## The solution set of the inequality ({log _2}left( {{x^2} + 3} right) – {log _2}x + {x^2} – 4x + 1 le 0) is – Math Book

Solution set of the inequality $${\log _2}\left( {{x^2} + 3} \right) – {\log _2}x + {x^2} – 4x + 1 \le 0$$ To be

A. $$S = \left( { – \infty ;1} \right]\cup \left[ {3; + \infty } \right)$$. B. $$\left[ { – 1;3} \right]$$. C. $$\left[ {1; + \infty } \right)$$. D. $$S = \left[ {1;3} \right]$$.
Condition: $$x > 0$$.
We have $${\log _2}\left( {{x^2} + 3} \right) – {\log _2}x + {x^2} – 4x + 1 \le 0 \Leftrightarrow {\log _2 }\left( {{x^2} + 3} \right) + {x^2} + 3 \le {\log _2}4x + 4x\,\,\,\left( * \right)$$.
Consider the function $$f\left( t \right) = {\log _2}t + t$$ on $$D = \left( {0;\,\, + \infty } \right)$$. We have
$$f’\left( t \right) = \frac{1}{{t\ln 2}} + 1 > 0\,\,\,\forall t \in D \Rightarrow$$function $$f$$ covariates on $$D$$.
Derive $$\left( * \right) \Leftrightarrow f\left( {{x^2} + 3} \right) \le f\left( {4x} \right) \Leftrightarrow {x^2} + 3 \le 4x \Leftrightarrow 1 \le x \le 3$$.