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Q.
The least integer a, for which $ 1 + log_{5} (x^{2} + 1) \leq log_{5} (ax^{2} + 4x + a )$ is true for all $x \in R$ is
Linear Inequalities
Solution:
For the validity of inequality $ax^{2} + 4x + a > 0$,
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}\left(x^{2}+1\right)\le \log_{5}\left(ax^{2}+4x +a\right) $
$\Rightarrow 5\left(x^{2}+1\right)\le ax^{2}+4x+a $
$\Rightarrow \left(a- 5\right)x^{2} + 4x + \left(a -5\right)\ge0$
It holds if a - 5 > 0 and $16 - 4(a - 5)^2 \leq 0$
$\Rightarrow \:\: a > 5$ ....(1) and $a \leq 3$ or $a \geq 7 \Rightarrow a \geq 7 $
.....(2)
Combining the results of (1) and (2) for
common values, we get $ a \, \in [ 7 , \infty)$