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2019年5月22日 星期三

108年 警專38期甲組數學科詳解




:$$(A)d(P,x軸)=\sqrt{(-3)^2+4^2}=5\\(B)d(P,y軸)=\sqrt{2^2+4^2}=2\sqrt{5}\\(C)d(P,平面)= \left|\frac{2-6+4-1}{\sqrt{1^2+2^2+1^2}}=\right|=\frac{1}{\sqrt{6}} \\ (D)d(P,點)=\sqrt{(2-4)^2+(-3-2)^2+(4-3)^2}=\sqrt{30}$$
故選\(\bbox[red,2pt]{(D)}\)



:由題意可知\((3x-1)及(3x-1)^2\)皆是\(f(x)\)的因式;此外\(2x+2=2(x+1)\),所以\(2x+2\)也是\(f(x)\)的因式,故選\(\bbox[red,2pt]{(D)}\)


:$$k=-4\Rightarrow \begin{cases} x^{ 2 }+k=x^{ 2 }-4=(x+2)(x-2) \\ x^{ 2 }-4x-k=x^{ 2 }-4x+4=(x-2)^{ 2 } \end{cases}\Rightarrow 有一次公因式 \;x-2$$,故選\(\bbox[red,2pt]{(A)}\)


:\(\left|x+5\right|+\left|x-3\right|\)相當於求\(x\)至(-5)的距離加上\(x\)至3的距離,只要\(x\)介於-5與3之間,該值等於\(5-(-3)=8\)為最小,故選\(\bbox[red,2pt]{(D)}\)



:$$\sin{\theta}=\frac{4}{5}且\theta為第二象限角\Rightarrow \cos{\theta}=-\frac{3}{5} \Rightarrow \sin{2\theta}=2\sin{\theta}\cos{\theta}=2\times\frac{4}{5}\times\frac{-3}{5}=-\frac{24}{25} $$故選\(\bbox[red,2pt]{(B)}\)


:$$\log _{ 3}{ x }+\log _{ 3}{ x^3 } =12\Rightarrow \log _{ 3}{ x }+3\log _{ 3}{ x }=4\log _{ 3}{ x }=12\Rightarrow \log _{ 3}{ x }=3 \Rightarrow x=3^3=27$$故選\(\bbox[red,2pt]{(B)}\)



:$$x=-\frac{2}{3}\Rightarrow (3x+2)^2+\frac { 2 }{ 5}有最小值\frac{2}{5} \Rightarrow f(x)有最小值 32^{\frac{2}{5}}=\sqrt[5]{32^2} =\sqrt[5]{4^5}=4$$故選\(\bbox[red,2pt]{(B)}\)



:$$\frac{2a+b}{2}\ge\sqrt{2a\cdot b}\Rightarrow \frac{8}{2}\ge\sqrt{2ab} \Rightarrow 4\ge\sqrt{2ab}\Rightarrow 16\ge 2ab \Rightarrow 8\ge ab,故選\bbox[red,2pt]{(C)}。 $$



:$$35^{100}為155位數\Rightarrow 154\le\log{35^{100}}<155 \Rightarrow 154\le 100\log{35}<155 \Rightarrow 1.54\le \log{35}<1.55\\ \Rightarrow 18\times 1.54\le 18\log{35}< 18\times 1.55 \Rightarrow 27.72\le\log{35^{18}}<27.9 \Rightarrow 35^{18}為28位數,故選\bbox[red,2pt]{(A)}。$$




7男6女相間排列,最左及最右一定都是男生;
先將7男任意排列,有7!排法;再將6女插入兩男之間,剛好有6個空位,6女任排有6!排法;
因此總共有7!6!排法,故選\(\bbox[red,2pt]{(B)}\)



:$$H^4_7=C^{10}_7=120$$,本題\(\bbox[red,2pt]{(送分)}\)



:取出2白球有\(C^3_2=3\)種情形、取出2黑球有\(C^4_2=6\)種情形、取出2紅球有\(C^5_2=10\)種情形,從12個球中取出2球有\(C^{12}_2=66\)種情形,因此取出兩同色球的機率為\((3+6+10)/66=19/66\),故選\(\bbox[red,2pt]{(A)}\)



:$$利用正弦定理:\frac{4}{\sin{30^\circ}}=2R\Rightarrow R=4 ,故選\bbox[red,2pt]{(A)}。$$



:$$\left( \vec { a } -\vec { b } \right) \cdot \left( \vec { a } -\vec { b } \right) =\left| \vec { a } -\vec { b } \right| ^{ 2 }\Rightarrow \left| \vec { a } \right| ^{ 2 }-2\vec { a } \cdot \vec { b } +\left| \vec { b } \right| ^{ 2 }=\left| \vec { a } -\vec { b } \right| ^{ 2 }\\ 將\left| \vec { a } -\vec { b } \right| =\left| \vec { b } \right| =\left| \vec { a } \right| 代入上式\Rightarrow \left| \vec { a } \right| ^{ 2 }-2\vec { a } \cdot \vec { b } +\left| \vec { a } \right| ^{ 2 }=\left| \vec { a } \right| ^{ 2 }\\ \Rightarrow \vec { a } \cdot \vec { b } =\frac { 1 }{ 2 } \left| \vec { a } \right| ^{ 2 }再代入餘弦定理\vec { a } \cdot \vec { b } =\left| \vec { a } \right| \left| \vec { b } \right| \cos { \theta } \\ \Rightarrow \frac { 1 }{ 2 } \left| \vec { a } \right| ^{ 2 }=\left| \vec { a } \right| ^{ 2 }\cos { \theta } \Rightarrow \cos { \theta } =\frac { 1 }{ 2 } \Rightarrow \theta =60°,故選\bbox[red,2pt]{(B)}。$$



:$$\vec{a}\bot\vec{b}\Rightarrow \vec{a}\cdot\vec{b}=0 \Rightarrow (2,t)\cdot (-3,6) = -6+6t=0 \Rightarrow t=1,故選\bbox[red,2pt]{(B)}$$


:$$\begin{cases} \vec { a } =(-1,1) \\ \vec { b } =(7,-1) \end{cases}\Rightarrow \begin{cases} \left| \vec { a } \right| =\sqrt { 2 } \\ \left| \vec { b } \right| =5\sqrt { 2 } \end{cases} \Rightarrow 5\left| \vec { a } \right|  =\left| \vec { b } \right| \Rightarrow 5\vec { a } +\vec { b } =(-5,5)+(7,-1)=(2,4)=2(1,2)\\ \Rightarrow (1,2)為角平分向量,故選\bbox[red,2pt]{(D)}$$


:直線與圓相切代表圓心至直線的矩離為圓半徑長;$$\left| \frac{-3+8-10}{\sqrt{3^2+4^2}}\right| =\frac{5}{5}=1,故選\bbox[red,2pt]{(D)}$$



:$$\begin{cases} A=(1,2,3) \\ B=(2,0,-2) \\ C=(4,2,0) \end{cases}\Rightarrow \begin{cases} \overrightarrow { AB } =(1,-2,-5) \\ \overrightarrow { AC } =(3,0,-3) \end{cases}\Rightarrow \vec { n } =\overrightarrow { AB } \times \overrightarrow { AC } =(6,-12,6)\\ \Rightarrow 經過A且法向量為\vec { n } 的平面方程式6(x-1)-12(y-2)+6(z-3)=0\\ \Rightarrow (x-1)-2(y-2)+(z-3)=0\Rightarrow x-2y+z=0,故選\bbox[red,2pt]{(B)}$$


:$$\begin{cases} x-y+z=1 \\ x+2y-3z=2 \end{cases}\Rightarrow \begin{cases} x=\frac { z+4 }{ 3 } \\ y=\frac { 4z+1 }{ 3 } \end{cases}\Rightarrow \left( x,y,z \right) =\left( \frac { z+4 }{ 3 } ,\frac { 4z+1 }{ 3 } ,z \right) \\ \Rightarrow 方向向量為\left( \frac { 1 }{ 3 } ,\frac { 4 }{ 3 } ,1 \right) =\left( 1,4,3 \right) ,故選\bbox[red,2pt]{(A)}$$


:$$y=\sin { \theta  } -\sqrt { 3 } \cos { \theta  } -3=2\left( \frac { 1 }{ 2 } \sin { \theta  } -\frac { \sqrt { 3 }  }{ 2 } \cos { \theta  }  \right) -3=2\left( \cos { \alpha  } \sin { \theta  } -\sin { \alpha  } \cos { \theta  }  \right) -3\\ =2\sin { \left( \theta -\alpha  \right)  } -3\le 2-3=-1,故選\bbox[red,2pt]{(A)}$$



:$$\begin{cases} A=(0,0,0) \\ B=(2,-2,1) \\ C=(1,1,2) \end{cases}\Rightarrow \begin{cases} \vec { u } =\overrightarrow { AB } =(2,-2,1) \\ \vec { v } =\overrightarrow { AC } =(1,1,2) \end{cases}\Rightarrow \frac { 1 }{ 2 } \sqrt { \left| \vec { u } \right| ^{ 2 }\left| \vec { v } \right| ^{ 2 }-\left( \vec { u } \cdot \vec { v } \right) ^{ 2 } }\\ =\frac { 1 }{ 2 } \sqrt { (4+4+1)(1+1+4)-(2-2+2)^{ 2 } } \\ =\frac { 1 }{ 2 } \sqrt { 54-4 } =\frac { 1 }{ 2 } \sqrt { 50 } =\frac { 5 }{ 2 } \sqrt { 2 } ,本題\bbox[red,2pt]{(送分)}$$



:$$對直線x=y鏡射相當於x與y對調,因此3x=4y鏡射成 3y=4x,故選\bbox[red,2pt]{(C)}$$


:$$z=1+i\Rightarrow \bar { z } =1-i=\sqrt { 2 } \left( \frac { 1 }{ \sqrt { 2 } } -\frac { 1 }{ \sqrt { 2 } } i \right) =\sqrt { 2 } \left( \cos { 315° } +i\sin { 315° } \right) ,故選\bbox[red,2pt]{(D)}。$$

:$$g\left( x \right) =f\left( f\left( x \right) \right) \Rightarrow g\left( 1 \right) =f\left( f\left( 1 \right) \right) \Rightarrow g\left( 1 \right) =f\left( 1-2+6-3 \right) =f\left( 2 \right) =8-8+12-3=9\\,故選\bbox[red,2pt]{(C)}$$


:$$f\left( x \right) =\left( x^{ 2 }-3x+2 \right) ^{ 2 }\Rightarrow f'\left( x \right) =2\left( x^{ 2 }-3x+2 \right) \left( 2x-3 \right) \Rightarrow f'\left( 0 \right) =2\times 2\times \left( -3 \right) =-12\\,故選\bbox[red,2pt]{(B)}$$



出現1次的機率\(C^5_1\times\frac{1}{6}\times\frac{5^4}{6^4}=\frac{5^5}{6^5}\);
出現2次的機率\(C^5_2\times\frac{1}{6^2}\times\frac{5^3}{6^3}=\frac{10\times 5^3}{6^5}\);
出現3次的機率\(C^5_3\times\frac{1}{6^3}\times\frac{5^2}{6^2}=\frac{10\times 5^2}{6^5}\);
出現4次的機率\(C^5_4\times\frac{1}{6^4}\times\frac{5}{6}=\frac{25}{6^5}\);
出現5次的機率\(C^5_5\times\frac{1}{6^5}\);
期望值為\(\frac{1}{6^5}(5^5+20\times 5^3+30\times 5^2+100+5)=\frac{1620}{1944}=\frac{5}{6}\),故選\(\bbox[red,2pt]{(B)}\)

:$$f\left( x \right) =\int _{ 0 }^{ x }{ \left( 4t^{ 2 }-t-2 \right) \, dt } \Rightarrow f'\left( x \right) =4x^{ 2 }-x-2\Rightarrow f'\left( 1 \right) =4-1-2=1,故選\bbox[red,2pt]{(B)}$$



:$$y=x^2-3x+2\Rightarrow y'=2x-3\Rightarrow y'(2)=4-3=1\Rightarrow 切線斜率為1\\\Rightarrow 切線方程式為y=x-2,故選\bbox[red,2pt]{(D)}$$

:$$f\left( x \right) =(x-4)^{ 3 }\Rightarrow \begin{cases} f\left( x \right) \ge 0 & x\ge 4 \\ f\left( x \right) \le 0 & x\le 4 \end{cases}\Rightarrow 面積=\int _{ 4 }^{ 2 }{ f\left( x \right) dx } +\int _{ 4 }^{ 6 }{ f\left( x \right) dx } \\ =\left. \left[ \frac { 1 }{ 4 } (x-4)^{ 4 } \right] \right| ^{ 2 }_{ 4 }+\left. \left[ \frac { 1 }{ 4 } (x-4)^{ 4 } \right] \right| ^{ 6 }_{ 4 }=4+4=8,故選\bbox[red,2pt]{(D)}$$



:$$f\left( x \right) =x^{ 3 }+x^{ 2 }-x+2\Rightarrow f'\left( x \right) =3x^{ 2 }+2x-1\Rightarrow f''\left( x \right) =6x+2\\ f''\left( x \right) =0\Rightarrow 6x+2=0\Rightarrow x=-\frac { 1 }{ 3 } ,故選\bbox[red,2pt]{(A)}$$



:$$A、B是轉移矩陣,則A^mB^n及B^mA^n 皆為轉移矩陣,故選\bbox[red,2pt]{(ACE)}$$



:$$x^2+y^2-2kx+2ky+9=0\Rightarrow (x^2-2kx+k^2)+(y^2+2ky+k^2)=2k^2-9\\ \Rightarrow (x-k)^2+(y+k)^2=2k^2-9 \Rightarrow 2k^2-9>0 \Rightarrow k^2>\frac{9}{2},故選\bbox[red,2pt]{(ADE)}$$


:$$(A)\bigcirc :\overrightarrow{AB}+3\overrightarrow{BC}\Rightarrow \overrightarrow{BA}= 3\overrightarrow{BC}\Rightarrow 共線\\ (B)\bigcirc :5\overrightarrow{OB}=3\overrightarrow{OA}+2\overrightarrow{OC} \Rightarrow \overrightarrow{OB}=\frac{3}{5}\overrightarrow{OA}+\frac{2}{5}\overrightarrow{OC}, 由於\frac{3}{5}+\frac{2}{5}=1\Rightarrow 共線\\ (C)\times :4\overrightarrow{OA}=3\overrightarrow{OB} -\overrightarrow{OC} \Rightarrow \overrightarrow{OA} =\frac{3}{4}\overrightarrow{OB}-\frac{1}{4}\overrightarrow{OC}\Rightarrow \frac{3}{4}-\frac{1}{4}\ne 1\Rightarrow 不共線\\ (D)\bigcirc :\overrightarrow{OB}=3\overrightarrow{OA}-\overrightarrow{OC} \Rightarrow 3-2=1 \Rightarrow 共線\\ (E)\times :\overrightarrow{OA}+ \overrightarrow{OB} +\overrightarrow{OC}=0 \Rightarrow \overrightarrow{OA}=-\overrightarrow{OB}-\overrightarrow{OC} \Rightarrow -1-1\ne 1 \Rightarrow 不共線\\故選\bbox[red,2pt]{(ABD)}$$



:$$x^2+y^2-2x+4y=0\Rightarrow (x-1)^2+(y+2)^2=(\sqrt{5})^2\Rightarrow 圓心O(1,-2), 半徑r=\sqrt{5}\\ 圓心至直線距離=\frac{2-2+5}{\sqrt{2^2+1^2}}=\sqrt{5}=r\Rightarrow 該直線為切線 \\\Rightarrow 0\le dist(P, 直線) \le 2r=2\sqrt{5}\approx 4.47,故選\bbox[red,2pt]{(ABC)}$$





$$x^3=8=8\left(1\pm 0i\right)=8\left(\cos{2\pi}\pm i\sin{2\pi}\right) \Rightarrow x=2\left(\cos{\frac{2\pi}{3}}\pm i\sin{\frac{2\pi}{3}}\right) =2\left(-\frac{1}{2} \pm i\frac{\sqrt{3}}{2}\right)\\=-1\pm\sqrt{3}i,故選\bbox[red,2pt]{(ACE)}$$


:$$\sin{x},\cos{x}及\sec{x}=\frac{1}{\cos{x}}週期皆為2\pi;\tan{x}週期為\pi \Rightarrow \tan{2x}的週期為\pi/2;\\\cos{2x}的週期為\pi;故選\bbox[red,2pt]{(ABD)}$$



:$$需滿足\left|x+3\right|<1,故選\bbox[red,2pt]{(AB)}$$


:$$(A)\bigcirc :\lim_{n\to\infty}{a_n}=0\\ (B)\times: \lim_{n\to\infty}{a_n}=\pm 1\ne 0\\(C) \times: \lim_{n\to\infty}{a_n}=\frac{1/2}{1-(1/2)}=1\ne 0\\(D)\times: \lim_{n\to\infty}{a_n}=\frac{1}{2}\ne 0 \\(E)\bigcirc: \lim_{n\to\infty}{a_n}=0\\,故選\bbox[red,2pt]{(AE)}$$


:$$\left( A \right) \left| \begin{bmatrix} -1 & 2 \\ 2 & -1 \end{bmatrix} \right| =\left| 1-4 \right| =3\neq 1\\ \left( B \right) \left| \begin{bmatrix} 0 & 1 \\ -1 & 2 \end{bmatrix} \right| =\left| 0+1 \right| =1\\ \left( C \right) \left| \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \right| =\left| 4-6 \right| =2\neq 1\\ \left( D \right) \left| \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \right| =\left| 4-3 \right| =1\\ \left( E \right) \left| \begin{bmatrix} \cos { \pi } & -\sin { \pi } \\ \sin { \pi } & \cos { \pi } \end{bmatrix} \right| =\left| \cos ^{ 2 }{ \pi } +\sin ^{ 2 }{ \pi } \right| =1\\,故選\bbox[red,2pt]{(BDE)}$$



(A)\(\bigcirc: \lim_{x\to 0}{\frac{x}{|x|}}=\pm 1\)
(B)\(\times: \lim_{x\to 0^+}{\frac{x}{|x|}}=1\)
(C)\(\bigcirc: \lim_{x\to 1}{x}=1或0\)
(D)\(\bigcirc: \lim_{x\to 1/2}{\frac{2x+1}{2x-1}}=\frac{2}{0}不存在\)
(E)\(\times: \lim_{x\to 1/2}{x[x]}=(1/2)\times 0=0\)
故選\(\bbox[red,2pt]{(ACD)}\)

解題僅供參考!  第9、11、17、21、35題的答案有疑慮!?
警專後來更正了答案!
有趣的是數學甲第35題應該是原答案CDE,更正為ACE,更正訊息寫錯了!!
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