Question

If for x ∈ R, $\frac{1}{3}$<$\frac{x^{2} - 2x + 4}{x^{2} + 2x + 4}$< 3, then $\frac{9 \cdot 3^{2x} - 6 \cdot 3^{x} + 4}{9 \cdot 3^{2x} + 6 \cdot 3^{x} + 4}$ lies between –

• A. $\frac{1}{2}$and 2
• B. $\frac{1}{3}$and 3
• C. 0 and 2
• D. None of these

Answer 1

Answer: (B)

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

Answer 2

$\frac{9 \cdot 3^{2x} - 6 \cdot 3^{x} + 4}{9 \cdot 3^{2x} + 6 \cdot 3^{x} + 4}$=$\frac{(3^{(x + 1)})^{2} - 2(3^{(x + 1)}) + 4}{(3^{(x + 1)})^{2} + 2(3^{(x + 1)}) + 4}$

= $\frac{t^{2} - 2t + 4}{t^{2} + 2t + 4}$ (where t = 3^{x + 1}) …..(1)

Since, $\frac{1}{3}$< $\frac{x^{2} - 2x + 4}{x^{2} + 2x + 4}$< 3

∴ From (1), the given expression lies between 1/3 and 3.