103學測試題線上作答

---

一 、 單 選 題 （ 占 3 0 分）

---

1.   請問下列哪一個選項等於 \(log(2^{(3^{5})})\)

(1)   \(5\, log\left ( 2^{3} \right )\)

(2)   \(3\times5\, log\, 2\)

(3)   \(5\, log\, 2\times log\, 3\)

(4)   \(5\left ( log\, 2+log\, 3 \right )\)

(5)   \(3^{5}\, log\, 2\)

記得嗎? 對數有下列運算性質:   \(log{_{a}}^{\left ( b^{c} \right )}=c\, log{_{a}}^{b}\)

記得嗎? 對數有下列運算性質:   \(log{_{a}}^{\left ( b^{c} \right )}=c\, log{_{a}}^{b}\)

如果底數 a 為10可省略不寫:

\(log{_{10}}^{b^{c}}=log\left ( b^{c} \right )=c\, log\, b\)



A: (5)

---

2.   令 \(A(5,0,12)\)、\(B(-5,0,12)\) 為坐標空間中之兩點，且令 \(P\) 為 \(xy\) 平面上滿足 \(\overline{PA}=\overline{PB}=13\) 的點。請問下列哪一個選項中的點可能為 \(P\) ？

(1)   \((5,0,0)\)

(2)   \((5,5,0)\)

(3)   \((0,12,0)\)

(4)   \((0,0,0)\)

(5)   \((0,0,24)\)

\(\require{color}\)因 \(P\) 點在 \(xy\) 平面上, 由此可見 \(P\) 點 \(z\) 座標值必為"0"

所以第(5)個選項 \((0,0,{\color{Red} 24})\) 當然是錯的啦

接下來看第(1)個選項,令 \((5,0,0)\) 為 \(Q\) 點

\(\overline{QA}=\left | \overrightarrow{QA} \right |=\left | (5-5,0-0,12-0) \right |=\left | (0,0,12) \right |={\color{Red} 12}\)

\(\overline{QA}\neq 13\) 由此可見(1)是錯的

接下來看第(4)個選項,令 \((0,0,0)\) 為 \(O\) 點

\(\overline{OA}=\left | \overrightarrow{OA} \right |=\left | (5-0,0-0,12-0) \right |=\left | (5,0,12) \right |=\sqrt{5^{2}+0^{2}+12^{2}}={\color{Green} 13}\)

\(\overline{OA}=13\) 由此可見 \(O\) 點有可能為\(P\) 點


A: (4)

---

3.   在坐標平面上, 以 \(\left ( 1,1 \right )\), \(\left ( -1,1 \right )\), \(\left ( -1,-1 \right )\) 及 \(\left ( 1,-1 \right )\) 等四個點為頂點的正方形, 與圓 \(x^{2}+y^{2}+2x+2y+1=0\) 有幾個交點？

(1)   1 個

(2)   2 個

(2)   3 個

(2)   4 個

(2)   0 個

你能在坐標平面上畫出正方形的四個頂點與圓嗎?

你能在坐標平面上畫出正方形的四個頂點與圓嗎?

將圓C:   \(x^{2}+y^{2}+2x+2y+1=0 \; \;\)標準化

你能在坐標平面上畫出正方形的四個頂點與圓嗎?

將圓C:   \(x^{2}+y^{2}+2x+2y+1=0\; \; \)標準化

\(\Rightarrow \;\; x^{2}+2x+1+y^{2}+2y=0\)

\(\Rightarrow \;\; x^{2}+2x+1+y^{2}+2y{\color{Green} \:\: +1}=0{\color{Green} \: +1}\)

\(\Rightarrow \; \; \left ( x+1 \right )^{2}+\left ( y+1 \right )^{2}=1\)

可看出圓C的圓心坐標為\({\color{Red} \left ( -1,-1 \right )}\),  半徑\({\color{Red} r=1}\)

  由圖可知圓與正方形交於兩點

---

4.   請問滿足絕對值不等式\(\, \left | 4x-12 \right |\leq 2x\,\)的實數\(\, x\,\)所形成的區間, 其長度為下列哪一個選項？

(1)  1

(2)  2

(3)  3

(4)  4

(5)  6

首先可以先將不等式化簡, 提出公因數再約分

首先可以先將不等式化簡, 提出公因數再約分

\(\, \left | 4x-12 \right |\leq 2x\,\)

\(\Rightarrow \, \; \; {\color{Green} 2}\left | 2x-6 \right |\leq{\color{Green} 2}x\) \(\Rightarrow \, \; \; \left | 2x-6 \right |\leq x\)

首先可以先將不等式化簡, 提出公因數再約分

\(\, \left | 4x-12 \right |\leq 2x\,\)

\(\Rightarrow \, \; \; {\color{Green} 2}\left | 2x-6 \right |\leq{\color{Green} 2}x\) \(\Rightarrow \, \; \; \left | 2x-6 \right |\leq x\)

因絕對值\(\, \left | 2x-6 \right |\,\)恆\(\, \geq 0\,\) , 且\(\, \left | 2x-6 \right |\leq x\)

故\(\, {\color{Red} x\geq 0}\)

因為不等式\(\, \left | 2x-6 \right |\leq x\,\)兩邊的數皆\(\, \geq 0\,\)

兩邊平方後不等式仍然成立

\(\Rightarrow \; \; \left ( 2x-6 \right )^{{\color{Green} 2}}\leq x^{{\color{Green} 2}}\)

因為不等式\(\, \left | 2x-6 \right |\leq x\,\)兩邊的數皆\(\, \geq 0\,\)

兩邊平方後不等式仍然成立

\(\Rightarrow \; \; \left ( 2x-6 \right )^{{\color{Green} 2}}\leq x^{{\color{Green} 2}}\)   \(\Rightarrow \; \; 4x^{2}-24x+36\leq x^{2}\)

\(\Rightarrow \; 3x^{2}-24x+36\leq 0\)   \(\Rightarrow \; x^{2}-8x+12\leq 0\)

因為不等式\(\, \left | 2x-6 \right |\leq x\,\)兩邊的數皆\(\, \geq 0\,\)

兩邊平方後不等式仍然成立

\(\Rightarrow \; \; \left ( 2x-6 \right )^{{\color{Green} 2}}\leq x^{{\color{Green} 2}}\)   \(\Rightarrow \; \; 4x^{2}-24x+36\leq x^{2}\)

\(\Rightarrow \; 3x^{2}-24x+36\leq 0\)   \(\Rightarrow \; x^{2}-8x+12\leq 0\)

因式分解 \(\Rightarrow \; \left ( x-2 \right )\left ( x-6 \right )\leq 0\)

因為不等式\(\, \left | 2x-6 \right |\leq x\,\)兩邊的數皆\(\, \geq 0\,\)

兩邊平方後不等式仍然成立

\(\Rightarrow \; \; \left ( 2x-6 \right )^{{\color{Green} 2}}\leq x^{{\color{Green} 2}}\)   \(\Rightarrow \; \; 4x^{2}-24x+36\leq x^{2}\)

\(\Rightarrow \; 3x^{2}-24x+36\leq 0\)   \(\Rightarrow \; x^{2}-8x+12\leq 0\)

因式分解 \(\Rightarrow \; \left ( x-2 \right )\left ( x-6 \right )\leq 0\)

你能找出不等式 \(\left ( x-2 \right )\left ( x-6 \right )\leq 0\)的解區間嗎?

首先令 \(y=\left ( x-2 \right )\left ( x-6 \right )\)

首先令 \(y=\left ( x-2 \right )\left ( x-6 \right )\)

你知道這是什麼圖形嗎?

沒錯, 開口向上的拋物線,
與x軸交於2, 6

現在你能找出不等式
\(\left ( x-2 \right )\left ( x-6 \right )\leq 0\)
的解區間嗎?

不等式
\(\left ( x-2 \right )\left ( x-6 \right )\leq 0\)
的解區間為

\(2\leq x\leq 6\)

長度為4

---

5.   設\(\; \left ( 1+\sqrt{2} \right )^{6}=a+b\sqrt{2}\;\)， 其中a,b 為整數。請問b 等於下列哪一個選項？

(1)   \(C_{\, 0}^{\, 6}+2\, C_{\, 2}^{\, 6}+2^{2}\, C_{\, 4}^{\, 6}+2^{3}\, C_{\, 6}^{\, 6}\)

(2)   \(C_{\, 1}^{\, 6}+2\, C_{\, 3}^{\, 6}+2^{2}\, C_{\, 5}^{\, 6}\)

(3)   \(C_{\, 0}^{\, 6}+2\, C_{\, 1}^{\, 6}+2^{2}\, C_{\, 2}^{\, 6}+2^{3}\, C_{\, 3}^{\, 6}+2^{4}\, C_{\, 4}^{\, 6}+2^{5}\, C_{\, 5}^{\, 6}+2^{6}\, C_{\, 6}^{\, 6}\)

(4)   \(2\, C_{\, 1}^{\, 6}+2^{2}\, C_{\, 3}^{\, 6}+2^{3}\, C_{\, 5}^{\, 6}\)

(5)   \(C_{\, 0}^{\, 6}+2^{2}\, C_{\, 2}^{\, 6}+2^{4}\, C_{\, 4}^{\, 6}+2^{6}\, C_{\, 6}^{\, 6}\)

還記得"二項式定理"嗎?

\(\Rightarrow \; \left ( x+y \right )^{6}=C_{\, 0}^{\, 6}\, x^{6}+C_{\, 1}^{\, 6}\, x^{5}y^{1}+C_{\, 2}^{\, 6}\, x^{4}y^{2}+C_{\, 3}^{\, 6}\, x^{3}y^{3}+C_{\, 4}^{\, 6}\, x^{1}y^{5}+C_{\, 6}^{\, 6}\, y^{6}\)

還記得"二項式定理"嗎?

\(\Rightarrow \; \left ( x+y \right )^{6}=C_{\, 0}^{\, 6}\, x^{6}+C_{\, 1}^{\, 6}\, x^{5}y^{1}+C_{\, 2}^{\, 6}\, x^{4}y^{2}+C_{\, 3}^{\, 6}\, x^{3}y^{3}+C_{\, 4}^{\, 6}\, x^{1}y^{5}+C_{\, 6}^{\, 6}\, y^{6}\)

令\(\; {\color{Red} x=1},\; {\color{Green} y=\sqrt{2}}\;\)代入

\(\Rightarrow \; \left ( {\color{Red} 1}+{\color{Green} \sqrt{2}} \right )^{6}=C_{\, 0}^{\, 6}\, {\color{Red} 1^{6}}+C_{\, 1}^{\, 6}\, {\color{Red} 1^{5}}{\color{Green} \left ( \sqrt{2} \right )^{1}}+C_{\, 2}^{\, 6}\, {\color{Red} 1^{4}}{\color{Green} \left ( \sqrt{2} \right )^{2}}+C_{\, 3}^{\, 6}\, {\color{Red} 1^{3}}{\color{Green} \left ( \sqrt{2} \right )^{3}}+\) \(C_{\, 4}^{\, 6}\, {\color{Red} 1^{2}}{\color{Green} \left ( \sqrt{2} \right )^{4}}+C_{\, 5}^{\, 6}\, {\color{Red} 1^{1}}{\color{Green} \left ( \sqrt{2} \right )^{5}}+C_{\, 6}^{\, 6}\, {\color{Green} \left ( \sqrt{2} \right )^{6}}\)

\(\Rightarrow \; \left ( {\color{Red} 1}+{\color{Green} \sqrt{2}} \right )^{6}=C_{\, 0}^{\, 6}\, {\color{Red} 1^{6}}+C_{\, 1}^{\, 6}\, {\color{Red} 1^{5}}{\color{Green} \left ( \sqrt{2} \right )^{1}}+C_{\, 2}^{\, 6}\, {\color{Red} 1^{4}}{\color{Green} \left ( \sqrt{2} \right )^{2}}+C_{\, 3}^{\, 6}\, {\color{Red} 1^{3}}{\color{Green} \left ( \sqrt{2} \right )^{3}}+\) \(C_{\, 4}^{\, 6}\, {\color{Red} 1^{2}}{\color{Green} \left ( \sqrt{2} \right )^{4}}+C_{\, 5}^{\, 6}\, {\color{Red} 1^{1}}{\color{Green} \left ( \sqrt{2} \right )^{5}}+C_{\, 6}^{\, 6}\, {\color{Green} \left ( \sqrt{2} \right )^{6}}\)

\(=C_{\, 0}^{\, 6}+C_{\, 1}^{\, 6}{\color{Green} \left ( \sqrt{2} \right )^{1}}+C_{\, 2}^{\, 6}\, {\color{Green} \left ( \sqrt{2} \right )^{2}}+C_{\, 3}^{\, 6}\, {\color{Green} \left ( \sqrt{2} \right )^{3}}+C_{\, 4}^{\, 6}\, {\color{Green} \left ( \sqrt{2} \right )^{4}}+C_{\, 5}^{\, 6}\, {\color{Green} \left ( \sqrt{2} \right )^{5}}+C_{\, 6}^{\, 6}\, {\color{Green} \left ( \sqrt{2} \right )^{6}}\)

\(\Rightarrow \; \left ( {\color{Red} 1}+{\color{Green} \sqrt{2}} \right )^{6}=C_{\, 0}^{\, 6}\, {\color{Red} 1^{6}}+C_{\, 1}^{\, 6}\, {\color{Red} 1^{5}}{\color{Green} \left ( \sqrt{2} \right )^{1}}+C_{\, 2}^{\, 6}\, {\color{Red} 1^{4}}{\color{Green} \left ( \sqrt{2} \right )^{2}}+C_{\, 3}^{\, 6}\, {\color{Red} 1^{3}}{\color{Green} \left ( \sqrt{2} \right )^{3}}+\) \(C_{\, 4}^{\, 6}\, {\color{Red} 1^{2}}{\color{Green} \left ( \sqrt{2} \right )^{4}}+C_{\, 5}^{\, 6}\, {\color{Red} 1^{1}}{\color{Green} \left ( \sqrt{2} \right )^{5}}+C_{\, 6}^{\, 6}\, {\color{Green} \left ( \sqrt{2} \right )^{6}}\)

\(=C_{\, 0}^{\, 6}+C_{\, 1}^{\, 6}{\color{Green} \left ( \sqrt{2} \right )^{1}}+C_{\, 2}^{\, 6}\, {\color{Green} \left ( \sqrt{2} \right )^{2}}+C_{\, 3}^{\, 6}\, {\color{Green} \left ( \sqrt{2} \right )^{3}}+C_{\, 4}^{\, 6}\, {\color{Green} \left ( \sqrt{2} \right )^{4}}+C_{\, 5}^{\, 6}\, {\color{Green} \left ( \sqrt{2} \right )^{5}}+C_{\, 6}^{\, 6}\, {\color{Green} \left ( \sqrt{2} \right )^{6}}\)

\(=C_{\, 0}^{\, 6}+C_{\, 1}^{\, 6}{\color{Green} \left ( \sqrt{2} \right )}+C_{\, 2}^{\, 6}\, \times 2+C_{\, 3}^{\, 6}\, {\color{Green} \left ( 2\sqrt{2} \right )}+C_{\, 4}^{\, 6}\, \times 4+C_{\, 5}^{\, 6}\, {\color{Green} \left ( 2^{2}\sqrt{2} \right )}+C_{\, 6}^{\, 6}\, \times 8\)

\(=C_{\, 0}^{\, 6}+2\, C_{\, 2}^{\, 6}+4\, C_{\, 4}^{\, 6}+8\, C_{\, 6}^{\, 6}+\left ( C_{\, 1}^{\, 6}+2\, C_{\, 3}^{\, 6}\, +2^{2}\, C_{\, 5}^{\, 6}\, \right ){\color{Green} \left ( \sqrt{2} \right )}\)