Question

If $x$ satisfies the inequality $(x^2-x-1)(x^2-x-7)<-5,$ then which of the following statements is (are) TRUE?

• A. Number of integral values of $x$ satisfying the given inequality is $2$
• B. $x^2+1\in (2,5),$ if $x<0$
• C. $x^2-1\in (3,8),$ if $x>0$
• D. Number of integral values of $x$ satisfying the given inequality is $0$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $x^2+1\in (2,5),$ if $x<0$
C $x^2-1\in (3,8),$ if $x>0$
D Number of integral values of $x$ satisfying the given inequality is $0$

$(x^2-x-1)(x^2-x-7)<-5$
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 $\left(1\right)$ and $\left(2\right)$,
$x\in (-2,-1)\cup (2,3)$
There are no integral values of $x$ in the solution set.
For $x<0,x\in (-2,-1)$
$\Rightarrow x^2\in (1,4)$
$\Rightarrow x^2+1\in (2,5)$
For $x>0,x\in (2,3)$
$\Rightarrow x^2\in (4,9)$
$\Rightarrow x^2-1\in (3,8)$