Question

The solution set of the inequality \[ 3^x + 3^{1-x} - 4 <0 \] contained in \( \mathbb{R} \) is:

• A. \( (1,2) \)
• B. \( (1,3) \)
• C. \( (0,2) \)
• D. \( (0,1) \)
  \

Hint:

For inequalities involving exponential expressions, use substitution to convert them into quadratic inequalities. Solve and revert back to the original variable.

Answer 1

The Correct Option is D

We need to solve the inequality: \[ 3^x + 3^{1-x} - 4<0. \] Step 1: Introduce a Substitution
Let \( y = 3^x \), then \( 3^{-x} = \frac{1}{3^x} = \frac{1}{y} \). Rewriting the inequality: \[ y + \frac{3}{y} - 4<0. \] Step 2: Multiply by \( y \) (Positive for \( y>0 \))
\[ y^2 - 4y + 3<0. \] Step 3: Solve the Quadratic Inequality
Factorizing: \[ (y - 3)(y - 1)<0. \] Using the sign analysis method, the inequality holds for: \[ 1<y<3. \] Step 4: Convert Back to \( x \)
Since \( y = 3^x \), we take logarithms: \[ 1<3^x<3. \] Taking the logarithm base 3: \[ 0<x<1. \] Step 5: Conclusion
Thus, the solution set is: \[ (0,1). \] \bigskip