Question Number 20051 by Tinkutara last updated on 22/Aug/17
![If x ∈ R then ((x^2 + 2x + a)/(x^2 + 4x + 3a)) can take all real values if (1) a ∈ (0, 2) (2) a ∈ [0, 1] (3) a ∈ [−1, 1] (4) None of these](https://www.tinkutara.com/question/Q20051.png)
$$\mathrm{If}\:{x}\:\in\:{R}\:\mathrm{then}\:\frac{{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:+\:{a}}{{x}^{\mathrm{2}} \:+\:\mathrm{4}{x}\:+\:\mathrm{3}{a}}\:\mathrm{can}\:\mathrm{take}\:\mathrm{all} \\ $$$$\mathrm{real}\:\mathrm{values}\:\mathrm{if} \\ $$$$\left(\mathrm{1}\right)\:{a}\:\in\:\left(\mathrm{0},\:\mathrm{2}\right) \\ $$$$\left(\mathrm{2}\right)\:{a}\:\in\:\left[\mathrm{0},\:\mathrm{1}\right] \\ $$$$\left(\mathrm{3}\right)\:{a}\:\in\:\left[−\mathrm{1},\:\mathrm{1}\right] \\ $$$$\left(\mathrm{4}\right)\:\mathrm{None}\:\mathrm{of}\:\mathrm{these} \\ $$
Answered by ajfour last updated on 22/Aug/17
![let y=((x^2 +2x+a)/(x^2 +4x+3a)) y(x^2 +4x+3a)=x^2 +2x+a (y−1)x^2 +2(2y−1)x+a(3y−1)=0 y= every real must yield at least one value of real x. So D≥0 ⇒ 4(2y−1)^2 ≥ 4a(y−1)(3y−1) (2y−1)^2 ≥ a(y−1)(3y−1) 4y^2 −4y+1 ≥ a(3y^2 −4y+1) (3a−4)y^2 −4(a−1)y+a−1 ≤ 0 ⇒ 3a−4 <0 or a< (4/3) and discriminant ≤ 0 ⇒ 16(a−1)^2 ≤ 4(3a−4)(a−1) 4(a−1)^2 −(3a−4)(a−1) ≤0 4a^2 −8a+4−3a^2 +7a−4 ≤ 0 a^2 −a ≤ 0 a(a−1) ≤ 0 ⇒ a ∈ [0, 1] .](https://www.tinkutara.com/question/Q20153.png)
$${let}\:{y}=\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}+{a}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}{a}} \\ $$$$\:\:{y}\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}{a}\right)={x}^{\mathrm{2}} +\mathrm{2}{x}+{a} \\ $$$$\:\left({y}−\mathrm{1}\right){x}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{2}{y}−\mathrm{1}\right){x}+{a}\left(\mathrm{3}{y}−\mathrm{1}\right)=\mathrm{0} \\ $$$${y}=\:{every}\:{real}\:{must}\:{yield}\:{at}\:{least}\: \\ $$$${one}\:{value}\:{of}\:{real}\:{x}.\:{So}\:{D}\geqslant\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{4}\left(\mathrm{2}{y}−\mathrm{1}\right)^{\mathrm{2}} \:\geqslant\:\mathrm{4}{a}\left({y}−\mathrm{1}\right)\left(\mathrm{3}{y}−\mathrm{1}\right) \\ $$$$\:\:\left(\mathrm{2}{y}−\mathrm{1}\right)^{\mathrm{2}} \:\geqslant\:{a}\left({y}−\mathrm{1}\right)\left(\mathrm{3}{y}−\mathrm{1}\right) \\ $$$$\:\:\:\mathrm{4}{y}^{\mathrm{2}} −\mathrm{4}{y}+\mathrm{1}\:\geqslant\:{a}\left(\mathrm{3}{y}^{\mathrm{2}} −\mathrm{4}{y}+\mathrm{1}\right) \\ $$$$\:\:\:\left(\mathrm{3}{a}−\mathrm{4}\right){y}^{\mathrm{2}} −\mathrm{4}\left({a}−\mathrm{1}\right){y}+{a}−\mathrm{1}\:\leqslant\:\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{3}{a}−\mathrm{4}\:<\mathrm{0}\:\:{or}\:\:\:{a}<\:\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${and}\:{discriminant}\:\leqslant\:\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{16}\left({a}−\mathrm{1}\right)^{\mathrm{2}} \:\leqslant\:\mathrm{4}\left(\mathrm{3}{a}−\mathrm{4}\right)\left({a}−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{4}\left({a}−\mathrm{1}\right)^{\mathrm{2}} \:−\left(\mathrm{3}{a}−\mathrm{4}\right)\left({a}−\mathrm{1}\right)\:\leqslant\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{4}{a}^{\mathrm{2}} −\mathrm{8}{a}+\mathrm{4}−\mathrm{3}{a}^{\mathrm{2}} +\mathrm{7}{a}−\mathrm{4}\:\leqslant\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:{a}^{\mathrm{2}} −{a}\:\leqslant\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:{a}\left({a}−\mathrm{1}\right)\:\leqslant\:\mathrm{0} \\ $$$$\:\:\:\:\Rightarrow\:\:\:\boldsymbol{{a}}\:\in\:\left[\mathrm{0},\:\mathrm{1}\right]\:. \\ $$
Commented by Tinkutara last updated on 23/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$