Question

If $y=3^{x-1}+3^{-x-1}$ ( $x$ real), then the least value of y is

• A. 2
• B. 6
• C. 2/3
• D. none of these

Answer 1

Answer: (C)

Key Idea: For any two numbers a, b, $\frac{a+b}{2}\ge \sqrt{ab}$ Given $y=3^{x-1}+3^{-x-1}$ $\Rightarrow$ $y=\frac{3^x}{3}+\frac{3^{-x}}{3}=\frac{3^x+3^{-x}}{3}$ ??(i) Now, $\frac{3^x+3^{-x}}{2}\le \sqrt{3^x{.3}^{-x}}$ $\Rightarrow$ $3^x+3^{-x}\ge 2$ $\Rightarrow$ $\frac{3^x+3^{-x}}{3}\ge \frac{2}{3}$ $\Rightarrow$ $y\ge \frac{2}{3}$ [from(i)]