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-

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 \dots .

Commented by SOMEDAVONG last updated on 26/Aug/21

Thanks sir

Commented by 1549442205PVT last updated on 26/Aug/21

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