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For real numbers \(x,y\) satisfying \(0 \le x,y \le 1\) and \({\log _2}\frac{{x + y}}{{2 – xy}} + 2\left( {x + 1} \right)\left( {y + 1} \right) – 6 = 0\). Find the maximum value of \(P = 3x + y\).
A. \(3\). B. \(\frac{7}{2}\). C. \(4\). D. \(\frac{5}{2}\).
The answer
With the condition \(\left\{ \begin{array}{l}0 \le x,y \le 1\\x + y > 0\end{array} \right.\) we have: \({\) log _2}\frac{{x + y}}{{2 – xy}} + 2\left( {x + 1} \right)\left( {y + 1} \right) – 6 = 0\)
\( \Leftrightarrow {\log _2}\left( {x + y} \right) – {\log _2}\left( {2 – xy} \right) + 2xy + 2x + 2y – 4 = 0\)
\( \Leftrightarrow {\log _2}\left( {x + y} \right) + 2\left( {x + y} \right) = {\log _2}\left( {2 – xy} \right) + 2\left( {2 – xy} \right)\left( * \right)\).
Consider the function \(f\left( t \right) = {\log _2}t + 2t\) on \(\left( {0;2} \right)\).
\(f’\left( t \right) = \frac{1}{{t\ln 2}} + 2 > 0,\forall t \in \left( {0;2} \right)\) so the function the number \(f\left( t \right)\) covariates on the interval \(\left( {0;2} \right)\).
So from \(\left( * \right)\) we have \(x + y = 2 – xy \Leftrightarrow y\left( {1 + x} \right) = 2 – x \Leftrightarrow y = \frac{ {2 – x}}{{1 + x}}\).
\(P = 3x + y = 3x + \frac{{2 – x}}{{1 + x}}\). Consider the function \(g\left( x \right) = 3x + \frac{{2 – x}}{{1 + x}},\,\,x \in \left[ {0;1} \right]\).
\(g’\left( x \right) = 3 – \frac{3}{{{{\left( {1 + x} \right)}^2}}} = \frac{{3x\left( { x + 2} \right)}}{{{\left( {1 + x} \right)}^2}}} \ge 0,\forall x \in \left[ {0;1} \right]\). It follows that \(MaxP = \frac{7}{2}\) is obtained when \(x = 1,y = \frac{1}{2}\).
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These are VD-VDC sentences in the topic REVIEW OF CODE FUNCTIONS – LOGARIT.