Q. If for $x \in R,\frac{1}{3} < \frac{x^{2} - 2x + 4}{x^{2} + 2x + 4} < 3,$then$\frac{{9 \bullet 3}^{2x} - {6 \bullet 3}^{x} + 4}{9 \bullet 3^{2x} + {6 \bullet 3}^{x} + 4}\$lies between
A) $\frac{1}{2}\$and$\ 2$
B) $\frac{1}{3}\$and$\ 3$
C) 0 and 2
D) None of these
Explanation
(b)
$\frac{{9 \bullet 3}^{2x} - {6 \bullet 3}^{x} + 4}{{9 \bullet 3}^{2x} + {6 \bullet 3}^{x} + 4} = \frac{\left( 3^{(x + 1)} \right)^{2} - 2(3^{(x + 1)}) + 4}{\left( 3^{(x + 1)} \right)^{2} + 2{(3}^{(x + 1)}) + 4}$
$= \frac{t^{2} - 2t + 4}{t^{2} + 2t + 4}\ \ \ ($where$\ t = 3^{x + 1})\ \$…(i)
Since, $\frac{1}{3} < \frac{x^{2} - 2x + 4}{x^{2} + 2x + 4} < 3$
$\therefore\$From Eq.(i), the given expression lies between 1/3 and 3
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.