The maximum real number, which most exceeds its cube, is

Options

A. `1/2`
B. `1/(sqrt(3))`
C. `1/(sqrt(2))`
D. `2/(sqrt(3))`

Answer

B

Explanation

Let number be `x` then cube `=x^(3)`
Now `f(x)=x-x^(3)`
On differentiating both sides w.r.t `x` we get
`f'(x)=1-3x^(2)`
For maximum or minimum put `f'(x)=0`
`:.1-3x^(2)=0impliesx=+-1/(sqrt(3))`
For maximum `f'(x)` should be negative
`:.f''(x)-6x=-ve` at `x=1/(sqrt(3))`