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The minimum value of $4^x+4^{1-x},\,x∈R$ is |
2 4 1 none of these |
4 |
A.M ≥ G.M $∴\frac{4^x+\frac{4}{4^x}}{2}≥\sqrt{4^x.\frac{4}{4^x}}$ $\Rightarrow 4^x+4^{1-x}\ge 4$ $\Rightarrow \text{Minimum value of } 4^x+4^{1-x} \text { is 4 }.$ |
