(left( P right):,,y = frac{1}{2}{x^2}).
TXĐ : (D = R). Ta có (y’ = x).
Giả sử (Aleft( {{x_1};frac{1}{2}x_1^2} right);,,Bleft( {{x_2};frac{1}{2}x_2^2} right) in left( P right),,left( {{x_1} ne {x_2}} right)).
Phương trình tiếp tuyến tại điểm A của (left( P right)) là (y = {x_1}left( {x – {x_1}} right) + frac{1}{2}x_1^2 Leftrightarrow y = {x_1}x – frac{1}{2}x_1^2,,left( {{d_1}} right))
Phương trình tiếp tuyến tại điểm B của (left( P right)) là (y = {x_2}left( {x – {x_2}} right) + frac{1}{2}x_2^2 Leftrightarrow y = {x_2}x – frac{1}{2}x_2^2,,left( {{d_2}} right))
Do (left( {{d_1}} right) bot left( {{d_2}} right)) nên ta có ({x_1}{x_2} = – 1 Leftrightarrow {x_2} = frac{{ – 1}}{{{x_1}}}).
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Phương trình đường thẳng AB :
(begin{array}{l}frac{{x – {x_1}}}{{{x_2} – {x_1}}} = frac{{y – frac{1}{2}x_1^2}}{{frac{1}{2}x_2^2 – frac{1}{2}x_1^2}} Leftrightarrow frac{1}{2}left( {x – {x_1}} right)left( {x_2^2 – x_1^2} right) = left( {y – frac{1}{2}x_1^2} right)left( {{x_2} – {x_1}} right)\ Leftrightarrow left( {x – {x_1}} right)left( {{x_2} + {x_1}} right) = 2y – x_1^2 Leftrightarrow left( {{x_1} + {x_2}} right)x – 2y – {x_1}{x_2} = 0\ Leftrightarrow y = frac{1}{2}left[ {left( {{x_1} + {x_2}} right)x – {x_1}{x_2}} right] = frac{1}{2}left[ {left( {{x_1} + {x_2}} right)x + 1} right]end{array})
Do đó diện tích hình phẳng giới hạn bởi (AB,,,left( P right)) là :
(begin{array}{l}S = frac{1}{2}intlimits_{{x_1}}^{{x_2}} {left( {left( {{x_1} + {x_2}} right)x + 1 – {x^2}} right)dx} \ Leftrightarrow frac{9}{4} = frac{1}{2}left. {left( {left( {{x_1} + {x_2}} right)frac{{{x^2}}}{2} + x – frac{{{x^3}}}{3}} right)} right|_{{x_1}}^{{x_2}}\ Leftrightarrow frac{9}{4} = frac{1}{2}left[ {left( {{x_1} + {x_2}} right)left( {frac{{x_2^2}}{2} – frac{{x_1^2}}{2}} right) + left( {{x_2} – {x_1}} right) – frac{{x_2^3 – x_1^3}}{3}} right]\ Leftrightarrow frac{9}{2} = frac{1}{2}left( {{x_1} + {x_2}} right)left( {x_2^2 – x_1^2} right) + left( {{x_2} – {x_1}} right) – frac{{x_2^3 – x_1^3}}{3}\ Leftrightarrow 27 = 3left( {{x_1}x_2^2 – x_1^3 + x_2^3 – x_1^2{x_2}} right) + 6left( {{x_2} – {x_1}} right) – 2x_2^3 + 2x_1^3\ Leftrightarrow 27 = 3{x_1}x_2^2 – 3x_1^2{x_2} + x_2^3 – x_1^3 + 6left( {{x_2} – {x_1}} right)\ Leftrightarrow 27 = – 3left( {{x_2} – {x_1}} right) + left( {{x_2} – {x_1}} right)left( {x_1^2 + x_2^2 – 1} right) + 6left( {{x_2} – {x_1}} right)\ Leftrightarrow 27 = 3left( {{x_2} – {x_1}} right) + left( {{x_2} – {x_1}} right)left( {x_1^2 + x_2^2 – 1} right)\ Leftrightarrow 27 = left( {{x_2} – {x_1}} right)left( {x_1^2 + x_2^2 + 2} right)\ Leftrightarrow 27 = left( {{x_2} – {x_1}} right)left( {x_1^2 + x_2^2 – 2{x_1}{x_2}} right)\ Leftrightarrow 27 = left( {{x_2} – {x_1}} right){left( {{x_2} – {x_1}} right)^2} = {left( {{x_2} – {x_1}} right)^3}\ Leftrightarrow {x_2} – {x_1} = 3end{array})
Thay ({x_2} = frac{{ – 1}}{{{x_1}}}) ta có : (frac{{ – 1}}{{{x_1}}} – {x_1} = 3 Leftrightarrow – 1 – x_1^2 – 3{x_1} = 0 Leftrightarrow left[ begin{array}{l}{x_1} = frac{{ – 3 – sqrt 5 }}{2} Rightarrow {x_2} = frac{2}{{3 + sqrt 5 }}\{x_1} = frac{{ – 3 + sqrt 5 }}{2} Rightarrow {x_2} = frac{{ – 2}}{{ – 3 + sqrt 5 }}end{array} right. Leftrightarrow {left( {{x_1} + {x_2}} right)^2} = 5).
Chọn B.