Statement-I: e^{pi} pi^e.Statement-II: The | vedclass

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(A) Consider the function $f(x) = x^{1/x}$ for $x > 0$.Taking the natural logarithm,we get $\ln(f(x)) = \frac{1}{x} \ln(x)$.Differentiating with respe

Vedclass · Accepted Answer

Statement-$I$ is true,Statement-$II$ is true; Statement-$II$ is a correct explanation for Statement-$I$.

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(A) Consider the function $f(x) = x^{1/x}$ for $x > 0$.Taking the natural logarithm,we get $\ln(f(x)) = \frac{1}{x} \ln(x)$.Differentiating with respect to $x$,we get $\frac{f'(x)}{f(x)} = \frac{x(\frac{1}{x}) - \ln(x)(1)}{x^2} = \frac{1 - \ln(x)}{x^2}$.Setting $f'(x) = 0$,we find $1 - \ln(x) = 0$,which implies $\ln(x) = 1$,so $x = e$.For $x  0$ and for $x > e$,$f'(x) Since $f(x)$ has a global maximum at $x = e$,for any $x 
eq e$,$f(e) > f(x)$.Substituting $x = \pi$,we have $e^{1/e} > \pi^{1/\pi}$.Raising both sides to the power of $e\pi$,we get $(e^{1/e})^{e\pi} > (\pi^{1/\pi})^{e\pi}$,which simplifies to $e^{\pi} > \pi^e$. This proves Statement-$I$ is true.Since Statement-$II$ directly leads to the inequality in Statement-$I$,Statement-$II$ is the correct explanation for Statement-$I$.

Statement-$I$: $e^{\pi} > \pi^e$.
Statement-$II$: The function $f(x) = x^{1/x}$ attains its global maximum at $x = e$.

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Statement-$I$: $e^{\pi} > \pi^e$.Statement-$II$: The function $f(x) = x^{1/x}$ attains its global maximum at $x = e$.