109學年度身心障礙學生升學大專校院甄試試題
甄試類(群)組別：四技二專組
考試科目(編號)：數學(C)
單選題，共 20 題，每題 5 分
$$\cases{A(0,-1) \\ B(a,2) \\ C(4,-3) \\ D(-2,b)} \Rightarrow \cases{ \overrightarrow{AB} =(a,3) \\ \overrightarrow{AC} = (4,-2) \\ \overrightarrow{AD} =(-2,b+1) } \Rightarrow \cases{{a\over 4}={3 \over -2} \\ {4\over -2}={ -2\over b+1}} \Rightarrow \cases{a=-6 \\ b=0} \Rightarrow a+b=-6\\，故選\bbox[red,2pt]{(B)} $$
解：
$$(\sin 75^\circ - \sin 15^\circ)^2 + (\cos 75^\circ + \cos 15^\circ)^2 \\=\sin^2 75^\circ -2\sin 75^\circ\sin 15^\circ +\sin^2 15^\circ +\cos^2 75^\circ +2\cos 75^\circ \cos 15^\circ + \cos^2 15^\circ \\ =(\sin^2 75^\circ+\cos^2 75^\circ) +(\sin^2 15^\circ +\cos^2 15^\circ)+2(\cos 75^\circ \cos 15^\circ -\sin 75^\circ\sin 15^\circ)\\ =1+1-2\cos(75^\circ+ 15^\circ) = 2-2\cos 90^\circ = 2-0=2，故選\bbox[red,2pt]{(C)}$$
解：$$(a+b+c)(a+b-c) = (a+b)^2-c^2 = a^2+b^2-c^2+2ab = 3ab \Rightarrow a^2+b^2-c^2 = ab\\ \Rightarrow \cos \angle C = \cfrac{a^2+b^2-c^2} {2ab} =\cfrac{ab}{2ab} =\cfrac{1}{2} \Rightarrow \angle C=60^\circ，故選\bbox[red,2pt]{(C)}$$
解：
$$f(x)= (x-1)^3 +2(x-1)^2+3(x-1)+4 = a(x+2)^3 +b(x+2)^2+c(x+2)+d \\ \Rightarrow f(-1) =-8+8-6+4= a+b+c+d \Rightarrow a+b+c+d =-2，故選\bbox[red,2pt]{(B)}$$
解：$$(\sqrt 2+1)^2 = 3+2\sqrt 2 \Rightarrow \sqrt{3+\sqrt 2} =\sqrt 2+1 \Rightarrow \cases{a=2 \\ b=\sqrt 2-1} \Rightarrow a+{1\over b} = 2+{1\over \sqrt 2-1}\\ =2 + \sqrt 2+1 =3+\sqrt 2，故選\bbox[red,2pt]{(C)}$$
解：$$\cases{3x+5y-z=-1 \cdots(1)\\ x-y+4z=11 \cdots(2) \\4x+2y-23z=k\cdots (3)}, 由(1)及(2)\Rightarrow \cases{x=(54-19z)/8 \\ y=(13z-34)/8}代入(3) \\\Rightarrow \cfrac{54-19z}{2} +\cfrac{65z-170}{2}-23z=k \Rightarrow k=-58，故選\bbox[red,2pt]{(A)}$$
解：$$-2 < x <5 \Rightarrow (x+2)(x-5)< 0 \Rightarrow x^2-3x-10<10 \Rightarrow \cases{a=-3 \\b=-10} \\ \Rightarrow x^2+3ax-2b >0 \Rightarrow x^2-9x+20 >0 \Rightarrow (x-4)(x-5)>0 \Rightarrow x>5或x<4，故選\bbox[red,2pt]{(D)}$$
解：$$11^{19} = (10+1)^{19} = \sum_{i=0}^{19}C^{19}_i10^i \Rightarrow 11^{19}除以100的餘數與\sum_{i=0}^{1}C^{19}_i10^i除以100的餘數相同;\\ 而\sum_{i=0}^{19}C^{19}_i10^i =1+19\times 10 =191 除以100的餘數為91，故選\bbox[red,2pt]{( D)}$$
解：
$$\cfrac{1+2+3+\cdots +n}{n-1} =\cases{45/8\approx 5.6 &n=9 \\ 55/9 \approx 6.1 & n=10\\ 66/10=6.6 & n=11 \\ 78/11\approx 7.1 & n=12 \\ 91/12 \approx 7.6 & n=13 \\ 105/13 \approx 8.1 & n=14 \\ 120/14 \approx 8.6 & n=15 \\ 136/15 = 9.1 & n=16} \\ \Rightarrow \cfrac{1+2+3+\cdots +n}{n-1}\ne 6,7,8,9，故題目有\bbox[red,2pt]{(疑慮)} $$
解：
$$假設第1次點數為a,第2次點數為b,則\begin{array}{c|l|c} a+b & (a,b) & 機率\\\hline 2 & (1,1) & 1/36 \\ 3& (1,2),(2,1) & 2/36 \\ \cdots & \cdots \\6 & (1,5),(2,4),\dots,(5,1) & 5/36 \\ 7 & (1,6),(2,5),\dots ,(6,1) & 6/36\\ 8 & (2,6),(3,5),\dots,(6,2) & 5/36\\ \cdots & \cdots \\11 & (5,6),(6,5) & 2/36 \\ 12 & (6,6) & 1/36\\ \hline\end{array} \\ \Rightarrow 點數和的期望值為 {1\over 36}(2\cdot 1+ \cdots +6\cdot 5+7\cdot 6 +8\cdot 5+ \cdots +12\cdot 1)\\ = {1\over 36}(2+6+12+20+30 +42+ 40 +36+ 30 + 22+ 12) =252/36=7，故選\bbox[red,2pt]{(B)}$$
解：
$$\cases{\overrightarrow{AB} =(2,-5) \\ \overrightarrow{BC}=(4,3) } \Rightarrow \overrightarrow{AC} = \overrightarrow{AB} +\overrightarrow{BC} =(2,-5)+(4,3) = (6,-2) \Rightarrow \cases{A(m,n) \\ C(m+6,n-2)} \\ \Rightarrow O({m+m+6 \over 2}, {n+n-2 \over 2}) =(2,5) \Rightarrow \cases{m=-1 \\n=6} \Rightarrow \cases{A(-1,6) \\ C(5,4)};\\ 假設D(p,q),由 \overrightarrow{AD} = \overrightarrow{BC} \Rightarrow (p+1,q-6) = (4,3) \Rightarrow \cases{p=3\\ q=9} \Rightarrow D(3,9)，故選\bbox[red,2pt]{(C)}$$
解：
$$(\vec a+\vec b+\vec c)\cdot (\vec a-\vec b+\vec c) = |\vec a|^2 -\vec a\cdot \vec b+\vec a\cdot \vec c+\vec b\cdot \vec a- |\vec b|^2 +\vec b\cdot \vec c+\vec c\cdot \vec a-\vec c\cdot b+|\vec c|^2 \\ =|\vec a|^2 +\vec a\cdot \vec c- |\vec b|^2 +\vec c\cdot \vec a+|\vec c|^2 = 4+3-9+3+16 = 17，故選\bbox[red,2pt]{(B)}$$
解：
$$(1+i)^2= 2i \Rightarrow (1+i)^4 = (2i)^2 =-4 \Rightarrow (1+i)^5 =-4(1+i) =-4-4i =a+bi \\ \Rightarrow \cases{a=-4 \\b=-4} \Rightarrow a^2+b^2 = 16+16=32， 故選\bbox[red, 2pt]{(D)}$$
解：
$$\cases{a_2+a_4+a_6 =45 \\ a_1+a_3+a_5 =33} \Rightarrow \cases{(a_1+d) +(a_1+3d)+ (a_1+5d) =45 \\ a_1+(a_1+2d) +(a_1 +4d) =33} \Rightarrow \cases{3a_1+ 9d =45 \\ 3a_1+ 6d =33} \\ \Rightarrow \cases{ a_1=3 \\d=4}，故選\bbox[red,2pt]{(A)}$$
解：
$$\log_2 3\cdot \log_3 5 \cdot \log_5 8 = \cfrac{\log 3}{\log 2} \cdot\cfrac{\log 5} {\log 3} \cdot\cfrac{\log 8}{\log 5} =\cfrac{\log 8}{\log 2} =\cfrac{3\log 2}{\log 2} =3 ，故選\bbox[red,2pt]{(C)}$$
解：
$$(A)\cases{2^{3^5} =2^{243} \\(2^3)^5 =2^{15}} \Rightarrow 2^{3^5}> (2^3)^5 \\(B)\cases{2^{3^5} =2^{243} \\2^{3\cdot 5} =2^{15}} \Rightarrow 2^{3^5}> 2^{3\cdot 5} \\(C) \cases{(2^3)^5 = 2^{15} \\ 2^{(3+5)} =2^8} \Rightarrow (2^3)^5 > 2^{(3+5)} \\(D) \cases{2^{(3+5)} =2^8 \\ 2^3\cdot 2^5 =2^8} \Rightarrow 2^{(3+5)} =2^3\cdot 2^5\\，故選\bbox[red,2pt]{(D)}$$
解：
$$\cases{(x-3)^2+(y-4)^2=25 \\y=3x } \Rightarrow (x-3)^2+(3x-4)^2=25 \Rightarrow 10x^2-30x=0 \Rightarrow 10x(x-3)=0 \\ \Rightarrow \cases{x=0 \\x=3} \Rightarrow \cases{y=0 \\ y=9} \Rightarrow \cases{A(0,0) \\B(3,9)} \Rightarrow A,B中點坐標({3\over 2},{9\over 2})，故選\bbox[red,2pt]{(B)}$$
解：
$$\cfrac{x^2}{9}- \cfrac{y^2}{4}=1 \Rightarrow \cases{a=3 \\b=2} \Rightarrow c^2=a^2+b^2 =9+4=13 \Rightarrow c=\sqrt{13}\\ \Rightarrow 兩焦點距離=2c =2\sqrt{13}，故選\bbox[red,2pt]{(A)}$$
解：
$$ 令k=3h \Rightarrow \lim_{h\to 0}\cfrac{g(1+3h)-g(1)}{h} =\lim_{k\to 0}\cfrac{g(1+k)-g(1)}{k/3} = 3\lim_{k\to 0}\cfrac{g(1+k)-g(1)}{k} \\ =3g'(1) =9 \Rightarrow g'(1)=3; \\又h(x)= f(x)\cdot g(x) \Rightarrow h'(x)=f'(x)g(x)\cdot f(x)g'(x) \Rightarrow h'(1)=f'(1)g(1)+f(1)g'(1)\\ = 4\times (-3)+5 \times 3=3，故選\bbox[red,2pt]{(C)}$$
解：$$f(x)=x^3-3x \Rightarrow f'(x)=3x^2-3 \Rightarrow f''(x)=6x \\f'(x)=0 \Rightarrow 3(x^2-1)=0 \Rightarrow x=\pm 1 有極值 \Rightarrow \cases{f''(1)=6>0 \\ f''(-1)= -6<0} \Rightarrow \cases{f(1)為極小值\\ f(-1)為極大值} \\ \Rightarrow f(-1)=-1+3=2 為最大值,x\in [-3,1]，故選\bbox[red,2pt]{(B)}$$
解題僅供參考
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