Question

Given that, for all real $x,$ the expression $\frac{x^2-2x+4}{x^2+2x+4}$ lies between $\frac{1}{3}$ and $3$. The values between which the expression $\frac{9.3^{2x}+6.3^x+4}{9.3^{2x}-6.3^x+4}$ lies are

• A. $0$ and $2$
• B. $-1$ and $1$
• C. $-1$ and $0$
• D. $\frac{1}{3}$ and $3$

Answer 1

The correct option is D $\frac{1}{3}$ and $3$

Given,$\frac{1}{3}<\frac{x^2-2x+4}{x^2+2x+4}<3$ for all $x\in R.$

$\Rightarrow \frac{1}{3}<\frac{x^2+2x+4}{x^2-2x+4}<3$ for all $x\in R.$ ...(1)

Let $3^{x+1}=y$

Then,$y\in R$for all $x\in R.$

$\therefore \frac{9.3^{2x}+6.3^x+4}{9.3^{2x}-6.3^x+4}=\frac{3^{2x+2}+2.3^{x+1}+4}{3^{2x+2}-2.3^{x+1}+4}=\frac{y^2+2y+4}{y^2-2y+4}$

From (1),$\frac{1}{3}<\frac{y^2+2y+4}{y^2-2y+4}<3$

$\therefore \frac{1}{3}<\frac{9.3^{2x}+6.3^x+4}{9.3^{2x}-6.3^x+4}<3.$