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

Answer 1

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

The first expression lies between $-\frac{1}{3}$ and $3\Rightarrow xϵR.$\\ \therefore The second expression is a real quantity.\\ Now the second expression can be written as $\frac{{\left(3^{x+1}\right)}^2+2.3^{x+1}+4}{{\left(3^{x+1}\right)}^2-2.3^{x+1}+4}=\frac{y^2+2y+4}{y^2-2y+4}[writing{3}^{x+1}=y]LetitsvaluebepwhenpϵR\therefore \frac{y^2+2y+4}{y^2-2y+4}=p\Rightarrow y^2(1-p)+(2+2p)y+4-4p=0\Rightarrow △={(2+2p)}^2-4(1-p)(4-4p)=4(3-p)(3p-1)Nowytobereal△>0\therefore (3-p)(3p-1)>0\Rightarrow p=\frac{1}{3},3Ans-OptionA$