Solve the following inequalities for x:

$\frac{2x-3}{(x-2)(x-4)}≤0$

$\left(−∞,\frac{3}{2}\right]∪(2,4)$

The correct answer is Option (1) → $\left(−∞,\frac{3}{2}\right]∪(2,4)$

Given $\frac{2x-3}{(x-2)(x-4)}≤0$. First, we note that $x≠ 2, 4$.

Since $(x-2)^2 (x-4)^2 > 0$ for all $x ∈ R, x≠ 2, 4$.

$\frac{2x-3}{(x-2)(x-4)}≤0⇒(2x-3)(x-2)(x-4)≤0$

$⇒2(x-\frac{3}{2})(x − 2) (x − 4) ≤ 0 ⇒ (x-\frac{3}{2}) (x − 2) (x −4) ≤ 0$   ...(1)

Mark the numbers $\frac{3}{2},2$ and $4$ on the number line.

By the method of intervals, the inequality (1) is satisfied when $x ≤\frac{3}{2}$ or $2≤x≤4$ but $x≠ 2, 4$.

∴ The solution set is $\left(−∞,\frac{3}{2}\right]∪(2,4)$.