If a moneylender charges 'interest' at the rate of 10 rupees per 100 rupees per half year, payable in advance, then the effective rate of interest…

To find the maximum value of $f(x) = \left(\frac{1}{x}\right)^x$f(x)=(x1​)x, we use calculus to find where the derivative equals zero.

Rewrite the function in a simpler form:

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

$$f(x) = x^{-x}$$f(x)=x−x

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Taking the natural log of both sides:

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

$$\ln f(x) = -x \ln x$$lnf(x)=−xlnx

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Differentiating both sides:

$$\frac{f'(x)}{f(x)} = -\ln x - 1$$f(x)f′(x)​=−lnx−1

$$f'(x) = f(x)(-\ln x - 1)$$f′(x)=f(x)(−lnx−1)

$$f'(x) = x^{-x}(-\ln x - 1)$$f′(x)=x−x(−lnx−1)

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For critical points, set $f'(x) = 0$f′(x)=0:

$$x^{-x}(-\ln x - 1) = 0$$x−x(−lnx−1)=0

Since $x^{-x}$x−x is always positive for $x > 0$x>0:

$$-\ln x - 1 = 0$$−lnx−1=0

$$\ln x = -1$$lnx=−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 function:

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

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

Therefore, the maximum value of $f(x) = \left(\frac{1}{x}\right)^x$f(x)=(x1​)x is $e^{1/e}$e1/e.