111年身障生升四技二專-數學(C)詳解

111 學年度身心障礙學生升學大專校院甄試

甄試類(群)組別：四技二專組-數學(C)

單選題，共 20 題，每題 5 分

解答：$$\cases{圖形有最低點(凹向上) \Rightarrow a\gt 0\\ f(0)=b=-4 \Rightarrow b\lt 0} \Rightarrow (a\gt 0,b\lt 0)在第四象限，故選\bbox[red, 2pt]{(D)}$$

解答：$$(\tan{\pi\over 7} +\cot {\pi\over 7})^2-(\sec {\pi\over 7}-\cos {\pi\over 7})^2 -\cot^2{\pi\over 7}+\cos^2 {\pi\over 7}\\ =\tan^2{\pi\over 7}+2 +\cot^2{\pi\over 7}- (\sec^2 {\pi\over 7}-2+\cos^2{\pi\over 7})-\cot^2{\pi\over 7}+\cos^2 {\pi\over 7}\\ =\tan^2{\pi\over 7}-\sec^2 {\pi\over 7}+4 =-1+4=3，故選\bbox[red, 2pt]{(C)}$$

解答：$$\sec A=56  \Rightarrow \cos A={1\over 56} ={4^2+7^2- \overline{BC}^2 \over 2\cdot 4\cdot 7} ={65- \overline{BC}^2 \over 56} \Rightarrow  \overline{BC}^2 =64 \Rightarrow  \overline{BC}=8，故選\bbox[red, 2pt]{(A)}$$

解答：$$\vec u\parallel (3,7) \Rightarrow \vec u=(3k,7k),k\in \mathbb{R} \Rightarrow  (7,-3)\cdot \vec u=21k-21k=0 \Rightarrow (7,-3)\bot \vec u，故選\bbox[red, 2pt]{(A)}$$

解答：$$\cases{A(3\sqrt 3,4) \\ B(2\sqrt 3,5)\\ C(4\sqrt 3,5)} \Rightarrow \cases{\overrightarrow{AB} =(-\sqrt 3,1)\\ \overrightarrow{AC} =(\sqrt 3,1)} \Rightarrow \cos \theta ={\overrightarrow{AB} \cdot \overrightarrow{AC} \over |\overrightarrow{AB}|| \overrightarrow{AC}|} ={-2\over 2\cdot 2} =-{1\over 2} \Rightarrow \theta ={2\over 3}\pi，故選\bbox[red, 2pt]{(D)}$$

解答：$$(2x^a-x+1)(x^3-1) = 2x^5+bx^4 +x^3-2x^2+cx-1\\\cases{ 左式最高項次數=a+3\\ 右式最高項次數=5} \Rightarrow a=2 \Rightarrow (2x^2-x+1)(x^3-1) =2x^5-x^4+x^3-2x^2+x-1\\ \Rightarrow \cases{b=-1\\ c=1} \Rightarrow a+b+c =2-1+1=2，故選\bbox[red, 2pt]{(B)}$$

解答：$$(4+2i)(3-7i)+ {5+2i\over 3-i} =12-28i +6i+14+ {(5+2i)(3+i)\over (3-i)(3+i)} =26-22i+{13+11i\over 10}\\ =26+ {13\over 10}+(-22+{11\over 10})i ={273\over 10}-{ 209\over 10}i，故選\bbox[red, 2pt]{(D)}$$

解答：$$圓x^2-2ax+y^2+4y = 5-a^2 \Rightarrow (x^2-2ax+a^2) +(y^2+4y+4)=9\\ \Rightarrow (x-a)^2+(y+2)^2 =3^2 \Rightarrow \cases{圓心P(a,-2) \\ 半徑r=3};\\ 直線L:3x-4y=2與圓相切 \Rightarrow d(P,L)=r \Rightarrow {3a+8-2\over 5} =3 \Rightarrow a=3，故選\bbox[red, 2pt]{(B)}$$

解答：$$\cases{a_1 +a_2+\cdots +a_{2n-1} =260\\ a_n=20} \Rightarrow \cases{{(2a_1+(2n-2)d)(2n-1)\over 2}= 260 \Rightarrow (a_1+(n-1)d)(2n-1)=260 \cdots(1)\\ a_1+(n-1)d= 20 \cdots(2)} \\ 將(2)代入(1) \Rightarrow 20(2n-1)=260 \Rightarrow 2n-1=13\Rightarrow n=7，故選\bbox[red, 2pt]{(D)}$$

解答：$$\sec \theta ={\sqrt{173}\over 13} \Rightarrow \cos \theta = {13\over \sqrt{173}} \Rightarrow \sin \theta ={ 2\over \sqrt{173} } \Rightarrow \cases{\sin (2\theta) =2\sin\theta \cos\theta =52/173\\ \cos(2\theta)= \cos^2\theta -\sin^2\theta =165/173} \\ \Rightarrow \sin(2\theta)+ \cos (2\theta)={217\over 173}，故選\bbox[red, 2pt]{(B)}$$

解答：$$f(x)= 2x^3-4x^2+1 \Rightarrow f'(x)=6x^2-8x \Rightarrow \cases{f(1)=-1\\ f'(1)=-2} \\ \Rightarrow 切線L通過(1,f(1))= (1,-1) 且斜率為-2 \Rightarrow L: y=-2(x-1)-1 \Rightarrow 2x+y=1\\ 將各選項坐標代入，可得(2,-3)在L上，故選\bbox[red, 2pt]{(A)}$$

解答：$$\cases{任意四位數:5\times 5\times 4\times 3 = 300個\\ 個位數=5的四位數: 4\times 4\times 3=48個\\ 個位數=0的四位數: 5\times 4\times 3= 60個} \Rightarrow 不是5的倍數=300-48-60=192個，故選\bbox[red, 2pt]{(B)}$$

解答：$$\int_0^1 3x^2-4x+6-(-1)\;dx= \int_0^1 3x^2-4x+7\;dx= \left.\left[ x^3-2x^2+7x \right]\right|_0^1 =6，故選\bbox[red, 2pt]{(D)}$$

解答：$${(x-1)^2 \over a^2} +{(y+2)^2 \over b^2} =1 \Rightarrow 中心點O(1,-2) \Rightarrow d(O,(6,-2))=a \Rightarrow a=5\\ 又正焦弦長{2b^2\over a} =6 \Rightarrow b^2= 3a= 15 \Rightarrow b=\sqrt{15} \Rightarrow c=\sqrt{a^2-b^2} =\sqrt{10} \Rightarrow \overline{F_1 F_2} =2c= 2\sqrt{10}\\，故選\bbox[red, 2pt]{(B)}$$

解答：

$$封閉區域\cases{2x+y\le 10\\ x-y\ge -1\\ x\ge 0\\ y\ge 0}各頂點坐標\cases{A(3,4)\\ B(0,1)\\ C(0,0)\\ D(5,0)  }，如上圖；\\令f(x,y)=3x+2y \Rightarrow \cases{f(A)=17\\ f(B)=2\\ f(C)=0\\ f(D)=15} \Rightarrow 最大值=17，故選\bbox[red, 2pt]{(C)}$$

解答：$$\cases{x-2y+3z=1\\ 2x-y+az= -3\\ 3x-y+4z= -2} \Rightarrow \begin{bmatrix} 1& -2 & 3\\ 2& -1 & a\\ 3 & -1 & 4 \end{bmatrix} \begin{bmatrix} x\\ y\\z\end{bmatrix}= \begin{bmatrix} 1\\ -3\\ 2\end{bmatrix} 無解\Rightarrow \begin{vmatrix} 1& -2 & 3\\ 2& -1 & a\\ 3 & -1 & 4 \end{vmatrix}=0 \\ \Rightarrow -4-6-6a+9+16+a=0 \Rightarrow 5a=15 \Rightarrow a=3，故選\bbox[red, 2pt]{(C)}$$

解答：$$\cases{A(1,-1,2)\\ B(2,0,-3) \\C(-3,2,4)} \Rightarrow \cases{\overrightarrow{AB} =(1,1,-5) \\ \overrightarrow{AC} =(-4,3,2) } \Rightarrow E之法向量 =\overrightarrow{AB} \times \overrightarrow{AC} =(17,18,7)，故選\bbox[red, 2pt]{(A)}$$

解答：$$\int_0^1 x\sqrt{x^2+3} \;dx =\left. \left[{1\over 3}(x^2+3)^{3/2} \right]\right|_0^1 ={1\over 3}(4^{3/2} -3^{3/2}) ={1\over 3}(8 -3\sqrt 3)={8\over 3}- \sqrt 3，故選\bbox[red, 2pt]{(C)}$$

解答：$$6\log_3 2-2\log_3 4+{1\over 3}\log_3 8 =6\log_3 2-2\log_3 2^2+{1\over 3}\log_3 2^3 =6\log_3 2-4\log_3 2+ \log_3 2 \\ =3 \log_3 2，故選\bbox[red, 2pt]{(A)}$$

解答：$$\cases{A= \begin{bmatrix} 3& -2 \\ -1 & 1\end{bmatrix} \\[1ex] B= \begin{bmatrix} 1& -1 \\ 0 & 2\end{bmatrix}} \Rightarrow \cases{X=A-B =\begin{bmatrix} 2& -1 \\ -1 & -1\end{bmatrix} \\[1ex] Y=-A+2B =\begin{bmatrix} -1& 0 \\ 1 & 3 \end{bmatrix} }\\ \Rightarrow C=XY =\begin{bmatrix} 2& -1 \\ -1 & -1\end{bmatrix}\begin{bmatrix} -1& 0 \\ 1 & 3 \end{bmatrix} =\begin{bmatrix} -3 & -3 \\ 0 & -3 \end{bmatrix}，故選\bbox[red, 2pt]{(C)}$$ 

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