Cho hình chóp S.ABCD  có cạnh SA=x còn tất c...

D

- Hướng dẫn giải

Đáp án D

$$\begin{gathered} V_{S.ABCD}=V_{S.ABC}+V_{S.ADC} \\ =2V_{S.ABC} \\ =2.\frac{1}{3}d\left(B;\left(SAC\right)\right)S_{SAC} \\ =\frac{2}{3}xBO \end{gathered}$$

$\begin{array}{l}OC=\frac{1}{2}AC=\frac{1}{2}\sqrt{S{A}^2+S{C}^2}\\ =\frac{1}{2}\sqrt{x^2+4}\\ BO=\sqrt{B{C}^2-O{C}^2}\\ =\sqrt{4-\frac{x^2}{4}-1}=\sqrt{3-\frac{x^2}{4}}\\ V_{S.ABCD}=\frac{2}{3}x\sqrt{3-\frac{x^2}{4}}\end{array}$

Đặt $f\left(x\right)=x\sqrt{3-\frac{x^2}{4}},x\in (0;2\sqrt{3}\text{]}$

$\begin{array}{l}\text{f'(x)=}\sqrt{3-\frac{x^2}{4}}+x\frac{-\frac{x}{2}}{2\sqrt{3-\frac{x^2}{4}}}\\ =\frac{6-x^2}{2\sqrt{3-\frac{x^2}{4}}}\\ f\text{'}\left(x\right)=0\iff x=\sqrt{6}\end{array}$

Bảng biến thiên

Vậy $V_{\max }=\frac{2}{3}.\underset{(0;2\sqrt{3}\text{]}}{\max }f\left(x\right)=\frac{2}{3}3=2$