Đáp án đúng: D
Phương pháp giải:
Đánh giá, chứng minh: \({\log _2}\left( {1 - {a^2} - {b^2} + 2b} \right) = \dfrac{{{2^a}}}{{{4^a} + 1}} + \dfrac{{{4^a}}}{{{2^a} + 1}} + \dfrac{1}{{{2^a} + {4^a}}} - \dfrac{1}{2} = 1\).
Giải chi tiết:Ta có: \({\log _2}\left( {1 - {a^2} - {b^2} + 2b} \right) = {\log _2}\left( {2 - {a^2} - {{\left( {b - 1} \right)}^2}} \right) \le {\log _2}2 = 1\) (1)
\(\dfrac{{{2^a}}}{{{4^a} + 1}} + \dfrac{{{4^a}}}{{{2^a} + 1}} + \dfrac{1}{{{2^a} + {4^a}}} - \dfrac{1}{2} \ge 1\) (2)
Thật vậy, xét hàm số \(f\left( t \right) = \dfrac{t}{{{t^2} + 1}} + \dfrac{{{t^2}}}{{t + 1}} + \dfrac{1}{{t + {t^2}}} - \dfrac{1}{2},\,\,\left( {\dfrac{1}{2} < t < 2} \right)\) có:
\(\begin{array}{l}f'\left( t \right) = \dfrac{{1 - {t^2}}}{{{{\left( {{t^2} + 1} \right)}^2}}} + \dfrac{{{t^2} + 2t}}{{{{\left( {t + 1} \right)}^2}}} - \dfrac{{1 + 2t}}{{{{\left( {t + {t^2}} \right)}^2}}}\\\,\,\,\,\,\,\,\,\,\,\, = - \dfrac{{\left( {t - 1} \right)\left( {t + 1} \right)}}{{{{\left( {{t^2} + 1} \right)}^2}}} + \dfrac{{{t^4} + 2{t^3} - 1 - 2t}}{{{t^2}{{\left( {t + 1} \right)}^2}}}\\\,\,\,\,\,\,\,\,\,\,\, = - \dfrac{{\left( {t - 1} \right)\left( {t + 1} \right)}}{{{{\left( {{t^2} + 1} \right)}^2}}} + \dfrac{{\left( {t - 1} \right)\left( {t + 1} \right)}}{{{t^2}}}\\\,\,\,\,\,\,\,\,\,\,\, = \left( {t - 1} \right)\left( {t + 1} \right)\left[ {\dfrac{1}{{{t^2}}} - \dfrac{1}{{{{\left( {{t^2} + 1} \right)}^2}}}} \right]\\\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{\left( {t - 1} \right)\left( {t + 1} \right)\left( {{t^2} - t + 1} \right)\left( {{t^2} + t + 1} \right)}}{{{t^2}{{\left( {{t^2} + 1} \right)}^2}}}\end{array}\)
Ta có: \(f'\left( t \right) = 0 \Rightarrow \left[ \begin{array}{l}t = 1\,\,\left( {tm} \right)\\t = - 1\,\,\left( {ktm} \right)\end{array} \right.\)
BBT:
Từ (1), (2), suy ra: \({\log _2}\left( {1 - {a^2} - {b^2} + 2b} \right) = \dfrac{{{2^a}}}{{{4^a} + 1}} + \dfrac{{{4^a}}}{{{2^a} + 1}} + \dfrac{1}{{{2^a} + {4^a}}} - \dfrac{1}{2} = 1\) \( \Leftrightarrow \left\{ \begin{array}{l}a = 0\\b - 1 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 0\\b = 1\end{array} \right.\)
Vậy có 1 giá trị của a thỏa mãn.
Chọn D.
