Cho hàm số \(y=f(x)\) có bảng biến thiên như sau
Câu hỏi: Cho hàm số \(y=f(x)\) có bảng biến thiên như sau![](data:image/png;base64,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)
A. e4
B. e3
C. \({e^{\frac{{15}}{{13}}}}\)
D. e5
Đáp án
A
- Hướng dẫn giải
Ta có: \({e^{2{f^3}\left( x \right) - \frac{{13}}{2}{f^2}\left( x \right) + 7f\left( x \right) + \frac{3}{2}}} = m \Leftrightarrow 2{f^3}\left( x \right) - \frac{{13}}{2}{f^2}\left( x \right) + 7f\left( x \right) + \frac{3}{2} = \ln m\)
Xét \(g\left( x \right) = 2{f^3}\left( x \right) - \frac{{13}}{2}{f^2}\left( x \right) + 7f\left( x \right) + \frac{3}{2}\) có
\(g'\left( x \right) = 6{f^2}\left( x \right).f'\left( x \right) - 13f\left( x \right).f'\left( x \right) + 7f'\left( x \right) = f'\left( x \right)\left[ {6{f^2}\left( x \right) - 13f\left( x \right) + 7} \right]\)
Suy ra \(g'\left( x \right) = 0 \Leftrightarrow \left[ \begin{array}{l}
f'\left( x \right) = 0\\
6{f^2}\left( x \right) - 13f\left( x \right) + 7 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
f'\left( x \right) = 0\\
f\left( x \right) = 1\\
f\left( x \right) = \frac{7}{6}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1;x = 3\\
x = 1;x = {x_1} > 3\\
x = {x_2} < 1
\end{array} \right.\)
Xét g(x) trên đoạn [0;2].
+ Trong khoảng (0;1) thì \(f'\left( x \right) < 0,f\left( x \right) > 1,f\left( x \right) < \frac{7}{6}\) nên \(f'\left( x \right)\left( {f\left( x \right) - 1} \right)\left( {f\left( x \right) - \frac{7}{6}} \right) > 0\) hay g'(x) > 0
+ Trong khoảng (1;2) thì \(f'\left( x \right) > 0,f\left( x \right) > 1,f\left( x \right) < \frac{7}{6}\) nên \(f'\left( x \right)\left( {f\left( x \right) - 1} \right)\left( {f\left( x \right) - \frac{7}{6}} \right) < 0\) hay g'(x) < 0
Từ đó ta có bảng biến thiên của g(x) như sau:
Từ bảng biến thiên ta thấy \(\mathop {\max }\limits_{[0;2]} g\left( x \right) = 4.\)
Vậy yêu cầu bài toán thỏa nếu và chỉ nếu \(\ln m \le 4 \Leftrightarrow m \le {e^4}\) hay giá trị lớn nhất của m là m = e4
Câu hỏi trên thuộc đề trắc nghiệm
Đề thi thử THPT QG năm 2019 môn Toán Sở GD & ĐT Bắc Ninh lần 2