The minimum value of

Question:

The minimum value of 4x + 41 – xx ∈ R, is

(a) 2

(b) 4

(c) 1

(d) 0

Solution:

For $4^{x}+4^{1-x}: x \in R$

4x and 41 – x are 2 terms say x and respectively

Since arithmetic mean ≥ Geometric mean of x and y

i. e $\frac{x+4}{2} \geq \sqrt{x y}$

i. e $\frac{4^{x}+4^{1-x}}{2} \geq \sqrt{4^{x} 4^{1-x}}$

i. e $4^{x}+4^{1-x} \geq 2 \sqrt{4^{x} 4.4^{-x}}$

i. e $4^{x}+4^{1-x} \geq 2 \sqrt{4}$

i. e $4 x+4^{1-x} \geq 4$

∴ minimum value of 4x + 41 – x is 4

Hence, the correct answer is option B.

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