Question

The fraction exceeding its $p^{th}$ power by the greatest number possible, where $p\ge 2$, is

• A. ${\left(\frac{1}{p}\right)}^{1/(p-1)}$
• B. ${\left(\frac{1}{p}\right)}^{(p-1)}$
• C. $p^{1/(p-1)}$
• D. None of these

Answer 1

The correct option is A ${\left(\frac{1}{p}\right)}^{1/(p-1)}$

Let $y=x-x^p$, where $x$ is the fraction

$\Rightarrow \frac{dy}{dx}=1-p{x}^{p-1}$

For maximum or minimum, $\frac{dy}{dx}=0$

$\Rightarrow 1-p{x}^{p-1}=0\Rightarrow x={\left(\frac{1}{p}\right)}^{1/(p-1)}$

Now, $\frac{d^{2}y}{{dx}^2}=-p(p-1)x^{p-2}$

$\therefore {\left[\frac{d^{2}y}{{dx}^2}\right]}_{x={\left(\frac{1}{p}\right)}^{1/(p-1)}}=-p(p-1){\left(\frac{1}{p}\right)}^{(p-2)/(p-1)}<0$

$\therefore y$ is maximum at $x={\left(\frac{1}{p}\right)}^{1/(p-1)}$