Question

The set of values of $x$ satisfying the inequalities $(x-1)(x-2)<0$ and $(3x-7)(2x-3)>0$ is

• A. $(1,2)$
• B. $(2,7/3)$
• C. $(1,7/3)$
• D. $(1,3/2)$

Answer 1

The correct option is D $(1,3/2)$

Given that,

$(x-1)(x-2)<0$ and $(3x-7)(2x-3)>0$

$(3x-7)(2x-3)>0$ can be written as

$\Rightarrow 3(x-\frac{7}{3})2(x-\frac{3}{2})>0$

$\Rightarrow (x-\frac{7}{3})(x-\frac{3}{2})>0$

So, according to this equation we have,

$\Rightarrow x<\frac{3}{2}$ and $x>\frac{7}{3}$

Now consider $(x-1)(x-2)<0$

So, according to this equation we have,

$\Rightarrow 1<x<2$

Considering these constraints we have

$\Rightarrow x$ lies in $(1,\frac{3}{2})$