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Question Number 7413 by Tawakalitu. last updated on 28/Aug/16

Answered by sandy_suhendra last updated on 28/Aug/16

9) the curve through the point (−1,0), (3,0) and (−2,−10) f(x)=a(x−x_1 )(x−x_2 ) which x_1 =−1 and x_2 =3 −10=a(−2+1)(−2−3) −10=5a ⇒ a=−2 f(x)=−2(x+1)(x−3) =−2x^2 +4x+6

$$\left.\mathrm{9}\right)\:{the}\:{curve}\:{through}\:{the}\:{point}\:\left(−\mathrm{1},\mathrm{0}\right),\:\left(\mathrm{3},\mathrm{0}\right)\:{and}\:\left(−\mathrm{2},−\mathrm{10}\right) \\ $$$$\:\:\:\:\:{f}\left({x}\right)={a}\left({x}−{x}_{\mathrm{1}} \right)\left({x}−{x}_{\mathrm{2}} \right)\:{which}\:{x}_{\mathrm{1}} =−\mathrm{1}\:{and}\:{x}_{\mathrm{2}} =\mathrm{3} \\ $$$$\:\:\:\:\:\:−\mathrm{10}={a}\left(−\mathrm{2}+\mathrm{1}\right)\left(−\mathrm{2}−\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:−\mathrm{10}=\mathrm{5}{a}\:\Rightarrow\:{a}=−\mathrm{2} \\ $$$${f}\left({x}\right)=−\mathrm{2}\left({x}+\mathrm{1}\right)\left({x}−\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{6} \\ $$

Commented by Tawakalitu. last updated on 28/Aug/16

Thanks so much sir

$${Thanks}\:{so}\:{much}\:{sir} \\ $$

Answered by sandy_suhendra last updated on 28/Aug/16

10a) AB^(→) =OB^(→) − OA^(→) =(3i+4j)−(2i−3j)=i+7j b) ∣AB^(→) ∣=(√(1^2 +7^2 ))=(√(50))=5(√2) c) ∣OA^(→) ∣=(√(2^2 +(−3)^2 ))=(√(13)) ∣OB^(→) ∣=(√(3^2 +4^2 ))=(√(25))=5 cos ∠AOB = ((OA^(→) ∙OB^(→) )/(∣OA^(→) ∣ ∣OB^(→) ∣)) = ((2×3+(−3)×4)/(((√(13)))×5)) = ((−6)/(5(√(13)))) ∠AOB = arc cos ((−6)/(5(√(13))))

$$\left.\mathrm{10}{a}\right)\:\overset{\rightarrow} {{AB}}\:=\overset{\rightarrow} {{OB}}\:−\:\overset{\rightarrow} {{OA}}=\left(\mathrm{3}{i}+\mathrm{4}{j}\right)−\left(\mathrm{2}{i}−\mathrm{3}{j}\right)={i}+\mathrm{7}{j} \\ $$$$\left.\:\:\:\:\:{b}\right)\:\mid\overset{\rightarrow} {{AB}}\mid=\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} }=\sqrt{\mathrm{50}}=\mathrm{5}\sqrt{\mathrm{2}} \\ $$$$\left.\:\:\:\:\:{c}\right)\:\mid\overset{\rightarrow} {{OA}}\mid=\sqrt{\mathrm{2}^{\mathrm{2}} +\left(−\mathrm{3}\right)^{\mathrm{2}} }=\sqrt{\mathrm{13}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mid\overset{\rightarrow} {{OB}}\mid=\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }=\sqrt{\mathrm{25}}=\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:{cos}\:\angle{AOB}\:=\:\frac{\overset{\rightarrow} {{OA}}\centerdot\overset{\rightarrow} {{OB}}}{\mid\overset{\rightarrow} {{OA}}\mid\:\mid\overset{\rightarrow} {{OB}}\mid}\:=\:\frac{\mathrm{2}×\mathrm{3}+\left(−\mathrm{3}\right)×\mathrm{4}}{\left(\sqrt{\mathrm{13}}\right)×\mathrm{5}}\:=\:\frac{−\mathrm{6}}{\mathrm{5}\sqrt{\mathrm{13}}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\angle{AOB}\:=\:{arc}\:{cos}\:\frac{−\mathrm{6}}{\mathrm{5}\sqrt{\mathrm{13}}} \\ $$

Commented by Tawakalitu. last updated on 28/Aug/16

Thanks so much sir

$${Thanks}\:{so}\:{much}\:{sir} \\ $$

Answered by sandy_suhendra last updated on 28/Aug/16

8) f(x)=x^2 +2kx+5 ⇒ a=1 ; b=2k ; c=5 always positive ⇒ a>0 and D<0 a=1 so a>0 D=b^2 −4ac <0 (2k^2 )−4×1×5<0 4k^2 −20<0 k^2 −5<0 (k+(√5))(k−(√5))<0 −(√5) < k < (√5)

$$\left.\mathrm{8}\right)\:{f}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{2}{kx}+\mathrm{5}\:\Rightarrow\:{a}=\mathrm{1}\:;\:{b}=\mathrm{2}{k}\:;\:{c}=\mathrm{5} \\ $$$$\:\:\:\:\:\:{always}\:{positive}\:\Rightarrow\:{a}>\mathrm{0}\:{and}\:{D}<\mathrm{0} \\ $$$$\:\:\:\:\:\:{a}=\mathrm{1}\:{so}\:{a}>\mathrm{0} \\ $$$$\:\:\:\:\:\:{D}={b}^{\mathrm{2}} −\mathrm{4}{ac}\:<\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}{k}^{\mathrm{2}} \right)−\mathrm{4}×\mathrm{1}×\mathrm{5}<\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}{k}^{\mathrm{2}} −\mathrm{20}<\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{k}^{\mathrm{2}} −\mathrm{5}<\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({k}+\sqrt{\mathrm{5}}\right)\left({k}−\sqrt{\mathrm{5}}\right)<\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\sqrt{\mathrm{5}}\:\:<\:{k}\:<\:\sqrt{\mathrm{5}} \\ $$$$ \\ $$

Commented by nume1114 last updated on 28/Aug/16

I think when a>0 and D<0, f(x) is always positive. So, D=b^2 −4ac<0 (2k)^2 −4×1×5<0 (k+(√5))(k−(√5))<0 −(√5)<k<(√5)

$${I}\:{think}\:{when}\:{a}>\mathrm{0}\:{and}\:{D}<\mathrm{0}, \\ $$$${f}\left({x}\right)\:{is}\:{always}\:{positive}. \\ $$$${So}, \\ $$$$\:\:\:\:{D}={b}^{\mathrm{2}} −\mathrm{4}{ac}<\mathrm{0} \\ $$$$\:\:\:\:\left(\mathrm{2}{k}\right)^{\mathrm{2}} −\mathrm{4}×\mathrm{1}×\mathrm{5}<\mathrm{0} \\ $$$$\:\:\:\:\left({k}+\sqrt{\mathrm{5}}\right)\left({k}−\sqrt{\mathrm{5}}\right)<\mathrm{0} \\ $$$$\:\:\:\:−\sqrt{\mathrm{5}}<{k}<\sqrt{\mathrm{5}} \\ $$

Commented by Rasheed Soomro last updated on 28/Aug/16

f(x)=x^2 +6x+8 f(−3)=(−3)^2 +6(−3)+8=9−18+8=−1<0 D=4>0 and a=1>0

$${f}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{8} \\ $$$${f}\left(−\mathrm{3}\right)=\left(−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{6}\left(−\mathrm{3}\right)+\mathrm{8}=\mathrm{9}−\mathrm{18}+\mathrm{8}=−\mathrm{1}<\mathrm{0} \\ $$$${D}=\mathrm{4}>\mathrm{0}\:{and}\:{a}=\mathrm{1}>\mathrm{0} \\ $$

Commented by Tawakalitu. last updated on 28/Aug/16

Thanks so much sir

$${Thanks}\:{so}\:{much}\:{sir} \\ $$

Commented by sandy_suhendra last updated on 28/Aug/16

That′s right. Sorry I have made a mistake. I′ve corrected it

$${That}'{s}\:{right}.\:{Sorry}\:{I}\:{have}\:{made}\:{a}\:{mistake}.\:{I}'{ve}\:{corrected}\:{it} \\ $$

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