最新高一数学期中考试测试题(必修一含答案)优秀名师资料.doc
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1、高一数学期中考试测试题(必修一含答案),x与的图象是 5(当0,a,1时,在同一坐标系中,函数yx,logya,高一年级上学期期中考试数学试题 a一、选择题(本大题共12小题,每小题5分,共60分。给出的四个选项中,只有一项是符合题目要求的) 1(设全集U=,1,2,3,4,5,,集合A=,1,2,,B=,2,3,,则A?CUB A( B( C ( D( 45,23,12,2(下列表示错误的是 0,(A) (B),12, , 0.766(令,则三个数a、b、c的大小顺序是 abc,6,0.7,log6210xy,,0.7(,)3,4xy,ABA,(C) (D)若则 35xy,AB,A(b,c,
2、a B(b,a,c C(c,a,b 3(下列四组函数,表示同一函数的是 D(c,b,a 22fxx()ln,7(函数的零点所在的大致区间是 A(f(x)=,g(x)=x xx21,xe,,,A(1,2) B(2,3) C(和(3,4) D( 1,,B(f(x)=x,g(x)= ,ex,xx239,xlog31,8(若,则的值为 C( fxlnxgxx(),()2ln,253x3A(6 B(3 C( fxaagxx()log(),(),D( a2x,112,2,x,fx(),D( 4(设则f ( f (2) )的值为 2log(1),2.xx,23A(0 B(1 C(2 D(3 1,29(若函数
3、y = f(x)的定义域为,则yfx,,(1)的定义域为 ,1 A( B( C( 2,30,1,1,0, 第?卷(非选择题共90分) D( ,3,2,二、填空题(本大题共4小题,每小题4分,共16分。把答案填在答题卡对应题号后的横线上() 10(已知是偶函数,当x,0时,则当x,0时, fx()fxxx()(1),,fx(),x,3恒过定点 。 13(函数ya,,3A( B( C xx(1),xx(1)xx(1),38a414(计算 6()_,3D( ,,xx(1)125b253,m11(设为偶函数,且在0,,,上是增函数,则、 15(幂函数在x,,,0,时为减函数,则fxxR()(),fx(
4、)f(2),ymmx,(1),,m 。 、的大小顺序是 f(),f(3)2x,3,316(函数,其中,则该函数的值域yxx,4,A( B( fff()(3)(2),fff()(2)(3),为 。 三、解答题(本大题共6小题,共74分(解答应写出文字说明,证明过程C( D( fff()(2),,,,,fff()(2)(3),或演算步骤() 12 已知函数f(x)的图象是连续不断的,x与f(x)的对应关系见下表,则函数17(本题满分10分) f(x)在区间1,6 2UaaAaCA,,,1,2,23,|2|,2,0已知全集,求a,U上的零点至少有 X 1 2 3 4 5 6 的值. Y 123.56
5、 21.45 -7.82 11.57 -53.76 18(每小题6分,共12分)不用计算器求下列各式的值。 (A) 2 (B) 3 (C) 4 (D) 5 21,13,0232,,(2)(9.6)(3)(1.5)(1); 482 427log27 。 (2)loglg25lg47,33 32x19(本题满分12分)已知函数在上述减函数,(,),yxbxc,,2a,2122(本题满分14分)设是R上的奇函数。 fx(),2x12,3在上述增函数,且两个零点满足,求二次函数(,),,,xx,2xx,1212(1)求实数a的值; 2的解析式。 (2)判定在R上的单调性。 fx()20(本题满分12分
6、)已知。 fxxaa()log(1)(0,1),a (1)求得定义域; fx() (2)求使成立的x的取值范围。 fx()0, 21(本题满分12分) 我国是水资源匮乏的国家为鼓励节约用水,某市打算出台一项水费政策措 施,规定:每一季度每人用水量不超过5吨时,每吨水费收基本价1(3元; 若超过5吨而不超过6吨时,超过部分水费加收200%;若超过6吨而不超 过7吨时,超过部分的水费加收400%,如果某人本季度实际用水量为 吨,应交水费为。 xx(07),fx()高一数学试题参考答案 (1)求f(4)、(f5.5)、(f6.5)的值; 一、CCDCC DBABA AB (2)试求出函数fx()的解
7、析式。 3 4322,4ab25 15(2 16( 二、13(3,4) 14( ,4,213,9(2)原式,,loglg(254)233三、17解 分 0,U由得1224分 4aa,,230? ,,log3lg10231151,A由得 ,,,22448分 a,21?12分 3b32,aa,,230b,619(解:由已知得:对称轴,所以得3,x,解得422,a,21,分 a,110分 ?2 故 fxxxc()26,, 21, 又,是的两个零点 xxfx()937312,232,,18(1)原式()1()() 4822260xxc,, 所以x,x是方程的两个12 根4分 21,,32333,232
8、,,()1()()3分 ?,,xx3 ,22212c333,22,,xgx,1()() 6分 1222221 所以, 22xxxxxx,,,()4922c,8分 1212126分 4 得xx01,511分 c,12分 2故21(解:(1)521f(4)41.35.2,,,12分 fxxx()26,,2分 10,x20(解:(1)依题意得1分 解得3f(5.5)51.30.53.98.45,,,x,12分 故所求定义域为分 4分 xx,1,5分 f(6.5)51.313.90.56.513.65,,,(2)由,0 fx()05,x(2)当时,得7分 fxxx()1.31.3,,,log(1)lo
9、g1,x6分 aa56,x 当时,a,111,x 当时,即9分 fxxx()1.35(5)3.93.913,,,,,x,08分 01,a011,x67,x 当时,即 当时,01,x10分 11分 fxxx()1.3513.9(6)6.56.528,,,,,.6a,101,axx,0 综上,当时,x的取值范围是,当时,, 故x的取值范围是5 xx1.3(05)xx,12,。 得2,fxxx()3.913(56),12分 ,xx12,2(22),6.528.6(67)xx, 则,即fxfx()(),012xx21(21)(21),22(1)法一:函数定义域是R,因为是奇函数, fx()fxfx()
10、(),12所以,即fxfx()(),所以说增函fx()xxx,12212,aaa数。14分 2分 ,xxx,121212,x,212xfx, 法二:由(1)可知,由于在R()12xxxx?,122aa解得 ,2121上是增函数, a,16分 22法二:由是奇函数,所以,故fx()f(0)0,?,在R上是减函数,在R上是增函数, xx,2121a,1,3分 是R上的增函?fx()x,21数。14分 a,1fx,()再由,验证,来确定的fxfx()(),x,12 合理性6分 (2)增函fx()数7分 x,21,Rfx,()xxxx, 法一:因为,设设,且,1212x,x126 (英文版 ) eas
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17、scipline a ruler, often the control inspection, and consciously in the ideological red line to draw the row Ming Good accumulation is indeed the bottom line, so that the heart has fear, said to have quit, the line has ended. Attached: indifferent to heart, calmly to the table in our life, there are
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19、 things. 1, life can be complex, can also be simple. Want simple life of precipitation, to have enough time to reflect, to make Become more perfect. Life is the most important thing is not to win, but the struggle; not to have conquered, but to have fought well. 2, the plain is the background of lif
20、e. Live a plain life, give up on themselves is not a coward, but the wise answers; not disillusioned after the heart, such as ashes, but experience the storm after the enlightenment; not unrewarding perfunctorily, but calm attitude of life of unrestrained self-confidence. Plain living, there is no n
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