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105指考數學乙試題-06


<多選題>設 \( a = 10^{1 – \frac{\sqrt{2}}{2}} \),\( b = a^{\sqrt{2}} \)。請選出正確的選項。
(1) \( 1 \lt a \)
(2) \( a \lt \sqrt{3} \)
(3) \( a^2 \lt b^{\sqrt{3}} \)
(4) \( 10^{0.4} \lt b \lt 10^{0.5} \)
(5) \((ab)^{\sqrt{2}} \lt 10\)

答案

\( a = 10^{1 - \frac{\sqrt{2}}{2}} \),\( b = a^{\sqrt{2}} = 10^{\sqrt{2} - 1} \)。
(1) \(1 - \frac{\sqrt{2}}{2} \approx 1 - 0.707 = 0.293 \gt 0\),所以 \(a \gt 10^0 = 1\),正確。
(2) \(a = 10^{0.293} \approx 1.96\),\(\sqrt{3} \approx 1.732\),所以 \(a \gt \sqrt{3}\),錯誤。
(3) \(a^2 = 10^{2 - \sqrt{2}} \approx 10^{0.586}\),\(b^{\sqrt{3}} = 10^{(\sqrt{2}-1)\sqrt{3}} = 10^{\sqrt{6} - \sqrt{3}} \approx 10^{2.449 - 1.732} = 10^{0.717}\),所以 \(a^2 \lt b^{\sqrt{3}}\),正確。
(4) \(b = 10^{\sqrt{2}-1} \approx 10^{0.414}\),\(10^{0.4} \lt b \lt 10^{0.5}\) 成立,正確。
(5) \(ab = 10^{1 - \frac{\sqrt{2}}{2} + \sqrt{2} - 1} = 10^{\frac{\sqrt{2}}{2}}\),\((ab)^{\sqrt{2}} = 10^{\frac{\sqrt{2}}{2} \cdot \sqrt{2}} = 10^1 = 10\),所以等於10,錯誤。
答案為 (1)(3)(4)。 報錯
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