Question

The minimum value of $\frac{9\cdot 3^{2x}+6\cdot 3^x+4}{9\cdot 3^{2x}-6\cdot 3^x+4}$ is

• A. $-1$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $\frac{1}{3}$

Answer 1

Answer: (D)

Let $y=\frac{9\cdot 3^{2x}+6\cdot 3^x+4}{9\cdot 3^{2x}-6\cdot 3^x+4}$
Put, $3^x=t$
$\therefore y=\frac{9\cdot t^2+6t+4}{9t^2-6t+4}$
$\Rightarrow y\left(9t^2-6t+4\right)=9t^2+6t+4$
$\Rightarrow 9t^2(y-1)-6t(y+1)+(4y-4)=0$
$\Rightarrow (y-1)t^2\frac{-6}{9}(y+1)t+\frac{4}{9}(y-1)=0$
$\Rightarrow (y-1)t^2-\frac{2}{3}(y+1)t+\frac{4}{9}(y-1)=0$
$\because t$ is real
$\therefore D\ge 0$
$\Rightarrow \frac{4}{9}(y+1{)}^2-4(y-1)\frac{4}{9}(y-1)\ge 0$
$\Rightarrow \frac{4}{9}(y+1{)}^2-\frac{16}{9}(y-1{)}^2\ge 0$
$\Rightarrow (y+1{)}^2-4(y-1{)}^2\ge 0$
$\Rightarrow y^2+1+2y-4y^2-4+8y\ge 0$
$\Rightarrow -3y^2+10y-3\ge 0$
$\Rightarrow 3y^2-10y+3\le 0$
$\Rightarrow (y-3)(3y-1)\le 0$
$\Rightarrow \frac{1}{3}\le y\le 3$
Hence, minimum value $=\frac{1}{3}$