Question

Prove that $\frac{2n^2-3n-2}{n^2}$ can not exceed $3\frac{1}{8}$.

Answer 1

Proof by contradiction:-

If possible let $\frac{2n^2-3n-2}{n^2}$ exceeds $3\frac{1}{8}$.

Then,

$\frac{2n^2-3n-2}{n^2}>$ $3\frac{1}{8}$

or, $\frac{2n^2-3n-2}{n^2}>$ $\frac{25}{8}$

or, $2-\frac{3n+2}{n^2}>$$\frac{25}{8}$

or, $-\frac{3n+2}{n^2}>$$\frac{9}{8}$

or, $9n^2+24n+16<0$

or, $(3n+4{)}^2<0$.

Which is not possible as square of any real number is always $\ge 0$.

So, $\frac{2n^2-3n-2}{n^2}$ can not exceed $3\frac{1}{8}$