CUET UG 2025 — Mathematics Calculus

To find the maximum value of $\left(\frac{1}{x}\right)^x$(x1​)x for $x > 0$x>0, rewrite the expression in a simpler form:

$$\left(\frac{1}{x}\right)^x = \frac{1^x}{x^x}$$(x1​)x=xx1x​

$$= \frac{1}{x^x}$$=xx1​

$$= x^{-x}$$=x−x

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Let $f(x) = x^{-x}$f(x)=x−x. To find the maximum, take the natural logarithm:

$$\ln(f(x)) = \ln(x^{-x})$$ln(f(x))=ln(x−x)

$$= -x \ln(x)$$=−xln(x)

Let $g(x) = -x \ln(x)$g(x)=−xln(x)

Since $\ln$ln is an increasing function, the maximum of $g(x)$g(x) occurs at the same point as the maximum of $f(x)$f(x).

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To find the maximum, take the derivative using the product rule:

$$g'(x) = -\left[\ln(x) + x \cdot \frac{1}{x}\right]$$g′(x)=−[ln(x)+x⋅x1​]

$$= -[\ln(x) + 1]$$=−[ln(x)+1]

$$= -\ln(x) - 1$$=−ln(x)−1

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Setting the derivative equal to zero:

$$-\ln(x) - 1 = 0$$−ln(x)−1=0

$$\ln(x) = -1$$ln(x)=−1

$$x = e^{-1}$$x=e−1

$$x = \frac{1}{e}$$x=e1​

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Substituting $x = \frac{1}{e}$x=e1​ into the original expression:

$$f\left(\frac{1}{e}\right) = \left(\frac{1}{\frac{1}{e}}\right)^{\frac{1}{e}}$$f(e1​)=(e1​1​)e1​

$$= e^{\frac{1}{e}}$$=ee1​

Therefore, the maximum value is $e^{1/e}$e1/e.