Equations-And-Inequalities Question 58
Question: The least integer a, for which $ 1+{\log_5}(x^{2}+1)\le log_5(ax^{2}+4x+a) $ is true for all $ x\in R $ is
Options:
A) 6
B) 7
C) 10
D) 1
Correct Answer: B $ \Rightarrow 5(x^{2}+1)\le ax^{2}+4x+a $ $ \Rightarrow (a-5)x^{2}+4x+(a-5)\ge 0. $ $ \Rightarrow a>5 $ And $ a\le 3 $ or $ a\ge 7\Rightarrow a\ge 7 $ ? (2)
Combining the results of (1) and (2) for common values, we get $ a\in [7,\infty ) $
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Answer:
Solution:
which is possible if a>0 and $ 16-4a^{2}<0 $
$ \Rightarrow a>2 $ ?. (1)
Further, the inequality can be rewritten as
$ {\log_5}5+{\log_5}(x^{2}+1)\le log_5(ax^{2}+4x+a) $
It holds if $ a-5>0 $ and $ 16-4{{(a-5)}^{2}}\le 0 $
BETA
