Question

Solution of inequality $\frac{{\left(x-2\right)}^2\left(x-4\right)}{{\left(x-7\right)}^3}\le 0$ is equal to

• A. $\left(-\infty ,2\right]\cup \left[7,\infty \right)$
• B. $\left(-\infty ,2\right]\cup \left[5,7\right)$
• C. $\left[4,7\right)\cup \left\{2\right\}$
• D. $\left[4,7\right)$

Answer 1

The correct option is C $\left[4,7\right)\cup \left\{2\right\}$

Solving this question by wavy curve method we get:

Power of expression $\Rightarrow$ even $\Rightarrow$ no change

odd $\Rightarrow$ change

$\therefore$ Ans $[4,7)\cup \left\{2\right\}$