The positive number which exceeds its cube by the greatest possible a…

The correct answer is Option (1) → \(\displaystyle\frac{1}{\sqrt{3}}\)

Let x be the number. The amount by which it exceeds its cube is

    \( f(x) = x - x^{3} \)

Setting the derivative equal to zero:

    \( f'(x) = 1 - 3x^{2} = 0 \)

    \( \displaystyle x^{2} = \frac{1}{3} \)

    \( \displaystyle x = \frac{1}{\sqrt{3}} \) (x is positive)

Also \(f''(x) = -6x \lt 0\) for x > 0, so this is a local maximum, and it is the only critical point for x > 0. So the required number is \(\frac{1}{\sqrt{3}}\).