Question

Given that for all real values of $x,$ the expression $\frac{x^2+2x+4}{x^2-2x+4}$ lies between $\frac{1}{3}$ and $3.$ Then the values of the expression $\frac{4^x+2^x+1}{4^x-2^x+1}$ lie between

• A. $\frac{1}{3}$ and $3$
• B. $-1$ and $0$
• C. $-1$ and $1$
• D. $0$ and $7$

Answer 1

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

$\frac{4^x+2^x+1}{4^x-2^x+1}$
$=\frac{(2^x{)}^2+2^x+1}{(2^x{)}^2-2^x+1}$
$=\frac{4(2^x{)}^2+4(2^x)+4}{4(2^x{)}^2-4(2^x)+4}=\frac{(2^{x+1}{)}^2+2(2^{x+1})+4}{(2^{x+1}{)}^2-2(2^{x+1})+4}$

Let $2^{x+1}=t$
Then, $\frac{(2^{x+1}{)}^2+2(2^{x+1})+4}{(2^{x+1}{)}^2-2(2^{x+1})+4}=\frac{t^2+2t+4}{t^2-2t+4}$
So, given expression has values lying in between $\frac{1}{3}$ and $3$.