113學年度中區縣市政府教師甄選策略聯盟
【科目名稱:國中數學】
選擇題【共50題,每題2分,共100分】請以2B鉛筆於答案卡上作答,單選題;答錯不倒扣。
解答:x2(1×2)/22(2×3)/22(3×4)/22(4×5)/22(5×6)/22(6×7)/22(7×8)/22(8×9)/2mod721121121⇒餘數為2,1,1,循環,共99項,餘數總和=(2+1+1)×33=132=6mod7,故選(D)
解答:

有兩塊區域,面積=∫10(x−x2)dx+∫21(x2−x)dx=[12x2−13x3]|10+[13x3−12x2]|21=16+56=1,故選(C)
解答:(A)×:f(x)=(−1)n⇒limx→a|f(x)|=1,但limx→af(x)不存在(B)×:{f(x)=1/(x−a)g(x)=−1/(x−a)⇒limx→a(f(x)+g(x))=0存在,但limx→af(x)不存在(C)×:{f(x)=x2g(x)=1/x2⇒{limx→2f(x)g(x)=24=16=αlimx→21f(x)=14=β⇒limx→2g(x)=14≠164=βα,故選(D)
解答:
解答:(A)×:f(x)=(−1)n⇒limx→a|f(x)|=1,但limx→af(x)不存在(B)×:{f(x)=1/(x−a)g(x)=−1/(x−a)⇒limx→a(f(x)+g(x))=0存在,但limx→af(x)不存在(C)×:{f(x)=x2g(x)=1/x2⇒{limx→2f(x)g(x)=24=16=αlimx→21f(x)=14=β⇒limx→2g(x)=14≠164=βα,故選(D)
解答:
¯AB與¯CD均為直徑⇒O為圓心,假設半徑為r,¯CE=a△DOF∼△DEC(AAA)⇒¯OD¯FD=¯DE¯CD⇒r6=82r⇒r=2√6直角△CDE:¯CD2=¯DE2+¯CE2⇒4r2=82+a2⇒a=4√2,故選(D)
解答:{α+β=pαβ=−276p⇒(β−α)2=(β+α)2−4αβ=p2+1104p⇒β−α=√p2+1104p=√p2+24⋅3⋅23⋅p=√232⋅72 (if p=23)=23⋅7=161,故選(D)
解答:{a>0b>0⇒{2a+b=10a=12⇒b<0矛盾;同理,{a<0b>0或{a<0b<0皆矛盾{a>0b<0⇒{2a+b=10a−2b=12⇒{a=32/5b=−14/5⇒a+b=185,故選(A)
解答:cosA=cos120∘=−12=42+22−¯BC22⋅2⋅4⇒¯BC=2√7¯AD為∠A的角平分線⇒¯AB¯AC=¯BD¯DC⇒{¯BC=4√7/3¯CD=2√7/3cos∠CAD=cos60∘=12=42+¯AD2−112/98¯AD⇒¯AD=43,83(不合),故選(B)
解答:A=[024224448082010263291]R1/2→R1,R2/4→R2→[012111112082010263291]R3−8R2→R3,R4−6R2→R4→[01211111200−6−8−620−3−4−31]R3+6R1→R3,R4+3R1→R4→[01211111200040800204]R1↔R2→[11120012110040800204]R1−R2→R1,R3/4→R3→[10−11−1012110010200204]R1+R3→R1,R2−2R3→R2,R4−2R2→R4→[100110101−30010200000]⇒rank(A)=3,故選(C)
解答:sin3θsinθ=−4sin3θ+3sinθsinθ=−4sin2θ+3=73⇒sin2θ=16⇒sinθ=1√6⇒cosθ=√5√6⇒cos3θcosθ=4cos3θ−3cosθcosθ=4cos2θ−3=4⋅56−3=13,故選(A)
解答:
解答:{α+β=pαβ=−276p⇒(β−α)2=(β+α)2−4αβ=p2+1104p⇒β−α=√p2+1104p=√p2+24⋅3⋅23⋅p=√232⋅72 (if p=23)=23⋅7=161,故選(D)
解答:{a>0b>0⇒{2a+b=10a=12⇒b<0矛盾;同理,{a<0b>0或{a<0b<0皆矛盾{a>0b<0⇒{2a+b=10a−2b=12⇒{a=32/5b=−14/5⇒a+b=185,故選(A)
解答:cosA=cos120∘=−12=42+22−¯BC22⋅2⋅4⇒¯BC=2√7¯AD為∠A的角平分線⇒¯AB¯AC=¯BD¯DC⇒{¯BC=4√7/3¯CD=2√7/3cos∠CAD=cos60∘=12=42+¯AD2−112/98¯AD⇒¯AD=43,83(不合),故選(B)
解答:A=[024224448082010263291]R1/2→R1,R2/4→R2→[012111112082010263291]R3−8R2→R3,R4−6R2→R4→[01211111200−6−8−620−3−4−31]R3+6R1→R3,R4+3R1→R4→[01211111200040800204]R1↔R2→[11120012110040800204]R1−R2→R1,R3/4→R3→[10−11−1012110010200204]R1+R3→R1,R2−2R3→R2,R4−2R2→R4→[100110101−30010200000]⇒rank(A)=3,故選(C)
解答:sin3θsinθ=−4sin3θ+3sinθsinθ=−4sin2θ+3=73⇒sin2θ=16⇒sinθ=1√6⇒cosθ=√5√6⇒cos3θcosθ=4cos3θ−3cosθcosθ=4cos2θ−3=4⋅56−3=13,故選(A)

¯AE為直徑⇒∠ACE=90∘⇒¯CE=√122−(4√5)2=8假設小圓半徑r⇒¯AD=12−2r⇒¯CD2=¯AC2−¯AD2=¯CE2−¯DE2⇒(4√5)2−(12−2r)2=82−(2r)2⇒r=83,故選(C)

解答:

∠APB=∠APC=∠CPB=120∘⇒P為費馬點(Fermat point)因此作正△ACD,則D,P,B三點共線,且¯PA+¯PB+¯PC=¯DBcos∠APC=cos120∘=−12=82+62−¯AC22⋅6⋅8⇒¯AC=2√37假設{C(0,0)E=¯PE∩¯AC,由於∠CPA=120∘⇒∠APE=∠EPC=60∘⇒¯EP是∠APC的角平分線⇒¯AE¯EC=¯PA¯PC=43⇒¯EC=2√37⋅33+4=67√37⇒E(0,67√37)¯AC=2√37⇒D(−√111,√37)⇒L=↔DE:y=−√378√111x+67√37⇒B(6√111,0)⇒¯DB=74⇒¯PB=74−6−8=60,故選(D)
解答:b=c⇒R=ab22b=a2⇒a=2R(直徑)⇒等腰直角三角形,故選(A)
解答:f(x)=xg(x),其中g(x)=(x−1)(x−2)⋯(x−2024)(x+1)(x+2)⋯(x+2024)⇒f′(x)=g(x)+xg′(x)⇒f′(0)=g(0)=2024!2024!=1,故選(C)
解答:f(x)=x4+2x3−3x2+2x−1⇒f′(x)=4x3+6x2−6x+2limh→0f(1+3h)−f(1−2h)h=limh→0(f(1+3h)−f(1−2h))′(h)′=limh→03f′(1+3h)+2f′(1−2h)1=3f″
解答:依均值定理,故選\bbox[red, 2pt]{(A)}
解答:\int_0^{\pi/2} \cos x\,dx =1 \Rightarrow 平均值={1\over \pi/2}= {2\over \pi},故選\bbox[red, 2pt]{(A)}
解答:\cases{\tan \alpha+\tan\beta=p\\ \tan \alpha\tan\beta =q} \Rightarrow \cases{{1\over \cot \alpha}+{1\over \cot\beta} =p\\ {1\over \cot \alpha \cot \beta}=q} \Rightarrow \cases{ \cot \alpha+\cot \beta=p/q \\\cot \alpha \cot \beta=1/q} \\ \Rightarrow rs=(\cot \alpha+\cot \beta)(\cot \alpha \cot \beta)={ p\over q} \cdot {1\over q} ={p\over q^2},故選\bbox[red, 2pt]{(C)}
解答:{(\log_a x)\cdot (\log_x{b\over a}) \over (\log_x b)(\log_{ab} x)} ={(\log_a x)\cdot (\log_x b-\log_x a) \over (\log_x b)({1\over \log_x ab })} = {\log_a b-1\over \log_{ab}b} \\={\log_a b-1\over {1\over \log_b a+1}} =(\log_b a+1)(\log_a b-1)=1-\log_ba+\log_a b-1=-{1\over 10}+10 ={99\over10},故選\bbox[red, 2pt]{(B)}
解答:{1+2\sqrt 3+\sqrt 5\over (1+\sqrt 3) (\sqrt 3+\sqrt 5)} +{3+2\sqrt 7+ \sqrt 5\over (\sqrt 5+\sqrt 7) (\sqrt 7+3)} ={(1+\sqrt 3)+(\sqrt 3+\sqrt 5)\over (1+\sqrt 3) (\sqrt 3+\sqrt 5)} +{(3+\sqrt 7)+ (\sqrt 7+\sqrt 5)\over (\sqrt 5+\sqrt 7) (\sqrt 7+3)} \\={1\over \sqrt 3+\sqrt 5} +{1\over 1+\sqrt 3} +{1\over \sqrt 5+\sqrt 7} +{1\over \sqrt 7+3}={1\over 2}\left(\sqrt 5-\sqrt 3+\sqrt 3-1+\sqrt 7-\sqrt 5+3-\sqrt 7 \right) \\=1,故選\bbox[red, 2pt]{(B)}
解答:\sum_{n=1}^\infty {\sqrt{n+1}-\sqrt n\over \sqrt{n^2+n}} =\sum_{n=1}^\infty \left( {1\over \sqrt{n}} -{1\over \sqrt{n+1}}\right) =1-{1\over \sqrt 2}+{1\over \sqrt 2}-{1\over \sqrt 3}+{1\over \sqrt 3}-{1\over \sqrt 4}+\cdots \\=1,故選\bbox[red, 2pt]{(A)}
解答:a=\sqrt 2+b \Rightarrow a^2=\sqrt 2a+ab \Rightarrow 2ab=2a^2-2\sqrt 2a \Rightarrow 2a^2-2\sqrt 2a+2\sqrt 2 c^2+1=0 \\ \Rightarrow 2(a-{\sqrt 2\over 2})^2+2\sqrt 2c^2=0 \Rightarrow \cases{a=\sqrt 2/2\\ c=0} \Rightarrow b=a-\sqrt 2=-{\sqrt 2 \over 2} \\ \Rightarrow \cases{a=\sqrt 2/2\\ b=-\sqrt 2/2\\ c=0} \Rightarrow a+b+c=0,故選\bbox[red, 2pt]{(D)}
解答:\cos x={1\over 2}(e^{ix}+e^{-ix}) \Rightarrow \int_0^\infty e^{-x}\cos x\,dx ={1 \over 2} \int_0^\infty \left(e^{(i-1)x}+e^{-(i+1)x} \right)\,dx \\={1\over 2} \left. \left[ {1\over i-1}e^{(i-1)x}-{1\over i+1}e^{-(i+1)x}\right] \right|_0^\infty ={1\over 2}\left({1\over i+1}-{1\over i-1} \right)={1\over 2}\cdot 1={1\over 2},故選\bbox[red, 2pt]{(C)}
解答:凸n邊形內角和=f(n)=180(n-2) \\\Rightarrow \cases{f(12)=1800\lt 2024,不合\\ f(13) =1980\lt2024,不合\\ f(14) =2160 \Rightarrow 另一角度數=2160-2024=136度\\ f(15)=2340 \Rightarrow 另一角度數=2340-2024=316度,不合},故選\bbox[red, 2pt]{(C)}
解答:\begin{vmatrix} 1& 4& 1& -3\\ 2& 10 & 0 &1 \\ 0 & 0& 2& 2\\ 0 & 0 &-2& 1\end{vmatrix} \xrightarrow{R_2-2R_1\to R_2} \begin{vmatrix} 1& 4& 1& -3\\ 0& 2 & -2 &7 \\ 0 & 0& 2& 2\\ 0 & 0 &-2& 1\end{vmatrix} =\begin{vmatrix} 2 & -2 &7 \\ 0& 2& 2\\ 0 &-2& 1\end{vmatrix} =2 \begin{vmatrix} 2& 2\\ -2& 1\end{vmatrix}=12,故選\bbox[red, 2pt]{(D)}
解答:(A) \bigcirc:\cases{\overrightarrow{AB} =(1,2,-1) \\ \overrightarrow{AC} =(-2,2,2) } \Rightarrow \overrightarrow{AB} \cdot \overrightarrow{AC} =0 \Rightarrow \angle BAC={\pi\over 2} \\(B)\bigcirc:\overrightarrow{AB} =(2,1,-1)-(1,-1,0)=(1,2,-1) \\(C)\bigcirc: \overrightarrow{AB} \times \overrightarrow{AC} =\begin{vmatrix}\vec i& \vec j&\vec k \\ 1& 2& -1\\ -2& 2& 2\end{vmatrix} =6\vec i+0\vec j+6\vec k \\(D)\times: {1\over 2}\Vert\overrightarrow{AB} \times \overrightarrow{AC} \Vert = 3\sqrt 2\ne 2\sqrt 2\\,故選\bbox[red, 2pt]{(D)}
解答:7^{10} = 1 \text{ mod }11 \Rightarrow 7^{223} =(7^{10})^{22}\cdot 7^3 =7^3 \text{ mod }11 =2\text{ mod }11 ,故選\bbox[red, 2pt]{(A)}
解答:0.\bar 3 ={3\over 9}={1\over 3},故選\bbox[red, 2pt]{(B)}
解答:\sqrt 5\approx2.236 \Rightarrow {1+\sqrt 5\over 4} \approx 0.81,故選\bbox[red, 2pt]{(C)}
解答:3f(1)+5f(1)=2+1 \Rightarrow f(1)={3\over 8}=0.375,故選\bbox[red, 2pt]{(D)}
解答:\int_0^1 {x^2\over 1+x^2}\,dx =\int_0^1 \left( 1-{1\over 1+x^2}\right)\,dx =\left. \left[ x-\tan^{-1}x\right] \right|_0^1 =1-{\pi\over 4},故選\bbox[red, 2pt]{(A)}
解答:只有四種可能\cases{1,1,2,3\\ 1,2,2,3\\ 1,2,3,3}排列數皆為{4!\over 2!}=12,因此共有12\times 3=36種,故選\bbox[red, 2pt]{(D)}
解答:f(x)=xg(x),其中g(x)=(x−1)(x−2)⋯(x−2024)(x+1)(x+2)⋯(x+2024)⇒f′(x)=g(x)+xg′(x)⇒f′(0)=g(0)=2024!2024!=1,故選(C)
解答:f(x)=x4+2x3−3x2+2x−1⇒f′(x)=4x3+6x2−6x+2limh→0f(1+3h)−f(1−2h)h=limh→0(f(1+3h)−f(1−2h))′(h)′=limh→03f′(1+3h)+2f′(1−2h)1=3f″
解答:依均值定理,故選\bbox[red, 2pt]{(A)}
解答:\int_0^{\pi/2} \cos x\,dx =1 \Rightarrow 平均值={1\over \pi/2}= {2\over \pi},故選\bbox[red, 2pt]{(A)}
解答:{(\log_a x)\cdot (\log_x{b\over a}) \over (\log_x b)(\log_{ab} x)} ={(\log_a x)\cdot (\log_x b-\log_x a) \over (\log_x b)({1\over \log_x ab })} = {\log_a b-1\over \log_{ab}b} \\={\log_a b-1\over {1\over \log_b a+1}} =(\log_b a+1)(\log_a b-1)=1-\log_ba+\log_a b-1=-{1\over 10}+10 ={99\over10},故選\bbox[red, 2pt]{(B)}
解答:{1+2\sqrt 3+\sqrt 5\over (1+\sqrt 3) (\sqrt 3+\sqrt 5)} +{3+2\sqrt 7+ \sqrt 5\over (\sqrt 5+\sqrt 7) (\sqrt 7+3)} ={(1+\sqrt 3)+(\sqrt 3+\sqrt 5)\over (1+\sqrt 3) (\sqrt 3+\sqrt 5)} +{(3+\sqrt 7)+ (\sqrt 7+\sqrt 5)\over (\sqrt 5+\sqrt 7) (\sqrt 7+3)} \\={1\over \sqrt 3+\sqrt 5} +{1\over 1+\sqrt 3} +{1\over \sqrt 5+\sqrt 7} +{1\over \sqrt 7+3}={1\over 2}\left(\sqrt 5-\sqrt 3+\sqrt 3-1+\sqrt 7-\sqrt 5+3-\sqrt 7 \right) \\=1,故選\bbox[red, 2pt]{(B)}
解答:\sum_{n=1}^\infty {\sqrt{n+1}-\sqrt n\over \sqrt{n^2+n}} =\sum_{n=1}^\infty \left( {1\over \sqrt{n}} -{1\over \sqrt{n+1}}\right) =1-{1\over \sqrt 2}+{1\over \sqrt 2}-{1\over \sqrt 3}+{1\over \sqrt 3}-{1\over \sqrt 4}+\cdots \\=1,故選\bbox[red, 2pt]{(A)}
解答:a=\sqrt 2+b \Rightarrow a^2=\sqrt 2a+ab \Rightarrow 2ab=2a^2-2\sqrt 2a \Rightarrow 2a^2-2\sqrt 2a+2\sqrt 2 c^2+1=0 \\ \Rightarrow 2(a-{\sqrt 2\over 2})^2+2\sqrt 2c^2=0 \Rightarrow \cases{a=\sqrt 2/2\\ c=0} \Rightarrow b=a-\sqrt 2=-{\sqrt 2 \over 2} \\ \Rightarrow \cases{a=\sqrt 2/2\\ b=-\sqrt 2/2\\ c=0} \Rightarrow a+b+c=0,故選\bbox[red, 2pt]{(D)}
解答:\cos x={1\over 2}(e^{ix}+e^{-ix}) \Rightarrow \int_0^\infty e^{-x}\cos x\,dx ={1 \over 2} \int_0^\infty \left(e^{(i-1)x}+e^{-(i+1)x} \right)\,dx \\={1\over 2} \left. \left[ {1\over i-1}e^{(i-1)x}-{1\over i+1}e^{-(i+1)x}\right] \right|_0^\infty ={1\over 2}\left({1\over i+1}-{1\over i-1} \right)={1\over 2}\cdot 1={1\over 2},故選\bbox[red, 2pt]{(C)}
解答:凸n邊形內角和=f(n)=180(n-2) \\\Rightarrow \cases{f(12)=1800\lt 2024,不合\\ f(13) =1980\lt2024,不合\\ f(14) =2160 \Rightarrow 另一角度數=2160-2024=136度\\ f(15)=2340 \Rightarrow 另一角度數=2340-2024=316度,不合},故選\bbox[red, 2pt]{(C)}
解答:\begin{vmatrix} 1& 4& 1& -3\\ 2& 10 & 0 &1 \\ 0 & 0& 2& 2\\ 0 & 0 &-2& 1\end{vmatrix} \xrightarrow{R_2-2R_1\to R_2} \begin{vmatrix} 1& 4& 1& -3\\ 0& 2 & -2 &7 \\ 0 & 0& 2& 2\\ 0 & 0 &-2& 1\end{vmatrix} =\begin{vmatrix} 2 & -2 &7 \\ 0& 2& 2\\ 0 &-2& 1\end{vmatrix} =2 \begin{vmatrix} 2& 2\\ -2& 1\end{vmatrix}=12,故選\bbox[red, 2pt]{(D)}
解答:(A) \bigcirc:\cases{\overrightarrow{AB} =(1,2,-1) \\ \overrightarrow{AC} =(-2,2,2) } \Rightarrow \overrightarrow{AB} \cdot \overrightarrow{AC} =0 \Rightarrow \angle BAC={\pi\over 2} \\(B)\bigcirc:\overrightarrow{AB} =(2,1,-1)-(1,-1,0)=(1,2,-1) \\(C)\bigcirc: \overrightarrow{AB} \times \overrightarrow{AC} =\begin{vmatrix}\vec i& \vec j&\vec k \\ 1& 2& -1\\ -2& 2& 2\end{vmatrix} =6\vec i+0\vec j+6\vec k \\(D)\times: {1\over 2}\Vert\overrightarrow{AB} \times \overrightarrow{AC} \Vert = 3\sqrt 2\ne 2\sqrt 2\\,故選\bbox[red, 2pt]{(D)}
解答:7^{10} = 1 \text{ mod }11 \Rightarrow 7^{223} =(7^{10})^{22}\cdot 7^3 =7^3 \text{ mod }11 =2\text{ mod }11 ,故選\bbox[red, 2pt]{(A)}
解答:0.\bar 3 ={3\over 9}={1\over 3},故選\bbox[red, 2pt]{(B)}
解答:\sqrt 5\approx2.236 \Rightarrow {1+\sqrt 5\over 4} \approx 0.81,故選\bbox[red, 2pt]{(C)}
解答:3f(1)+5f(1)=2+1 \Rightarrow f(1)={3\over 8}=0.375,故選\bbox[red, 2pt]{(D)}
解答:\int_0^1 {x^2\over 1+x^2}\,dx =\int_0^1 \left( 1-{1\over 1+x^2}\right)\,dx =\left. \left[ x-\tan^{-1}x\right] \right|_0^1 =1-{\pi\over 4},故選\bbox[red, 2pt]{(A)}
解答:只有四種可能\cases{1,1,2,3\\ 1,2,2,3\\ 1,2,3,3}排列數皆為{4!\over 2!}=12,因此共有12\times 3=36種,故選\bbox[red, 2pt]{(D)}

解答:假設\cases{\vec u=(2,4,3,1)\\ \vec v=(-1,-1,2,0)}\\(A)\times: (2,-1,0,0)\cdot \vec v=3\ne 0\\(B)\times:(1,1,-1,0)\cdot \vec u=3\ne 0 \\ (C)\bigcirc: \cases{ (2,0,1,-7)\cdot \vec u=0,(2,0,1,-7)\cdot \vec v=0\\ (1,-1,0,2)\cdot \vec u=0,(1,-1,0,2)\cdot \vec v=0} \\(D)\times: (0,1,-2,2)\cdot \vec v=-5\ne 0\\,故選\bbox[red, 2pt]{(C)}
解答:28人任排有28!排法,其中一半小明在小華前面,另一半小明在小華後面,因此有28!/2,故選\bbox[red, 2pt]{(D)}
解答:\cases{7= 7 \text{ mod }10 \\7^2= 9 \text{ mod }10 \\7^3= 3 \text{ mod }10 \\ 7^4= 1 \text{ mod }10 \\7^5= 7 \text{ mod }10 } \Rightarrow 循環數=4, 又\cases{7= 3 \text{ mod }4 \\7^2= 1 \text{ mod }4 \\7^3= 3 \text{ mod }4 \\7^4= 1 \text{ mod }4 } \Rightarrow 循環數2\\ 7^{7^7} =7^{3 \text{ mod }4} =3 \text{ mod }10,故選\bbox[red, 2pt]{(B)}
解答:abc=1 \Rightarrow 取a=b=c=1 \Rightarrow {1\over 1+1+1}+ {1\over 1+1+1}+ {1\over 1+1+1}=1,故選\bbox[red, 2pt]{(C)}\\\bbox[red, 2pt]{另解}\;{a\over 1+a+ab}+ {b\over 1+b+bc}+{c\over 1+c+ac} ={a\over 1+a+ab}+ {ab\over a+ab+abc}+{abc\over ab+abc+a^2bc} \\={a\over 1+a+ab}+ {ab\over a+ab+1}+{1\over ab+1+a}={a+ab+1\over 1+a+ab} =1
解答:A=\begin{bmatrix}2 & -3 & 6 \\0 & 3 & -4 \\ 0 & 2 & -3 \end{bmatrix} \Rightarrow \det(A-\lambda I)=-\lambda^3+2\lambda^2 +\lambda-2 =-(\lambda-1)(\lambda-2)(\lambda+1)=0 \\ \Rightarrow 特徵值\lambda=1,2,-1,故選\bbox[red, 2pt]{(B)}
解答:(A)x={\sqrt 2\over 4} \Rightarrow \sqrt{1-x^2} =\sqrt{7\over 8},不可能\\ (B)x={\sqrt 2\over 3} \Rightarrow \sqrt{1-x^2} =\sqrt{7\over 9},不可能\\(C)x={\sqrt 2\over 2} \Rightarrow \sqrt{1-x^2} ={\sqrt 2\over 2} \Rightarrow 2\sqrt 2\cdot {1\over 2}+{\sqrt 2\over 2}-{\sqrt 2\over 2} =\sqrt 2,符合題意\\ (D)x=\sqrt 2 \Rightarrow 1-x^2 \lt 0,不可能\\,故選\bbox[red, 2pt]{(C)}
解答:\triangle ABC :\cases{\angle C=90^\circ\\ 2\overline{BC}=\overline{AB}} \Rightarrow \cases{\angle A=30^\circ \\\angle ABC=60^\circ} \\ \Rightarrow \tan \angle DEC=\tan 60^\circ ={\overline{DC} \over \overline{EC}} \Rightarrow \sqrt 3={3\over \overline{EC}} \Rightarrow \overline{EC}=\sqrt 3,故選\bbox[red, 2pt]{(A)}
解答:\left( {\sqrt 3\over 2}-{1\over 2}i\right)^{2022} =\left( \cos{11\over 6}\pi+ i\sin {11\over 6}\pi \right)^{2022} =(e^{11\pi i/6})^{2022} =e^{11\cdot 337 \pi i} \\=e^{3707\pi i} =e^{(2 \cdot 1853+1)\pi i} =e^{\pi i} =-i,故選\bbox[red, 2pt]{(D)}
解答:\cases{A \begin{bmatrix}1 \\2\\3 \end{bmatrix} =\begin{bmatrix}1 \\1\\ 0 \end{bmatrix} \\A \begin{bmatrix}2 \\2\\4 \end{bmatrix} =\begin{bmatrix}0 \\1\\ 1 \end{bmatrix} } \Rightarrow \cases{A^{-1}\begin{bmatrix}1 \\1\\ 0 \end{bmatrix} = \begin{bmatrix}1 \\2\\3 \end{bmatrix} \\A^{-1}\begin{bmatrix}0 \\1\\ 1 \end{bmatrix} = \begin{bmatrix}2 \\2\\4 \end{bmatrix}} \Rightarrow 3A^{-1}\begin{bmatrix}1 \\1\\ 0 \end{bmatrix}-2 A^{-1}\begin{bmatrix}0 \\1\\ 1 \end{bmatrix} =3\begin{bmatrix}1 \\2\\3 \end{bmatrix}-2 \begin{bmatrix}2 \\2\\4 \end{bmatrix} \\ \Rightarrow A^{-1} \left( \begin{bmatrix}3 \\3\\ 0 \end{bmatrix}+ \begin{bmatrix}0 \\-2\\ -2 \end{bmatrix}\right) = \begin{bmatrix}3 \\6\\9 \end{bmatrix}+ \begin{bmatrix}-4 \\-4\\-8 \end{bmatrix} \Rightarrow A^{-1}\begin{bmatrix}3 \\1\\ -2 \end{bmatrix} =\begin{bmatrix}-1 \\2\\ 1 \end{bmatrix} \\ \Rightarrow A\begin{bmatrix}-1 \\2\\ 1 \end{bmatrix}= \begin{bmatrix}3 \\1\\ -2 \end{bmatrix} \Rightarrow \mathbf X=\begin{bmatrix}-1 \\2\\ 1 \end{bmatrix},故選\bbox[red, 2pt]{()}
解答:\cases{A(4,0,0) \\B(5,4,2) \\C(0,4,0) \\D(1,3,1)} \Rightarrow L=\overleftrightarrow{CD} :{x\over 1} ={y-4\over -1}={z\over 1} \Rightarrow P(t,-t+4,t),t\in \mathbb R \Rightarrow \cases{\overrightarrow{PA} =(t-4,-t+4,t) \\ \overrightarrow{PB} =(t-5,-t,t-2)}\\ \Rightarrow \overrightarrow{PA} \cdot \overrightarrow{PB} =(t-4)(t-5)-t(-t+4)+t(t-2) =3(t-{5\over 2})^2+{5\over 4} \\ \Rightarrow 最小值={5\over 4}=1.25,故選\bbox[red, 2pt]{(B)}

解答:L_1\bot L_2 \Rightarrow x=4 \Rightarrow \cases{Q(2,0)\\ R(1,0)} \\ 由於\angle RPQ=90^\circ \Rightarrow \overline{QR}為外接圓直徑=2R \Rightarrow R={1\over 2}\overline{QR} ={1\over 2},故選\bbox[red, 2pt]{(D)}
解答:{3+8-1\over \sqrt{3^2+4^2}} =2,故選\bbox[red, 2pt]{(D)}
解答:\sqrt{{1\over 3^2}+{1\over 4^2}+1} =\sqrt{1+({5\over 12}^2)}\\ 假設\tan \theta={5\over 12} \Rightarrow \sqrt{1+({5\over 12}^2)} =\sqrt{1+\tan^2 \theta} =\sec \theta={13\over 12} =1.08,故選\bbox[red, 2pt]{(B)}
解答:\cases{兩根之和6-m\ge 2\\ 兩根之積m-1\ge 1} \Rightarrow 2\le m\le 4 \Rightarrow \cases{m=2 \Rightarrow x^2-4x+1=0兩根非正整數\\ m=3 \Rightarrow x^2-3x+2=0\Rightarrow x=2,1},故選\bbox[red, 2pt]{(D)}
解答:(a^x+a^{-x})^2=a^{2x}+a^{-2x}+2=4 \Rightarrow a^{2x}+a^{-2x}=2\\ \Rightarrow a^{3x}+ a^{-3x} =(a^x+a^{-x})(a^{2x}-1+a^{-2x}) =2\cdot (2-1)=2,故選\bbox[red, 2pt]{(B)}
解答:{1\over 2}\log{125\over 16} +\log {3\sqrt 8\over 125} -\log {3\over 5} ={1\over 2}(\log 125-\log 16) +\log 3\sqrt 8-\log 125-\log 3+\log 5 \\={3\over 2}\log 5-2\log 2+\log 3+{3\over 2} \log 2-3\log 5-\log 3+\log 5 =-{1\over 2}\log 5-{1\over 2}\log 2 \\=-{1\over 2}(1-\log 2)-{1\over 2}\log 2=-{1\over 2},故選\bbox[red, 2pt]{(A)}
解答:由於限制在區間內,因此至少有兩個極值,故選\bbox[red, 2pt]{(C)}
=========== END ================
解題僅供參考,教甄其他歷年試題及詳解
謝謝,你的方法比較 【完備】!
回覆刪除老師您好,想請教第39題如何得知角C為90度?謝謝您
回覆刪除嚴格地講,題目並沒有標示角C是90度, 只不過這樣畫圖應該是90度吧!!?
刪除謝謝您🙂
刪除感謝提供!
回覆刪除