Question

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