Cho hình chóp S.ABCD có đáy ABC...

B

- Hướng dẫn giải

Đáp án B

Ta có: $\frac{V_{S.MNP}}{V_{S.ABC}}=\frac{2V_{S.MNP}}{V_{S.ABC}}=\frac{SM}{SA}.\frac{SN}{SB}.\frac{SP}{SC}=\frac{1}{3}.\frac{SP}{SC}$ 

Tương tự $\frac{V_{S.MPQ}}{V_{S.ACD}}=\frac{2V_{S.MPQ}}{V_{S.ABCD}}=\frac{1}{2}.\frac{SP}{SC}.\frac{SQ}{SD}$ 

Do đó $\frac{2V_{S.MNPQ}}{V_{S.ABCD}}=\frac{1}{3}\frac{SP}{SC}+\frac{1}{2}.\frac{SP}{SC}.\frac{SQ}{SD}$

Đặt $\frac{SP}{SC}=x\left(0<x\le 1\right)$, ta chứng minh được $\frac{SA}{SM}+\frac{SC}{SP}=\frac{SB}{SN}+\frac{SD}{SQ}=2\frac{SO}{SI}$

Do đó $\frac{SD}{SQ}=\frac{1}{x}+\frac{1}{2}\Rightarrow 2k=x\left(\frac{1}{3}+\frac{x}{x+2}\right)=\frac{2}{3}$ 

Do $0<x\le 1$nên ${\left(2k\right)}_{m\text{ax}}=f\left(1\right)=\frac{2}{3}\Rightarrow k=\frac{1}{3}.$