Question

The real number which most exceeds its cube is

• A. $\frac{1}{2}$
• B. $\frac{1}{\sqrt{3}}$
• C. $\frac{1}{\sqrt{2}}$
• D. None of these

Answer 1

The correct option is B

$\frac{1}{\sqrt{3}}$

Compute the required number:

Let the number be $x$, then the cube of the number is $x^3$.

$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}-{\mathrm{x}}^3$

Differentiate with respect to $x$,

$\mathrm{f}\prime \left(\mathrm{x}\right)=1-3{\mathrm{x}}^2$

Put $\mathrm{f}\prime \left(\mathrm{x}\right)=0$

$$\begin{gathered} 1-3x^2=0 \\ \Rightarrow x^2=\frac{1}{3} \\ \Rightarrow x=\pm \frac{1}{\sqrt{3}} \end{gathered}$$

Differentiate $\mathrm{f}\text{'}\left(\mathrm{x}\right)$with respect to $x$,

$\mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right)=-6\mathrm{x}$

Substitute the value of $x$,

$\mathrm{f}\text{'}\text{'}\left(\frac{1}{\sqrt{3}}\right)=-\frac{6}{\sqrt{3}}$

$\mathrm{f}\text{'}\text{'}\left(-\frac{1}{\sqrt{3}}\right)=+\frac{6}{\sqrt{3}}$

Since, for $x=\frac{1}{\sqrt{3}}$, $\mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right)$ is negative.

Therefore, $\frac{1}{\sqrt{3}}$ is the real number that most exceeds its cube.

Hence, option (B) is the correct option.