#4
設小水晶球半徑\(r\)
立方體的斜對角線\(8\sqrt 3 = 4 \cdot 2 + 2r\sqrt 3 + 2r\)
移項即可解出\(r\)
這題考出來算是秒殺題了
關鍵在斜對角線與內接球半徑的關係
110.2.28補充
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#8
設切點P為\((\sqrt t ,t)\),可算出切線交y軸於A點\((0, - t)\)
\(\Delta PAB = \frac{1}{2} \cdot 2t \cdot \sqrt t = t\sqrt t \)
拋物線與y軸、\(\overline {PB} \)所夾面積\(\int_0^{\sqrt t } {(t - {x^2})} dx = \frac{2}{3}t\sqrt t \)
故所求比值為\(\frac{2}{3}\)
#10
\(f'(x) = 6{x^2} + 6ax + 6(a - 1) = 0\)解得\(x = - 1,1 - a\)
若\(1 - a > - 1\),即\(a < 2\)
\(f(1 - a) = 0\)解得\(a = 2 - \sqrt 3 \)
若\(1 - a < - 1\),即\(a > 2\)
\(f(-1) = 0\),此時\(a\)無解
故答案只有一個