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
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
Show 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.
Updated On: Jul 9, 2026
- \( \dfrac{1}{3} \) and \( 3 \)
- \( \dfrac{1}{2} \) and \( 2 \)
- \( 0 \) and \( 2 \)
- \( 3 \) and \( 9 \) \bigskip
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
When an expression is given to satisfy an inequality for all real values, try to transform the required expression into the same form by an appropriate substitution.
Step 1: Rewrite the given expression. Let \[ t=3^x. \] Since \[ 3^x>0, \] we have \[ t>0. \] The given expression becomes \[ E= \frac{9t^2+6t+4}{9t^2-6t+4}. \]
Step 2: Express \(E\) in the form given in the question. Put \[ u=3t. \] Then \(u>0\) and \[ E= \frac{u^2+2u+4}{u^2-2u+4}. \] Now let \[ x=-u. \] Then \[ E= \frac{x^2-2x+4}{x^2+2x+4}. \]
Step 3: Use the given inequality. The question states that for all real \(x\), \[ \frac{1}{3} < \frac{x^2-2x+4}{x^2+2x+4} <3. \] Since \(E\) has exactly the same form, \[ \frac{1}{3} < E <3. \] Therefore, \[ \frac{1}{3} < \frac{9\cdot 3^{2x}+6\cdot 3^x+4} {9\cdot 3^{2x}-6\cdot 3^x+4} <3. \]
Step 4: Write the final answer. \[ \boxed{\frac{1}{3}<E<3} \] Hence, the values lie between \[ \boxed{\frac{1}{3}\ \text{and}\ 3}. \]
Step 1: Rewrite the given expression. Let \[ t=3^x. \] Since \[ 3^x>0, \] we have \[ t>0. \] The given expression becomes \[ E= \frac{9t^2+6t+4}{9t^2-6t+4}. \]
Step 2: Express \(E\) in the form given in the question. Put \[ u=3t. \] Then \(u>0\) and \[ E= \frac{u^2+2u+4}{u^2-2u+4}. \] Now let \[ x=-u. \] Then \[ E= \frac{x^2-2x+4}{x^2+2x+4}. \]
Step 3: Use the given inequality. The question states that for all real \(x\), \[ \frac{1}{3} < \frac{x^2-2x+4}{x^2+2x+4} <3. \] Since \(E\) has exactly the same form, \[ \frac{1}{3} < E <3. \] Therefore, \[ \frac{1}{3} < \frac{9\cdot 3^{2x}+6\cdot 3^x+4} {9\cdot 3^{2x}-6\cdot 3^x+4} <3. \]
Step 4: Write the final answer. \[ \boxed{\frac{1}{3}<E<3} \] Hence, the values lie between \[ \boxed{\frac{1}{3}\ \text{and}\ 3}. \]
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