Question

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

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

Answer: (B)

$\frac{x^2(3x-4{)}^3(x-2{)}^4}{(x-5{)}^5(2x-7{)}^6}\le 0$
$\Rightarrow x=0,\frac{4}{3},2,3x-4<0,x-5>0$
or $3x-4>0,x-5>0$
$\left[\because x^2,{\left(x-2\right)}^4,{\left(2x-7\right)}^6>0\right]$
$\Rightarrow x=0,\frac{4}{3},2,x<\frac{4}{3},x>5$ or $x>\frac{4}{3},x<5$
$\Rightarrow x=0,2$ and integral value between
$\frac{4}{3}<x<5$ i.e., $x=2,3,4$.
Hence positive integral solutions are $2$, $3$, $4$.