Theo đề tham khảo Toán 2021
ĐỀ BÀI:
Cho hàm số bậc ba \(y = f\left( x \right) = {x^3} + b{x^2} + cx + d\)có đồ thị như hình vẽ. \(\left| {f\left( {{x_1}} \right)} \right| = 2,5\left| {{x_1}} \right|\). Xác định tỉ lệ \(\frac{{{S_1}}}{{{S_2}}}\).
A. \(\frac{8}{{27}}\).
B. \(\frac{2}{{27}}\).
C. \(\frac{{16}}{{27}}\).
D. \(\frac{8}{{27}}\).
LỜI GIẢI CHI TIẾT
Ta có\(y = f\left( x \right) = {x^3} + b{x^2} + cx + d\)
\(f\left( 0 \right) = – 1 \Rightarrow d = – 1\)
\(f’\left( x \right) = 3{x^2} + 2bx + c\); \(f’\left( 0 \right) = 0 \Rightarrow c = 0\)
\(f’\left( x \right) = 3{x^2} + 2bx = 0 \Leftrightarrow \left[ \begin{array}{l}x = 0\\x = – \frac{{2b}}{3}\end{array} \right. \Rightarrow {x_1} = – \frac{{2b}}{3} > 1 \Rightarrow b < – \frac{3}{2}\).
Ta có \(y = f\left( x \right) = {x^3} + b{x^2} – 1\)\(,\left| {f\left( {{x_1}} \right)} \right| = 2,5\left| {{x_1}} \right| \Leftrightarrow f\left( {\frac{{ – 2b}}{3}} \right) = \frac{5}{3}b\)\( \Leftrightarrow – \frac{8}{{27}}{b^3} + \frac{4}{9}{b^3} – 1 = \frac{5}{3}b\)
\( \Leftrightarrow 4{b^3} – 35b – 27 = 0 \Leftrightarrow \left( {b + 3} \right)\left( {4{b^2} – 12b – 9} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}b = – 3\\b = \frac{{3 + 2\sqrt 2 }}{2}\,(l)\\b = \frac{{3 – 2\sqrt 2 }}{2}\,(l)\end{array} \right.\) \( \Rightarrow {x_1} = 2 \Rightarrow b = – 3\)
\( \Rightarrow f\left( x \right) = {x^3} – 3{x^2} – 1\)
Xét phương trình: \({x^3} – 3{x^2} – 1 = – 1 \Leftrightarrow {x^3} – 3{x^2} = 0 \Leftrightarrow \left[ \begin{array}{l}x = 0\\x = 3\end{array} \right.\)
\( \Rightarrow {S_2} = \int\limits_0^3 {\left| { – 1 – \left( {{x^3} – 3{x^2} – 1} \right)} \right|} {\rm{d}}x = \int\limits_0^2 {\left( { – {x^3} + 3{x^2}} \right)} {\rm{d}}x = \frac{{27}}{4}\)
\( \Rightarrow {S_1} = \int\limits_0^2 {\left| {{x^3} – 3{x^2} – 1 – \left( { – 5} \right)} \right|} {\rm{d}}x = \int\limits_0^2 {\left( {{x^3} – 3{x^2} + 4} \right)} {\rm{d}}x = 4\)
\( \Rightarrow \frac{{{S_1}}}{{{S_2}}} = \frac{{16}}{{27}}\).
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