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For positive real numbers \(a\), \(b\) such that \(\ln \frac{{2 – 2ab}}{{a + b}} = 2ab + a + b – 2\). Find the smallest value \({P_{\min }}\) of \(P = a + 2b\).
A. \({P_{\min }} = \frac{{2\sqrt {10} – 3}}{2}\). B. \({P_{\min }} = \frac{{3\sqrt {10} – 7}}{2}\). C. \({P_{\min }} = \frac{{2\sqrt {10} – 1}}{2}\). D. \({P_{\min }} = \frac{{2\sqrt {10} – 5}}{2}\).
The answer
Condition: \(ab 0,\forall t > 0\). Infer that the function \(f\left( t \right)\) covariates on the interval \(\left( {0; + \infty } \right)\).
Therefore, \(\left( * \right) \Leftrightarrow f\left[ {2\left( {1 – ab} \right)} \right] = f\left( {a + b} \right)\)\( \Leftrightarrow 2\left( {1 – ab} \right) = a + b\)\( \Leftrightarrow a\left( {2b + 1} \right) = 2 – b\)\( \Leftrightarrow a = \frac{{ – b + 2}}{{2b + 1}}\).
We have \(P = a + 2b = \frac{{ – b + 2}}{{2b + 1}} + 2b = g\left( b \right)\).
\(g’\left( b \right) = \frac{{ – 5}}{{{{\left( {2b + 1} \right)}^2}}} + 2 = 0\)\( \ Leftrightarrow {\left( {2b + 1} \right)^2} = \frac{5}{2}\)\( \Leftrightarrow 2b + 1 = \frac{{\sqrt {10} }}{2}\ )\( \Leftrightarrow b = \frac{{\sqrt {10} – 2}}{4}\) .
Variation table:
From the variation table deduce \({P_{\min }} = g\left( {\frac{{\sqrt {10} – 2}}{4}} \right) = \frac{{2\sqrt { 10} – 3}}{2}\).
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These are VD-VDC sentences in the topic REVIEW OF CODE FUNCTIONS – LOGARIT.
