Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 96712 by john santu last updated on 04/Jun/20

Answered by bobhans last updated on 04/Jun/20

since f(x) is defined on (0,+∞), from { ((2a^2 +a+1=2(a+(1/4))^2 +(7/8)>0)),((3a^2 −4a+1=(3a−1)(a−1)>0)) :} we get a<(1/3) ∪ a>1...(i) since f(x) is a decreasing function on (0, +∞) so 2a^2 +a+1 >3a^2 −4a+1 ⇒a^2 −5a < 0 , thus 0<a<5...(ii) combining this with (i) we have 0<a<(1/3) ∪ 1<a<5

$$\mathrm{since}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{defined}\:\mathrm{on}\:\left(\mathrm{0},+\infty\right),\:\mathrm{from} \\ $$$$\begin{cases}{\mathrm{2}{a}^{\mathrm{2}} +{a}+\mathrm{1}=\mathrm{2}\left({a}+\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{7}}{\mathrm{8}}>\mathrm{0}}\\{\mathrm{3}{a}^{\mathrm{2}} −\mathrm{4}{a}+\mathrm{1}=\left(\mathrm{3}{a}−\mathrm{1}\right)\left({a}−\mathrm{1}\right)>\mathrm{0}}\end{cases} \\ $$$$\mathrm{we}\:\mathrm{get}\:{a}<\frac{\mathrm{1}}{\mathrm{3}}\:\cup\:{a}>\mathrm{1}...\left({i}\right) \\ $$$${since}\:{f}\left({x}\right)\:{is}\:{a}\:{decreasing}\:{function} \\ $$$${on}\:\left(\mathrm{0},\:+\infty\right)\:\mathrm{so}\:\mathrm{2}{a}^{\mathrm{2}} +{a}+\mathrm{1}\:>\mathrm{3}{a}^{\mathrm{2}} −\mathrm{4}{a}+\mathrm{1} \\ $$$$\Rightarrow{a}^{\mathrm{2}} −\mathrm{5}{a}\:<\:\mathrm{0}\:,\:\mathrm{thus}\:\mathrm{0}<{a}<\mathrm{5}...\left({ii}\right) \\ $$$${combining}\:{this}\:{with}\:\left({i}\right)\:\mathrm{we}\:\mathrm{have}\: \\ $$$$\mathrm{0}<{a}<\frac{\mathrm{1}}{\mathrm{3}}\:\cup\:\mathrm{1}<{a}<\mathrm{5}\: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com