Question

If $\left(x-1\right)\left(x^2-5x+7\right)<\left(x-1\right)$ then $x$ belongs to

• A. $\left(1,2\right)\cup \left(3,\infty \right)$
• B. $\left(-\infty ,1\right)\cup \left(2,3\right)$
• C. $\left(2,3\right)$
• D. None of the above.

Answer 1

The correct option is B

$\left(-\infty ,1\right)\cup \left(2,3\right)$

Explanation for the correct option.

Step 1. Simplify the inequality.

In the inequality $\left(x-1\right)\left(x^2-5x+7\right)<\left(x-1\right)$, gather all the terms in the left hand side and simplify the inequality.

$$\begin{gathered} \left(x-1\right)\left(x^2-5x+7\right)-\left(x-1\right)<0 \\ \Rightarrow \left(x-1\right)\left(x^2-5x+7-1\right)<0 \\ \Rightarrow \left(x-1\right)\left(x^2-2x-3x+6\right)<0 \\ \Rightarrow \left(x-1\right)\left(x\left(x-2\right)-3\left(x-2\right)\right)<0 \\ \Rightarrow \left(x-1\right)\left(x-2\right)\left(x-3\right)<0 \end{gathered}$$

Step 2. Find the solution of inequality.

The inequality $\left(x-1\right)\left(x-2\right)\left(x-3\right)<0$ divides the number line into four parts they are $\left(-\infty ,1\right),\left(1,2\right),\left(2,3\right),\left(3,\infty \right)$.

For the region $\left(-\infty ,1\right)$, check the value of $\left(x-1\right)\left(x-2\right)\left(x-3\right)$ at $x=0$.

$\begin{array}{rcl}\left(x-1\right)\left(x-2\right)\left(x-3\right)& =& \left(0-1\right)\left(0-2\right)\left(0-3\right)\\ & =& \left(-1\right)\left(-2\right)\left(-3\right)\\ & =& -6\end{array}$

As the value is less than $0$, so the region $\left(-\infty ,1\right)$ is in the solution set of $\left(x-1\right)\left(x-2\right)\left(x-3\right)<0$.

For the region $\left(1,2\right)$, check the value of $\left(x-1\right)\left(x-2\right)\left(x-3\right)$ at $x=1.5$.

$\begin{array}{rcl}\left(x-1\right)\left(x-2\right)\left(x-3\right)& =& \left(1.5-1\right)\left(1.5-2\right)\left(1.5-3\right)\\ & =& \left(0.5\right)\left(-0.5\right)\left(-1.5\right)\\ & =& 0.375\end{array}$

As the value is greater than $0$, so the region $\left(1,2\right)$ is not in the solution set of $\left(x-1\right)\left(x-2\right)\left(x-3\right)<0$.

Step 3:For the region $\left(2,3\right)$, check the value of $\left(x-1\right)\left(x-2\right)\left(x-3\right)$ at $x=2.5$.

$\begin{array}{rcl}\left(x-1\right)\left(x-2\right)\left(x-3\right)& =& \left(2.5-1\right)\left(2.5-2\right)\left(2.5-3\right)\\ & =& \left(1.5\right)\left(0.5\right)\left(-0.5\right)\\ & =& -0.375\end{array}$

As the value is less than $0$, so the region $\left(2,3\right)$ is in the solution set of $\left(x-1\right)\left(x-2\right)\left(x-3\right)<0$.

For the region $\left(3,\infty \right)$, check the value of $\left(x-1\right)\left(x-2\right)\left(x-3\right)$ at $x=4$.

$\begin{array}{rcl}\left(x-1\right)\left(x-2\right)\left(x-3\right)& =& \left(4-1\right)\left(4-2\right)\left(4-3\right)\\ & =& \left(3\right)\left(2\right)\left(1\right)\\ & =& 6\end{array}$

As the value is greater than $0$, so the region $\left(3,\infty \right)$ is not in the solution set of $\left(x-1\right)\left(x-2\right)\left(x-3\right)<0$.

So the solution set of inequality $\left(x-1\right)\left(x-2\right)\left(x-3\right)<0$ is $\left(-\infty ,1\right)$and $\left(2,3\right)$.

Thus, $x$ belongs to $\left(-\infty ,1\right)\cup \left(2,3\right)$.

Hence, the correct option is B.