The maximum value of left(frac{1}{x}right)^{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $ y = \left(\frac{1}{x}\right)^{x} $. Taking the natural logarithm on both sides,we get $ \ln y = x \ln \left(\frac{1}{x}\right) = -x \ln x $.

Vedclass · Accepted Answer

$ e^{1/e} $

Vedclass · Answer

(C) Let $ y = \left(\frac{1}{x}\right)^{x} $. Taking the natural logarithm on both sides,we get $ \ln y = x \ln \left(\frac{1}{x}\right) = -x \ln x $. Differentiating with respect to $ x $,we have $ \frac{1}{y} \frac{dy}{dx} = - (1 \cdot \ln x + x \cdot \frac{1}{x}) = -(1 + \ln x) $. Thus,$ \frac{dy}{dx} = -y(1 + \ln x) $. Setting $ \frac{dy}{dx} = 0 $,we get $ 1 + \ln x = 0 $,which implies $ \ln x = -1 $,so $ x = e^{-1} = \frac{1}{e} $. To check for the maximum,we find the second derivative $ \frac{d^2y}{dx^2} = -\frac{dy}{dx}(1 + \ln x) - y(\frac{1}{x}) $. At $ x = \frac{1}{e} $,$ \ln x = -1 $,so $ \frac{d^2y}{dx^2} = 0 - y(e) = -y e The maximum value is $ y = \left(\frac{1}{1/e}\right)^{1/e} = e^{1/e} $.

The maximum value of $ \left(\frac{1}{x}\right)^{x} $ is

Similar Questions

In a seconds pendulum,the mass of the bob is $30\, g$. If it is replaced by a $90\, g$ mass,then its time period will be ... $\sec$.

The equation of the hyperbola whose directrix is $2x + y = 1$,focus is $(1, 1)$,and eccentricity $e = \sqrt{3}$ is:

If $\alpha$ and $\beta$ are imaginary cube roots of unity,then the value of $\alpha^4 + \beta^{28} + \frac{1}{\alpha\beta}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module