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1-for-x-gt-0-find-the-value-of-m-to-1-log-5-x-2-1-log-5-mx-2-4x-m-verify-x-

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Question Number 152208 by SOMEDAVONG last updated on 26/Aug/21
1.for ∀x>0.find the value of m to 1+log_5 (x^2 +1)≥log_5 (mx^2 +4x+m) verify ∀x.
$$\mathrm{1}.\mathrm{for}\:\forall\mathrm{x}>\mathrm{0}.\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{m}\:\mathrm{to}\: \\ $$$$\mathrm{1}+\mathrm{log}_{\mathrm{5}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\geqslant\mathrm{log}_{\mathrm{5}} \left(\mathrm{mx}^{\mathrm{2}} +\mathrm{4x}+\mathrm{m}\right)\:\mathrm{verify}\:\forall\mathrm{x}. \\ $$
Answered by Rasheed.Sindhi last updated on 26/Aug/21
1+log_5 (x^2 +1)≥log_5 (mx^2 +4x+m) log_5 5 +log_5 (x^2 +1)≥log_5 (mx^2 +4x+m) log_5 ( 5(x^2 +1) )≥log_5 (mx^2 +4x+m) 5(x^2 +1)≥mx^2 +4x+m (5−m)x^2 −4x+5−m≥0 Continue....
$$\mathrm{1}+\mathrm{log}_{\mathrm{5}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\geqslant\mathrm{log}_{\mathrm{5}} \left(\mathrm{mx}^{\mathrm{2}} +\mathrm{4x}+\mathrm{m}\right)\: \\ $$$$\mathrm{log}_{\mathrm{5}} \mathrm{5}\:+\mathrm{log}_{\mathrm{5}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\geqslant\mathrm{log}_{\mathrm{5}} \left(\mathrm{mx}^{\mathrm{2}} +\mathrm{4x}+\mathrm{m}\right)\: \\ $$$$\mathrm{log}_{\mathrm{5}} \left(\:\mathrm{5}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\:\right)\geqslant\mathrm{log}_{\mathrm{5}} \left(\mathrm{mx}^{\mathrm{2}} +\mathrm{4x}+\mathrm{m}\right)\: \\ $$$$\mathrm{5}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\geqslant\mathrm{mx}^{\mathrm{2}} +\mathrm{4x}+\mathrm{m} \\ $$$$\left(\mathrm{5}−\mathrm{m}\right)\mathrm{x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{5}−\mathrm{m}\geqslant\mathrm{0} \\ $$$$\mathrm{Continue}…. \\ $$
Commented by SOMEDAVONG last updated on 26/Aug/21
Thanks sir
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$
Commented by 1549442205PVT last updated on 26/Aug/21
Put f(x)=(5−m)x^2 −4x+5−m i)For m=5 we get −4x≥0⇔x≤0,so it is rejected ii)For m<5 f(x)≥0∀x>0 if and only if [_(f(x)have has two roots x_1 ≤x_2 ≤0(2)) ^(△′=4+m(5−m)<0(1)) (1)⇔−m^2 +5m+4<0⇔m^2 −5m−4>0 ⇔m∈(−∞,((5−(√(41)))/2))∪(((5+(√(41)))/2),∞) Combining to the condition m<5 we get m∈(−∞,((5−(√(41)))/2)) (2)⇔ { ((△′≥0)),((((x_1 +x_2 )/2)=(2/(5−m))≤0)) :}⇔ { ((m^2 −5m−4≤0)),((m≥5 this is contradiction to the hypothesis that m<5)) :} .Hence this case don′t occur ii)For m>5 don′t exist m
$${Put}\:{f}\left({x}\right)=\left(\mathrm{5}−{m}\right){x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}−{m} \\ $$$$\left.{i}\right){For}\:{m}=\mathrm{5}\:{we}\:{get}\:−\mathrm{4}{x}\geqslant\mathrm{0}\Leftrightarrow{x}\leqslant\mathrm{0},{so}\:{it}\:{is}\:{rejected} \\ $$$$\left.{ii}\right){For}\:{m}<\mathrm{5}\:{f}\left({x}\right)\geqslant\mathrm{0}\forall{x}>\mathrm{0}\:{if}\:{and}\:{only}\:{if}\: \\ $$$$\:\left[_{{f}\left({x}\right){have}\:{has}\:{two}\:{roots}\:{x}_{\mathrm{1}} \leqslant{x}_{\mathrm{2}} \leqslant\mathrm{0}\left(\mathrm{2}\right)} ^{\bigtriangleup'=\mathrm{4}+{m}\left(\mathrm{5}−{m}\right)<\mathrm{0}\left(\mathrm{1}\right)} \right. \\ $$$$\left(\mathrm{1}\right)\Leftrightarrow−{m}^{\mathrm{2}} +\mathrm{5}{m}+\mathrm{4}<\mathrm{0}\Leftrightarrow{m}^{\mathrm{2}} −\mathrm{5}{m}−\mathrm{4}>\mathrm{0} \\ $$$$\Leftrightarrow{m}\in\left(−\infty,\frac{\mathrm{5}−\sqrt{\mathrm{41}}}{\mathrm{2}}\right)\cup\left(\frac{\mathrm{5}+\sqrt{\mathrm{41}}}{\mathrm{2}},\infty\right) \\ $$$${Combining}\:{to}\:{the}\:{condition}\:{m}<\mathrm{5}\:{we}\:{get} \\ $$$${m}\in\left(−\infty,\frac{\mathrm{5}−\sqrt{\mathrm{41}}}{\mathrm{2}}\right) \\ $$$$\left(\mathrm{2}\right)\Leftrightarrow\begin{cases}{\bigtriangleup'\geqslant\mathrm{0}}\\{\frac{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{2}}{\mathrm{5}−{m}}\leqslant\mathrm{0}}\end{cases}\Leftrightarrow\begin{cases}{{m}^{\mathrm{2}} −\mathrm{5}{m}−\mathrm{4}\leqslant\mathrm{0}}\\{{m}\geqslant\mathrm{5}\:{this}\:{is}\:{contradiction}\:{to}\:{the}\:{hypothesis}\:{that}\:{m}<\mathrm{5}}\end{cases} \\ $$$$.{Hence}\:{this}\:{case}\:{don}'{t}\:{occur} \\ $$$$\left.{ii}\right){For}\:{m}>\mathrm{5}\:{don}'{t}\:{exist}\:{m} \\ $$

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