Question

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

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

Answer 1

Answer: (D)

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