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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. \(S = \left[ {1;3} \right]\).
The answer
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 \(\left( * \right) \Leftrightarrow f\left( {{x^2} + 3} \right) \le f\left( {4x} \right) \Leftrightarrow {x^2} + 3 \le 4x \Leftrightarrow 1 \le x \le 3\).
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These are VD-VDC sentences in the topic REVIEW OF CODE FUNCTIONS – LOGARIT.
