高一数学上期末复习精简.ppt

2 已知集合A1,2，Bx|mx10，若ABB，则则所有实实数m的 值组值组 成的集合是( ) A1,2 B1， C1,0， D1,0， 解析：ABB，即BA，若m0，BA； 若m0，Bx|x ；由BA得： 1或 2，m1或m .综上选C. 答案：C Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. (改编题)已知集合Ax|2x25x20，By|y2xa，xR，若 ABA，求a的取值值范围围 解答：A ，B(a，)，由ABA即AB得a2，因此a 的取值值范围围是(，2). 变式2. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 3在R上定义义运算 ：x yx(1y)若不等式(xa) (xa)1对对任意实实数x 成立，则则( ) A1a1 B0a2 C a D a 解析：(xa) (xa)1对任意实数x成立，即(xa)(1xa)1对任意实数 x成立x2xa2a10恒成立 14(a2a1)0， a . 答案：C 解析：原不等式等价于 10 0x21的解集是_ Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 此类问题类问题 是解不等式的逆向思维问题维问题 ，要在熟练练掌握不等式解法的基础础上 进进行 求解 【例2】 (1)关于x的不等式 1的解集为为x|x1或x2，则实则实 数a _. (2)若不等式ax2bxc0的解集是 ，则则不等式cx2bxa0 的 解集是_ Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 解析：(1)原不等式可化为 0. 解集为x|x1或x2，a10且 2.a . (2)由已知条件知a0，且 ，即b ，c ， 不等式cx2bxa0即x2x60，其解集为(3,2) 答案：(1) (2)(3,2) Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 解析：(a2)x24xa10恒成立， 由得a2，由得a3或a2. 答案：2，) 变式2. 若xR，ax24xa2x21恒成立，则a的范围是_ Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 变式3. 已知不等式 0(aR) (1)解这这个关于x的不等式；(2)若xa时时不等式成立，求a的取值值范围围 解：(1)原不等式等价于(ax1)(x1)0. 当a0时时，由(x1)0，得x1； 当a0时时，不等式化为为 (x1)0，解得x1或x ； 当a0时时，不等式化为为 (x1)0； 当 1，即1a0，则则 x1； Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 若 1，即a1，则则不等式解集为为空集； 若 1，即a1，则则1x . 综综上所述，a1时时，解集为为 ； a1时时，原不等式无解；1a0时时，解集为为 ； a0时时，解集为为x|x1； a0时时，解集为为 . (2)xa时时不等式成立， 0，即a10，a1，即a的取值值范围为围为 a1. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 1已知p是r的充分条件而不是必要条件，q是r的充分条件，s是r的必要条件， q是s的必要条件现现有下列命题题： s是q的充要条件； p是q的充分条件，而不是必要条件； r是q的必要条件， 而不是充分条件； 綈p是綈s的必要条件， 而不是充分条件； r是s的充分条件，而不是必要条件 则则正确命题题的序号是( ) A B C D Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 解析：由已知条件可知： 因此为为正确命题题 答案：B Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【例1】 已知c0，设设p：函数ycx在R上递递减；q：不等式x|x2c|1的解 集为为 R，如果“p或q”为为真，且“p且q”为为假，求c的范围围 解答：由p01c ，“p或q”为为真，且“p且q”为为假， p真q假或p假q真， 若p真q假，则则c的范围围是(0,1)(， (0， ； 若p假q真，则则c的范围围是(，01，)( ，)1，)， 因此c的范围围是(0， 1，). Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 2.(2009安徽)“acbd”是“ab且cd”的( ) A必要不充分条件 B充分不必要条件 C充分必要条件 D既不充分也不必要条件 解析：由ab且cd知，ab0且cd0，(ac)(bd)(ab)(c d)0，因此acbd，即 acbd，若a10，c1，b6， d2，acbd，/ ab，cd.综上可知，“acbd”是“ab 且cd”的必要不充分条件 答案：A Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【例1】 已知三个不等式：ab0； ；bcad以其中两个作条件 ，余下一个作结论，则可组成_个正确命题 答案：3 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【例2】设abc，求证证：bc2ca2ab2b2cc2aa2b. 证明：(bc2ca2ab2)(b2cc2aa2b)(ba)c2(a2b2)cab2a2b (ba)c2(ab)cab(ba)(ca)(cb) abc，ba0，ca0，cb0.(bc2ca2ab2)(b2cc2a a2b)0， 即bc2ca2ab2b2cc2aa2b. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 利用比较较法可证证明函数的单调单调 性和凸凹性等问题问题 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 2给给出下列四个不等式，其中正确不等式的个数是( ) x232x(xR) a5b5a3b2a2b3(a，bR) a2b22(ab1)(a，bR) (m0) A1 B2 C3 D4 解析：其中不等式一定成立 答案：B Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 3.设设a0，b0，则则以下不等式中不恒成立的是( ) A(ab)( )4 Ba3b32ab2 Ca2b222a2b D. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 解法二：取a ，b ，则a3b32ab2.故选B项 答案：B Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【例1】 已知函数f(x)，g(x)分别由下表给出： 则fg(1)的值为值为 _；满满足fg(x)gf(x)的x值值是_ 解析：fg(1)f(3)1，fg(2)f(2)3，fg(3)f(1)1， gf(1)g(1)3，gf(2)g(3)1，gf(3)g(1)3， 满足fg(x)gf(x)的x值是2. 答案：1 2 x123 f(x)131 x123 g(x)321 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. (2)已知f(x)为为一次函数，且f f f (x)8x7，求f(x)； (3)已知f(x)2f( )2x1，求f(x) 解答：(1)方法一：设x1t，则则xt1，代入f(x1)的解析式， 得f(t)(t1)24(t1)1t22t2，f(x)x22x2. 方法二：f(x1)x24x1(x22x1)2(x1)2 (x1)22(x1)2.用x替代x1，得f(x)x22x2. 【例2】 (1)已知f(x1)x24x1，求f(x)； Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. (2)设设f(x)axb(a0)，所以fff(x)ff(axb)fa(axb)b aa(axb)bba3xa2babb8x7， 所以 解得 所以f(x)2x1. (3)由已知得 消去f( )， 得f(x) . Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. A2 B2 C D. 解析：f(4x) ，依题意 x，解得x . 答案：D 变式2.(1)若f(x) ，则方程f(4x)x的根是( ) Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 1函数的定义义域通常由问题问题 的实际实际 背景确定如果只给给出解析式yf(x)， 而没有指明它的定义义域，那么函数的定义义域就是指能使这这个式子有意义义的实实数x 的集合 2常见基本初等函数的定义域 (1)一次函数f(x)axb(a0)的定义义域为为 ； (2)二次函数f(x)ax2bxc(a0)的定义义域为为 ； (3)反比例函数f(x) (k0)的定义义域为为 ； R R x|x0 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 如果函数yf(x)的定义义域为为A，那么函数的值值域为为y|yf(x)，xA 3函数的值域 一般地，设设函数yf(x)的定义义域为为I，如果存在实实数M满满足： (1)对对于任意的xI，都有f(x)M；(2)存在x0I，使得f(x0)M. 那么，我们们称M是函数yf(x)的最大值值(maximum value) 思考：你能仿照函数最大值的定义，给出函数yf(x)的最小值(minimum value)的定义吗？ 4函数最大值与最小值的含义 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. A1,1 B(1,1 C1,1) D(，11，) 解析：由y 得：x2 0，解得：15，则m5. 答案：(，5 5(2009湖南)若x0，则则x 的最小值为值为 _ Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【例2】 求下列函数的值值域： Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. (1) 解法：分离常数法 (2)解法一：配方法 原函数的值值域为为 ，1) Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 由y ，得(y1)x2(1y)xy0.y1时时，x， y1，又xR，必须须(1y)24y(y1)0. y1.y1，函数的值值域为为 ，1) (3)解法一：单调单调 性法 解法二：判别式法 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 解法二：换换元法 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. (1)当a 时，求函数f(x)的最小值； (2)若对任意x1，)，f(x)0恒成立，试求实数a的取值范围 【例3】 已知函数f(x) ，x1，)， Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. (2)若对对任意x1，)，f(x)0恒成立，即 0， x22xa0对对于一切x1，)恒成立； 又x22xa(x1)2a13a， 由3a0得a3. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 2已知f(x)为为R上的减函数，则满则满 足f(| |)1，不等式等价于 ，解得10)在(2，)上递递增，求实实数a的取值值范围围 解答：解法一：设设2a恒成立又x1x24，则则00)的递递增区间间是 (， )，( ，)， 根据已知条件 2，解得00时时，f(x)x2x1，则则f(x)_. 解析：当x0时，f(0)f(0)，即f(0)0.当xx|x|0知f(x)ln(x )的定义域为R， 又f(x)ln(x )ln ln(x )f(x) ， 则f(x)为奇函数； f(x) 的定义域为R，又f(x) f(x)， 则f(x)为奇函数； 答案：C Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【例2】已知f(x)x( )(x0)(1)判断f(x)的奇偶性；(2)证明：f(x)0. 解答：(1)解法一：f(x)的定义域是(，0)(0，) 是偶函数 解法二：f(x)的定义域是(，0)(0，)，f(1) ，f(1) ， f(x)不是奇函数 1)x(11)0，f(x)f(x)，f(x)是偶函数 (2)证证明：当x0时时，2x1,2x10，所以f(x)x( )0.当x0 时时，x0，所以f(x)0，又f(x)是偶函数，f(x)f(x)，所以f(x)0. 综综上，均有f(x)0. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. (本题满分4分)对于函数f(x) (其中a为实数，x1)，给出下列命题： 当a1时，f(x)在定义域上为单调函数； f(x)的图象关于点(1，a)对称； 对任意aR，f(x)都不是奇函数； 当a1时，f(x)为偶函数； 当a2时，对于满足条件2x1x2的所有x1、x2总有f(x1)f(x2)3(x2 x1) 其中正确命题的序号为_. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 解析：(1)当a1时，f(x) 的定义域为(，1)(1，)， 又f(x)1 ，函数的两个递减区间分别为(，1)、(1，)， 命题错误 的图象关于点(1，a)对称，命题正确； (3)f(0)1，因此f(x)不是奇函数，是正确命题； (4)当a1时，f(x) 1(x1) 因此f(x)不是偶函数，命题不正确 【答题模板】 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 1右图是指数函数(1)yax，(2)ybx，(3)ycx，(4)ydx的图图象，则则a，b，c ，d与1的大小关系是( ) Aad1a1b1，b0且a1) (1)求f(x)的定义义域和值值域；(2)讨论讨论 f(x)的奇偶性；(3)讨论讨论 f(x)的单调单调 性 解答：(1)易得f(x)的定义义域为为x|xR设设y ，解得ax ax0，当且仅仅当 0时时，方程有解解得11时时，ax1为为增函数，且ax10. Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 1函数y1 的图图象是( ) 答案：B Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 3. (2009重庆模拟)已知图中的图图象对应对应 的函数为为yf(x)，则图则图 的图图象对对 应应的函数为为( ) Ayf(|x|) By|f(x)| Cyf(|x|) Dyf(|x|) 答案：C Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【例1】作出下列函数的图象： 解答：(1)解法一：函数的定义义域为为(，1)(1,1)(1，)，且函数为为偶 函数，函数的递递增区间为间为 (，1)，(1,0)，递递减区间为间为 (0,1)，(1，) 可根据以上性质质取值值列表：Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 在直角坐标系中描出上表对应点并用光滑的曲线连结起来再根据y 是偶函数，把所作图象关于y轴对称到y轴左侧后，就得到y 的图象(如图1) 当x0且x1时，y ，它的图象可由y 的图象向右平移一个单位后得 到(仅要y轴及其右侧部分) 当x0且x1时，y ，它的图象可由y 的图象先关于x轴对称后 ，再向左平移一个单位后得到(仅要y轴左侧部分)，把上述两次得到的图象合 在一起就得到函数y 的图象(如图1) 图1 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. (3)若x2，原式为y x(x4)， 若xVn，即当h愈大时，相等高度增加 的水量愈少，其图象呈“上凸”形状，故选A. (2)时间t愈大，该学生离学校的距离d愈小，d是t的减函数，答案应为C、 D中的一个，由于前一段时间速度快，后一段时间速度慢，即 的值前大后小， 故选D. 答案：(1)A (2)D Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 变式2.如下图图所示，向高为为h的水瓶A、B、C、D同时时以等速注水，注满为满为 止 (1)若水量V与水深h函数图图象是下图图的(a)，则则水瓶的形状是_； (2)若水深h与注水时间时间 t的函数图图象是下图图的(b)，则则水瓶的形状是_； (3)若注水时间时间 t与水深h的函数图图象是下图图的(c)，则则水瓶的形状是_； (4)若水深h与注水时间时间 t的函数的图图象是图图中的(d)，则则水瓶的形状是_ Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 答案：(1)A (2)D (3)B (4)C Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 【例3】已知二次函数yf1(x)的图图象以原点为顶为顶 点且过过点(1,1)，反比例函数y f2(x) 的图图象与直线线yx的两个交点间间的距离为为8，f(x)f1(x)f2(x) (1)求函数f(x)的表达式； (2)证证明：当a3时时，关于x的方程f(x)f(a)有三个实实数解Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile .Created with Aspose.Slides for .NET 3.5 Client Profile . Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd. 解答：(1)由已知，设设f1(x)ax2(a0)，由f1(1)1，得a1，f1(x)x2. 设设 f2(x) (k0)，它的图图象与直线线yx的交点分别为别为 A( ， ) B( ， )，由|AB|8，得k8，f2(x) .故f(x)x2 . (2)证证明：证证法一：由f(x)f(a)，得x2 a2 ，即 x2a2 .在同 一坐标标系内作出f2(x) 和f3(x)x2a2 的大致图图象，其中f2(x)的图图象是以 坐标轴为渐标轴为渐 近线线，且位于第一、三象限的双曲线线，f3(x)的图图象是(0，a2 )为为 顶顶点，开口向下的抛物线线因此，f2(x)与f3(x)的图图象在第三象限有一个交点，即 f(x)f(a)有一个负负数解 Ev