A) 6
B) 7
C) 10
D) 1
Correct Answer: B
Solution :
[b] For the validity of inequality \[a{{x}^{2}}+4x+a>0,\] |
which is possible if a>0 and \[16-4{{a}^{2}}<0\]\[\Rightarrow a>2\] ?. (1) |
Further, the inequality can be rewritten as |
\[{{\log }_{5}}5+{{\log }_{5}}({{x}^{2}}+1)\le lo{{g}_{5}}(a{{x}^{2}}+4x+a)\] |
\[\Rightarrow 5({{x}^{2}}+1)\le a{{x}^{2}}+4x+a\] |
\[\Rightarrow (a-5){{x}^{2}}+4x+(a-5)\ge 0.\] |
It holds if \[a-5>0\] and \[16-4{{(a-5)}^{2}}\le 0\] |
\[\Rightarrow a>5\] 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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