Question

The greatest of the numbers $1,2^{1/2},3^{1/3},4^{1/4},5^{1/5},6^{1/6}$ and $7^{1/7}$ is

• A. $2^{1/2}$
• B. $3^{1/3}$
• C. $7^{1/7}$
• D. $6^{1/6}$

Answer 1

The correct option is B $3^{1/3}$

Let $f\left(x\right)=x^{1/x}$

For $x>0$, $f\left(x\right)$ is positive and attains a maximum at $f^{\prime }\left(x\right)=0$

Hence differentiating $f\left(x\right)$ with respect to $x$ we get

$y=x^{1/x}$

$x\ln y=\ln x$

Differentiating both sides with respect to x, we get

$\ln y+\frac{x{y}^{\prime }}{y}=\frac{1}{x}$

$\frac{x{y}^{\prime }}{y}=\frac{1}{x}-\ln \left(x^{1/x}\right)$

$x{y}^{\prime }=\left(x^{1/x}\right)\left(\frac{1-\ln x}{x}\right)$

$y^{\prime }=\left(x^{1/x}\right)\left(\frac{1-\ln x}{x^2}\right)$

Since $\frac{x^{1/x}}{x^2}$ will not be equal to $0$,

For $y^{\prime }=0$, we get

$1-\ln x=0$

$\ln x=1$

$\therefore x=e$

As $y^{\prime }$ changes sign from negative to positive, $x=e$ is a point of maximum.

Now, $2<e<3$. Hence the maximum values of the function occurs between $2$ and $3$. Hence, let us compare $f\left(2\right)$ and $f\left(3\right)$

$f\left(2\right)=\sqrt{2}$

$f\left(3\right)=3^{1/3}$

$f(2{)}^6=8$

$f(3{)}^6=9$

$\Rightarrow f(3{)}^6>f(2{)}^6$

$\Rightarrow f\left(3\right)>f\left(2\right)$

Hence, $f\left(3\right)$ is the greatest among the given terms.