Question

Consider the logarithmic inequality $1+{\log }_5(x^2+1)\ge {\log }_5(a{x}^2+4x+a)$ for all real values of $x.$ The number of integers which $a$ cannot take, is

Answer 1

$1+{\log }_5(x^2+1)\ge {\log }_5(a{x}^2+4x+a)$
$\Rightarrow {\log }_{5}5(x^2+1)\ge {\log }_5(a{x}^2+4x+a)...\left(1\right)$
Now, ${\log }_5(a{x}^2+4x+a)$ is defined when
$a{x}^2+4x+a>0$
$\Rightarrow D<0(\text{assuming}a\ne 0)$
$\Rightarrow 16-4a^2<0$
$\Rightarrow a^2-4>0$
$\Rightarrow a\in (-\infty ,-2)\cup (2,\infty )...\left(2\right)$

From equation $\left(1\right),$
$5(x^2+1)\ge a{x}^2+4x+a$
$\Rightarrow (5-a)x^2-4x+(5-a)\ge 0$
$\Rightarrow D\le 0(\text{assuming}a\ne 5)$
$\Rightarrow 16-4(a-5{)}^2\le 0$
$\Rightarrow a^2-10a+21\ge 0$
$\Rightarrow (a-7)(a-3)\ge 0$
$\Rightarrow a\in (-\infty ,3]\cup [7,\infty )...\left(3\right)$

From equation $\left(2\right)$ and $\left(3\right),$
$a\in (-\infty ,-2)\cup (2,3]\cup [7,\infty )$
So, $a$ cannot take the values $-2,-1,0,1,2,4,5,6$