Question

If for all \(x\in\mathbb{R}\), \[ \frac{1}{3} < \frac{x^2-2x+4}{x^2+2x+4} <3, \] then the values of \[ \frac{9\cdot 3^{2x}+6\cdot 3^x+4} {9\cdot 3^{2x}-6\cdot 3^x+4} \] lie between

• A. \( \dfrac{1}{3} \) and \( 3 \)
• B. \( \dfrac{1}{2} \) and \( 2 \)
• C. \( 0 \) and \( 2 \)
• D. \( 3 \) and \( 9 \) \bigskip

Hint:

Whenever an inequality involving a rational expression is already provided, try to convert the required expression into the same form using a suitable substitution. This avoids lengthy calculations.

Answer 1

The Correct Option is A

Concept: Transform the given expression into the standard form for which the inequality is already known.

Step 1: Let \(t=3^x>0\). Then \(E=\dfrac{9t^2+6t+4}{9t^2-6t+4}\). Putting \(u=3t\), we get \(E=\dfrac{u^2+2u+4}{u^2-2u+4}\).

Step 2: Replacing \(u\) by \(-x\) gives \(E=\dfrac{x^2-2x+4}{x^2+2x+4}\), which matches the given inequality.

Step 3: Therefore, \(\boxed{\frac13<E<3}\).