114學年度四技二專統測--數學(C)詳解

114 學年度科技校院四年制與專科學校二年制
統 一 入 學 測 驗-數學(C)

解答: $$(1+2i)(2-i)=2-i+4i+2=4+3i =a+bi\Rightarrow \cases{ a=4\\ b=3} \Rightarrow a+b=7，故選\bbox[red, 2pt]{(D)}$$

解答: $$\overrightarrow{AB} \bot \overrightarrow{AC} \Rightarrow \overrightarrow{AB} \cdot \overrightarrow{AC}=0，故選\bbox[red, 2pt]{(C)}$$

解答: $$\int x^3+2x-5 \,dx ={1\over 4}x^4+x^2-5x+C ，故選\bbox[red, 2pt]{(C)}$$

解答: $$2025^\circ =360^\circ\times 5+225^\circ \Rightarrow \tan 2025^\circ+ \sec 2025^\circ=\tan 225^\circ +\sec 225^\circ =\tan 45^\circ -\sec 45^\circ \\=1-\sqrt 2，故選\bbox[red, 2pt]{(B)}$$

解答: $$3+2\sqrt 2={a\over \sqrt 2-1}+b(\sqrt 2-1) \Rightarrow (3+2\sqrt 2)(\sqrt 2-1)=a+b(\sqrt 2-1)^2 \\ \Rightarrow 1+\sqrt 2=a+3b-2b\sqrt 2 \Rightarrow \cases{a+3b=1\\ -2b=1} \Rightarrow \cases{a=5/2\\ b=-1/2} \Rightarrow a^2-b^2={24\over 4}=6，故選\bbox[red, 2pt]{(A)}$$

解答: $$\cases{-\log H^+=7 \\ -\log H^+=4} \Rightarrow \cases{濃度最小值a=10^{-7} \\ 濃度最大值=10^{-4}} \Rightarrow 10^{-4} =10^3\times 10^{-7} =1000a，故選\bbox[red, 2pt]{(D)}$$

解答: $$假設公差為d \Rightarrow -2+3d=7 \Rightarrow d=3 \Rightarrow \cases{a=-2+d=1\\ b=-2+2d =4} \Rightarrow 1,4,p,q成等比數列\\ \Rightarrow 公比r={4\over 1} =4 \Rightarrow \cases{p=4r=16\\ q=4r^2=64} \Rightarrow p+q=16+64=80，故選\bbox[red, 2pt]{(C)}$$

解答: $$\cases{\sin 155^\circ =\sin (180^\circ-155^\circ) =\sin 25^\circ \Rightarrow \sin 155^\circ -\sin 55^\circ \lt 0\\ \cases{\cos 222^\circ =-\cos(222^\circ-180^\circ)= -\cos 42^\circ \\ \cos122^\circ =-\cos(180^\circ-122^\circ) =-\cos 58^\circ} \Rightarrow \cos 222^\circ-\cos 122^\circ =\cos 58^\circ-\cos42^\circ \lt 0} \\ \Rightarrow (\sin 155^\circ-\sin 55^\circ, \cos 222^\circ-\cos 122^\circ) 在第三象限，故選\bbox[red, 2pt]{(C)}$$

解答: $$x^2+kx-k+3=0有兩個共軛虛根\Rightarrow k^2-4(3-k)\lt 0 \Rightarrow k^2+4k-12\lt 0 \\ \Rightarrow (k+6)(k-2) \lt 0 \Rightarrow k\in (-6,2) =(a,b) \Rightarrow b-a=2-(-6)=8，故選\bbox[red, 2pt]{(C)}$$

解答: $$\cases{\star在L_1左側,即2x-y\le 6\\ \star 在L_2右側,即x+y\ge 8}，故選\bbox[red, 2pt]{(B)}$$

解答: $$\Gamma'正焦弦長=4 \Rightarrow \Gamma':y^2=4x \Rightarrow c=1\Rightarrow \cases{(A)\bigcirc:焦點(c,0)=(1,0) \\ (B)\times : y^2=4x\\ (C)\times: c=1\ne 2\\ (D)\times:準線y=-1}，故選\bbox[red, 2pt]{(A)}$$

解答: $$f(x)=2x^3-3x^2-12x+1 \Rightarrow f'(x)=6x^2-6x-12 \Rightarrow f''(x)=12x-6\\ f'(x)=0 \Rightarrow 6(x-2)(x+1)=0 \Rightarrow x=2,-1 \Rightarrow \cases{f''(2)=18 \gt 0\\ f''(-1)=-18 \lt 0} \\ \Rightarrow 遞增區域(-\infty,-1),(2, \infty)，故選\bbox[red, 2pt]{(D)}$$

解答: $$\cases{3x+4y=114\\ 4x+5y=2025} \Rightarrow \begin{bmatrix} 3& 4\\ 4& 5\end{bmatrix} \begin{bmatrix} x\\ y\end{bmatrix}= \begin{bmatrix}114\\ 2025 \end{bmatrix} \Rightarrow \begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix} a\\ b\end{bmatrix} =\begin{bmatrix} 3& 4\\ 4& 5\end{bmatrix}^{-1} \begin{bmatrix}114\\ 2025 \end{bmatrix}\\= \left[ \begin{matrix} -5 & 4 \\4 & -3 \end{matrix} \right]\begin{bmatrix}114\\ 2025 \end{bmatrix}，故選\bbox[red, 2pt]{(B)}$$

解答: $$假設兩個奇數為a_1,a_2,兩個偶數為b_1,b_2，則可能的四位數為:a_1b_1a_2b_2,b_1a_1b_2a_2, \\及a_1與a_2互換或b_1與b_2互換，因此共有8種可能\\ 1到9有5個奇數、4個偶數，因此符合要求的四位數有C^5_2C^4_2\times 8=480個，故選\bbox[red, 2pt]{(B)}$$

解答: $$\cases{(A)\bigcirc:AB \Rightarrow n=k\\ (B)\times:A+B \Rightarrow k=3,n=4\\ (C)\times:n=k\\ (D)\times:k=3,n=4}，故選\bbox[red, 2pt]{(A)}$$

解答: $$f(x)=a^x \Rightarrow f(T)=a^T={1\over 2} \\ (A)\times: f(4T)=a^{4T} =\left({1\over 2} \right)^4={1\over 16}\lt {1\over 10} \\(B)\times: f(6T) = \left({1\over 2} \right)^6= {1\over 64} \lt {1\over 50} \\(C)\times: f(8T) =\left({1\over 2} \right)^8={1\over 256} \gt {1\over 500} \\(D)\bigcirc: f(10T) =\left({1\over 2} \right)^{10}= {1\over 1024} \lt {1\over 1000}\\，故選\bbox[red, 2pt]{(D)}$$

解答: $$\cases{A(1,4) \\B(-3,-4) \\C(5,2)} \Rightarrow \cases{\overline{AB} =4\sqrt 5\\ \overline{BC} =10 \\ \overline{CA} =2\sqrt 5} \Rightarrow 10^2=(4\sqrt 5)^2+(2\sqrt 5)^2 \Rightarrow \angle A=90^\circ \ \\\Rightarrow \cases{直徑=2r=\overline{BC} =10 \Rightarrow r=5\\ 圓心O=(B+C)/2=(1,-1)} \Rightarrow 外接圓:(x-1)^2+ (y+1)^2=5^2，故選\bbox[red, 2pt]{(D)}$$

解答: $${\triangle OAP\over \triangle OBP} ={2\over 7} ={\overline{PA} \over \overline{PB}} \Rightarrow P=(7A+2B)/9=\left({1\over 9},{22\over 9} \right)，故選\bbox[red, 2pt]{(A)}$$

解答: $$\cases{\overline{AB}=2\\ \overline{AC} =k\\ \overline{BC}=3k} \Rightarrow \cos \angle C={k^2+9k^2-4\over 2\cdot k\cdot 3k} \Rightarrow {1\over 2} ={10k^2-4\over 6k^2} \Rightarrow k^2={4\over 7 } \\ \Rightarrow a\triangle ABC ={1\over 2}\overline{AC}\cdot \overline{BC} \sin \angle C={1\over 2}\cdot k\cdot 3k\cdot {\sqrt 3\over 2} ={3\sqrt 3\over 4}k^2={3\sqrt 3\over 4}\cdot {4\over 7}={3\sqrt 3\over 7}，故選\bbox[red, 2pt]{(B)}$$

解答: $$\overline{AB}=10\times \sin 60^\circ\times 2=10\sqrt 3，故選\bbox[red, 2pt]{(C)}$$

解答: $$L:y+1=m(x-1) \Rightarrow x截距={1\over m}+1=a \\(A) \bigcirc: y截距=-m-1\gt 0 \Rightarrow m\lt -1 \Rightarrow 0\lt a<1 \\(B)\times: -m-1\lt 0 \Rightarrow m\gt -1 \Rightarrow a=1+{1\over m }\gt 0\\ (C)\times: m=5\Rightarrow a=1+{1\over m}\not \gt 2 \\ (D)\times: m=-1 \Rightarrow a=1+{1\over m} =0 \not \lt 0\\，故選\bbox[red, 2pt]{(A)}$$

解答: $$A(0,0,0) \to B(0,0,10) \to C(5,0,10) \to D(5,-7,10) \to E(5,-7,6) \\ \Rightarrow \overline{AE}=\sqrt{25+49+36}=\sqrt{110}，故選\bbox[red, 2pt]{(B)}$$

解答: $$f(x)=x^2 \Rightarrow f'(x)=2x \Rightarrow f'(4)=8 \Rightarrow L:y=8(x-4)+16 \Rightarrow x={y\over 8}+2  \\ \Rightarrow 陰影面積=\int_0^{16} {y\over 8}+2-\sqrt y\,dy = \left. \left[ {1\over 16}y^2 +2y-{2\over 3}y^{3/2}\right] \right|_0^{16} ={16\over 3}，故選\bbox[red, 2pt]{(D)}$$

解答: $$假設C(0,0,0) \Rightarrow \cases{A(1,-3,-4) \\ B(x,y,z) \\D(-2,1,-2) } \Rightarrow \cases{\overrightarrow{AB }=(x-1,y+3,z+4) \\ \overrightarrow{DC}=(-2,1,-2)} \Rightarrow \overrightarrow{AB} \parallel \overrightarrow{DC} \\ \Rightarrow {x-1\over -2}=y+3={z+4\over -2} \Rightarrow \cases{x=-2y-5 \\ z=-2y-10} \Rightarrow B(-2y-5,y,-2y-10) \\ \Rightarrow \cases{\overrightarrow{AB}=(-2y-6,y+3,-2y-6) \\ \overrightarrow{CB}=(-2y-5,y,-2y-10)} \Rightarrow \overrightarrow{AB} \bot \overrightarrow{CB} \Rightarrow (-2y-6,y+3,-2y-6)\cdot (-2y-5,y,-2y-10)=0\\ \Rightarrow 9y^2+57y+90=0 \Rightarrow (3y+9)(3y+10)=0 \Rightarrow \cases{y=-3 \Rightarrow B(1,-3,-4)=A,不合\\ y=-10/3} \\ \Rightarrow \overline{AB} =\sqrt{(-2y-6)^2+(y+3)^2+(-2y-6)^2} =\sqrt{9y^2+54y+81} =\sqrt{-3y-9} =1，故選\bbox[red, 2pt]{(A)}$$

解答: $$體積=f(a) =(3-2a)(2-4a)a =8a^3-16a^2+6a \Rightarrow f'(a)=24a^2-32a+6=0\\ \Rightarrow a={16-4\sqrt 7\over 24} ={4-\sqrt 7\over 6}，故選\bbox[red, 2pt]{(B)}$$

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