Question

The range of values of $x$ satisfying the inequality $\frac{(x-2)(x-4)(x-7)}{(x+2)(x+4)(x+7)}>1$ is

• A. $(-\infty ,-7)\cup (-4,-2)$
• B. $(-\infty ,-7]\cup (-4,-2)$
• C. $(-\infty ,-7]\cup [-4,-2)$
• D. $(-\infty ,-7)\cup [-4,-2]$

Answer 1

The correct option is A $(-\infty ,-7)\cup (-4,-2)$

$\frac{(x-2)(x-4)(x-7)}{(x+2)(x+4)(x+7)}>1$

$\frac{(x-2)(x-4)(x-7)}{(x+2)(x+4)(x+7)}-1>0$

$\frac{(x-2)(x-4)(x-7)-(x+2)(x+4)(x+7)}{(x+2)(x+4)(x+7)}>0$

$\Rightarrow \frac{-26x^2-112}{(x+2)(x+4)(x+7)}>0$

$\Rightarrow \frac{-2(13x^2+56)}{(x+2)(x+4)(x+7)}>0$

$\Rightarrow (x+2)(x+4)(x+7)<0$
$\Rightarrow x\in (-\infty ,-7)\cup (-4,-2)$
Alternative Method :
$\frac{(x-2)(x-4)(x-7)}{(x+2)(x+4)(x+7)}>1$
Clearly, the function f(x) does not exists at $x=-2,-4,-7$
So, the solution set of the inequality does not contain these points.
From the options , only option A does not contain these points.