Question 30
The maximum value of (1/π₯)^π₯ is:
(A) e (B) ee
(C) π^(1/π) (D) γ1/πγ^(1/π)
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)) = π log (π/π)
Differentiating wrt π₯
π/(π(π)) fβ(x) = 1βlog (π/π) + π Γ (π/(π/π)) Γ ((βπ)/π^π )
1/(π(π₯)) fβ(x) = log (1/π₯) + π₯ Γ (π₯) Γ ((β1)/π₯^2 )
1/(π(π₯)) fβ(x) = log (1/π₯) + π₯^2 Γ ((β1)/π₯^2 )
1/(π(π₯)) fβ(x) = log (1/π₯) β 1
fβ(x) = f(x) [logβ‘γ(1/π₯)β1γ ]
Putting f (π₯) =(1/π₯)^π₯
fβ(x) = (π/π)^π (π₯π¨π β‘γ(π/π)βπγ )
Putting fβ(x) = 0
(1/π₯)^π₯ (logβ‘γ(1/π₯)β1γ ) = 0
Since, there is only one critical point, so it will be point of maxima
Either
(π/π)^π = 0
Since, it is an exponential function
It can never be zero.
Or
(πππβ‘γ(π/π)βπγ ) = 0
log 1/π₯ = 1
Taking exponential on both sides
e^logβ‘γ1/xγ = π^1
1/π₯ = e
π = π^(βπ)
Putting π₯ = 1/π in f (x)
f (π/π) = (1/(1/π))^(1/π)
f (π/π) = π^(π/π)
So, the correct answer is (C)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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