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2022年8月4日 星期四

91年大學學測(補考)-數學詳解

大學入學考試中心
九十一學年度學科能力測驗(補考)

第一部分:選擇題  

壹、單一選擇題 

解答231=3×77235=5×47237=3×79233239(2)
解答x4+2x21=(x2+1)22=0(x2+1)2=2x2+1=2x=±(21)2(3)
解答

(1)(3)(4)
解答y=12x2x2=412yc=12F(0.12)(1)

解答3+3+1+1=88%12×8%=9600(2)
解答{ABCRcABDRdABERe{¯ACsinB=2Rc¯ADsinB=2Rd¯AEsinB=2Re¯AC=¯AD>¯AE2Rc=2Rd>2Rec=d>e(5)

貳、多重選擇題

解答(1):(±x)2y2=1(2)×:(1,0)(0,1)(3):x2y2=0x=±y(4)×:{a=1b=1c=2(±2,0)(5):a=1(±1,0)(135)
解答(3)×:{b=0.9a=0.1ba=0.8>0(4)×:{a=0.8b=0.1a/b=8
解答(1)\times: \angle C=90^\circ \Rightarrow \cos C=0,但{h\over a}+{h\over b}\gt 0\\ (2)\times: 反例如(1)\\ (3)\times: \cos(A+B)= \cos(180^\circ -C) =-\cos C \ne \cos C\\ (4)\bigcirc: 餘弦定理\\ (5)\bigcirc: \cases{h^2= a^2-x^2\\ h^2=b^2-y^2} \Rightarrow 2h^2=a^2+b^2-(x^2+y^2) =a^2+b^2-(x+y)^2+2xy\\\qquad =a^2+b^2-c^2 +2xy \Rightarrow h^2-xy={a^2+b^2-c^2\over 2} \Rightarrow \cos C ={a^2+b^2-c^2\over 2ab} ={h^2-xy\over ab}\\,故選\bbox[red, 2pt]{(45)}
解答
(1)(2)
(4)由(2)之圖例可知m_2\not \le 0

其餘皆正確,故選\bbox[red, 2pt]{(35)}
解答(1)\bigcirc: f(x)={1\over 2}(\cos 10x-\cos 12x)={1\over 2}(-2)\sin{10+12\over 2}x\sin {10-12\over 2}x \\\qquad \quad =-\sin 11x\sin(-x)=\sin 11x\sin x \\(2) \bigcirc:\cases{|\sin 11x|\le 1 \\|\sin x|\le 1} \Rightarrow |f(x)|=|\sin 11x\sin x| =|\sin 11x|| \sin x| \le 1 \\(3) \times: f(x)=\sin 11x\sin x=1 \Rightarrow \cases{\sin x=1 \Rightarrow  x=2k\pi +{\pi \over 2}\\ \sin 11x =1 \Rightarrow x=2t\pi +{\pi\over 22}},兩者無交集 \Rightarrow f(x)\ne 1\\ (4) \bigcirc: x={\pi\over 2} \Rightarrow \cases{\sin x=1\\ \sin 11x= -1} \Rightarrow f(x)=-1 \\(5)\bigcirc: x=k\pi,k\in \mathbb{Z} \Rightarrow f(x)=0\\,故選\bbox[red, 2pt]{(1245)}
解答
(1)三交線交於一點↓
 

(2)三交線相互平行
其他皆不可能,故選\bbox[red, 2pt]{(345)}

第二部分:填充題

解答11^{15}=(11^3)^5 = 1331^5 \equiv 31^5 \mod 100 \equiv 31^2 \times 31^2 \times 31 \mod 100 \\ \equiv 61\times 61\times 31 \mod 100 \equiv 21 \times 31 \mod 100 \equiv \bbox[red, 2pt]{51} \mod 100
解答\cases{z= 2\left( \cos{\pi\over 7}+ i\sin {\pi\over 7}\right) \\i=\cos{\pi\over 2} +i \sin {\pi\over 2}} \Rightarrow zi= 2\left( \cos({\pi\over 7}+{\pi\over 2})+ i\sin ({\pi\over 7}+{\pi\over 2})\right) =2(\cos {9\pi \over 14} +i\sin{9\pi\over 14})\\ \Rightarrow a=\bbox[red, 2pt]{9\over 14}
解答年利率4\% \Rightarrow 半年利率2\% \Rightarrow Q=10000(1+2\%)^2 = \bbox[red, 2pt]{10404}
解答

作\overline{MR}\parallel \overline{AD}、\overline{NQ} \parallel \overline{AB},並令\angle MPR=\theta,則\angle NQS=\theta,見上圖;\\ \cases{\triangle PMR: \sin\theta =a/3\\ \triangle NQS: \cos\theta =a/4} \Rightarrow ({a\over 3})^2 +({a\over 4})^2 =1 \Rightarrow a^2= {3^2\times 4^2\over 25} \Rightarrow a=\bbox[red, 2pt]{12\over 5}
解答\cases{5x+4y-4z=kx\\ 4x+5y+2z=ky \\ x+y+z=0} \Rightarrow \cases{(5-k)x+4y-4z=0\\ 4x+(5-k)y+2z=0 \\ x+y+z=0} \Rightarrow \left|\begin{matrix}5-k & 4 & -4 \\4 & 5-k & 2 \\1 & 1 & 1\end{matrix}\right| =0\\ \Rightarrow k^2-12k+11=0 \Rightarrow (k-1)(k-11)=0 \Rightarrow k=\bbox[red, 2pt]{1} (k=11不合,違反k\lt 10)
解答\cases{邊長為1的正方形有6\times 4=24個\\ 邊長為2的正方形有5\times 3=15個\\ 邊長為3的正方形有4\times 2= 8個\\ 邊長為4的正方形有3 \times 1=3個 } \Rightarrow 共有24+15+8+3= \bbox[red, 2pt]{50}個正方形
解答{4\over 3}\pi (12^3-5^3)= {4\over 3} \cdot \pi \cdot 2^3 \cdot a \Rightarrow a={12^3-5^3\over 2^3} =200{3\over 8} \Rightarrow 最多\bbox[red, 2pt]{200}顆
解答\cases{猜對3題的機率:1/8,期望值:5\cdot (1/8)\\ 猜對2題的機率:C^3_2/8=3/8,期望值:2.5\cdot(3/8)\\ 猜對1題的期望值:0}\\ \Rightarrow 期望值={5\over 8}+{7.5\over 8}= \bbox[red,2pt]{1 +{9\over 16}}

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解答僅供參考,其他歷屆試題及詳解

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