1.for ∀x>0.find the value of m to 1+log_5 (x^2 +1)≥log


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Question Number 152208 by SOMEDAVONG last updated on 26/Aug/21

$1 . for \forall x > 0 . find the value of m to$ $1 + \log_{5} (x^{2} + 1) ⩾ \log_{5} ({mx}^{2} + 4 x + m) verify \forall x .$
	Answered by Rasheed.Sindhi last updated on 26/Aug/21

	$1 + \log_{5} (x^{2} + 1) ⩾ \log_{5} ({mx}^{2} + 4 x + m)$ $\log_{5} 5 + \log_{5} (x^{2} + 1) ⩾ \log_{5} ({mx}^{2} + 4 x + m)$ $\log_{5} (5 (x^{2} + 1)) ⩾ \log_{5} ({mx}^{2} + 4 x + m)$ $5 (x^{2} + 1) ⩾ {mx}^{2} + 4 x + m$ $(5 - m) x^{2} - 4 x + 5 - m ⩾ 0$ $Continue . . . .$
	Commented by SOMEDAVONG last updated on 26/Aug/21

	$Thanks sir$
	Commented by 1549442205PVT last updated on 26/Aug/21
	$Put f(x)=(5−m)x^2 −4x+5−m i)For m=5 we get −4x≥0⇔x≤0,so it is rejected ii)For m<5 f(x)≥0∀x>0 if and only if [_(f(x)have has two roots x_1 ≤x_2 ≤0(2)) ^(△′=4+m(5−m)<0(1)) (1)⇔−m^2 +5m+4<0⇔m^2 −5m−4>0 ⇔m∈(−∞,((5−(√(41)))/2))∪(((5+(√(41)))/2),∞) Combining to the condition m<5 we get m∈(−∞,((5−(√(41)))/2)) (2)⇔ { ((△′≥0)),((((x_1 +x_2 )/2)=(2/(5−m))≤0)) :}⇔ { ((m^2 −5m−4≤0)),((m≥5 this is contradiction to the hypothesis that m<5)) :} .Hence this case don′t occur ii)For m>5 don′t exist m$
	$P u t f (x) = (5 - m) x^{2} - 4 x + 5 - m$ $i) F o r m = 5 w e g e t - 4 x ⩾ 0 \Leftrightarrow x ⩽ 0, s o i t i s r e j e c t e d$ $i i) F o r m < 5 f (x) ⩾ 0 \forall x > 0 i f a n d o n l y i f$ $[_{f (x) h a v e h a s t w o r o o t s x_{1} ⩽ x_{2} ⩽ 0 (2)}^{△^{'} = 4 + m (5 - m) < 0 (1)}$ $(1) \Leftrightarrow - m^{2} + 5 m + 4 < 0 \Leftrightarrow m^{2} - 5 m - 4 > 0$ $\Leftrightarrow m \in (- \infty, \frac{5 - \sqrt{41}}{2}) \cup (\frac{5 + \sqrt{41}}{2}, \infty)$ $C o m b i n i n g t o t h e c o n d i t i o n m < 5 w e g e t$ $m \in (- \infty, \frac{5 - \sqrt{41}}{2})$ $(2) \Leftrightarrow {\begin{cases} △^{'} ⩾ 0 \\ \frac{x_{1} + x_{2}}{2} = \frac{2}{5 - m} ⩽ 0 \end{cases} \Leftrightarrow {\begin{cases} m^{2} - 5 m - 4 ⩽ 0 \\ m ⩾ 5 t h i s i s c o n t r a d i c t i o n t o t h e h y p o t h e s i s t h a t m < 5 \end{cases}$ $. H e n c e t h i s c a s e {d o n}^{'} t o c c u r$ $i i) F o r m > 5 {d o n}^{'} t e x i s t m$
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