Question

Solve the inequality: $ 1 \leqslant \dfrac{{3{x^2} - 7x + 8}}{{{x^2} + 1}} \leqslant 2 $ .

Answer 1

Hint: In the given question, we have to solve the inequality given to us in the problem itself. So, we have made use of the algebraic and simplification rules in order to simplify the inequality and find a solution set to the inequality. One must know that we can multiply a positive number without changing the sign of inequality.

Complete step by step solution:
In the given question, we are given an inequality $ 1 \leqslant \dfrac{{3{x^2} - 7x + 8}}{{{x^2} + 1}} \leqslant 2 $ .
Now, we know that $ \left( {{x^2} + 1} \right) $ is a compulsorily positive number. So, multiplying both sides of the inequality by $ \left( {{x^2} + 1} \right) $ without changing the inequality signs, we get,
 $ \Rightarrow {x^2} + 1 \leqslant 3{x^2} - 7x + 8 \leqslant 2\left( {{x^2} + 1} \right) $
Opening the brackets, we get,
 $ \Rightarrow {x^2} + 1 \leqslant 3{x^2} - 7x + 8 \leqslant 2{x^2} + 2 $
Now, we can separate the inequality into two parts,
 $ {x^2} + 1 \leqslant 3{x^2} - 7x + 8 $ and $ 3{x^2} - 7x + 8 \leqslant 2{x^2} + 2 $
 $ \Rightarrow 2{x^2} - 7x + 7 \geqslant 0 $ and $ {x^2} - 7x + 6 \leqslant 0 $
Now, we have to first solve both the inequalities individually and then take the intersection of the solutions of both the inequalities.
So, we have, $ 2{x^2} - 7x + 7 \geqslant 0 $
The discriminant of the quadratic expression $ \left( {2{x^2} - 7x + 7} \right) $ is negative. Also, the coefficient of the $ {x^2} $ is a positive integer. So, the quadratic expression is always positive for any value of x. So, we get the solution set of the inequality $ 2{x^2} - 7x + 7 \geqslant 0 $ as the set of real numbers.
Now, we have, $ {x^2} - 7x + 6 \leqslant 0 $
Splitting the middle term into two parts, we get,
Taking x common in first two terms and $ - 6 $ common in last two terms, we get,
 $ \Rightarrow {x^2} - x - 6x + 6 \leqslant 0 $
 $ \Rightarrow x\left( {x - 1} \right) - 6\left( {x - 1} \right) \leqslant 0 $
Now, taking the factor $ \left( {x - 1} \right) $ common, we get,
 $ \Rightarrow \left( {x - 6} \right)\left( {x - 1} \right) \leqslant 0 $
Now, the product $ \left( {x - 6} \right)\left( {x - 1} \right) $ must be non-positive. So, one of the two terms should be negative and the other one should be positive.
So, we get the solution set of the inequality $ \left( {x - 6} \right)\left( {x - 1} \right) \leqslant 0 $ as $ \left[ {1,6} \right] $ .
Therefore, taking the intersection of solutions sets of both the inequalities $ 2{x^2} - 7x + 7 \geqslant 0 $ and $ {x^2} - 7x + 6 \leqslant 0 $ , we get the solution set for the inequality $ 1 \leqslant \dfrac{{3{x^2} - 7x + 8}}{{{x^2} + 1}} \leqslant 2 $ as $ \left[ {1,6} \right] $ .

Note: One must know that when multiplying both sides of the inequality by a positive integer does not change the sign of inequality but when we multiply both sides of the inequality by a negative number, the signs of inequality change. One should take care of the calculations and should recheck them to verify the final answer