Question

If $x$ satisfies the inequality $(x^2-x-1)(x^2-x-7)<-5,$ then number of integral value(s) of $x$ satisfying the given inequality is

• A. $4$
• B. $5$
• C. $0$
• D. $1$

Answer 1

The correct option is C $0$

$(x^2-x-1)(x^2-x-7)<-5$
Let $x^2-x=t$
Then, $(t-1)(t-7)<-5$
$\Rightarrow t^2-8t+12<0$
$\Rightarrow (t-2)(t-6)<0$
$\Rightarrow t\in (2,6)$
$\Rightarrow 2<x^2-x<6$

Case $\text{I}:x^2-x>2$
$\therefore x^2-x-2>0$
$\Rightarrow (x+1)(x-2)>0$
$\therefore x\in (-\infty ,-1)\cup (2,\infty )\cdots \left(1\right)$

Case $\text{II}:x^2-x<6$
$\therefore x^2-x-6<0$
$\Rightarrow (x+2)(x-3)<0$
$\therefore x\in (-2,3)\cdots \left(2\right)$

From $\left(1\right)\cap \left(2\right),$
$x\in (-2,-1)\cup (2,3)$
There is no integral value of $x$ in the solution set.