阅读小说网站建设,白羊影院,商务网站开发考题,专做坏消息的网站前言赋值法是高中数学中比较常用的一种方法#xff0c;使用“赋值法”的数学素材和知识点#xff0c;散落在高中数学的几乎各个章节中#xff0c;现对其进行整理#xff0c;以便于学习。比如学习函数时可以赋值法给出单调性#xff0c;奇偶性#xff0c;周期性等#xf…前言赋值法是高中数学中比较常用的一种方法使用“赋值法”的数学素材和知识点散落在高中数学的几乎各个章节中现对其进行整理以便于学习。比如学习函数时可以赋值法给出单调性奇偶性周期性等求函数的值。理论依据比如给定\(R\)上的函数\(f(x)\)满足条件\(f(xy)f(x)f(y)\)则我们就可以令\(x1y1\)得到\(f(2)f(1)f(1)2f(1)\)当然也可以令\(x1y0\)得到\(f(1)f(1)f(0)\)从而得到\(f(0)0\)实际上由于是\(R\)上的函数\(f(x)\)那么我们给\(x、y\)任意赋值都是合理的这样就能得到无穷个等式但是我们解题不需要这么多只需要有针对性的一两个所以我们其实是有目的的赋值的那么到底该怎么赋值呢这要结合具体题目来分析需要一定的数学素养。以下为涉及到的数学素材函数性质【单调性】如定义在\(R\)上的函数\(f(x)\)满足\(f(xy)f(x)f(y)\)且\(x 0\)时\(f(x)0\)判定函数单调性。分析令\(x_1 x_2\)则\(x_1-x_20\)故\(f(x_1-x_2)0\)则有$ f(x_1) f(x_1-x_2x_2) f(x_1-x_2)f (x_2) 故函数\(f(x)\)在\(R\)上单调递减。【奇偶性】已知函数\(f(x)\)满足\(f(1)\cfrac{1}{2}\)且\(f(xy)f(x-y)2f(x)f(y)\)判断函数的奇偶性分析令\(xy0\)则有\(2f(0)2f^2(0)\)得到\(f(0)0或f(0)1\)再令\(x1y0\)则有\(2f(1)2f(1)f(0)\)得到\(f(0)1\)又题目已知\(f(1)\cfrac{1}{2}\)若令\(x0\)则得到\(f(y)f(-y)2f(0)f(y)2f(y)\)所以\(f(-y)f(y)\)可知函数是偶函数。【周期性】比如已知定义在\(R\)上的函数满足\(f(x6)f(x)f(3)\)且函数\(f(x)\)是偶函数试判断函数的周期性。分析由于\(f(x6)f(x)f(3)\)令\(x-3\)则\(f(-36)f(-3)f(3)\)即\(f(-3)0\)又由于\(f(-x)f(x)\)则有\(f(-3)f(3)0\)那么原式变为\(f(x6)f(x)\)即\(T6\)引申已知\(f(x6)f(x)nf(3)(n\in N^*)\)再加上\(f(x)\)是偶函数\(\Longrightarrow T6\)提示用赋值法令\(x-3\)\(f(-36)f(-3)nf(3)\)推出\(f(3)0\)从而\(f(x6)f(x)\)故\(T6\)。【对称性】已知函数\(f(x)\cfrac{2^x}{1a\cdot 2^x}(a\in R)\)的图像关于点\((0\cfrac{1}{2})\)对称则\(a\)_______。分析由题目可知\(f(x)f(-x)1\)法1定义法对定义域内的任意\(x\)必须恒有\(f(x)f(-x)1\)由\((a-1)[2^{2x}(a-1)2^x1]0\)恒成立故必须\(a-10\)从而得到\(a1\)法2赋值法比如\(f(0)f(-0)1\)变形为\(\cfrac{2}{1a}1\)得到\(a1\)当然这个方法要注意定义域。函数求值【2019届高三理科函数的奇偶性周期性课时作业第13题】设定义在\(R\)上的函数\(f(x)\)同时满足以下条件①\(f(x)f(-x)0\)②\(f(x)f(x2)\)③当\(0\leq x1\)时\(f(x)2^x-1\)则\(f(\cfrac{1}{2})f(1)f(\cfrac{3}{2})f(2)f(\cfrac{5}{2})\)的值是_________。分析由①知函数为奇函数在利用③先做出\([01)\)上的图像再利用奇函数做出\((-10]\)上的图像一个周期基本完成就差端点值\(f(-1)\)和\(f(1)\)的值未确定难点是求\(f(1)\)的值可以通过以下几个思路求解法1图像法假设\(f(1)\cfrac{1}{2}\)则\(f(-1)-\cfrac{1}{2}\)奇偶性是说的通的但是周期性不满足因为向右平移一个周期后元素\(1\)对应\(\cfrac{1}{2}\)还对应\(-\cfrac{1}{2}\)出现了一对多不是函数了故只能有\(f(1)0\)即也有\(f(-1)0\)这样在一个周期上奇偶性和周期性都是满足的。法2题中没有明确告诉但是由①②可知\(f(x2)-f(-x)\)即\(f(x2)f(-x)0\)即对称中心是\((10)\)这时要么函数在\((10)\)处没有定义这个不满足题意要么必有\(f(1)0\)则\(f(-1)0\)其余就好处理了。法3赋值法由\(f(x)f(-x)0\)令\(x1\)得到\(f(1)f(-1)0\)①令\(x-1\)由\(f(x)f(x2)\)得到\(f(-1)f(1)\)②故有\(f(1)f(-1)0\)在此基础上做出函数的大致图像可知\(f(1)f(2)f(0)0\)\(f(\cfrac{3}{2})f(\cfrac{5}{2})0\)\(f(\cfrac{1}{2})\sqrt{2}-1\)故\(f(\cfrac{1}{2})f(1)f(\cfrac{3}{2})f(2)f(\cfrac{5}{2})\sqrt{2}-1\)。周期性给定\(f(x2)f(x)\)则可知函数\(f(x)\)的周期为\(T2\)那么给定\(f(x-1)f(x1)\)给其中的\(x\)赋值\(x1\)变换得到\(f(x)f(x2)\)也说明\(f(x-1)f(x1)\)刻画的是周期性且周期为\(T2\)练习①\(f(x1)f(x5)\)则\(T4\)说明给其中的\(x\)赋值\(x-1\)得到\(f(x)f(x4)\)则\(T4\)②\(f(2-x)f(4-x)\)则\(T2\)说明给其中的\(x\)赋值\(-x\)得到\(f(x2)f(x4)\)给其中的\(x\)赋值\(x-2\)得到\(f(x)f(x2)\)则\(T2\)③\(f(xa)f(xb)(a求解析式已知定义在\(R\)上的函数\(f(x)\)满足条件\(f(x-y)f(x)-y(2x-y1)\)且\(f(0)1\)求\(f(x)\)的解析式分析令\(yx\)代入原式得到\(f(x-x)f(x)-x(2x-x1)\)即\(f(0)f(x)-x(x1)\)即\(f(x)x^2x1\)函数\(f(x)\)对一切实数\(x、y\)均有\(f(xy)-f(y)x(x2y1)\)成立且\(f(1)0\)求函数\(f(x)\)的解析式分析注意到\(f(1)0\)故令\(y1\)代入原式得到\(f(x1)-f(1)x(x2\times 1)x^23x\)即\(f(x1)x^23x\)令\(x1t\)则\(xt-1\)代入上式得到\(f(t)(t-1)^23(t-1)t^2t-2\)即\(f(x)x^2x-2\)。已知函数\(f(x)x^22f(2)\cdot x1\)求函数的解析式\(f(x)\).分析给原式两边同时求导可得\(f(x)2x2f(2)\)再令\(x2\)得到\(f(2)42f(2)\)解得\(f(2)-4\)可知\(f(x)x^2-8x1\)。已知函数\(f(x)1f(\cfrac{1}{2})\cdot log_2x\)求函数\(f(x)\)的解析式及\(f(2)\)的值。分析令\(x\cfrac{1}{2}\)则\(f(\cfrac{1}{2})1f(\cfrac{1}{2})\cdot log_2\cfrac{1}{2}\)即\(f(\cfrac{1}{2})1-f(\cfrac{1}{2})\)解得\(f(\cfrac{1}{2})\cfrac{1}{2}\)故所求解析式为\(f(x)1\cfrac{1}{2}log_2x\)则\(f(2)1\cfrac{1}{2}\cfrac{3}{2}\)。通项公式如\(a_{nm}a_n\cdot a_m\)\(a_12\)求通项公式\(a_n\)令\(m1\)得到\(a_{n1}a_n\cdot a_1a_1\cdot a_n\)即\(\cfrac{a_{n1}}{a_n}a_12\)不就是等比数列嘛故\(a_n2\cdot 2^{n-1}2^n\)如\(a_{nm}a_n a_m\)\(a_12\)求通项公式\(a_n\)令\(m1\)得到\(a_{n1}a_n a_1\)即\(a_{n1}-a_na_12\)不就是等差数列嘛故\(a_n2(n-1)\times 22n\)等差数列\(\{a_n\}\)\(a_n0\)且数列\(\{\cfrac{1}{a_na_{n1}}\}\)的前\(n\)项和为\(\cfrac{n}{2(n2)}\)求数列\(\{a_n\}\)的通项公式分析当\(n1\)时\(\cfrac{1}{a_1a_2}\cfrac{1}{2\times3}\cfrac{1}{6}\)①当\(n2\)时\(\cfrac{1}{a_1a_2}\cfrac{1}{a_2a_3}\cfrac{2}{2\times4}\cfrac{1}{4}\)②①-②得到\(\cfrac{1}{a_2a_3}\cfrac{1}{12}\)则有\(a_1\cdot (a_1d)6\)③\((a_1d)(a_12d)12\)④由③④解得\(a_12\)\(d1\)故\(a_nn1\)[引申]如\(a_{nm}a_na_mmn\)\(a_12\)求通项公式\(a_n\)令\(m1\)得到\(a_{n1}a_na_1na_nn2\)即\(a_{n1}-a_nn2\)使用累加法即可求解通项公式【高三理科用题】已知\(f(x)\)是定义在\(R\)上不恒为零的函数对于任意的\(xy\in R\)都有\(f(xy)\)\(xf(y)\)\(yf(x)\)成立数列\(\{a_n\}\)满足\(a_nf(2^n)(n\in N^*)\)且\(a_12\)则数列\(\{a_n\}\)的通项公式\(a_n\)______________.法1【迭代递推】\(a_1f(2^1)2\)即\(f(2)2\)\(a_nf(2^n)f(2\cdot2^{n-1})2f(2^{n-1})2^{n-1}f(2)\)\(2^1\cdot f(2^{n-1})2^n\cdot 12[2f(2^{n-2})2^{n-2}f(2)]2^n\cdot 1\)\(2^2\cdot f(2^{n-2})2^n\cdot 2\)\(2^3\cdot f(2^{n-3})2^n\cdot 3\)\(2^4\cdot f(2^{n-4})2^n\cdot 4\)\(2^{n-1}\cdot f(2^1)2^n \cdot (n-1)n\cdot 2^n\)法2【赋值法】由题目\(a_nf(2^n)\)可知\(a_{n1}f(2^{n1})\)且\(a_1f(2)2\)由于对任意的\(xy\in R\)都有\(f(xy)\)\(xf(y)\)\(yf(x)\)成立令\(x2^n\)\(y2\)则有\(f(2^{n1})f(2^n\cdot 2)2^nf(2)2f(2^n)\)即\(a_{n1}2a_n2\times 2^n\)即\(a_{n1}2a_n2^{n1}\)接下来两边同时除以\(2^{n1}\)得到\(\cfrac{a_{n1}}{2^{n1}}\cfrac{a_{n}}{2^{n}}1\)则数列\(\{\cfrac{a_n}{2^n}\}\)是首项为\(1\)公差为\(1\)的等差数列则有\(\cfrac{a_n}{2^n}1(n-1)\times 1n\)即所求通项公式为\(a_nn\cdot 2^n\)。大小比较设\(0 b a 1\)则下列不等式成立的是【】$A.ab b^2 1$ $B.log_\frac{1}{2}b log_\frac{1}{2}a 0$ $C.2b 2a 2$ $D.a^2 ab 1$【法1】不等式性质法此处略。【法2】赋值法由题设令\(b\cfrac{1}{4}\)\(a\cfrac{1}{2}\)对选项A而言\(ab\cfrac{1}{2}\times \cfrac{1}{4}\cfrac{1}{8}\)\(b^2\cfrac{1}{4}\times \cfrac{1}{4}\cfrac{1}{16}\)\(\cfrac{1}{8}\cfrac{1}{16}\) 故A错对选项B而言\(log_\frac{1}{2}blog_\frac{1}{2} \cfrac{1}{4}20\)故B错对选项D而言\(a^2(\cfrac{1}{2})^2\cfrac{1}{4}\) $ ab\cfrac{1}{2}\times \cfrac{1}{4}\cfrac{1}{8} a^2$故D错故选C。若\(x\in (e^{-1}1)\)\(alnx\)\(b(\cfrac{1}{2})^{lnx}\)\(ce^{lnx}\)则其大小关系为__________。分析借助赋值法令\(x\cfrac{1}{2}\)则可知\(b(\cfrac{1}{2})^{lnx}1\)\(alnx0\)\(ce^{lnx}\cfrac{1}{2}\)故大小关系为\(bca\)系数求和【二项式系数求和】若\((1x)(a-x)^6a_0a_1xa_2x^2\cdotsa_7x^7\)其中\(a\int_0^{\pi}(sinx-cosx)dx\)则\(a_0a_1a_2\cdotsa_6\)的值为__________.分析先求得\(a(-cosx-sinx)|_0^{\pi}2\)代入已知表达式再赋值\(x1\)得到\(a_0a_1a_2\cdotsa_6a_7(11)(2-1)^62\)又\(C_6^6\cdot 2^0\cdot (-x)^6\cdot xx^7\)故\(a_71\)从而解得\(a_0a_1a_2\cdotsa_61\).二项式定理将赋值法的使用发挥到了极致。关系判断【集合的关系判断】已知集合\(M\{x\mid x\cfrac{k}{2}\cfrac{1}{4}k\in Z \}\)\(N\{x\mid x\cfrac{k}{4}\cfrac{1}{2}k\in Z \}\)则两个集合的关系是【】$A. N\subsetneqq M$ $B. M\subsetneqq N$ $C. M N$ $D. M\cap N\varnothing $分析【法1】赋值法由于\(k\in Z\)故我们给\(k\)赋值\(\cdots-2-1012\cdots\)这样就分别得到了两个无限集合了。\(M\{\cdots-\cfrac{3}{4}-\cfrac{1}{4}\cfrac{1}{4}\cfrac{3}{4}\cfrac{5}{4}\cfrac{7}{4}\cdots\}\)元素之间的间隔为\(\cfrac{2}{4}\cfrac{1}{2}\)元素少\(N\{\cdots-\cfrac{1}{4}0\cfrac{1}{4}\cfrac{2}{4}\cfrac{3}{4}\cfrac{4}{4}\cdots\}\)元素之间的间隔为\(\cfrac{1}{4}\)元素多这样我们就很容易发现关系为\(M\subsetneqq N\)故选\(B\)。【法2】数列法如果我们经常将数列当成函数来理解那么我们可以看到两个无限集合可以看成数列集合\(M\)的元素可以看成公差为\(\cfrac{1}{2}\)其中某一项为\(\cfrac{1}{4}\)的无穷等差数列合\(N\)的元素可以看成公差为\(\cfrac{1}{4}\)其中某一项为\(\cfrac{1}{2}\)的无穷等差数列我们画两条平行的数轴在上面依照两个集合的元素取点时就会发现关系为\(M\subsetneqq N\)故选\(B\)。三角函数已知\(f(x)2Acos^2(\omega x\phi)(A0,\omega0,0分析由\(f(x)2Acos^2(\omega x\phi)A[cos2(\omega x\phi)1]-AAcos(2\omega x2\phi)\)故其周期为\(T\cfrac{2\pi}{2\omega}\cfrac{\pi}{\omega}\)又由题目可知\(\cfrac{T}{4}\cfrac{\pi}{3}-\cfrac{\pi}{12}\cfrac{\pi}{4}\)则\(T\pi\cfrac{\pi}{\omega}\)故\(\omega1\)则函数简化为\(f(x)Acos(2x2\phi)\)再利用直线\(x\cfrac{\pi}{3}\)是函数\(f(x)\)图象上的一条对称轴故\(2\times \cfrac{\pi}{3}2\phik\pi,(k\in Z)\),解得\(\phi\cfrac{k\pi}{2}-\cfrac{\pi}{3}\)令\(k1\)则\(\phi\cfrac{\pi}{6}\in (0,\cfrac{\pi}{2})\)满足题意故\(f(x)Acos(2x2\phi)Acos(2x\cfrac{\pi}{3})\).抽象不等式【2019届宝鸡中学高三文科第一次月考第22题】设函数\(f(x)\)是增函数对于任意\(xy\in R\)都有\(f(xy)f(x)f(y)\)(1)求\(f(0)\)分析考查赋值法令\(xy0\)得到\(f(00)f(0)f(0)\)即\(f(0)0\)。(2)证明函数\(f(x)\)是奇函数分析由题目可知定义域关于原点对称令\(y-x\)代入已知得到\(f(x-x)f(x)f(-x)\)即\(f(x)f(-x)0\)即\(f(-x)-f(x)\)故函数\(f(x)\)是奇函数(3)解不等式\(\cfrac{1}{2}f(x^2)-f(1-x) 分析先将已知变形为\(f(x^2)-2f(1-x) f(3x)\)再变形为\(f(x^2)-f(3x) 2f(1-x)\)(提示上式变形的最终形式应该是\(f(M) f(N)\)的形式为此需要将\(-f(3x)\)变形需要将\(2f(1-x)\)变形)由于任意\(xy\in R\)都有\(f(xy)f(x)f(y)\)令\(xy\)得到\(f(2x)f(x)f(x)2f(x)\)应用到题目中有\(2f(1-x)f(2-2x)\)又\(-f(x)f(-x)\)应用到题目中有\(-f(3x)f(-3x)\)故\(f(x^2)-f(3x)2f(1-x)\)可以再次变形得到\(f(x^2)f(-3x) f(2-2x)\)即\(f(x^2-3x) f(2-2x)\)由于函数\(f(x)\)是\(R\)上的增函数故由单调性有\(x^2-3x 2-2x\)即\(x^2-x-20\)解得\(-1 x 2\)即解集为\(x\in (-12)\)。【2019届高三理科教学资料用题】函数\(f(x)\)对任意的\(mn\in R\)都有\(f(mn)f(m)f(n)-1\)并且\(x0\)恒有\(f(x)1\)。(1)求证\(f(x)\)在\(R\)上是增函数证明设\(x_1x_2\in R\)且\(x_1 x_2\)则\(x_2-x_1 0\)由题目当\(x 0\)恒有\(f(x) 1\)则\(f(x_2-x_1)1\)\(f(x_2)f[(x_2-x_1)x_1]f(x_2-x_1)f(x_1)-1\)则\(f(x_2)-f(x_1)f(x_2-x_1)-10\)故\(f(x_1) f(x_2)\)即\(f(x)\)在\(R\)上是增函数(2)若\(f(3)4\)解不等式\(f(a^2a-5)2\)。分析\(mn\in R\)都有\(f(mn)f(m)f(n)-1\)令\(mn1\)则\(f(11)f(1)f(1)-1\)即\(f(2)2f(1)-1\)又由已知\(f(3)4\)即\(4f(21)f(2)f(1)-1\)即\(3f(1)-24\)即\(f(1)2\)也即\(2f(1)\)故\(f(a^2a-5)2f(1)\)又\(f(x)\)在\(R\)上是增函数则有\(a^2a-51\)解得\(a\in (-32)\)。【解后反思】解抽象函数不等式的一般步骤①(定性)确定函数\(f(x)\)在给定区间上的单调性②(转化)将抽象函数不等式转化为\(f(M) f(N)\)的形式③(脱去\(f\))利用单调性去掉函数符号\(\large{f}\)转化为一般的不等式(组)④(求解)求解上述的不等式组⑤(反思)反思回顾查看关键点易错点及解题规范。OK!公式证明①由\(cos(\alpha-\beta)cos\alpha\cdot cos\betasin\alpha\cdot sin\beta\)用\(-\beta\)替换\(\beta\)或者给\(\beta\)赋值\(-\beta\)得到\(cos(\alpha\beta)cos\alpha\cdot cos\beta-sin\alpha\cdot sin\beta\)②由\(sin(\alpha\beta)\)推导\(sin(\alpha-\beta)\)③由\(f(x4)f(-x)\)等价于\(f(x3)f(-x1)\)④由\(a^3b^3\)推导\(a^3-b^3\)正态分布【2015·高考湖北卷】设\(X\sim N(\mu_1\sigma_1^2)\)\(Y\sim N(\mu_2\sigma_2^2)\)这两个正态分布密度曲线如图所示下列结论中正确的是【】$A.P(Y\ge \mu_2)\ge P(Y\ge \mu_1)$$B.P(X\leq \sigma_2)\ge P(X\leq \sigma_1)$$C$.对任意实数$t$$P(X\leq t)\ge P(Y\leq t)$$D$.对任意实数$t$$P(X\ge t)\ge P(Y\ge t)$分析根据正态密度曲线可知\(\mu_1则有\(P(Y\ge \mu_2) P(Y\ge \mu_1)\)故\(A\)错 且有\(P(X\leq \sigma_2) P(X\leq \sigma_1)\)故\(B\)错对\(C\)选项而言不妨赋值设\(t\mu_1\)由图可知必有\(P(X\leq t)\ge P(Y\leq t)\)故\(C\)正确对\(D\)选项而言不妨赋值设\(t\mu_1\)由图可知必有\(P(X\ge t) P(Y\ge t)\)故\(D\)错误综上所述选\(C\)。
