The fraction exceeding its nth power by the greatest possible number, where n ≥ 2, is
Answer & explanation
Correct answer: option 1
Let x be the fraction and let $y=x-x^n$. Then,
$\frac{d y}{d x}=1-n x^{n-1}$ and $\frac{d^2 y}{d x^2}=-n(n-1) x^{n-2}$
Now, $\frac{d y}{d x}=0 \Rightarrow x=\left(\frac{1}{n}\right)^{\frac{1}{n-1}}$
Clearly, $\frac{d^2 y}{d x^2}<0$ for $x=\left(\frac{1}{n}\right)^{\frac{1}{n-1}}$
Hence, y is maximum when $x=\left(\frac{1}{n}\right)^{n-1}$