Question Number 161854 by mnjuly1970 last updated on 23/Dec/21
Commented by cortano last updated on 23/Dec/21
$$\:{yx}^{\mathrm{2}} −{yx}−\mathrm{6}{y}={x}^{\mathrm{2}} −{ax}+{a} \\ $$$$\:\left({y}−\mathrm{1}\right){x}^{\mathrm{2}} +\left({a}−{y}\right){x}−\left(\mathrm{6}{y}+{a}\right)=\mathrm{0} \\ $$$$\:\Delta\geqslant\mathrm{0} \\ $$$$\:\left({a}−{y}\right)^{\mathrm{2}} +\mathrm{4}\left({y}−\mathrm{1}\right)\left(\mathrm{6}{y}+{a}\right)\geqslant\mathrm{0} \\ $$$$\:{y}^{\mathrm{2}} −\mathrm{2}{ay}+{a}^{\mathrm{2}} +\mathrm{4}\left(\mathrm{6}{y}^{\mathrm{2}} +\left({a}−\mathrm{6}\right){y}−{a}\right)\geqslant\mathrm{0} \\ $$$$\:\mathrm{25}{y}^{\mathrm{2}} +\left(\mathrm{2}{a}−\mathrm{24}\right){y}+{a}^{\mathrm{2}} −\mathrm{4}{a}\geqslant\mathrm{0} \\ $$$$\:{since}\:{R}_{{f}} ={R}\:{then} \\ $$$$\:\Rightarrow\left(\mathrm{2}{a}−\mathrm{24}\right)^{\mathrm{2}} −\mathrm{4}.\mathrm{25}\left({a}^{\mathrm{2}} −\mathrm{4}{a}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{2}} −\mathrm{24}{a}+\mathrm{144}−\mathrm{25}{a}^{\mathrm{2}} +\mathrm{100}{a}\geqslant\mathrm{0} \\ $$$$\Rightarrow−\mathrm{24}{a}^{\mathrm{2}} +\mathrm{76}{a}+\mathrm{144}\geqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{6}{a}^{\mathrm{2}} −\mathrm{19}{a}−\mathrm{36}\leqslant\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{2}{x}−\mathrm{9}\right)\left(\mathrm{3}{x}+\mathrm{4}\right)\leqslant\mathrm{0} \\ $$$$\Rightarrow−\frac{\mathrm{4}}{\mathrm{3}}\leqslant{x}\leqslant\frac{\mathrm{9}}{\mathrm{2}} \\ $$
Answered by mr W last updated on 23/Dec/21
![y=((x^2 −ax+a)/(x^2 −x−6)) (y−1)x^2 −(y−a)x−(6y+a)=0 (y−a)^2 +4(y−1)(6y+a)≥0 ⇒25y^2 +2(a−12)y+a^2 −4a≥0 (a−12)^2 −25(a^2 −4a)≤0 ⇒(2a−9)(3a+4)≥0 a≤−(4/3) a≥(9/2) i.e. a∈(−∞,−(4/3)] ∧ [(9/2),+∞)](https://www.tinkutara.com/question/Q161858.png)
$${y}=\frac{{x}^{\mathrm{2}} −{ax}+{a}}{{x}^{\mathrm{2}} −{x}−\mathrm{6}} \\ $$$$\left({y}−\mathrm{1}\right){x}^{\mathrm{2}} −\left({y}−{a}\right){x}−\left(\mathrm{6}{y}+{a}\right)=\mathrm{0} \\ $$$$\left({y}−{a}\right)^{\mathrm{2}} +\mathrm{4}\left({y}−\mathrm{1}\right)\left(\mathrm{6}{y}+{a}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{25}{y}^{\mathrm{2}} +\mathrm{2}\left({a}−\mathrm{12}\right){y}+{a}^{\mathrm{2}} −\mathrm{4}{a}\geqslant\mathrm{0} \\ $$$$\left({a}−\mathrm{12}\right)^{\mathrm{2}} −\mathrm{25}\left({a}^{\mathrm{2}} −\mathrm{4}{a}\right)\leqslant\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{2}{a}−\mathrm{9}\right)\left(\mathrm{3}{a}+\mathrm{4}\right)\geqslant\mathrm{0} \\ $$$${a}\leqslant−\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${a}\geqslant\frac{\mathrm{9}}{\mathrm{2}} \\ $$$${i}.{e}.\:{a}\in\left(−\infty,−\frac{\mathrm{4}}{\mathrm{3}}\right]\:\wedge\:\left[\frac{\mathrm{9}}{\mathrm{2}},+\infty\right) \\ $$
Commented by mnjuly1970 last updated on 23/Dec/21
$$\:\:\:{grateful}\:{mr}\:\mathrm{W} \\ $$
Commented by Tawa11 last updated on 23/Dec/21
$$\mathrm{Am}\:\mathrm{sorry}\:\mathrm{sir}. \\ $$