Solve $1<\left(\dfrac{3x^2-7x+8}{x^2+1}\right)\leq 2,\ \ x\in\mathbb{R}$

Question

Solve $1<\left(\dfrac{3x^2-7x+8}{x^2+1}\right)\leq 2,\ \ x\in\mathbb{R}$

options

$a.)\ 1<x<6\\ b.)\ 1 \leq x<6\\ c.)\ 1<x\leq 6\\ \color{green}{d.)\ 1\leq x \leq 6}$

I simplified it in $0<2x^2-7x+7\leq x^2+1$

I don't know how to move further.

Meanwhile using options I found that it is option $d.)$

But I am looking for a short and simple method without options.

I have studied maths up to $12$th grade.

Answer 1

As $x^2+1>0$, this is equivalent to $x^2+1<3x^2-7x+8\le 2x^2+2$, i.e. $$\begin{cases} \begin{aligned} \!2x^2-7x+7&>0\\ \!x^2-7x+6&\le 0 \end{aligned} \end{cases}$$ The first polynomial has no real root (its discriminant is $-7$), hence the first inequation is always satisfied.

The second polynomial has $1$ as a root, hence the other root is $6$. The solutions to $x^2-7x+6=(x-1)(x-6)\le 0$ (and the original inequations) is finally: $$1\le x\le 6.$$

Answer 2

You don't even really need to do much math for this question, as the 4 choices you're given mean you can skip all derivations entirely.

Note in particular that the choices your given basically ask you to figure out whether or not the equality holds at $x=6$ and $x=1$. There's not a "None of the above" option so you don't even need to check if it holds for $x$ between 1 and 6.

So checking at $x=1$:

Is $1<\frac{3-7+8}{1+1}=2\leq2$? Yes. So it can't be a) or c).

Checking at $x=6$:

Is $1<\frac{108-42+8}{36+1}=\frac{74}{37}=2\leq2$? Yes, so it can't be b). Therefore the answer is d) (process of elimination)