Question

The number of positive and integral solutions of $\frac{x^2(5x-4{)}^3(x-3{)}^4}{(x-6{)}^5(3x-7{)}^6}\le 0$ is

Answer 1

$\frac{x^2(5x-4{)}^3(x-3{)}^4}{(x-6{)}^5(3x-7{)}^6}\le 0$
$x\ne 6,\frac{7}{3}$
We have,
$\frac{(x-3{)}^4}{(3x-7{)}^6}\ge 0$ (Ratio of two squares)
$\Rightarrow \frac{(5x-4{)}^3}{(x-6{)}^5}\le 0$
Hence,
$\frac{4}{5}\le x<6$
The integral values that $x$ can take to satisfy the inequality are $1,2,3,4,5$. Hence, the number of positive integral solutions $=5$