Statement-I: Among the numbers 1, 2^{1/2}, 3^ | vedclass

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

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Statement-$I$ is true,Statement-$II$ is true. Statement-$II$ is the 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,$\ln(f(x)) = \frac{1}{x} \ln(x)$.Differentiating with respect to $x$,$\frac{f'(x)}{f(x)} = \frac{x(\frac{1}{x}) - \ln(x)(1)}{x^2} = \frac{1 - \ln(x)}{x^2}$.Thus,$f'(x) = x^{1/x} \left( \frac{1 - \ln(x)}{x^2} \right)$.$f'(x) > 0$ when $1 - \ln(x) > 0$,i.e.,$\ln(x) $f'(x)  1$,which means $x > e$.So,Statement-$II$ is true.Now,$e \approx 2.718$. Since $2 Comparing $2^{1/2}$ and $3^{1/3}$: Since $2 Comparing $3^{1/3}$ and $4^{1/4}$: Since $3 Since $f(x)$ decreases for $x > e$,$f(3) > f(4) > f(5) > f(6) > f(7)$.Thus,$3^{1/3}$ is indeed the maximum value among the given numbers.Therefore,Statement-$I$ is true and Statement-$II$ is the correct explanation for Statement-$I$.

Statement-$I$: Among the numbers $1, 2^{1/2}, 3^{1/3}, 4^{1/4}, 5^{1/5}, 6^{1/6}, 7^{1/7}$,the maximum is $3^{1/3}$.
Statement-$II$: The function $f(x) = x^{1/x}$ increases for $0 < x < e$ and decreases for $x > e$.

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Statement-$I$: Among the numbers $1, 2^{1/2}, 3^{1/3}, 4^{1/4}, 5^{1/5}, 6^{1/6}, 7^{1/7}$,the maximum is $3^{1/3}$.Statement-$II$: The function $f(x) = x^{1/x}$ increases for $0 < x < e$ and decreases for $x > e$.