A) \[1\]
B) \[2\]
C) \[3\]
D) \[4\]
Correct Answer: D
Solution :
We know that,\[AM\ge GM\] Therefore, for\[{{4}^{x}}\]and\[{{4}^{1-x}}\], we have \[\frac{{{4}^{x}}+{{4}^{1-x}}}{2}\ge \sqrt{{{4}^{x}}\cdot {{4}^{1-x}}}\] \[\Rightarrow \] \[{{4}^{x}}+{{4}^{1-x}}\ge 2\sqrt{4}\] \[\Rightarrow \] \[{{4}^{x}}+{{4}^{1-x}}\ge 4\] Hence, minimum value of\[{{4}^{x}}+{{4}^{1-x}}\]is\[4\].
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