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Question-62676




Question Number 62676 by bshahid010@gmail.com last updated on 24/Jun/19
Commented by kaivan.ahmadi last updated on 24/Jun/19
if tanx=n>0⇒  n^2 +an+n+a−3<0⇒a(n+1)<3−n^2 −n⇒  a<((3−n−n^2 )/(n+1))
$${if}\:{tanx}={n}>\mathrm{0}\Rightarrow \\ $$$${n}^{\mathrm{2}} +{an}+{n}+{a}−\mathrm{3}<\mathrm{0}\Rightarrow{a}\left({n}+\mathrm{1}\right)<\mathrm{3}−{n}^{\mathrm{2}} −{n}\Rightarrow \\ $$$${a}<\frac{\mathrm{3}−{n}−{n}^{\mathrm{2}} }{{n}+\mathrm{1}} \\ $$
Answered by ajfour last updated on 24/Jun/19
If   t^2 +(a+1)t+(a−3)<0     for every t∈(0,∞)  ⇒ a∈{ }.
$${If}\:\:\:{t}^{\mathrm{2}} +\left({a}+\mathrm{1}\right){t}+\left({a}−\mathrm{3}\right)<\mathrm{0}\:\:\: \\ $$$${for}\:{every}\:{t}\in\left(\mathrm{0},\infty\right) \\ $$$$\Rightarrow\:{a}\in\left\{\:\right\}. \\ $$
Commented by kaivan.ahmadi last updated on 24/Jun/19
if t=1⇒ 1+a+1+a−3<0⇒2a<1⇒a<(1/2)
$${if}\:{t}=\mathrm{1}\Rightarrow\:\mathrm{1}+{a}+\mathrm{1}+{a}−\mathrm{3}<\mathrm{0}\Rightarrow\mathrm{2}{a}<\mathrm{1}\Rightarrow{a}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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