Check sibling questions

The maximum value of (1/x) x Β  is:

(A) eΒ  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  (B) ee

(C) e (1/e)Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β (D) 1/e (1/e)




Question 30 The maximum value of (1/π‘₯)^π‘₯ is: (A) e (B) ee (C) 𝑒^(1/𝑒) (D) γ€–1/𝑒〗^(1/𝑒) Let f (π‘₯) = (1/π‘₯)^π‘₯ To find maximum value, we need to differentiate f(x) For differentiating f (π‘₯), we use logarithmic differentiation f (π‘₯) = (1/π‘₯)^π‘₯ log (f(x)) = 𝒙 log (𝟏/𝒙) Differentiating wrt π‘₯ 𝟏/(𝒇(𝒙)) f’(x) = 1βˆ™log (𝟏/𝒙) + 𝒙 Γ— (𝟏/(𝟏/𝒙)) Γ— ((βˆ’πŸ)/𝒙^𝟐 ) 1/(𝑓(π‘₯)) f’(x) = log (1/π‘₯) + π‘₯ Γ— (π‘₯) Γ— ((βˆ’1)/π‘₯^2 ) 1/(𝑓(π‘₯)) f’(x) = log (1/π‘₯) + π‘₯^2 Γ— ((βˆ’1)/π‘₯^2 ) 1/(𝑓(π‘₯)) f’(x) = log (1/π‘₯) βˆ’ 1 f’(x) = f(x) [log⁑〖(1/π‘₯)βˆ’1γ€— ] Putting f (π‘₯) =(1/π‘₯)^π‘₯ f’(x) = (𝟏/𝒙)^𝒙 (π₯𝐨𝐠⁑〖(𝟏/𝒙)βˆ’πŸγ€— ) Putting f’(x) = 0 (1/π‘₯)^π‘₯ (log⁑〖(1/π‘₯)βˆ’1γ€— ) = 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 (π’π’π’ˆβ‘γ€–(𝟏/𝒙)βˆ’πŸγ€— ) = 0 log 1/π‘₯ = 1 Taking exponential on both sides e^log⁑〖1/xγ€— = 𝑒^1 1/π‘₯ = e 𝒙 = 𝒆^(βˆ’πŸ) Putting π‘₯ = 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)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.