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

Answer: (B)

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

Answer 2

Let number = x, then cube = $x^{3}$

Now $f(x)$ = $x - x^{3}$ (Maximum) ⇒ $f^{'}(x) = 1 - 3x^{2}$

Put $f^{'}(x) = 0$ ⇒ $1 - 3x^{2} = 0$ ⇒ $x = \pm \frac{1}{\sqrt{3}}$

Because $f^{''}(x) = - 6x = - ve$. when $x = + \frac{1}{\sqrt{3}}$.