Question

If we have the inequality as $\left( x-1 \right)\left( {{x}^{2}}-5x+7 \right)<\left( x-1 \right)$, then $x$ belongs to
$\left( 1 \right)\text{ }\left( 1,2 \right)\cup \left( 3,\infty \right)$
$\left( 2 \right)\text{ }\left( -\infty ,1 \right)\cup \left( 2,3 \right)$
$\left( 3 \right)\text{ }\left( 2,3 \right)$
$\left( 4 \right)\text{ None of the above}$

Answer 1

Hint: In this question we have been given with a polynomial expression for which we have to find the value of $x$ such that it satisfies the condition. We will solve this question by first rearranging the terms in the expression and then factoring the polynomial expression. We will then find the critical points of the value of $x$ such that the condition is satisfied and get the required domain.

Complete step-by-step solution:
We have the expression given to us as:
$\Rightarrow \left( x-1 \right)\left( {{x}^{2}}-5x+7 \right)<\left( x-1 \right)$
On transferring the term $\left( x-1 \right)$ from the right-hand side to the left-hand side, we get:
$\Rightarrow \left( x-1 \right)\left( {{x}^{2}}-5x+7 \right)-\left( x-1 \right)<0$
Now on taking the term$\left( x-1 \right)$ common from both the terms, we get:
$\Rightarrow \left( x-1 \right)\left[ \left( {{x}^{2}}-5x+7 \right)-1 \right]<0$
On opening the bracket and simplifying, we get:
$\Rightarrow \left( x-1 \right)\left( {{x}^{2}}-5x+6 \right)<0$
Now we will factorize the term $\left( {{x}^{2}}-5x+6 \right)$. On splitting the middle term, we get:
$\Rightarrow \left( x-1 \right)\left( {{x}^{2}}-3x-2x+6 \right)<0$
On taking the terms common, we get:
$\Rightarrow \left( x-1 \right)\left( x\left( x-3 \right)-2\left( x-3 \right) \right)<0$
Now on taking the term $\left( x-3 \right)$ common, we get:
$\Rightarrow \left( x-1 \right)\left( x-2 \right)\left( x-3 \right)<0$
Therefore, from the above terms, we can see that the critical points are $1,2,3$ and since the sign is strictly lesser than, all the values ranging from negative infinity till $1$ and then the numbers $2,3$ are the values which $x$ belongs to.
Therefore, we can write:
$x\in \left( -\infty ,1 \right)\cup \left( 2,3 \right)$, which is the required solution therefore, the correct option is $\left( 2 \right)$.

Note: It is to be remembered that whenever round brackets are used to write any interval, it is called an open interval as it does not include the endpoints. Therefore, $1,2,3$ are not a part of the values of $x$ because they would yield $0$. Whenever there is a closed interval, it includes the endpoints. It is denoted using square brackets.