高考數(shù)學(xué)離奇題目及答案

一、單項(xiàng)選擇題1.已知集合\(A=\{x|x^2-5x+6=0\}\)，\(B=\{x|mx-1=0\}\)，若\(B\subseteqA\)，則實(shí)數(shù)\(m\)組成的集合為（）A.\(\{\frac{1}{2},\frac{1}{3}\}\)B.\(\{\frac{1}{2},0,\frac{1}{3}\}\)C.\(\{\frac{1}{2},\frac{1}{3},0\}\)D.\(\{\frac{1}{3},0\}\)答案：B2.函數(shù)\(y=\log_2(x^2-4x+3)\)的單調(diào)遞減區(qū)間是（）A.\((-\infty,1)\)B.\((3,+\infty)\)C.\((-\infty,2)\)D.\((2,+\infty)\)答案：A3.已知向量\(\vec{a}=(1,2)\)，\(\vec=(m,-1)\)，若\(\vec{a}\parallel(\vec{a}+\vec)\)，則\(m\)的值為（）A.\(-\frac{1}{2}\)B.\(\frac{1}{2}\)C.\(2\)D.\(-2\)答案：A4.若\(\sin\alpha=\frac{3}{5}\)，\(\alpha\)是第二象限角，則\(\tan2\alpha\)的值為（）A.\(\frac{24}{7}\)B.\(-\frac{24}{7}\)C.\(\frac{7}{24}\)D.\(-\frac{7}{24}\)答案：B5.已知等差數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項(xiàng)和為\(S_n\)，若\(a_3+a_7=10\)，則\(S_9\)的值為（）A.\(45\)B.\(50\)C.\(90\)D.\(100\)答案：A6.雙曲線\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1(a\gt0,b\gt0)\)的一條漸近線方程為\(y=\frac{4}{3}x\)，則雙曲線的離心率為（）A.\(\frac{5}{3}\)B.\(\frac{4}{3}\)C.\(\frac{5}{4}\)D.\(\frac{3}{2}\)答案：A7.已知\(f(x)\)是定義在\(R\)上的奇函數(shù)，當(dāng)\(x\geq0\)時(shí)，\(f(x)=x^2-2x\)，則\(f(-1)\)的值為（）A.\(-3\)B.\(-1\)C.\(1\)D.\(3\)答案：C8.從\(1\)，\(2\)，\(3\)，\(4\)，\(5\)這\(5\)個(gè)數(shù)字中任取\(3\)個(gè)數(shù)字組成無重復(fù)數(shù)字的三位數(shù)，則這個(gè)三位數(shù)是偶數(shù)的概率為（）A.\(\frac{1}{5}\)B.\(\frac{2}{5}\)C.\(\frac{3}{5}\)D.\(\frac{4}{5}\)答案：B9.若函數(shù)\(y=f(x)\)的圖象上存在兩點(diǎn)，使得函數(shù)的圖象在這兩點(diǎn)處的切線互相垂直，則稱\(y=f(x)\)具有\(zhòng)(T\)性質(zhì)。下列函數(shù)中具有\(zhòng)(T\)性質(zhì)的是（）A.\(y=\sinx\)B.\(y=\lnx\)C.\(y=e^x\)D.\(y=x^3\)答案：A10.已知\(a=\log_32\)，\(b=\log_52\)，\(c=0.5^{a-1}\)，則（）A.\(a\ltc\ltb\)B.\(b\ltc\lta\)C.\(b\lta\ltc\)D.\(c\ltb\lta\)答案：C二、多項(xiàng)選擇題1.下列說法正確的是（）A.若\(a\gtb\)，\(c\gtd\)，則\(ac\gtbd\)B.若\(a\gtb\)，則\(a^3\gtb^3\)C.若\(a\gtb\gt0\)，\(m\gt0\)，則\(\frac{b+m}{a+m}\gt\frac{a}\)D.若\(a\gtb\)，\(c\gtd\)，則\(a-d\gtb-c\)答案：BCD2.已知函數(shù)\(f(x)=\sin(2x+\varphi)(0\lt\varphi\lt\pi)\)的圖象關(guān)于點(diǎn)\((\frac{\pi}{6},0)\)對稱，則下列選項(xiàng)正確的是（）A.\(\varphi=\frac{\pi}{3}\)B.\(f(x)\)在\([0,\frac{\pi}{6}]\)上單調(diào)遞減C.\(f(x)\)的圖象向左平移\(\frac{\pi}{6}\)個(gè)單位長度得到的圖象對應(yīng)的函數(shù)是奇函數(shù)D.\(f(x)\)在\([0,\frac{\pi}{2}]\)上的最小值為\(-\frac{\sqrt{3}}{2}\)答案：BCD3.已知拋物線\(y^2=2px(p\gt0)\)的焦點(diǎn)為\(F\)，準(zhǔn)線為\(l\)，過點(diǎn)\(F\)的直線與拋物線交于\(A\)，\(B\)兩點(diǎn)，過\(A\)，\(B\)兩點(diǎn)分別作準(zhǔn)線\(l\)的垂線，垂足分別為\(C\)，\(D\)，則（）A.\(AC\perpBD\)B.梯形\(ABDC\)的面積的最小值為\(p^2\)C.若\(|AF|=2|BF|\)，則直線\(AB\)的斜率為\(\pm2\sqrt{2}\)D.\(\overrightarrow{FA}\cdot\overrightarrow{FB}\lt0\)答案：CD4.已知函數(shù)\(f(x)=\begin{cases}x+1,x\leq0\\\lnx,x\gt0\end{cases}\)，則下列說法正確的是（）A.\(f(x)\)在\(R\)上單調(diào)遞增B.\(f(x)\)的值域?yàn)閈(R\)C.若\(f(x)=m\)有兩個(gè)不等實(shí)根，則\(m\leq1\)D.方程\(f(x)-\frac{1}{x}=0\)有兩個(gè)不同的解答案：BD5.已知\(a\)，\(b\)，\(c\)分別為\(\triangleABC\)內(nèi)角\(A\)，\(B\)，\(C\)的對邊，\(b^2=ac\)，且\(c=2a\)，則\(\cosB\)的值為（）A.\(\frac{1}{4}\)B.\(\frac{3}{4}\)C.\(\frac{\sqrt{2}}{4}\)D.\(\frac{\sqrt{15}}{4}\)答案：B6.已知\(\{a_n\}\)是等比數(shù)列，公比\(q\gt1\)，前\(n\)項(xiàng)和為\(S_n\)，若\(a_2+a_3=12\)，\(a_1a_4=27\)，則下列說法正確的是（）A.\(q=3\)B.\(a_n=3^{n-1}\)C.\(S_n=\frac{3^n-1}{2}\)D.\(\{a_n\}\)的前\(n\)項(xiàng)積\(T_n\)的最大值為\(3\)答案：ABC7.已知\(x\)，\(y\)滿足約束條件\(\begin{cases}x+y\geq2\\x-y\leq2\\y\leq2\end{cases}\)，則（）A.\(z=3x-y\)的最大值為\(6\)B.\(z=x+2y\)的最小值為\(2\)C.\(z=x^2+y^2\)的最大值為\(8\)D.\(z=\frac{y}{x}\)的最小值為\(-1\)答案：ABC8.已知函數(shù)\(f(x)=x^3-3x^2+2\)，則（）A.\(f(x)\)在\((-\infty,0)\)上單調(diào)遞增B.\(f(x)\)在\((0,2)\)上單調(diào)遞減C.\(f(x)\)的極大值為\(2\)D.\(f(x)\)的極小值為\(-2\)答案：ABD9.已知圓\(C\)：\((x-1)^2+(y-2)^2=25\)，直線\(l\)：\((2m+1)x+(m+1)y-7m-4=0\)，則（）A.直線\(l\)恒過定點(diǎn)\((3,1)\)B.直線\(l\)與圓\(C\)相交C.當(dāng)直線\(l\)被圓\(C\)截得的弦長最短時(shí)，直線\(l\)的方程為\(2x-y-5=0\)D.直線\(l\)被圓\(C\)截得的弦長最長時(shí)，直線\(l\)的方程為\(x+y-4=0\)答案：ABC10.已知函數(shù)\(f(x)\)是定義在\(R\)上的偶函數(shù)，且當(dāng)\(x\geq0\)時(shí)，\(f(x)=x^2-2x\)，則（）A.\(f(x)\)的解析式為\(f(x)=x^2-2|x|\)B.\(f(x)\)在\((-\infty,-1)\)上單調(diào)遞增C.不等式\(f(x)\lt3\)的解集為\((-3,3)\)D.方程\(f(x)-\log_2|x|=0\)有\(zhòng)(4\)個(gè)不同的實(shí)根答案：ABCD三、判斷題1.若\(a\gtb\)，則\(\frac{1}{a}\lt\frac{1}\)。（×）2.函數(shù)\(y=\sin^2x\)的最小正周期為\(\pi\)。（√）3.向量\(\vec{a}=(1,2)\)與向量\(\vec=(2,4)\)共線。（√）4.若直線\(l_1\)：\(A_1x+B_1y+C_1=0\)與直線\(l_2\)：\(A_2x+B_2y+C_2=0\)垂直，則\(A_1A_2+B_1B_2=0\)。（√）5.函數(shù)\(y=\log_a(x+1)(a\gt0,a\neq1)\)的圖象恒過定點(diǎn)\((0,0)\)。（√）6.若等比數(shù)列\(zhòng)(\{a_n\}\)的公比\(q\gt1\)，則數(shù)列\(zhòng)(\{a_n\}\)是遞增數(shù)列。（×）7.雙曲線\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1(a\gt0,b\gt0)\)的漸近線方程為\(y=\pm\frac{a}x\)。（√）8.若\(x\gt0\)，\(y\gt0\)，且\(x+y=1\)，則\(\frac{1}{x}+\frac{1}{y}\)的最小值為\(4\)。（√）9.函數(shù)\(y=2\sin(2x+\frac{\pi}{3})\)的圖象可以由函數(shù)\(y=2\sin2x\)的圖象向左平移\(\frac{\pi}{3}\)個(gè)單位長度得到。（×）10.已知\(f(x)\)是定義在\(R\)上的奇函數(shù)，若\(f(x)\)在\((0,+\infty)\)上單調(diào)遞增，則\(f(x)\)在\((-\infty,0)\)上也單調(diào)遞增。（√）四、簡答題1.已知函數(shù)\(f(x)=x^2-2x+3\)，求\(f(x)\)在區(qū)間\([0,3]\)上的最值。答案：對\(f(x)=x^2-2x+3\)進(jìn)行配方可得\(f(x)=(x-1)^2+2\)。對稱軸為\(x=1\)，開口向上。當(dāng)\(x=1\)時(shí)，\(f(x)\)取得最小值\(f(1)=2\)。\(f(0)=3\)，\(f(3)=9-6+3=6\)，比較\(f(0)\)與\(f(3)\)，\(f(3)\)更大，所以\(f(x)\)在\([0,3]\)上的最大值為\(6\)。2.已知\(\tan\alpha=2\)，求\(\frac{\sin\alpha-3\cos\alpha}{\sin\alpha+\cos\alpha}\)的值。答案：因?yàn)閈(\tan\alpha=2\)，即\(\frac{\sin\alpha}{\cos\alpha}=2\)，所以\(\sin\alpha=2\cos\alpha\)。將\(\sin\alpha=2\cos\alpha\)代入\(\frac{\sin\alpha-3\cos\alpha}{\sin\alpha+\cos\alpha}\)可得：\(\frac{2\cos\alpha-3\cos\alpha}{2\cos\alpha+\cos\alpha}=\frac{-\cos\alpha}{3\cos\alpha}=-\frac{1}{3}\)。3.已知等差數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項(xiàng)和為\(S_n\)，\(a_3=5\)，\(S_6=36\)，求數(shù)列\(zhòng)(\{a_n\}\)的通項(xiàng)公式。答案：設(shè)等差數(shù)列\(zhòng)(\{a_n\}\)的首項(xiàng)為\(a_1\)，公差為\(d\)。由\(a_3=5\)可得\(a_1+2d=5\)；由\(S_6=36\)可得\(6a_1+\frac{6\times5}{2}d=36\)，即\(6a_1+15d=36\)。將\(a_1=5-2d\)代入\(6a_1+15d=36\)，得\(6(5-2d)+