金門縣113學年度國民中學正式教師聯合甄選
解答:三根之和=a+b+c=4⇒log1/2(a+b+c)=log1/24=−2,故選(D)
解答:x2+4x+3=x+k⇒x2+3x+3−k=0交點只有一個⇒判別式9−4(3−k)=0⇒k=34⇒x2+3x+94=0⇒x=−32⇒y=x+k=−32+34=−34⇒交點為(−32,−34),故選(C)
解答:cosπ17⋅cos13π17cos3π17+cos5π17=12⋅cos14π17+cos12π17cos3π17+cos5π17=12⋅−cos3π17−cos15π17cos3π17+cos5π17=−12,故選(A)
解答:f(k)=k2⇒(k+3)(k−2)+7=k2⇒k+1=0⇒k=−1,故選(C)
解答:P在y=x2上⇒P(t,t2),t∈R⇒P至(0,1)的距離=√t2+(t2−1)2=√t4−t2+1=√(t2−12)2+34⇒當t2=12時,P至(0,1)的距離最近,此時{P(1/√2,1/2)Q(−1/√2,1/2)⇒¯PQ=√2,故選(B)
解答:{抽中白球的機率=11/35抽中綠球的機率=6/35⇒1135+635=1735,故選(C)
解答:σ(X)=σ(X+50)=25,故選((B)
解答:A=[cos60∘−sin60∘sin60∘cos60∘]為旋轉60度矩陣⇒A6=I,故選(C)
解答:f(θ)=1+tanθsecθ⇒f′(θ)=sec2θsecθ−(1+tanθ)secθtanθsec2θ=secθ−(1+tanθ)tanθsecθ⇒f′(π/4)=√2−2√2=0,故選(A)
解答:∫80√x+1dx=∫91√udu=[23u3/2]|91=23(27−1)=523,故選(A)
解答:{A(1,7,1)B(4,7,1)C(1,7,5)⇒{¯AB=3¯BC=5¯CA=4⇒內心坐標=5A+4B+3C5+4+3=112(5+16+3,35+28+21,5+4+15)=112(24,84,24)=(2,7,2),故選(B)
解答:x+y+3i=−4+xyi⇒{x+y=−4xy=3⇒x,y為a2+4a+3=0的兩根⇒(x,y)=(−3,−1),(−1,−3)⇒(√x−√y)2=((√3−1)i)2=−(4−2√3)=−4+2√3,故選(D)
解答:全部−六人同船=36−3=729−3=726,故選(B)
解答:limx→1x2+2x−3x−1=limx→1(x+3)(x−1)x−1=limx→1(x+3)=4,故選(D)
解答:11×4+12×5+13×6+⋯+1n×(n+3)+⋯=13(11−14+12−15+13−16+14−17+⋯+1n−1n+3+⋯)=13(11+12+13)=1118,故選(B)
解答:
兩圖形{y=x3/2y=2x交點{A(0,0)B(2,4)C(−2,−4)且皆為奇函數,因此所圍面積=2∫20(2x−x32)dx=2[x2−18x4]|20=2⋅2=4,故選(A)
解答:f(x)=x3+ax2+bx−3⇒f′(x)=3x2+2ax+b=3(x−1)(x+2)=3x2+3x−6⇒{a=3/2b=−6⇒a×b=−9,故選(A)
解答:f(x)=x2+x⇒f′(x)=2x+1⇒假設切點P(t,t2+t)⇒切線斜率2t+1=t2+t−1t−1⇒t2−2t=0⇒{t=0t=2⇒{P(0,0)P(2,6)⇒{切線¯PQ方程式:x=y切線¯PQ方程式:5x−y=4,故選(B)
解答:x+1(x+1)+1(x−1)+1x+1=x+2x+3⇒(x+1)(x+3)=(x+1)(x+2)+x+2(x−1)+1x+1=(x+1)(x+2)+x+2x2x+1⇒x+3=(1x2+1)(x+2)⇒x2−x−2=0⇒兩根之和=1,故選(A)
解答:{3x2+2xy+y2−2y−1=0⋯(1)x2+2xy−y+1=0⋯(2)⇒1=3x2+2xy+y2−2y=y−x2−2xy⇒4x2+4xy+y2=3y⇒y=13(2x+y)2代入(2)⇒(x−y)2=3⇒{x=y−√3⇒(y−√3)2+2(y−√3)y−y+1=0⇒3y2−(4√3+1)y+4=0⇒有二解x=y+√3⇒(y+√3)2+2(y+√3)y−y+1=0⇒3y2+(4√3−1)y+4=0⇒無實數解⇒有兩組解,故選(C),但公布的答案是(A)經電腦繪圖,兩圖形只有兩個交點,如下:
解答:f(x)=x3+ax2+bx−3⇒f′(x)=3x2+2ax+b=3(x−1)(x+2)=3x2+3x−6⇒{a=3/2b=−6⇒a×b=−9,故選(A)
解答:f(x)=x2+x⇒f′(x)=2x+1⇒假設切點P(t,t2+t)⇒切線斜率2t+1=t2+t−1t−1⇒t2−2t=0⇒{t=0t=2⇒{P(0,0)P(2,6)⇒{切線¯PQ方程式:x=y切線¯PQ方程式:5x−y=4,故選(B)
解答:x+1(x+1)+1(x−1)+1x+1=x+2x+3⇒(x+1)(x+3)=(x+1)(x+2)+x+2(x−1)+1x+1=(x+1)(x+2)+x+2x2x+1⇒x+3=(1x2+1)(x+2)⇒x2−x−2=0⇒兩根之和=1,故選(A)
解答:{3x2+2xy+y2−2y−1=0⋯(1)x2+2xy−y+1=0⋯(2)⇒1=3x2+2xy+y2−2y=y−x2−2xy⇒4x2+4xy+y2=3y⇒y=13(2x+y)2代入(2)⇒(x−y)2=3⇒{x=y−√3⇒(y−√3)2+2(y−√3)y−y+1=0⇒3y2−(4√3+1)y+4=0⇒有二解x=y+√3⇒(y+√3)2+2(y+√3)y−y+1=0⇒3y2+(4√3−1)y+4=0⇒無實數解⇒有兩組解,故選(C),但公布的答案是(A)經電腦繪圖,兩圖形只有兩個交點,如下:
解答:(√2+√3−√2−√3)2=4−2=2⇒(√2+√3−√2−√3)12=26=82⇒log8(√2+√3−√2−√3)12=log882=2,故選(C)
解答:{tanA=5/6cotB=11⇒{sinA=5/√61cosA=6/√61sinB=1/√122cosB=11/√122⇒sin(A+B)=sinAcosB+sinBcosA=55√61⋅√122+6√122⋅√61=1√2⇒A+B=45∘,故選(C)
解答:sin(sin−145+sin−1513)=45⋅1213+35⋅513=6365⇒cos(sin−145+sin−1513)=1665⇒sin(sin−11665)=cos(sin−145+sin−1513)=1665⇒sin−145+sin−1513+sin−11665=π2,故選(A)
解答:橢圓:(x−1)2+4(y−32)2=4⇒P(2cosθ+1,sinθ+32)⇒(2,2)為¯PQ中點⇒Q(−2cosθ+3,−sinθ+52)⇒(−2cosθ+2)2+4(−sinθ+1)2=4⇒cosθ+sinθ=1⇒θ=π2,2π⇒¯PQ斜率=2sinθ−14cosθ−2=−12,故選(C)
解答:M∈L:x−y+2=0⇒M(t,t+2)⇒¯MA−¯MB=f(t)=√(t−2)2+(t+3)2−√(t+1)2+(t−3)2⇒f′(t)=0⇒4t+2√2t2+2t+13=4t−4√2t2−4t+10⇒9t2−66t+21=0⇒(t−7)(9t−3)=0⇒{t=7⇒M(7,9)t=1/3⇒M(1/3,7/3)=¯AB∩L⇒f(t)為極小值⇒a+b=7+9=16,故選(D)
解答:x=12(√11−3)⇒{2x2=10−3√114x4=199−60√11⇒x⋅4x4+3⋅4x4+x⋅2x2+8⋅2x2+10x+5=(x+3)4x4+(x+8)2x2+19x+5=(x+3)(199−60√11)+(x+8)(19−3√11)+10x+5=(219−63√11)x+682−204√11=12(219−63√11)(√11−3)+682−204√11=7,故選(D)
解答:直接取c=0⇒{ab=−1a−3b=2√3⇒{a=√3b=−1/√3⇒ab+c=−1,故選(B)
解答:依題意公差1,假設首項a⇒an=a+n−1⇒S=n∑i=1ai=(2a+n−1)n2=1000⇒(2a+n−1)n=2000=2453⇒(2a+n−1)與n必須是2,5的倍數⇒{a=28n=25⇒ara1=5228=137⇒a+b=20,故選(D)
解答:t=ax+a−x⇒t2=a2x+a−2x+2⇒a2x+a−2x=t2−2又a3x+a−3x=(ax+a−x)(a2x+a−2x−1)=t(t2−3)=18⇒t3−3t−18=0⇒(t−3)(t2+3t+6)=0⇒t=3⇒a2x+a−2x=32−2=7,故選(A)
解答:√12+√128=√12+2√32=√8+√4=2√2+2⇒{x=4y=2√2−2⇒√y+2+√4y+y2y+2−√4y+y2=√2√2+√42√2−√4=√2√2+22√2−2=√12+8√24=√3+2√2=√2+1,故選(C)
解答:cosB=12=22+62−¯AC22⋅2⋅6⇒¯AC=2√7¯BD是∠的角平分線⇒¯AD¯DC=¯AB¯BC=26⇒¯AD=√72⇒cos∠ABD=cos30∘=√32=22+¯BD2−744¯BD⇒¯BD=3√32≈2.6,故選(B)
解答:200∑i=5ai=200∑i=5⌊logi⌋=9∑i=5⌊logi⌋+99∑i=10⌊logi⌋+200∑i=100⌊logi⌋=0+90×1+101×2=292,故無解,公布的答案是(D)
解答:y=3−sinx2+cosx⇒2y+ycosx=3−sinx⇒ycosx+sinx=3−2y⇒√y+1sin(x+α)=3−2y⇒sin(x+α)=3−2y√y+1⇒|3−2y√y+1|≤1⇒3y2−12y+8≤0⇒(y−2)2≤43⇒2−2√3≤y≤2+2√3⇒{a=2+2√3b=2−2√3⇒a+b=4,故選(A)
解答:用 Lagrange 算子來求:令F(a,b,c,λ,μ)=b2−4b+3−λ(a+2b+4c)−μ(a2+4b2+16c2−6)⇒{Fa=−λ−2aμ=0Fb=2b−4−2λ−8bμ=0Fc=−4λ−32μc=0⇒λ=−2μa=−8μc⇒a=4c⇒4c+2b+4c=0⇒c=−14b⇒{a=−bc=−b/4⇒(−b)2+4b2+16(−b/4)2=6⇒b=±1⇒{b=1⇒b2−4b+3=0b=−1⇒b2−4b+3=8⇒最小值=0,故選(B)
解答:{A(3,−1)B(2,5)C(−1,3)⇒{LAB:6x+y=17LBC:−2x+3y=11LAC:x+y=2⇒△ABC面積=12|3−11251−131|=10⇒{S△MAB=15/3S△MBC=10/3S△MCA=5/3⇒{‖3−11251xy1‖=|−6x−y+17|=10‖251−131xy1‖=|−2x+3y−11|=20/3‖3−11−131xy1‖=|4x+4y−8|=10/3⇒{x=5/6y=2,故選(C)
解答:{|→a|=1,|→b|=2,|→c|=3→a+→b+→c=0⇒{→b=2→a→c=−3→a⇒|2→a+3→b+5→c|=|2→a+6→a−15→a|=|−7→a|=7,故選(A)
解答:√(x−6)2+52+√(x+2)2+1=¯PQ+¯PR,其中{P(x,0)Q(6,5)R(−2,1)⇒R對稱x軸的對稱點R′(−2,−1)⇒L=↔QR′:4y=3x+2⇒P=L∩x軸=P(−23,0)⇒s=¯QR′=10⇒6r+s=−4+10=6,故選(B)
解答:{x+ky=10−2k⋯(1)kx−y=10k⋯(2)⇒(1)×k⇒kx+k2y=10k−2k2⋯(3)由(2)⇒kx=10k+y代入(3)⇒10k+y+k2y=10k−2k2⇒y=−2k2k2+1⇒y=−2+2k2+1為整數⇒k=0,±1,有三種可能,故選(C)
解答:M=abcde,其中{a為萬位數字b為千位數字c為百位數字d為十位數字e為個位數字,需滿足1≤a≤9,0≤b,c,d,e≤9⇒{a=9⇒H43a=8⇒H44⋯a=3⇒H49a=2⇒H410−4(b,c,d,e=10四種情形)a=1⇒H411−4−12(b,c,d,e=11,或(10,1,0,0)排列數⇒共有(11∑k=3H4k)−4−4−12=(1611∑k=3(k+1)(k+2)(k+3))−20=1350−20=1330,故選(A)
解答:112=121⇒113=1210+121=1331只考慮後三位→331⇒3310+331=3641⇒6410+641=7051⇒0510+051=561⇒5610+561=6171⇒1710+171=1881⇒8810+881=9691⇒6910+691=7601⇒601⇒6+0+1=7,故選(C)
解答:32+3+42+4−3+42+3+4=2245⇒b−a=45−22=23,故選(B)
解答:
¯BQ為∠B的角平分線⇒¯AB¯BC=¯AQ¯QC⇒35=2¯CQ⇒¯CQ=103同理可得¯BP=95,¯AR=4831又¯AP為∠A的角平分線⇒cos∠BAP=cos∠CAP⇒9+¯AP2−81256¯AP=2569+¯AP2−25625323¯AP⇒¯AP=165,又¯AM¯MP=¯AB¯BP=39/5=159⇒¯AM=1524⋅¯AP=1524⋅165=2,故選(C)
解答:
解答:
假設\cases{\overline{AB}=1\\ \overline{BP}=a\\ \overline{DQ}=b}, 由於\angle PAQ=45^\circ \Rightarrow \overline{PQ}=a+b \; \href{https://zhidao.baidu.com/question/193097272.html}{參考資料} \\ {ABCD\over \triangle APQ}={5\over 2} \Rightarrow S_{\triangle APQ}={2\over 5} \Rightarrow S_{\triangle ABP}+ S_{\triangle ADQ}+ S_{\triangle PQC}= {3\over 5} \\ \Rightarrow {1\over 2}(a+b+(1-a)(1-b))={1\over 2}(1+ab)={3\over 5} \Rightarrow ab={1\over 5}\\ 又\overline{PQ}^2=\overline{PC}^2 +\overline{QC}^2 \Rightarrow (a+ b)^2=(1-a)^2+(1-b)^2 \Rightarrow 2ab=-2(a+b)+2 \\ \Rightarrow a+b=1-{1\over 5}={4\over 5} \Rightarrow {\overline{AB} \over \overline{PQ}}={1\over 4/5}={5\over 4},故選\bbox[red, 2pt]{(B)}
解答:
解答:
\triangle ADF \cong \triangle ABE \;(RHS) \Rightarrow \overline{BE}=\overline{DF}=a \Rightarrow \overline{CE}=\overline{CF}= 1-a \\ \Rightarrow \cases{\overline{AE}^2= 1+a^2\\ \overline{EF}^2=2(1-a)^2} \Rightarrow \overline{AE}=\overline{EF} \Rightarrow 1+a^2 =2(1-a)^2 \Rightarrow a^2-4a+1=0 \\ \Rightarrow a=2-\sqrt 3 \;(a=2+\sqrt 3\gt 1不合) \Rightarrow \overline{AE}=\sqrt{1+a^2} =\sqrt{8-4\sqrt 3} =\sqrt{8-2\sqrt{12}} \\=\sqrt 6-\sqrt 2,故選\bbox[red, 2pt]{(B)}
解答:
解答:
\angle B= \angle D=90^\circ \Rightarrow ABCD共圓,因此假設\cases{ 圓心O\\ 半徑r} \Rightarrow \angle C=180^\circ- \angle A=120^\circ\\ \triangle BCD: \cos \angle C={\overline{BC}^2+\overline{CD}^2-\overline{BD}^2 \over 2\cdot \overline{BC} \cdot \overline{CD}} \Rightarrow \overline{BD}=\sqrt 7\\ \triangle OBD: \cos \angle DOB=\cos (2\angle A) = {2r^2-\overline{BD}^2\over 2r^2} \Rightarrow r=\sqrt{7 \over 3} \\ \Rightarrow \overline{AC}=2r= 2\sqrt{7\over 3} \Rightarrow \overline{AC}^2={28\over 3} \Rightarrow a+b=28+3=31,故選\bbox[red, 2pt]{(D)}
解答:x={1\over \sqrt 2-1} =\sqrt 2+1 \Rightarrow \cases{a=2\\ b=\sqrt 2-1} \Rightarrow -x=-\sqrt 2-1 \Rightarrow \cases{c=-a-1\\ d=2-\sqrt 2} \\ \Rightarrow b^3+d^3+3bd=13-9\sqrt 2-12+9\sqrt 2=1,故選\bbox[red, 2pt]{(B)}
解答:\cases{\sum_{i=1}^8 a_i=56\\ \sum_{i=2}^7 a_i=44} \Rightarrow a_1 +a_8=12 \Rightarrow \cases{(a_1,a_8)=(1,11) \Rightarrow 在2-10挑6個數總和為44 \\ (a_1,a_8)=(2,10) \Rightarrow 在3-9挑6個數總和為44 \\ (a_1,a_8)=(3,9) \times:4-8不足6個數} \\ \Rightarrow 1\lt 4\lt 6\lt 7\lt 8\lt 9\lt 10\lt 11 \Rightarrow \cases{a_2=4\\ a_7=10} \Rightarrow 2_2+a_7=18,故選\bbox[red, 2pt]{(A)}
解答:f(n)=a(2n+2)(2n+3)(2n+4) +b(2n+1)(2n+3)(2n+4) \\\qquad +c(2n+1)(2n+2)(2n+4)+ d(2n+1)(2n+2)(2n+3) =6 \\ \Rightarrow \cases{f(-1/2)=6a=6 \Rightarrow a=1\\ f(-1)=-2b=6 \Rightarrow b=-3 \\f(-3/2) =2c=6 \Rightarrow c=3\\ f(-2)=-6d=6 \Rightarrow d=-1} \Rightarrow 2a+b+2c+d=2-3+6-1=4,故選\bbox[red, 2pt]{(C)}
解答:S=\sum_{i=1}^{10} x_i \le 13 \Rightarrow S=10,11,12,13 \\ \Rightarrow \cases{S=10 \Rightarrow 10個1 \Rightarrow 1種情形\\ S=11 \Rightarrow 9個1,1個2 \Rightarrow 10種情形\\ S=12 \Rightarrow \cases{9個1,1個3 \Rightarrow 10種情形 \\ 8個1,2個2 \Rightarrow 45種情形} \\ S=13 \Rightarrow \cases{9個1,1個4\Rightarrow 10種情形\\ 8個1,1個2,1個3\Rightarrow 90種情形 \\ 7個1,3個2 \Rightarrow 120種情形}} \Rightarrow 合計286,故選\bbox[red, 2pt]{(C)}
解答:x={1\over \sqrt 2-1} =\sqrt 2+1 \Rightarrow \cases{a=2\\ b=\sqrt 2-1} \Rightarrow -x=-\sqrt 2-1 \Rightarrow \cases{c=-a-1\\ d=2-\sqrt 2} \\ \Rightarrow b^3+d^3+3bd=13-9\sqrt 2-12+9\sqrt 2=1,故選\bbox[red, 2pt]{(B)}
解答:\cases{\sum_{i=1}^8 a_i=56\\ \sum_{i=2}^7 a_i=44} \Rightarrow a_1 +a_8=12 \Rightarrow \cases{(a_1,a_8)=(1,11) \Rightarrow 在2-10挑6個數總和為44 \\ (a_1,a_8)=(2,10) \Rightarrow 在3-9挑6個數總和為44 \\ (a_1,a_8)=(3,9) \times:4-8不足6個數} \\ \Rightarrow 1\lt 4\lt 6\lt 7\lt 8\lt 9\lt 10\lt 11 \Rightarrow \cases{a_2=4\\ a_7=10} \Rightarrow 2_2+a_7=18,故選\bbox[red, 2pt]{(A)}
解答:f(n)=a(2n+2)(2n+3)(2n+4) +b(2n+1)(2n+3)(2n+4) \\\qquad +c(2n+1)(2n+2)(2n+4)+ d(2n+1)(2n+2)(2n+3) =6 \\ \Rightarrow \cases{f(-1/2)=6a=6 \Rightarrow a=1\\ f(-1)=-2b=6 \Rightarrow b=-3 \\f(-3/2) =2c=6 \Rightarrow c=3\\ f(-2)=-6d=6 \Rightarrow d=-1} \Rightarrow 2a+b+2c+d=2-3+6-1=4,故選\bbox[red, 2pt]{(C)}
解答:S=\sum_{i=1}^{10} x_i \le 13 \Rightarrow S=10,11,12,13 \\ \Rightarrow \cases{S=10 \Rightarrow 10個1 \Rightarrow 1種情形\\ S=11 \Rightarrow 9個1,1個2 \Rightarrow 10種情形\\ S=12 \Rightarrow \cases{9個1,1個3 \Rightarrow 10種情形 \\ 8個1,2個2 \Rightarrow 45種情形} \\ S=13 \Rightarrow \cases{9個1,1個4\Rightarrow 10種情形\\ 8個1,1個2,1個3\Rightarrow 90種情形 \\ 7個1,3個2 \Rightarrow 120種情形}} \Rightarrow 合計286,故選\bbox[red, 2pt]{(C)}
老師您好:第44題若解 ab=1/5及 a+b=4/5 算出來應該是無解的,謝謝老師
回覆刪除題目的確有疑慮,主要的問題在ABCD:APQ=5:2, 只能參考參考!!!
刪除