Question

The minimum value of 4^x + 4^{1 – x}, x ∈ R, is
(a) 2
(b) 4
(c) 1
(d) 0

Answer 1

For 4^x + 4^{1 – x} ; x ∈ R
4^x and 4^{1 – x} are 2 terms say x and y respectively
Since arithmetic mean ≥ Geometric mean of x and y
$$\begin{gathered} \mathrm{i}.\mathrm{e}\frac{x+4}{2}\ge \sqrt{xy} \\ \mathrm{i}.\mathrm{e}\frac{4^x+4^{1-x}}{2}\ge \sqrt{4^x{4}^{1-x}} \\ \mathrm{i}.\mathrm{e}{4}^x+4^{1-x}\ge 2\sqrt{4^{x}4.4^{-x}} \\ \mathrm{i}.\mathrm{e}{4}^x+4^{1-x}\ge 2\sqrt{4} \\ \mathrm{i}.\mathrm{e}4x+4^{1-x}\ge 4 \end{gathered}$$
∴ minimum value of 4^x + 4^{1 – x} is 4
Hence, the correct answer is option B.