc 14 Important
Misc 15 Important
Misc 16
Misc 17 Important
Misc 18 Important
Misc. 19 (MCQ) Deleted for CBSE Board 2022 Exams
Misc 20 (MCQ) Important
Misc 21 (MCQ) Important
Misc 22 (MCQ)
Misc. 23 (MCQ) Important
Misc 24 (MCQ) Important
Last updated at April 19, 2021 by Teachoo
Misc 2 Show that the function given by f(x) = log𝑥/𝑥 is maximum at x = e.Let f(𝑥) = log𝑥/𝑥 Finding f’(𝒙) f’(𝑥) = 𝑑/𝑑𝑥 (log𝑥/𝑥) f’(𝑥) = (𝑑(log𝑥 )/𝑑𝑥 " " . 𝑥 − 𝑑(𝑥)/𝑑𝑥 " . log " 𝑥)/𝑥2 f’(𝑥) = (1/𝑥 × 𝑥 − log𝑥)/𝑥2 f’(𝑥) = (1 − log𝑥)/𝑥2 Using quotient rule as (𝑢/𝑣)^′ = (𝑢^′ 𝑣−𝑣^′ 𝑢)/𝑣^2 Putting f’(𝒙) = 0 (1 − log𝑥)/𝑥2=0 1 – log 𝑥 = 0 log 𝑥 = 1 𝒙 = e Finding f’’(𝒙) f’(𝑥) = (1 − log𝑥)/𝑥2 Diff w.r.t. 𝑥 f’’(𝑥) = 𝑑/𝑑𝑥 ((1 − log𝑥)/𝑥2) log 𝑥=1 is only possible when 𝒙=𝒆 Using quotient rule as (𝑢/𝑣)^′=(𝑢^′ 𝑣 − 𝑣^′ 𝑢)/(𝑣2 .) f’’(𝑥) = (𝑑(1 − log𝑥 )/𝑑𝑥 . 𝑥2− 𝑑(𝑥2)/𝑑𝑥 . (1 − log𝑥 ))/(𝑥^2 )^2 = ((0 − 1/𝑥) . 𝑥2 − 2𝑥(1 − log𝑥 ))/𝑥4 = ((−1)/𝑥 × 𝑥2 − 2𝑥(1 − log𝑥 ))/𝑥^4 = (−𝑥 − 2𝑥(1 − log𝑥 ))/𝑥^4 = (−𝑥[1 + 2(1 − log𝑥 )])/𝑥4 = (−𝑥[3 − 2 log𝑥 ])/𝑥4 ∴ f’’(𝑥) = (−(3 − 2 log𝑥 ))/𝑥3 Putting 𝒙 = e f’’(𝑒) = (−(3 − 2 log𝑒 ))/𝑒3 = (−(3 − 2))/𝑒3 = (−1)/𝑒3 = –(1/𝑒3) < 0 (∵1/𝑒𝑥>0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥 ∈𝑹) Since f’’(𝑥) < 0 at 𝑥 = e . ∴ 𝑥 = e is point of maxima Hence, f(𝑥) is maximum at 𝒙 = e.