Question Number 24641 by math solver last updated on 23/Nov/17
Commented by math solver last updated on 23/Nov/17
$$? \\ $$
Answered by ajfour last updated on 23/Nov/17
![x^2 +ax+b=(x+(a/2))^2 +b−(a^2 /4) Case: I if −(a/2) < 0 f(x) is increasing in [0,2] so f(0)=b=2 and f(2)=4+2a+b=3 ⇒ a=−(3/2) ; b=2 Case II : if 2 < −(a/2) f(x) is decreasing in [0,2] So f(0)=b=3 and f(2)=4+2a+b=2 ⇒ a=−(5/2) ; b=3 Case III: if 0 < −(a/2) < 2 Then b−(a^2 /4) = 2 and either f(0)= b=3 or f(2)=4+2a+b=3 ⇒ case III(a): b=3 then (a^2 /4)=1 or a=±2 that is (a,b)≡(±2, 3) case III(b): 4+2a+b=3 ⇒ b=−1−2a so (a^2 /4)=b−2 = −1−2a−2 a^2 +8a+12=0 ⇒ (a+4)^2 =4 a=−2, −6 ⇒ b=−1−2a = 3 or 11 that is (a,b)≡(−2, 3) or ≡(−6, 11) .](https://www.tinkutara.com/question/Q24645.png)
$${x}^{\mathrm{2}} +{ax}+{b}=\left({x}+\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +{b}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\boldsymbol{{C}}{ase}:\:{I} \\ $$$${if}\:\:\:\:\:\:\:−\frac{{a}}{\mathrm{2}}\:\:<\:\mathrm{0} \\ $$$${f}\left({x}\right)\:{is}\:{increasing}\:{in}\:\left[\mathrm{0},\mathrm{2}\right] \\ $$$${so}\:\:\:\:{f}\left(\mathrm{0}\right)={b}=\mathrm{2} \\ $$$${and}\:\:{f}\left(\mathrm{2}\right)=\mathrm{4}+\mathrm{2}{a}+{b}=\mathrm{3} \\ $$$$\Rightarrow\:\:\:\boldsymbol{{a}}=−\frac{\mathrm{3}}{\mathrm{2}}\:\:;\:\:\boldsymbol{{b}}=\mathrm{2} \\ $$$$\boldsymbol{{C}}{ase}\:{II}\:: \\ $$$${if}\:\:\:\:\:\:\:\:\:\mathrm{2}\:<\:−\frac{{a}}{\mathrm{2}}\: \\ $$$${f}\left({x}\right)\:{is}\:{decreasing}\:{in}\:\left[\mathrm{0},\mathrm{2}\right] \\ $$$${So}\:\:\:\:\:\:{f}\left(\mathrm{0}\right)={b}=\mathrm{3}\:\:\:\:{and} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{f}\left(\mathrm{2}\right)=\mathrm{4}+\mathrm{2}{a}+{b}=\mathrm{2} \\ $$$$\Rightarrow\:\:\:\boldsymbol{{a}}=−\frac{\mathrm{5}}{\mathrm{2}}\:\:;\:\:\boldsymbol{{b}}=\mathrm{3} \\ $$$$\boldsymbol{{C}}{ase}\:{III}: \\ $$$${if}\:\:\:\:\:\:\:\:\:\mathrm{0}\:<\:−\frac{{a}}{\mathrm{2}}\:<\:\mathrm{2} \\ $$$${Then}\:\:\:\:\:\:{b}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\:=\:\mathrm{2} \\ $$$${and}\:\:\:{either}\:\:\:{f}\left(\mathrm{0}\right)=\:{b}=\mathrm{3}\:\:{or}\:\:\: \\ $$$$\:\:\:\:{f}\left(\mathrm{2}\right)=\mathrm{4}+\mathrm{2}{a}+{b}=\mathrm{3} \\ $$$$\Rightarrow\:\:\:\:{case}\:{III}\left({a}\right): \\ $$$$\:\:\:\:\:\:\:{b}=\mathrm{3}\:\:\:{then}\:\:\:\frac{{a}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{1}\:\:\:{or}\:\:{a}=\pm\mathrm{2} \\ $$$${that}\:{is}\:\:\:\:\left(\boldsymbol{{a}},\boldsymbol{{b}}\right)\equiv\left(\pm\mathrm{2},\:\mathrm{3}\right) \\ $$$${case}\:{III}\left({b}\right): \\ $$$$\mathrm{4}+\mathrm{2}{a}+{b}=\mathrm{3}\:\:\:\:\Rightarrow\:\:\:{b}=−\mathrm{1}−\mathrm{2}{a} \\ $$$${so}\:\:\:\:\frac{{a}^{\mathrm{2}} }{\mathrm{4}}={b}−\mathrm{2}\:=\:−\mathrm{1}−\mathrm{2}{a}−\mathrm{2} \\ $$$${a}^{\mathrm{2}} +\mathrm{8}{a}+\mathrm{12}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\left({a}+\mathrm{4}\right)^{\mathrm{2}} =\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:{a}=−\mathrm{2},\:−\mathrm{6} \\ $$$$\Rightarrow\:\:{b}=−\mathrm{1}−\mathrm{2}{a}\:=\:\mathrm{3}\:\:{or}\:\:\mathrm{11} \\ $$$${that}\:{is}\:\:\:\left(\boldsymbol{{a}},\boldsymbol{{b}}\right)\equiv\left(−\mathrm{2},\:\mathrm{3}\right)\:\:\:{or} \\ $$$$\:\:\:\:\:\:\:\:\equiv\left(−\mathrm{6},\:\mathrm{11}\right)\:. \\ $$