Question

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

• A. $0$
• B. $3$
• C. $5$
• D. $7$

Answer 1

The correct option is B $3$

$1+\log 5(x^2+1)\le \log 5({ax}^2+4x+a){\log }_{5}5+\log 5(x^2+1)\le \log 5({ax}^2+4x+a){5x}^2+5\le {ax}^2+4x+a(a-5)x^2+4x+(a-5)\ge 0D<016-4(a-5{)}^2<04(a-5{)}^2>16(a-5{)}^2>4|a-5|>2-2>a-5>2a\le 3a\ge 7a\in [3,7]$