Cho parabol (P): y=x^2 và hai điểm A,B

* Hướng dẫn giải

Đáp án A

Gọi $A\left(a;a^2\right),B\left(b;b^2\right)$ với a<b. Ta có $AB=2\iff {\left(b-a\right)}^2+{\left(b^2-a^2\right)}^2=4$.

$\begin{array}{l}AB:\frac{x-a}{b-a}=\frac{y-a^2}{b^2-a^2}\iff \frac{x-a}{1}=\frac{y-a^2}{b+a}\iff y=\left(a+b\right)\left(x-a\right)+a^2\iff y=\left(a+b\right)x-ab\\ S=\underset{a}{\overset{b}{\int }}\left(\left(a+b\right)x-ab-x^2\right)dx=\underset{a}{\overset{b}{\int }}\left(x-a\right)\left(b-x\right)dx\end{array}$

Đặt t=x-a.

Suy ra $S=\underset{0}{\overset{b-a}{\int }}t\left(b-a-t\right)dt=\underset{0}{\overset{b-a}{\int }}\left(\left(b-a\right)t-t^2\right)dt=\frac{\left(b-a\right)t^2}{2}\left|{}_0^{b-a}\right.-\frac{t^3}{3}\left|{}_0^{b-a}\right.=\frac{{\left(b-a\right)}^3}{6}$

Ta có: ${\left(b-a\right)}^2+{\left(b^2-a^2\right)}^2=4\iff {\left(b-a\right)}^2\left(1+{\left(b+a\right)}^2\right)=4\iff {\left(b-a\right)}^2=\frac{4}{1+{\left(a+b\right)}^2}\le 4$.

Suy ra $b-a\le 2\Rightarrow S=\frac{{\left(b-a\right)}^3}{6}\le \frac{2^3}{6}=\frac{4}{3}$.

Dấu “=” xảy ra khi và chỉ khi $\begin{array}{l}\left\{\begin{array}{l}a+b=0\\ b-a=2\end{array}\right.\iff \left\{\begin{array}{l}b=1\\ a=-1\end{array}\right.\iff A\left(-1;1\right),B\left(1;1\right)\\ \end{array}$.