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2020年6月6日 星期六

100年 警專30期乙組數學科詳解


臺灣警察專科學校專科警員班三十期(正期學生組)
新生入學考試乙組數學科試題
壹、單選題


:$$令\cases{a=mk \\b=nk},其中k為最大公因數且m與n為自然數且互質 \Rightarrow \cases{mnk=330\\ (m+n)k=176} \\(A)\times: k=11 \Rightarrow \cases{mn=30\\ m+n=16},無(m,n)符合此條件\\ (B)\bigcirc: k=22 \Rightarrow \cases{mn=15\\ m+n=8}\Rightarrow (m,n)=(7,8),(8,7)\\ (C)\times: k=33\Rightarrow  m+n=176/33 非整數\\ (D)\times: k=66\Rightarrow  m+n=176/66 非整數\\,故選\bbox[red,2pt]{(B)}$$


:$$\sqrt{13+4\sqrt 3} = \sqrt{13+2\sqrt 12} =\sqrt{(1+\sqrt{12})^2} =1+\sqrt{12} =1+2\sqrt{3} =1+2(1.732)\\,整數部份為4,故選\bbox[red,2pt]{(D)}$$


:$$\sum_{k=1}^n {1\over k(k+1)} = \sum_{k=1}^n ({1\over k}-{1\over k+1}) = 1-{1\over n+1}= {n\over n+1} ={20\over 21} \Rightarrow n=20,故選\bbox[red,2pt]{(B)}$$



$$\sum_{n=1}^\infty \left( {2x \over 1+x}\right)^n 收斂\Rightarrow \left| {2x \over 1+x}\right|<1 \Rightarrow \begin{cases} {2x \over 1+x}<1 & 如果x>0 或x< -1 \\ {-2x \over 1+x}<1 & 如果 -1 < x<0\end{cases} \\ \Rightarrow  \begin{cases} (x+1)(x-1)<0 & 如果x>0 或x< -1 \\ (x+1)(3x+1) >0 & 如果 -1 < x<0\end{cases} \Rightarrow \cases{0 < x< 1 \\ -{1\over 3} < x< 0} \Rightarrow -{1\over 3} < x < 1,故選\bbox[red,2pt]{(D)}$$



:$$利用長除法\Rightarrow f(x)=(3x^2-11x+9)(x-1)+6 =((3x-8)(x-1)+1)(x-1)+6 \\ = ((3(x-1)-5)(x-1)+1)(x-1)+6 = 3(x-1)^3-5(x-1)^2 +(x-1)+6 \\ \Rightarrow c=1,故選\bbox[red,2pt]{(A)}$$


:$$令y=\log x \Rightarrow \log(2x)\log(3x)= (\log 2+\log x)(\log 3+\log x)= (y+\log 2)(y+\log 3)=1 \\ \Rightarrow y^2 +(\log 6)y+\log 2\log 3-1=0 \Rightarrow \log \alpha +\log \beta =-\log 6 \Rightarrow \log \alpha\beta = \log {1\over 6} \\ \Rightarrow \alpha \beta={1\over 6},故選\bbox[red,2pt]{(A)}$$



:$$\log 25^{30} =\log 5^{60} = 60\log 5=60(1-\log 2)=60(1-0.301) =41.94\\ 又3\times 0.301=0.903= \log 2^3 <0.94 < \log 3^2=2\times 0.4771= 0.9542 \\\Rightarrow \cases{m=41+1=42 \\ n=8},故選\bbox[red,2pt]{(C)}$$



:$${\cos(90^\circ-\theta) \cot(180^\circ+\theta) \over \sin(270^\circ-\theta)} +{\tan (180^\circ+\theta) \over \tan(360^\circ -\theta)} ={\sin \theta \cot \theta \over -\cos \theta} +{\tan \theta \over -\tan \theta} ={\cos \theta  \over -\cos \theta} -1 \\=-1-1=-2,故選\bbox[red,2pt]{(A)}。 $$



:$$令s=(7+3+5)\div 2=15/2 \Rightarrow \triangle ABC面積= \sqrt{s(s-7)(s-3)(s-5)} =\sqrt{{15\over 2}\cdot {1\over 2}\cdot {9\over 2}\cdot {5\over 2}}\\ = {15\sqrt{3}\over 4},故選\bbox[red,2pt]{(D)}。$$





$$\angle A的內角平分線交\overline{BC}於D,並作\overline{DE}\bot \overline{AB}及\overline{DF}\bot \overline{AC},如上圖;\\令\overline{AD}=a \Rightarrow \overline{DE}=\overline{DF} ={\sqrt 3\over 2}a \Rightarrow \triangle ABC面積=\cases {{1\over 2}\overline{AB}\times \overline{AC}\sin \angle A \\ {1\over 2}(\overline{AB}\times \overline{DE}+\overline{AC}\times\overline{DF})} \\ =\cases{3\sin 120^\circ={3\sqrt 3\over 2} \\ {\sqrt 3\over 4}(2a+ 3a)={5\over 4}\sqrt 3a} \Rightarrow {3\sqrt 3\over 2}= {5\over 4}\sqrt 3a \Rightarrow a={6\over 5},故選\bbox[red, 2pt]{(B)}$$



:$$\cases{a=\sin1 =\sin {180^\circ \over \pi} (第1象限角)>0\\ b=\cos 2=\cos {360^\circ \over \pi} (第2象限角) <0 \\c=\tan 3=\tan {540^\circ \over \pi} (第2象限角)<0 \\ d=\sec 4=\sec {720^\circ \over \pi}(第3象限角) <0},故選\bbox[red, 2pt]{(A)}$$



:$$\cases{\sin \alpha ={\sqrt 2\over 2} \\\cos \beta=-{3\over 5}} \Rightarrow \cases{\cos \alpha ={\sqrt 2\over 2} \\\sin \beta={4\over 5}} \Rightarrow \cos(\alpha+ \beta)= \cos \alpha \cos \beta-\sin \alpha\sin\beta \\ ={\sqrt 2\over 2}\times {-3\over 5}-{\sqrt 2\over 2}\times {4\over 5} =-{7\sqrt 2\over 10},故選\bbox[red, 2pt]{(D)}$$




$$(\vec a+\vec b)\cdot (\vec a+\vec b) = |\vec a+\vec b|^2 =|\vec a|^2+2\vec a\cdot \vec b +|\vec b|^2 \Rightarrow (\sqrt{61})^2 =4^2 +2\vec a\cdot \vec b+5^2 \\ \Rightarrow \vec a\cdot \vec b=10 \Rightarrow \cos \theta ={\vec a \cdot \vec b \over |\vec a||\vec b|} ={10 \over 4\times 5} ={1\over 2} \Rightarrow \theta =60^\circ ,故選\bbox[red,2pt]{(B)}。$$





$$\overrightarrow{AP}= x\overrightarrow{AB} +y\overrightarrow{AC} =\cases{x\overrightarrow{AB} +y\cdot {4\over 3}\overrightarrow{AE} \\ x\cdot {3\over 2}\overrightarrow{AD} +y\overrightarrow{AC}} \Rightarrow \cases{x+{4\over 3}y=1 \\ {3\over 2}x+y=1} \Rightarrow \cases{x=1/3\\ y=1/2},故選\bbox[red,2pt]{(C)}。$$

15. 若一個正四體相鄰兩面的夾角為\(\theta\),則\(\sin \theta\)之值為何?
(A)\({1\over 3}\)  (B)\({\sqrt 2\over 3}\)  (C)\({2\sqrt 2\over 3}\)  (D)\({\sqrt 3\over 3}\)




$$\cases{ \overrightarrow{OD} =(1,0,1) \\\overrightarrow{OC} =(1,1,0) \\\overrightarrow{DE} =(-1,1,0) \\\overrightarrow{DC} =(0,1,-1) } \Rightarrow \cases{\vec u = \overrightarrow{OD}\times \overrightarrow{OC} =(-1,1,1) \\ \vec v =\overrightarrow{DE} \times \overrightarrow{DC}= (-1,-1,-1)} \Rightarrow \cos \theta ={\vec u\cdot \vec v \over |\vec u||\vec v|} = -{1\over 3}\\ \Rightarrow \sin \theta = {2\sqrt 2\over 3},故選\bbox[red,2pt]{(C)}$$


:$$\cases{A(1,1,1) \\B(2,3,3) \\C(3,2,3) } \Rightarrow \cases{\vec u=\overrightarrow{AB} =(1,2,2) \\\vec v=\overrightarrow{AC} =(2,1,2)} \Rightarrow \triangle ABC面積= {1\over 2}\sqrt{|\vec u|^2|\vec v|^2 -(\vec u\cdot \vec v)^2} \\={1\over 2}\sqrt {9\times 9-(2+2+4)^2} = {\sqrt{17} \over 2},故選\bbox[red,2pt]{(B)}$$




$$x^2+y^2-2x-4y-4=0 \Rightarrow (x-1)^2 +(y-2)^2=3^2 \Rightarrow \cases{圓心O(2,3)\\ 半徑r=3} \Rightarrow \overline{OP}= \sqrt {4^2+4^2} =4\sqrt 2 \\\Rightarrow 切線長\overline{AP}=\sqrt{\overline{OP}^2-r^2} = \sqrt {32-9} =\sqrt{23},故選\bbox[red, 2pt]{(C)}$$





$$ \cases{(x-1)^2 +(y-2)^2=5^2 \Rightarrow \cases{圓心O(2,3)\\ 半徑r=5} \\L:3x+4y+9=0}\Rightarrow \text{dist}(O,L) = \overline{OC}= \left|{ 3+8+9\over \sqrt{3^2+4^2}} \right| ={20\over 5}=4 \\ \Rightarrow \overline{AC} =\sqrt{r^2-\overline{OC}^2} =\sqrt{5^2-4^2} =3 \Rightarrow \overline{AB} = 2\overline{OA}=6,故選\bbox[red,2pt]{(D)}$$


:$$x^2-2x+4y+9=0 \Rightarrow (x-1)^2+4(y+2)=0 \Rightarrow (x-1)^2 =-4(y+2) \\\Rightarrow 正焦弦長=|-4|=4,故選\bbox[red,2pt]{(B)}$$



$$x^2 +4y^2+2x+8y+1=0 \Rightarrow (x+1)^2+4(y+1)^2 = 4 \Rightarrow {(x+1)^2 \over 2^2} +{(y+1)^2 \over 1^2}=1 \\ \Rightarrow a=2 \Rightarrow \overline{PF_1}+\overline{PF_2} =2a=4,故選\bbox[red,2pt]{(B)}$$



:$$令f(x)=(1+x)^3+(1+x)^4+ \cdots+(1+x)^{10},則f(x)的x^3係數即為所求;\\f(x)={(1+x)^3-(1+x)^{11}\over 1-(1+x)} = {g(x) \over -x} \Rightarrow g(x)的x^4係數為-C^{11}_4 \\\Rightarrow f(x)的x^3係數={-C^{11}_4 \over -1} =C^{11}_4,故選\bbox[red,2pt]{(D)}$$



:$$g(x)=\sum_{k=1}^{10} (1-x)^k = {(1-x)(1-(1-x)^{10}) \over 1-(1-x)} ={ (1-x)-(1-x)^{11}\over x}= {f(x) \over x} \\ \Rightarrow f(x)的x^3係數為C^{11}_3 = g(x)的x^2係數,故選\bbox[red,2pt]{(C)}$$

23. 某次考試全班數學成績不佳,總平均 55 分,標準差12分。老師決定全班每人數學加15 分,則全班加分後的標準差為多少分?
(A) 6 (B)12 (C)15 (D) 24 。

:$$\sigma(X)=12 \Rightarrow \sigma(X+15)=12 ,故選\bbox[red,2pt]{(B)}。$$

24. 若某校1000位學生的數學段考成績呈現常態分配,其中平均分數是 70 分,標準差是5分,則全校約有多少人數學成績低於 60 分?(設已知常態分配的資料約有 95% 的觀測值,落在距平均數左右各兩個標準差的範圍內)
(A) 25 (B)50 (C)100 (D) 200 。

:$$P(X<60) = P(X<\mu-2\sigma) = P(X<\mu)-{1\over 2}P(2\sigma < x < 2\sigma) =50\%-{1\over 2}\times 95\% = 2.5\% \\ \Rightarrow 1000\times P(X<60)=1000\times 0.025 =25,故選\bbox[red,2pt]{(A)}$$

25. 根據過去紀錄可知,某燈泡工廠檢驗其產品的過程中,將良品檢驗為不良品的機率為 0.1,將不良品檢驗為良品的機率為 0.2 。又知該產品中,良品占 90% ,不良品占10% 。若已知一件產品被檢驗為良品,則該產品實際上為不良品之機率為何?
(A)\({2\over 83}\)   (B)\({81\over 83}\)   (C)\({8\over 17}\)   (D)\({9\over 17}\)

:$${檢驗為良品但其實是不良品\over 檢驗為良品} = {10\%\times 0.2\over 90\%\times 0.9 + 10\%\times 0.2} ={ 0.02\over 0.81+0.02} ={0.02 \over 0.83} = {2\over 83}\\,故選\bbox[red,2pt]{(A)}$$


:$$迴歸直線斜率=r\times {S_Y \over S_X} = 0.88 \times {10\over 5} =1.76,故選\bbox[red, 2pt]{(A)}$$



:$$A=\left[\matrix{1 & 1 \\-1 & 1} \right] \Rightarrow A^2= \left[\matrix{1 & 1 \\-1 & 1} \right]\left[\matrix{1 & 1 \\-1 & 1} \right] =\left[\matrix{0 & 2 \\-2 & 0} \right] \\\Rightarrow A^4=\left[\matrix{0 & 2 \\-2 & 0} \right]\left[\matrix{0 & 2 \\-2 & 0} \right]=\left[\matrix{-4 & 0 \\0 & -4} \right] \\ \Rightarrow A^8=\left[\matrix{-4 & 0 \\0 & -4} \right]\left[\matrix{-4 & 0 \\0 & -4} \right] =\left[\matrix{16 & 0 \\0 & 16} \right] ,故選\bbox[red,2pt]{(A)}$$



:$$\begin{vmatrix} a_1 & b_1 & c_1 \\a_2 & b_2 & c_2 \\a_3 & b_3 & c_3  \end{vmatrix} =5 \Rightarrow \begin{vmatrix} 3a_1 & b_1 & c_1 \\3a_2 & b_2 & c_2 \\3a_3 & b_3 & c_3  \end{vmatrix} =5\times 3=15 \Rightarrow \begin{vmatrix} 3a_1+2b_1 & b_1 & c_1 \\3a_2+2b_2 & b_2 & c_2 \\3a_3+2b_3 & b_3 & c_3  \end{vmatrix} =15 \\ \Rightarrow \begin{vmatrix} 3a_1+2b_1 & 3b_1 & c_1 \\3a_2+2b_2 & 3b_2 & c_2 \\3a_3+2b_3 & 3b_3 & c_3  \end{vmatrix} =15\times 3=45 \Rightarrow \begin{vmatrix} 3a_1+2b_1 & 3b_1-5c_1 & c_1 \\3a_2+2b_2 & 3b_2-5c_2 & c_2 \\3a_3+2b_3 & 3b_3-5c_3 & c_3  \end{vmatrix} =45,故選\bbox[red,2pt]{(D)}$$



:$$x^2+y^2+z^2-4x-2y-2z-3=0 \Rightarrow (x-2)^2+(y-1)^2+(z-1)^2=9 \\ \Rightarrow ((x-2)^2+(y-1)^2+(z-1)^2)(1^2+(-2)^2+(-2)^2) \ge ((x-2)-2(y-1)-2(z-1))^2 \\ \Rightarrow 9\times 9 \ge (x-2y-2z+2)^2 \Rightarrow 9\ge x-2y-2z+2 \ge -9 \\ \Rightarrow 7 \ge x-2y-2z \ge -11 \Rightarrow 最大值為7,故選\bbox[red,2pt]{(C)}$$



:$$f(x)=x^2+4x-5 \Rightarrow g(x)=f(f(x)) = (x^2+4x-5)^2+4(x^2+4x-5)-5 \\ \Rightarrow g'(x)=2(x^2+4x-5)(2x+4)+4(2x+4) \Rightarrow g(x)=0 \Rightarrow 2(2x+4)((x^2+4x-5)+4)=0\\ \Rightarrow 4(x+2)(x^2+4x-1)=0 \Rightarrow x=-2,-2\pm \sqrt 5 (-2\pm \sqrt 5不在區間 [-3,3]內)\\ 又g''(x)=4(x^2+4x-1)+4(x+1)(2x+4) \Rightarrow g(-2)=4(4-8-1) < 0 \Rightarrow g(-2)為極大值 \\ \Rightarrow g(x)的極小值出現在區間端點\Rightarrow \cases{g(0)=(-5)^2+4\times (-5)-5=0\\g(-3)=(-8)^2+4\times (-8)-5=27} \Rightarrow 極小值為0\\難,故選\bbox[red,2pt]{(C)}$$

貳、多重選擇題


:$$(A) \bigcirc:\cases{a_5=4 \\ a_7=1 } \Rightarrow \cases{a_1+4d=4 \\ a_1+6d=1 } \Rightarrow \cases{a_1=10 \\d=-3/2} \\(B)\bigcirc: 理由同(A),d=-3/2 \\(C) \bigcirc:\cases{a_7= a_1+6d=10-6\times {3\over 2}=1 >0\\a_8=a_1+7d = 10-7\times {3\over 2}=-{1\over 2}<0 } \\(D) \times: a_8開始為負值,所以S_7最大,不是S_8\\ (E)\bigcirc: S_{21}= {21(a_1+a_{21})\over 2} = {21(2a_1+20d)\over 2} = 21(a_1+10d) = 21(10-15) = -105\\,故選\bbox[red,2pt]{(ABCE)}$$



:$$(A) \bigcirc: f(x)=(2x-1)Q(x)+r = 2(x-{1\over 2})Q(x)+r \Rightarrow 商為2Q(x),餘式為r\\ (B)\times: 有可能f(a)f(b)=0,不一定f(a)f(b)<0 \\(C) \bigcirc:奇次多項式圖形一定是左上右下或右上左下,與X軸一定有相交\\ (D)\times: f(x)必須是實係數,共軛根才會成對出現\\ (E)\bigcirc: f(x)為實係數,1\pm \sqrt 3皆為其根\\,故選\bbox[red,2pt]{(ACE)}$$


:$$(A)\times: \log_{1/4}{1\over 2} = {1\over 2} \not \gt 1\\ (E)\times: \log b> 0 \Rightarrow b> 10^0=1,不符b<1\\其餘皆正確,故選\bbox[red,2pt]{(BCD)}$$



:$$-160^\circ位於第3象限\Rightarrow \cases{\tan (-16 0^\circ)=k >0 \\ \sin (-160^\circ)<0 \\ \cos(-160^\circ) <0} \\(A)\times: \sin 20^\circ >0,但 {-k\over \sqrt{1+k^2}}< 0 \\(E) \times: \csc 20^\circ >0,但{\sqrt{k^2+1} \over -k} < 0\\其餘皆正確,故選\bbox[red,2pt]{(BCD)}$$





$$f(x)=2\sin(x+{\pi \over 6})-2\cos x = 2(\sin x\cos {\pi \over 6}+ \sin{\pi \over 6}\cos x)-2\cos x =2({\sqrt 3\over 2}\sin x+{1\over 2}\cos x)-2\cos x\\ =\sqrt 3\sin x-\cos x = 2))({\sqrt 3\over 2}\sin x-{1\over 2}\cos x) = 2(\cos {\pi \over 6}\sin x-\sin{\pi \over 6}\cos x) =2\sin(x-{\pi \over 6}) \\ \Rightarrow f(x)=\begin{cases}M=2\sin({\pi \over 2}-{\pi \over 6}) =2\sin {\pi \over 3} =\sqrt 3 & x=\pi/2=\alpha \\ m= 2\sin(0-{\pi\over 6}) =-1 & x=0=\beta\end{cases} \\ \Rightarrow \cases{M=\sqrt 3 \\ m=-1 \\ \alpha=\pi/2 \\ \beta=0 \\ M-m=\sqrt 3+1}\\,故選\bbox[red,2pt]{(AD)}$$


:$$(A) \bigcirc: \omega=\cos {2\pi \over 3} +i\sin {2\pi \over 3} \Rightarrow \omega^3= \cos 2\pi +i\sin 2\pi = 1 \Rightarrow \omega^{30}=1 \\(B)\bigcirc: x^3-1=0 \Rightarrow (x-1)(x^2+x+1)=0 \Rightarrow \cases{x=1\\ \omega^2+\omega+1=0} \\(C) \bigcirc: (1-\omega)(1-\omega^2)= 1-(\omega+\omega^2)+\omega^3=1-(-1)+1=3 \\(D)\times: (1+\omega)(1+\omega^2) = 1+\omega+\omega^2 +\omega^3 =1+(-1)+1=1 \ne 3 \\(E)\bigcirc: {1\over 1-\omega}+ {1\over 1-\omega^2} = {2+\omega \over 1-\omega^2} = {2+\omega \over 1-(-\omega-1)} ={2+\omega \over 2+\omega}=1\\故選\bbox[red,2pt]{(ABCE)}$$



:$$(A) \times: 可能平行、垂直...\\(E)\times: 有無限多個平面\\其它皆正確,故選\bbox[red,2pt]{(BCD)}$$


:$$(C)\times: 可能無解,也可能無限多解\\其他皆正確,故選\bbox[red,2pt]{(ABDE)}$$


:$$(C)\times:分成三堆,各堆不用再排列,因此有{C^9_5C^6_3C^3_3 \over 2!}種方法\\ (D)\times: 每個人是不同的,三個人要再排列,因此有C^9_5C^4_3C^1_1\times 3!種方法\\其餘正確,故選\bbox[red,2pt]{(ABE)}$$


:$$(A)\times: \cases{A=\begin{bmatrix}0 & 1\\ 0 & 1\end{bmatrix} \\ B=\begin{bmatrix}1 & 1\\ 0 & 0\end{bmatrix}} \Rightarrow \cases{AB=\begin{bmatrix}0 & 0\\ 0 & 0\end{bmatrix} \\BA= \begin{bmatrix}0 & 2\\ 0 & 0\end{bmatrix}} \Rightarrow AB\ne BA \\(B)\times: 除非AB=BA,否則(A+B)(A-B)\ne A^2-B^2\\ (C)\times: 由(A)知:A\ne 0且B\ne 0,但AB=0\\ (D)\times: A=\begin{bmatrix}1 & 0\\ 0 & -1\end{bmatrix} \Rightarrow A^2=\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}= I,但A\ne I,A\ne -I \\(E)\times: \cases{A=\begin{bmatrix}0 & 1\\ 0 & 1\end{bmatrix} \\ B=\begin{bmatrix}1 & 1\\ 0 & 0\end{bmatrix} \\C= \begin{bmatrix}0 & 0\\ 0 & 0\end{bmatrix}} \Rightarrow \cases{AB=\begin{bmatrix}0 & 0\\ 0 & 0\end{bmatrix} \\AC= \begin{bmatrix}0 & 0\\ 0 & 0\end{bmatrix}} \Rightarrow AB=AC,但B\ne C\\各選項均錯誤,故選\bbox[red, 2pt]{(ABCDE)}$$


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