大滿貫學測指考試題觀摩

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包含於大滿貫復習講義數學1~4冊

目錄

第1單元

第2單元

第3單元

學測觀摩站

壹、單一選擇題

1

2

3

4

貳、多重選擇題

參、填充題

1

\begin{align} & \left\{ \begin{align} & f\left( x \right)=x^{3}-17x^{2}+32x-30=0 \\ & f\left( \alpha \right)=f\left( \beta \right)=f\left( \gamma \right)=0 \\ & \alpha =a+i,\beta =1+bi,a\ne 0,b\ne 0,a,b\in R \\ \end{align} \right. \\ & \gamma =? \\ & \because \beta =\overline{\alpha }\therefore a=1,b=-1\Rightarrow \left\{ \begin{align} & \alpha =1+i \\ & \beta =1-i \\ \end{align} \right. \\ & f\left( \alpha \right)=f\left( \beta \right)=f\left( \gamma \right)=0\Rightarrow \left\{ \begin{align} & \alpha +\beta +\gamma =17 \\ & \alpha \beta +\alpha \gamma +\beta \gamma =32 \\ & \alpha \beta \gamma =30 \\ \end{align} \right. \\ & \alpha \beta \gamma =30\Rightarrow \gamma =\frac{30}{\alpha \beta }=\frac{30}{\left( 1+i \right)\cdot \left( 1-i \right)}=\frac{30}{2}=15\# \\ \end{align}

2

\begin{align} & \left\{ \begin{align} & f\left( x \right)=kx^{2}+7x+1,k\in N \\ & f\left( \alpha \right)=f\left( \beta \right)=0,\alpha ,\beta \in R \\ & \frac{5}{71}<\alpha \beta <\frac{6}{71} \\ \end{align} \right. \\ & k=? \\ & \left\{ \begin{align} & \alpha \beta =\frac{1}{k} \\ & \frac{5}{71}<\alpha \beta <\frac{6}{71} \\ \end{align} \right.\Rightarrow \frac{5}{71}<\frac{1}{k}<\frac{6}{71}\Rightarrow \frac{71}{5}>k>\frac{71}{6}\Rightarrow 14+\frac{1}{5}>k>11+\frac{5}{6} \\ & \left\{ \begin{align} & 14+\frac{1}{5}>k>11+\frac{5}{6} \\ & \Delta =7^{2}-4k>0\Rightarrow k<\frac{49}{4}=12+\frac{1}{4} \\ \end{align} \right.\Rightarrow k=12\# \\ \end{align}

3

\begin{align} & \left\{ \begin{align} & f\left( x \right)=\left( x+1 \right)^{6} \\ & R_{x^{2}+1}\left[ f\left( x \right) \right]=ax+b \\ \end{align} \right. \\ & a=?,b=? \\ & f\left( x \right)=\left( x+1 \right)^{6}=\left[ \left( x+1 \right)^{2} \right]^{3}=\left( x^{2}+2x+1 \right)^{3} \\ & g\left( x \right)=x^{2}+2x+1 \\ & \left\{ \begin{align} & R_{x^{2}+1}\left[ g\left( x \right) \right]=2x \\ & f\left( x \right)=\left[ g\left( x \right) \right]^{3} \\ \end{align} \right.\Rightarrow R_{x^{2}+1}\left[ f\left( x \right) \right]=R_{x^{2}+1}\left[ \left( 2x \right)^{3} \right]=R_{x^{2}+1}\left[ 8x^{3} \right] \\ & x^{2}+1\overset{8x}{\overline{\left){\begin{align} & 8x^{3} \\ & 8x^{3}+8x \\ & \overline{\text{ }-8x} \\ \end{align}}\right.}} \\ & \left\{ \begin{align} & R_{x^{2}+1}\left[ f\left( x \right) \right]=ax+b \\ & R_{x^{2}+1}\left[ f\left( x \right) \right]=-8x \\ \end{align} \right.\Rightarrow \left\{ \begin{align} & a=-8 \\ & b=0 \\ \end{align} \right.\# \\ \end{align}

4

5

6

7

8

指考觀摩站

壹、單一選擇題

\begin{align} & f\left( x \right)=x^{4}-4x^{3}-3x^{2}+x+1 \\ & \begin{matrix} x & -3 & -2 & -1 & 0 & 1 & 2 \\ f\left( x \right) & + & + & + & + & - & - \\ \end{matrix} \\ \end{align}

貳、多重選擇題

參、填充題

1

\begin{align} & \left\{ \begin{align} & f\left( x \right)=x^{2}+x+c,c\in R \\ & a,b\notin R,\left\{ \begin{align} & f\left( a \right)=f\left( b \right)=0 \\ & f\left( \frac{1}{a} \right)=f\left( \frac{1}{b} \right)=0 \\ \end{align} \right. \\ \end{align} \right. \\ & a^{2}+b^{2}=? \\ & f\left( a \right)=f\left( b \right)=0\Rightarrow \left\{ \begin{align} & a+b=-1 \\ & ab=c \\ \end{align} \right. \\ & f\left( \frac{1}{a} \right)=f\left( \frac{1}{b} \right)=0\Rightarrow \left\{ \begin{align} & \frac{1}{a}+\frac{1}{b}=-1 \\ & \frac{1}{a}\times \frac{1}{b}=c \\ \end{align} \right. \\ & \left\{ \begin{align} & ab=c \\ & \frac{1}{a}\times \frac{1}{b}=c \\ \end{align} \right.\Rightarrow ab\times \frac{1}{a}\times \frac{1}{b}=c\times c\Rightarrow c^{2}=1\Rightarrow c=\pm 1 \\ & \because a,b\notin R\therefore \Delta =1-4c<0\Rightarrow c>\frac{1}{4}\therefore c=1 \\ & a^{2}+b^{2}=\left( a+b \right)^{2}-2ab=\left( -1 \right)^{2}-2\times \left( 1 \right)=1-2=-1\# \\ \end{align}

2

第4單元

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