Question

The least integer a, for which $1+lo{g}_5(x^2+1)\le lo{g}_5(a{x}^2+4x+a)$ is true for all $x\in R$ is

• A. 6
• B. 7
• C. 10
• D. 1

Answer 1

Answer: (B)

For the validity of inequality $a{x}^2+4x+a>0$, which is possible if $a>0$ and $16-4a^2<0$
$\Rightarrow a>2$ .....(1)
Further, the inequality can be rewritten as
${\log }_{5}5+{\log }_5\left(x^2+1\right)\le {\log }_5\left(a{x}^2+4x+a\right)$
$\Rightarrow 5\left(x^2+1\right)\le a{x}^2+4x+a$
$\Rightarrow \left(a-5\right)x^2+4x+\left(a-5\right)\ge 0$
It holds if a - 5 > 0 and $16-4(a-5{)}^2\le 0$
$\Rightarrow a>5$ ....(1) 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 $a\in [7,\infty )$