Question

If $1+{\log }_5(x^2+1)\ge {\log }_5({ax}^2+4x+a)$ for all $x\in R,$ then $a$ can be equal to

• A. $3$
• B. $4$
• C. $2$
• D. $\frac{3}{2}$

Answer 1

The correct option is A $3$

We have, ${\log }_{5}5(x^2+1)\ge {\log }_5(a{x}^2+4x+a)$
$\Rightarrow 5(x^2+1)\ge a{x}^2+4x+a,\forall x\in R$
$\Rightarrow x^2(5-a)-4x+(5-a)\ge 0,\forall x\in R$

$5-a>0$
$\Rightarrow a<5\cdots \left(1\right)$
and $D\le 0$
$\Rightarrow 16-4(5-a{)}^2\le 0$
$\Rightarrow (2-(5-a\left)\right)(2+5-a)\le 0$
$\Rightarrow (a-3)(a-7)\ge 0$
$\Rightarrow a\in (-\infty ,3]\cup [7,\infty )\cdots \left(2\right)$

$\left(1\right)\cap \left(2\right),$ we get
$a\in (-\infty ,3]\cdots \left(3\right)$

But for the domain of ${\log }_5(a{x}^2+4x+a),$
we must have $a{x}^2+4x+a>0,\forall x\in R$
$\Rightarrow a>0$ and $16-4a^2<0$
$\Rightarrow a>0$ and $a\in (-\infty ,-2)\cup (2,\infty )$
$\Rightarrow a\in (2,\infty )\cdots \left(4\right)$

Thus, from both $\left(3\right)\cap \left(4\right),$ we get
$a\in (2,3]$