109年度自學進修普通型高級中等學校
畢業程度學力鑑定考試
解：
$$\vec a \bot \vec b \Rightarrow \vec a\cdot \vec b=0 \Rightarrow (t+4,2t+3) \cdot (-3,2)=-3t-12+4t+6=t-6=0\\ \Rightarrow t=6,故選\bbox[red,2pt]{(D)}$$
$$\log{5\over 9}-\log{3\over 14} +\log{27\over 7} =(\log 5-\log 9)-(\log 3-\log 14)+ (\log 27-\log 7) \\ =\log 5-2\log 3-\log 3+\log 2+\log 7+3\log 3-\log 7 =\log 5+\log 2= \log 10 =1,故選\bbox[red, 2pt]{(A)}$$
解：
$$每個選項都有2種選擇,因此共有2^4=16-1(全不選)=15種填寫方式,故選\bbox[red, 2pt]{(C)}$$
解：
$$x^2+y^2-4x+2y+k=0 \Rightarrow (x-2)^2 +(y+1)^2 = 5-k \Rightarrow 圓半徑為\sqrt{5-k} \\ \Rightarrow 5-k > 0 \Rightarrow 5 > k,故選\bbox[red, 2pt]{(D)}$$
$$(A) \times: 正弦定理: {7\over \sin A} = {8\over \sin B} = {5\over \sin C} \Rightarrow \sin A:\sin B: \sin C=7:8:5 \\(B) \times: \cos \angle C= {7^2+8^2 -5^2 \over 2\times 7\times 8} > 0 \Rightarrow \angle C < 90^\circ \\(C) \bigcirc: \cos \angle A = {5^2+8^2- 7^2 \over 2\times 5\times 8} = {40\over 80} ={1\over 2} \Rightarrow \angle A= 60^\circ \\ (D)\times: \triangle ABC 面積= {1\over 2}\overline{AB} \cdot \overline{AC} \sin \angle A ={1\over 2} \times 5\times 8 \times {\sqrt 3\over 2} =10\sqrt 3\\ 故選\bbox[red, 2pt]{(C)}$$
$$-15 \le x \le 9 \Rightarrow -12 \le x+3 \le 12 \Rightarrow |x+3| \le 12 \Rightarrow a=12,故選\bbox[red, 2pt]{(B)}$$
$$\cases{L_1: {x-1\over 2} ={y-2\over -2} ={z+2\over 1} \\ L_2: {x-3\over 4} ={y\over -4} ={z+1\over 2}} \Rightarrow \cases{L_1方向向量\vec u=(2,-2,1)且經過P(1,2,-2) \\L_2方向向量\vec v=(4,-4,2)且經過Q(3,0,-1)} \\ \Rightarrow \cases{\vec u=2\vec v \\ L_1經過Q \\ L_2經過P} \Rightarrow L_1,L_2重合,故選\bbox[red, 2pt]{(A)}$$
$$0.\bar 3={3\over 9}={1\over 3}為有理數,故選\bbox[red, 2pt]{(B)}$$
解：
$$(A)\times: A+B=\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix} +\begin{bmatrix}4 & 3 \\ 2 & 1\end{bmatrix} =\begin{bmatrix}5 & 5 \\ 5 & 5\end{bmatrix} \ne \begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix} \\(B) \bigcirc: AB=\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix} \begin{bmatrix}4 & 3 \\ 2 & 1\end{bmatrix} =\begin{bmatrix}8 & 5 \\ 20 & 13\end{bmatrix} \\(C)\times: det(A) = 4-6=-2\ne 10\\ (D)\times: \cases{\det(B)=4-6=-2 \\ \det(2B)=2^2 \det(B) =-8} \Rightarrow \det(2B)\ne 2\det(B)\\,故選\bbox[red, 2pt]{(B)}$$
解：
$$由散布圖可知:x越大,則y越大,因此r> 0,故選\bbox[red, 2pt]{(C)}$$
解：
$$(A)\bigcirc: \sqrt{4^2+5^2} =\sqrt{41} \\(B)\times: 5\ne 7 \\(C)\times: \sqrt{3^2+4^2+5^2} =5\sqrt 2 \ne 5 \\ (D)\times: (-3,4,5) \ne (3,-4,-5)\\,故選\bbox[red, 2pt]{(A)}$$
$$\lim_{n\to \infty} {n^2 \over 3n+1} =\infty \ne 0,故選\bbox[red, 2pt]{(D)}$$
$$40\pi = r^2\pi \times {144\over 360} \Rightarrow r^2 =40\times 360\div 144 = 100 \Rightarrow r=\bbox[red, 2pt]{10}$$
解：
$$\cases{f(5-3i)=\sqrt 2+3i\\ f(i)=1-\sqrt 2} \Rightarrow \cases{f(5+3i)=-\sqrt 2-3i\\ f(-i)=1+\sqrt 2} \Rightarrow f(5+3i) +f(-i) = \bbox[red, 2pt]{1-3i}$$
解：
$$\cases{\tan \theta = -4/3 \\ \sin \theta > 0} \Rightarrow \cases{\sin \theta = 4/5 \\ \cos \theta = -3/5} \Rightarrow {4\cos \theta+1 \over 3-2\sin \theta} = {-12/5+1 \over 3-8/5} ={-7/5 \over 7/5} = \bbox[red, 2pt]{-1}$$
解：$$6^3+7^3 +\cdots +15^3 = \sum_{k=1}^{15}k^3- \sum_{k=1}^{5}k^3 =\left({15\times 16\over 2} \right)^2 -\left({5\times 6\over 2} \right)^2 = 120^2-15^2 =\bbox[red, 2pt]{14175}$$
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