The maximum value of (1/x) x is:
(A) e (B) ee
(C) e (1/e) (D) 1/e (1/e)
NCERT Exemplar - MCQs
Question 2 Important
Question 3 Important
Question 4 Important
Question 5 Important
Question 6
Question 7 Important
Question 8 Important
Question 9 Important
Question 10
Question 11 Important
Question 12 Important
Question 13
Question 14
Question 15 Important
Question 16 Important
Question 17 Important You are here
Question 1 Important Deleted for CBSE Board 2024 Exams
Question 2 Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams
Question 4 Deleted for CBSE Board 2024 Exams
Question 5 Deleted for CBSE Board 2024 Exams
Question 6 Deleted for CBSE Board 2024 Exams
Question 7 Important Deleted for CBSE Board 2024 Exams
Question 8 Deleted for CBSE Board 2024 Exams
Question 9 Important Deleted for CBSE Board 2024 Exams
Question 10 Important Deleted for CBSE Board 2024 Exams
Question 11 Deleted for CBSE Board 2024 Exams
Question 12 Deleted for CBSE Board 2024 Exams
Question 13 Deleted for CBSE Board 2024 Exams
Last updated at April 16, 2024 by Teachoo
Question 17 The maximum value of (1/𝑥)^𝑥 is: (A) e (B) ee (C) 𝑒^(1/𝑒) (D) 〖1/𝑒〗^(1/𝑒) Let f (𝑥) = (1/𝑥)^𝑥 To find maximum value, we need to differentiate f(x) For differentiating f (𝑥), we use logarithmic differentiation f (𝑥) = (1/𝑥)^𝑥 log (f(x)) = 𝒙 log (𝟏/𝒙) Differentiating wrt 𝑥 𝟏/(𝒇(𝒙)) f’(x) = 1∙log (𝟏/𝒙) + 𝒙 × (𝟏/(𝟏/𝒙)) × ((−𝟏)/𝒙^𝟐 ) 1/(𝑓(𝑥)) f’(x) = log (1/𝑥) + 𝑥 × (𝑥) × ((−1)/𝑥^2 ) 1/(𝑓(𝑥)) f’(x) = log (1/𝑥) + 𝑥^2 × ((−1)/𝑥^2 ) 1/(𝑓(𝑥)) f’(x) = log (1/𝑥) − 1 f’(x) = f(x) [log〖(1/𝑥)−1〗 ] Putting f (𝑥) =(1/𝑥)^𝑥 f’(x) = (𝟏/𝒙)^𝒙 (𝐥𝐨𝐠〖(𝟏/𝒙)−𝟏〗 ) Putting f’(x) = 0 (1/𝑥)^𝑥 (log〖(1/𝑥)−1〗 ) = 0 Since, there is only one critical point, so it will be point of maxima Either (𝟏/𝒙)^𝒙 = 0 Since, it is an exponential function It can never be zero. Or (𝒍𝒐𝒈〖(𝟏/𝒙)−𝟏〗 ) = 0 log 1/𝑥 = 1 Taking exponential on both sides e^log〖1/x〗 = 𝑒^1 1/𝑥 = e 𝒙 = 𝒆^(−𝟏) Putting 𝑥 = 1/𝑒 in f (x) f (𝟏/𝒆) = (1/(1/𝑒))^(1/𝑒) f (𝟏/𝒆) = 𝒆^(𝟏/𝒆) So, the correct answer is (C)