A) 2
B) 6
C) 2/3
D) none of these
Correct Answer: C
Solution :
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)]
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