2025年日本數(shù)學(xué)奧林匹克（JMO）模擬試卷：代數(shù)方程與幾何變換競賽技巧與解析一、選擇題（共10題，每題3分，共30分）1.設(shè)函數(shù)$f(x)=x^3-3x+2$，則$f(x)$的零點(diǎn)個(gè)數(shù)是（）A.1B.2C.3D.無法確定2.已知等差數(shù)列$\{a_n\}$的公差$d=2$，且$a_1=3$，則$a_{10}=（）$A.17B.19C.21D.233.在$\triangleABC$中，若$AB=AC=4$，$\angleBAC=60^\circ$，則$\triangleABC$的面積是（）A.$4\sqrt{3}$B.$8\sqrt{3}$C.$16\sqrt{3}$D.$32\sqrt{3}$4.設(shè)$a,b,c$為等比數(shù)列的前三項(xiàng)，若$a+b+c=6$，$ab+bc+ac=12$，則$a^3+b^3+c^3=（）$A.27B.36C.48D.545.在直角坐標(biāo)系中，點(diǎn)$A(2,3)$關(guān)于直線$y=x$的對稱點(diǎn)$B$的坐標(biāo)是（）A.$(3,2)$B.$(2,3)$C.$(3,3)$D.$(2,2)$6.若$\log_2(3x-2)=3\log_2(2-x)$，則$x$的取值范圍是（）A.$1<x<2$B.$2<x<3$C.$3<x<4$D.$4<x<5$7.設(shè)$a,b$是方程$x^2-2x+a=0$的兩個(gè)實(shí)數(shù)根，則$ab$的取值范圍是（）A.$[0,1]$B.$[1,2]$C.$[2,3]$D.$[3,4]$8.已知函數(shù)$f(x)=x^2-2x+1$，則$f(-x)+f(x)=（）$A.2B.4C.6D.89.在$\triangleABC$中，若$AB=AC=5$，$BC=7$，則$\sinB$的取值范圍是（）A.$(0,\frac{\sqrt{3}}{2})$B.$(\frac{\sqrt{3}}{2},1)$C.$(\frac{1}{2},\frac{\sqrt{3}}{2})$D.$(1,\frac{2\sqrt{3}}{3})$10.若方程$ax^2+bx+c=0$的兩根為$m$和$n$，則$\frac{1}{m}+\frac{1}{n}=\frac{c}$的充分必要條件是（）A.$a\neq0$B.$b^2-4ac\geq0$C.$b^2-4ac=0$D.$b^2-4ac\leq0$二、填空題（共10題，每題4分，共40分）1.若方程$3x^2-2x-1=0$的兩根分別為$m$和$n$，則$m+n=\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\四、解答題（共2題，每題20分，共40分）4.已知函數(shù)$f(x)=x^3-6x^2+9x-1$，求證：對于任意實(shí)數(shù)$x$，都有$f(x)\geq-1$。五、解答題（共2題，每題20分，共40分）5.在平面直角坐標(biāo)系中，點(diǎn)$A(1,2)$，$B(3,4)$，$C(x,y)$。若$\triangleABC$是等腰三角形，且$AB=BC$，求點(diǎn)$C$的軌跡方程。六、解答題（共2題，每題20分，共40分）6.已知數(shù)列$\{a_n\}$滿足遞推關(guān)系$a_{n+1}=\frac{1}{2}a_n+\frac{1}{3}$，且$a_1=1$。求證：數(shù)列$\{a_n\}$的通項(xiàng)公式為$a_n=\frac{2}{3}-\frac{1}{3}\cdot\left(\frac{1}{2}\right)^{n-1}$。本次試卷答案如下：一、選擇題1.B解析：函數(shù)$f(x)=x^3-3x+2$的導(dǎo)數(shù)為$f'(x)=3x^2-3$，令$f'(x)=0$得$x=\pm1$，因此$f(x)$在$x=-1$和$x=1$處有極值。由于$f(-1)=2$，$f(1)=0$，且$f(x)$在$x\to\pm\infty$時(shí)趨于$\pm\infty$，故$f(x)$有兩個(gè)零點(diǎn)。2.A解析：等差數(shù)列的通項(xiàng)公式為$a_n=a_1+(n-1)d$，代入$a_1=3$，$d=2$，$n=10$，得$a_{10}=3+(10-1)\cdot2=19$。3.A解析：由海倫公式，$\triangleABC$的面積$S=\sqrt{s(s-a)(s-b)(s-c)}$，其中$s=\frac{a+b+c}{2}$。代入$AB=AC=4$，$BC=7$，得$s=\frac{4+4+7}{2}=7.5$，$S=\sqrt{7.5(7.5-4)(7.5-4)(7.5-7)}=4\sqrt{3}$。4.C解析：由等比數(shù)列的性質(zhì)，$a^2=bc$，$b^2=ac$，$c^2=ab$。代入$a+b+c=6$，$ab+bc+ac=12$，得$a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)=6(36-12)=144$。5.A解析：點(diǎn)$A(2,3)$關(guān)于直線$y=x$的對稱點(diǎn)$B$的坐標(biāo)為$(3,2)$，因?yàn)閷τ谌我庖稽c(diǎn)$(x,y)$，它關(guān)于直線$y=x$的對稱點(diǎn)坐標(biāo)為$(y,x)$。6.B解析：由對數(shù)函數(shù)的性質(zhì)，$\log_2(3x-2)=3\log_2(2-x)$可化為$3x-2=(2-x)^3$，解得$x=\frac{4}{5}$。7.A解析：由韋達(dá)定理，$a+b=2$，$ab=a$，因此$ab$的取值范圍是$[0,1]$。8.B解析：$f(-x)+f(x)=(-x)^2-2(-x)+1+x^2-2x+1=2$。9.C解析：由余弦定理，$\cosB=\frac{AB^2+BC^2-AC^2}{2\cdotAB\cdotBC}=\frac{25+49-25}{2\cdot5\cdot7}=\frac{1}{2}$，因此$\sinB=\sqrt{1-\cos^2B}=\frac{\sqrt{3}}{2}$。10.A解析：由韋達(dá)定理，$m+n=-\frac{a}$，$mn=\frac{c}{a}$，因此$\frac{1}{m}+\frac{1}{n}=\frac{m+n}{mn}=-\frac{c}$，所以$a\neq0$是充分必要條件。二、填空題1.2解析：由韋達(dá)定理，$m+n=-\frac{-2}{3}=2$。2.19解析：等差數(shù)列的通項(xiàng)公式$a_n=a_1+(n-1)d$，代入$a_1=3$，$d=2$，$n=10$，得$a_{10}=3+(10-1)\cdot2=19$。3.$4\sqrt{3}$解析：由海倫公式，$\triangleABC$的面積$S=\sqrt{s(s-a)(s-b)(s-c)}$，其中$s=\frac{a+b+c}{2}$。代入$AB=AC=4$，$BC=7$，得$s=\frac{4+4+7}{2}=7.5$，$S=\sqrt{7.5(7.5-4)(7.5-4)(7.5-7)}=4\sqrt{3}$。4.144解析：由等比數(shù)列的性質(zhì)，$a^2=bc$，$b^2=ac$，$c^2=ab$。代入$a+b+c=6$，$ab+bc+ac=12$，得$a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)=6(36-12)=144$。5.$(3,2)$解析：點(diǎn)$A(2,3)$關(guān)于直線$y=x$的對稱點(diǎn)$B$的坐標(biāo)為$(3,2)$，因?yàn)閷τ谌我庖稽c(diǎn)$(x,y)$，它關(guān)于直線$y=x$的對稱點(diǎn)坐標(biāo)為$(y,x)$。6.$\frac{4}{5}$解析：由對數(shù)函數(shù)的性質(zhì)，$\log_2(3x-2)=3\log_2(2-x)$可化為$3x-2=(2-x)^3$，解得$x=\frac{4}{5}$。7.[0,1]解析：由韋達(dá)定理，$a+b=2$，$ab=a$，因此$ab$的取值范圍是$[0,1]$。8.2解析：$f(-x)+f(x)=(-x)^2-2(-x)+1+x^2-2x+1=2$。9.$\frac{\sqrt{3}}{