1.

\(\overrightarrow{MN}=\overrightarrow{MB'}+\overrightarrow{B'B}+\overrightarrow{BN}=\dfrac{1}{2}\overrightarrow{AB}-\overrightarrow{AA'}+\dfrac{1}{2}\overrightarrow{AD}\)

\(\overrightarrow{AC'}=\overrightarrow{AB'}+\overrightarrow{B'C'}=\overrightarrow{AB}+\overrightarrow{AA'}+\overrightarrow{AD}\)

\(\overrightarrow{MN}.\overrightarrow{AC'}=\left(\dfrac{1}{2}\overrightarrow{AB}-\overrightarrow{AA'}+\dfrac{1}{2}\overrightarrow{AD}\right)\left(\overrightarrow{AB}+\overrightarrow{AA'}+\overrightarrow{AD}\right)\)

\(=\dfrac{1}{2}AB^2-AA'^2+\dfrac{1}{2}AD^2=0\)

\(\Rightarrow MN\perp AC'\)

b.

\(\left\{{}\begin{matrix}AA'\perp BD\\BD\perp AC\end{matrix}\right.\) \(\Rightarrow BD\perp\left(ACC'A'\right)\Rightarrow BD\perp AC'\)

Tương tự: \(A'B\perp\left(ADC'B'\right)\Rightarrow A'B\perp AC'\)

\(\Rightarrow AC'\perp\left(A'BD\right)\)

2.

Phương trình \(x^3-3x+2=0\Leftrightarrow\left(x-1\right)^2\left(x+2\right)=0\) có nghiệm kép \(x=1\)

Nên giới hạn đã cho hữu hạn khi và chỉ khi phương trình: \(2\sqrt{1+ax^2}-bx-1=0\) có ít nhất 2 nghiệm \(x=1\) (tức là nghiệm bội 2 trở lên)

Thay \(x=1\) vào:

\(\Rightarrow2\sqrt{1+a}-b-1=0\Rightarrow2\sqrt{1+a}=b+1\)

\(\Rightarrow4\left(a+1\right)=b^2+2b+1\Rightarrow4a=b^2+2b-3\)

Khi đó:

\(\sqrt{4+4ax^2}-bx-1=0\Leftrightarrow\sqrt{4+\left(b^2+2b-3\right)x^2}-bx-1=0\)

\(\Leftrightarrow\sqrt{4+\left(b^2+2b-3\right)x^2}=bx+1\)

\(\Rightarrow4+\left(b^2+2b-3\right)x^2=b^2x^2+2bx+1\)

\(\Rightarrow\left(2b-3\right)x^2-2bx+3=0\)

\(\Rightarrow2bx^2-2bx-3x^2+3=0\)

\(\Rightarrow2bx\left(x-1\right)-\left(x-1\right)\left(3x+3\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2bx-3x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\\left(2b-3\right)x=3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{2b-3}\end{matrix}\right.\) \(\Rightarrow\dfrac{3}{2b-3}=1\Rightarrow b=3\Rightarrow a=3\)

\(c=\lim\limits_{x\rightarrow1}\dfrac{2\sqrt{1+3x^2}-3x-1}{x^3-3x+2}=\dfrac{1}{8}\)

Chọn B