教育部受託辦理112學年度公立高級中等學校教師甄選
第一部分:選擇題( 共40分)
一、單選題( 每題3分, 共24分)
解答:n3+103n+11=n2−11n+121−1228n+11⇒取1228=n+11⇒n=1217,故選(A)
解答:x2+xy+y2=3⇒2x+y+xy′+2yy′=0⇒y′=−2x+yx+2y⇒y′(1,1)=−1⇒過(1,1)且斜率為−1的直線方程式:y=−(x−1)+1⇒x+y=2,故選(B)
解答:∞∑k=012kcos(120∘k)=∞∑k=0123kcos(120∘⋅3k)+123k+1cos(120∘⋅(3k+1))+123k+2cos(120∘⋅(3k+2))=11−18−141−18−181−18=87−27−17=57,故選(B)
解答:原式S=13910∑k=2k(k−1)C10k2k−2令f(x)=(1+x)10=10∑k=0C10kxk⇒f′(x)=10(1+x)9=10∑k=1kC10kxk−1⇒f″(x)=90(1+x)8=10∑k=2k(k−1)C10kxk−2⇒S=f″(2)39=90⋅3839=30,故選(A)
解答:ω=−1+√3i2=cos2π3+isin2π3=e2πi/3⇒ω3=1⇒(ω−1)(ω2+ω+1)=0⇒{ω3=1ω2+ω+1=0因此(ω5+ω4+5)(−3ω2+2ω+2)(3ω2−ω+3)=(ω2+ω+5)(−3(−ω−1)+2ω+2)(3(−ω−1)−ω+3)=4(5ω+5)(−4ω)=4(5(−ω2))(−4ω)=80ω3=80,故選(B)
解答:對於每種壽司{可能被拿手一盤:有C31=3種分配方式可能被拿走二盤:有C32=3種分配方式共有六種可能⇒10種壽司有610種可能,故選(B)
解答:log2n≥12⇒n≥12log2=120.301≈39.8⇒n=40,故選(A)
狀態1→狀態1:P(A2W→B2W→A)+P(A1R1W→B1R1W→A)=13+49=79狀態1→狀態2:P(A1R1W→B2W→A)=23⋅13=29狀態轉變圖如上(A)×:第一回合後仍在狀態一的機率=79(B)◯:第一回合後在狀態二的機率=29(C)×:P(S1→S1→S1)+P(S1→S2→S1)=(79)2+29⋅23>2081(D)◯:P(S1→S1→S1→S1)+P(S1→S1→S2→S1)+P(S1→S2→S2→S1)+P(S1→S2→S1→S1)=(79)3+79⋅29⋅23+29⋅13⋅23+29⋅23⋅79=547729,故選(BD)
解答:f(x)=(x−α)(x−β)(x−γ)(x−δ)=(x2−(α+β)x+αβ)(x2−(γ+δ)x+γδ)=(x2−(3−2i)x+4−i)(x2−(3+2i)x+4+i)=x4−6x3+21x2+28x+17(A)◯:a=−6(B)×:b=21(C)◯:c=−28(D)◯:d=17,故選(ACD)
解答:(A)◯:x+1503<x⇒x>75⇒原始分數大於75調整後分數會變低(B)×:相關係數都是1(C)×:∑z2i≠1(D)◯:作弊同學分數剛好是全班平均分數,因此移除該同學分數不改變全班平均,因此甲案平均仍高於乙案,故選(AD)
解答:令{→a=→OA→b=→OB→c=→OC→d=→OD(A)◯:O,A,D共線⇒→d→a=k→a+(2−k)→b→a=常數⇒k=2(B)◯:→CD=−→c+→d=−→a−→b+k→a+(2−k)→b=−(1−k)→a+(1−k)→b=(1−k)(−→a+→b)=(1−k)→AB⇒→CD∥→AB(C)◯:{→AB=(b1−a1,b2−a2)→AD=((k−1)a1+(2−k)b1,(k−1)a2+(2−k)b2)⇒△ABD面積=|b1−a1b2−a2(k−1)a1+(2−k)b1(k−1)a2+(2−k)b2)|=|a2b1−a1b2|與k無關(D)×:{→AC=→OB=(b1,b2)→AD=((k−1)a1+(2−k)b1,(k−1)a2+(2−k)b2)⇒△ACD=|b1b2(k−1)a1+(2−k)b1(k−1)a2+(2−k)b2)|=|(k−1)(a2b1−a1b2)|與k有關,故選(ABC)
解答:鏡射矩陣[cos2θsin2θsin2θ−cos2θ],其中{L1:x+y=0⇒θ=135∘L2:x=√3y⇒θ=30∘⇒{鏡射矩陣A=[0−1−10]鏡射矩陣B=[1/2√3/2√3/2−1/2]⇒Q=AP=[−1−1]⇒R=BQ=[(−1−√3)/2(1−√3)/2]⇒R=(−1−√32,1−√32)
解答:A=[1a−1−2]⇒A3=[1−a2+3aa−33a−8]=I⇒a=3⇒A=[13−1−2]⇒A2=[−2−311]⇒A2+A+I=0令B=A+2A2+⋯+20A20⇒AB=A2+2A3+⋯+20A21⇒B−AB=A+A2+⋯+A20−20A21=A19+A20−20A21⇒(I−A)B=A+A2−20I=−21I⇒B=−21I(I−A)−1=[−2100−21][0−31−3]=[−2100−21][11−1/30]=[−21−2170]
解答:X∼b(n=10,p=12)⇒{μ=E(X)=np=5σ=√np(1−p)=√10/2≈1.58⇒P(μ−σ≤X≤μ+σ)=P(5−1.58≤X≤5+1.58)=P(3.42≤X≤6.58)=P(X=4)+P(X=5)+P(X=6)=1210(C104+C105+C106)=6721024=2132
解答:6×24x−2×27x+3×16x−18x=2⋅3⋅3x⋅23x−2⋅33x+3⋅24x−2x⋅32x=23x+1⋅3x+1−2⋅33x+3⋅24x−2x⋅32x=(23x+1⋅3x+1+3⋅24x)−(2⋅33x+2x⋅32x)=6⋅23x(3x+2x−1)−2⋅32x(3x+2x−1)=(6⋅23x−2⋅32x)(3x+2x−1)=0⇒6⋅23x=2⋅32x⇒log6+3xlog2=log2+2xlog3⇒log3=(2log3−3log2)x⇒x=log32log3−3log2⇒2x−13x=3log22log3−3log2×2log3−3log23log3=log2log3=log32
解答:an=n∑k=1nn2+k2=n∑k=11/n1+(k/n)2⇒lim
解答:令f(k)={x\over k+4^2} +{y\over k+5^2} +{z\over k+6^2} -1,依題意:f(1^2)=f(2^2)=f(3^2)=0 \\ 因此(k-1)(k-4) (k-9) \\ \qquad =(k+4^2)(k+5^2)(k+6^2)-x(k+5^2)(k+6^2)-y(k+4^2)(k+6^2)-z(k+4^2)(k+5^2) \\ \Rightarrow \cases{左式k^2係數=-1-4-9=-14\\ 右式k^2係數=4^2+5^2+6^2-(x+y+z)} \Rightarrow x+y+z=77+14=\bbox[red, 2pt]{91}
解答:x^2f(x)=(3x-2)p(x)+r \Rightarrow x^3f(x) =(3x-2)xp(x)+rx =(3x-2)xp(x)+{r\over 3}(3x-2)+{2r\over 3} \\ \Rightarrow {2r\over 3}=6 \Rightarrow r=\bbox[red, 2pt]9
解答:Q=a^2-2a =(a-1)^2-1為有理數\Rightarrow a=1+ \sqrt q,其q為有理數,但\sqrt q為無理數\\ 又{P\over Q} ={a^3+3a^2-16a+6\over a^2-2a } =a+5+{-6a+6\over a^2-2a}=6+ \sqrt q+{ - 6\sqrt q\over q-1} 為有理數\\ \Rightarrow q-1=6 \Rightarrow q=7 \Rightarrow a=\bbox[red, 2pt]{1+\sqrt 7}
解答:\cases{A(2,2,1)\\ B(1,0,2)\\ C(0,1,2)} \Rightarrow \cases{\overrightarrow{AB} =(-1,-2,1)\\ \overrightarrow{AC} =(-2,-1,1)} \Rightarrow \vec n=\overrightarrow{AB} \times \overrightarrow{AC} =(-1,-1, -3) \\ \Rightarrow E=\triangle ABC:x+y+3z =7 \Rightarrow \cases{E\cap 直線(2,0,s) =\bbox[red,2pt]{(2,0,5/3)}\\ E\cap 直線(0,2,t)=\bbox[red,2pt]{(0,2,5/3)}}
解答:
解答:\cases{u=\ln x\\ dv =xdx} \Rightarrow \cases{du={dx\over x}\\ v={1\over 2}x^2} \Rightarrow \int x\ln x\,dx ={1\over 2}x^2\ln x -{1\over 2}\int xdx ={1\over 2}x^2 \ln x-{1\over 4}x^2+C\\ \Rightarrow \int_1^e x\ln x\,dx =\left. \left[{1\over 2}x^2 \ln x-{1\over 4}x^2 \right]\right|_1^e ={1\over 2}e^2-{1\over 4}e^2+{1\over 4} =\bbox[red, 2pt]{{1\over 4}e^2+{1\over 4}}
解答:f(x)=(x−α)(x−β)(x−γ)(x−δ)=(x2−(α+β)x+αβ)(x2−(γ+δ)x+γδ)=(x2−(3−2i)x+4−i)(x2−(3+2i)x+4+i)=x4−6x3+21x2+28x+17(A)◯:a=−6(B)×:b=21(C)◯:c=−28(D)◯:d=17,故選(ACD)
解答:(A)◯:x+1503<x⇒x>75⇒原始分數大於75調整後分數會變低(B)×:相關係數都是1(C)×:∑z2i≠1(D)◯:作弊同學分數剛好是全班平均分數,因此移除該同學分數不改變全班平均,因此甲案平均仍高於乙案,故選(AD)
解答:令{→a=→OA→b=→OB→c=→OC→d=→OD(A)◯:O,A,D共線⇒→d→a=k→a+(2−k)→b→a=常數⇒k=2(B)◯:→CD=−→c+→d=−→a−→b+k→a+(2−k)→b=−(1−k)→a+(1−k)→b=(1−k)(−→a+→b)=(1−k)→AB⇒→CD∥→AB(C)◯:{→AB=(b1−a1,b2−a2)→AD=((k−1)a1+(2−k)b1,(k−1)a2+(2−k)b2)⇒△ABD面積=|b1−a1b2−a2(k−1)a1+(2−k)b1(k−1)a2+(2−k)b2)|=|a2b1−a1b2|與k無關(D)×:{→AC=→OB=(b1,b2)→AD=((k−1)a1+(2−k)b1,(k−1)a2+(2−k)b2)⇒△ACD=|b1b2(k−1)a1+(2−k)b1(k−1)a2+(2−k)b2)|=|(k−1)(a2b1−a1b2)|與k有關,故選(ABC)
解答:鏡射矩陣[cos2θsin2θsin2θ−cos2θ],其中{L1:x+y=0⇒θ=135∘L2:x=√3y⇒θ=30∘⇒{鏡射矩陣A=[0−1−10]鏡射矩陣B=[1/2√3/2√3/2−1/2]⇒Q=AP=[−1−1]⇒R=BQ=[(−1−√3)/2(1−√3)/2]⇒R=(−1−√32,1−√32)
解答:A=[1a−1−2]⇒A3=[1−a2+3aa−33a−8]=I⇒a=3⇒A=[13−1−2]⇒A2=[−2−311]⇒A2+A+I=0令B=A+2A2+⋯+20A20⇒AB=A2+2A3+⋯+20A21⇒B−AB=A+A2+⋯+A20−20A21=A19+A20−20A21⇒(I−A)B=A+A2−20I=−21I⇒B=−21I(I−A)−1=[−2100−21][0−31−3]=[−2100−21][11−1/30]=[−21−2170]
解答:X∼b(n=10,p=12)⇒{μ=E(X)=np=5σ=√np(1−p)=√10/2≈1.58⇒P(μ−σ≤X≤μ+σ)=P(5−1.58≤X≤5+1.58)=P(3.42≤X≤6.58)=P(X=4)+P(X=5)+P(X=6)=1210(C104+C105+C106)=6721024=2132
解答:6×24x−2×27x+3×16x−18x=2⋅3⋅3x⋅23x−2⋅33x+3⋅24x−2x⋅32x=23x+1⋅3x+1−2⋅33x+3⋅24x−2x⋅32x=(23x+1⋅3x+1+3⋅24x)−(2⋅33x+2x⋅32x)=6⋅23x(3x+2x−1)−2⋅32x(3x+2x−1)=(6⋅23x−2⋅32x)(3x+2x−1)=0⇒6⋅23x=2⋅32x⇒log6+3xlog2=log2+2xlog3⇒log3=(2log3−3log2)x⇒x=log32log3−3log2⇒2x−13x=3log22log3−3log2×2log3−3log23log3=log2log3=log32
解答:an=n∑k=1nn2+k2=n∑k=11/n1+(k/n)2⇒lim
解答:令f(k)={x\over k+4^2} +{y\over k+5^2} +{z\over k+6^2} -1,依題意:f(1^2)=f(2^2)=f(3^2)=0 \\ 因此(k-1)(k-4) (k-9) \\ \qquad =(k+4^2)(k+5^2)(k+6^2)-x(k+5^2)(k+6^2)-y(k+4^2)(k+6^2)-z(k+4^2)(k+5^2) \\ \Rightarrow \cases{左式k^2係數=-1-4-9=-14\\ 右式k^2係數=4^2+5^2+6^2-(x+y+z)} \Rightarrow x+y+z=77+14=\bbox[red, 2pt]{91}
解答:x^2f(x)=(3x-2)p(x)+r \Rightarrow x^3f(x) =(3x-2)xp(x)+rx =(3x-2)xp(x)+{r\over 3}(3x-2)+{2r\over 3} \\ \Rightarrow {2r\over 3}=6 \Rightarrow r=\bbox[red, 2pt]9
解答:Q=a^2-2a =(a-1)^2-1為有理數\Rightarrow a=1+ \sqrt q,其q為有理數,但\sqrt q為無理數\\ 又{P\over Q} ={a^3+3a^2-16a+6\over a^2-2a } =a+5+{-6a+6\over a^2-2a}=6+ \sqrt q+{ - 6\sqrt q\over q-1} 為有理數\\ \Rightarrow q-1=6 \Rightarrow q=7 \Rightarrow a=\bbox[red, 2pt]{1+\sqrt 7}
解答:\cases{A(2,2,1)\\ B(1,0,2)\\ C(0,1,2)} \Rightarrow \cases{\overrightarrow{AB} =(-1,-2,1)\\ \overrightarrow{AC} =(-2,-1,1)} \Rightarrow \vec n=\overrightarrow{AB} \times \overrightarrow{AC} =(-1,-1, -3) \\ \Rightarrow E=\triangle ABC:x+y+3z =7 \Rightarrow \cases{E\cap 直線(2,0,s) =\bbox[red,2pt]{(2,0,5/3)}\\ E\cap 直線(0,2,t)=\bbox[red,2pt]{(0,2,5/3)}}
解答:
令\overline{MF} =\overline{FC}=a \Rightarrow 在直角\triangle BMF中,\overline{MF}^2 = \overline{MB}^2 + \overline{BF}^2 \Rightarrow a^2=120^2+(288-a)^2 \Rightarrow a=169\\ 令\overline{DE}= \overline{EG} =b \Rightarrow \overline{AE}=288-b \Rightarrow \cases{直角\triangle MAE: \overline{ME}^2 = \overline{MA}^2 + \overline{AE}^2 =120^2+(288-b)^2 \\ 直角\triangle EDC: \overline{EC}^2 = \overline{ED}^2 + \overline{DC}^2 =b^2+240^2}\\ 由於\overline{EM} =\overline{EC} \Rightarrow 120^2+(288-b)^2 =b^2+240^2 \Rightarrow b=69 \\ 直角\triangle FHE: \overline{EF}^2= \overline{EH}^2 + \overline{HF}^2 =240^2+ (a-b)^2=240^2+ (169-69)^2=260^2 \Rightarrow \overline{EF} =\bbox[red, 2pt]{260}
解答:\cases{u=\ln x\\ dv =xdx} \Rightarrow \cases{du={dx\over x}\\ v={1\over 2}x^2} \Rightarrow \int x\ln x\,dx ={1\over 2}x^2\ln x -{1\over 2}\int xdx ={1\over 2}x^2 \ln x-{1\over 4}x^2+C\\ \Rightarrow \int_1^e x\ln x\,dx =\left. \left[{1\over 2}x^2 \ln x-{1\over 4}x^2 \right]\right|_1^e ={1\over 2}e^2-{1\over 4}e^2+{1\over 4} =\bbox[red, 2pt]{{1\over 4}e^2+{1\over 4}}
解答:
假設\overline{AB} \bot \overline{BO} 且\overline{OB}\bot \overline{BC},則欲求之|\cos \theta | = |\cos \angle ABC|\\ 在直角\triangle OAB中,假設\overline{AB}=1再加上\angle AOB=30^\circ \Rightarrow \cases{\overline{OA}=2\\ \overline{OB}=\sqrt 3}\\ 同理,在直角\triangle OBC中,\angle BOC=45^\circ \Rightarrow \overline{BC}=\overline{OB}=\sqrt 3,且\overline{OC}=\sqrt 6\\ 在\triangle OAC中,\cos \angle AOC =\cos 60^\circ = {1\over 2} ={2^2+(\sqrt 6)^2 -\overline{AC}^2 \over 2\cdot 2\cdot \sqrt 6} \Rightarrow \overline{AC}^2 =10-2\sqrt 6\\ 在\triangle ABC中,\cos \angle ABC={1^2+(\sqrt 3)^2-(10-2\sqrt 6) \over 2\cdot 1\cdot \sqrt 3} = {2\sqrt 6-6\over 2\sqrt 3}=\sqrt 2-\sqrt 3\\ \Rightarrow |\cos \theta| =|\sqrt 2-\sqrt 3|= \bbox[red, 2pt]{\sqrt 3-\sqrt 2}
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解題僅供參考,其他試題及詳解
多選九D選項第一項應該都是轉移到S_1機率
回覆刪除已修正,謝謝
刪除請問複選題
回覆刪除9.
狀態1---->狀態1與狀態1---->狀態2的機率怎麼得到
狀態1->狀態1有兩種情形,情形1:A箱抽到2W(p=1/3)丟到B箱,再從B箱抽2W(p=1)丟回A箱,機率=1/3; 情形2:A箱抽到1R1W(p=2/3)丟到B箱,再從B箱抽1R1W(p=2/3)丟回A箱,機率=4/9; 兩情況機率和=1/3+4/9=7/9
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