Question

Given that, for all real ${}^{\prime }{x}^{\prime }$, 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. $-1$ and $1$
• B. $\frac{1}{3}$ and $3$
• C. $0$ and $2$
• D. $-2$ and $0$

Answer 1

The correct option is B

$\frac{1}{3}$ and $3$

As given in the question,
$\frac{1}{3}<\frac{x^2+2x+4}{x^2-2x+4}<3$ for all $x\in R$ . . . $\left(1\right)$

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

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

$\therefore \frac{9.3^{2x}+{6.3}^x+4}{9.3^{2x}-{6.3}^x+4}$

$\Rightarrow \frac{3^{2x+2}+{2.3}^{x+1}+4}{3^{2x+2}-{2.3}^{x+1}+4}$

$\Rightarrow \frac{y^2+2y+4}{y^2-2y+4}$

From $\left(1\right)$

$\frac{1}{3}<\frac{y^2+2y+4}{y^2-2y+4}<3$

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