高二必修一综合训练 2015.5.9

高一数学必修一综合测试卷一、选择题（本大题共12小题，每小题5分，在每小题给出的四个选项中，只有一项是符合题目要求的）1．若集合}1,1{-=A ，}1|{==mx x B ，且A B A =⋃，则m 的值为（ ） A ．1 B ．1- C ．1或1- D ．1或1-或02、函数1()(0)f x x x x =+≠是（ ）A 、奇函数，且在(0，1)上是增函数B 、奇函数，且在(0，1)上是减函数C 、偶函数，且在(0，1)上是增函数D 、偶函数，且在(0，1)上是减函数3. 已知b ax y x f B y A x R B A +=→∈∈==:,,,是从A 到B 的映射，若1和8的原象分别是3和10，则5在f 下的象是（ ）A .3B .4C .5D .64. 下列各组函数中表示同一函数的是（ ）⑴3)5)(3(1+-+=x x x y ，52-=x y ； ⑵111-+=x x y ， )1)(1(2-+=x x y ；⑶x x f =)(， 2)(x x g = ； ⑷x x f =)(，()g x =； ⑸21)52()(-=x x f ， 52)(2-=x x fA 、⑴、⑵B 、 ⑵、⑶C 、 ⑷D 、 ⑶、⑸5．若)(x f 是偶函数，其定义域为()+∞∞-,，且在[)+∞,0上是减函数，则)252()23(2++-a a f f 与的大小关系是（ ）A ．)23(-f >)252(2++a a f B ．)23(-f <)252(2++a a f C ．)23(-f ≥)252(2++a a f D ．)23(-f ≤)252(2++a a f6.设⎪⎩⎪⎨⎧-=-)1(log 2)(231x ex f x )2()2(≥<x x 则[])2(f f =（ ） A .2 B .3 C .9 D .187．函数1(0,1)x y a a a a=->≠的图象可能是（ ）8.给出以下结论：①11)(--+=x x x f 是奇函数；②221)(2-+-=x x x g 既不是奇函数也不是偶函数；③)()()(x f x f x F -= )(R x ∈是偶函数 ；④xxx h +-=11lg)(是奇函数.其中正确的有（ ）个 A .1个 B .2个 C .3个 D .4个9. 函数1)3(2)(2+-+=x a ax x f 在区间[)+∞-,2上递减，则实数a 的取值范围是（ ）A .(]3,-∞-B .[]0,3-C . [)0,3-D .[]0,2-10.函数33()11f x x x =++-,则下列坐标表示的点一定在函数f(x)图象上的是（ ）A ．(,())a f a --B ．(,())a f a -C ．(,())a f a -D ．(,())a f a ---11. 若函数a x x x f +-=24)(有4个零点，则实数a 的取值范围是（ ）A . []0,4- B. []4,0 C. )4,0( D. )0,4(-12. 设()f x 是奇函数，且在(0,)+∞内是增函数，又(3)0f -=，则()0x f x ⋅<的解集是（ ） A ．{}|303x x x -<<>或 B ．{}|303x x x <-<<或C ．{}|3003x x x -<<<<或D ．{}|33x x x <->或二、填空题（本大题共4小题，每小题5分）13．若函数2()(1)3f x kx k x =+-+是偶函数，则()f x 的递减区间是 ；14.已知函数11()()142x x y =-+的定义域为[3,2]-，则该函数的值域为 ；15. 函数()()R b a xbax x f ∈+-=,25，若()55=f ，则()=-5f ； 16.设函数()f x ＝x |x |＋b x ＋c ，给出下列四个命题： ①若()f x 是奇函数，则c ＝0③()f x 的图象关于(0，c )对称④若b ≠0，方程()f x ＝0必有三个实根 其中正确的命题是 (填序号)三、解答题（解答应写文字说明，证明过程或演算步骤）17.(10分)已知集合{}0652<--=x x x A ，集合{}01562≥+-=x x x B ，集合⎭⎬⎫⎩⎨⎧<---=09m x mx x C（1）求B A ⋂（2）若C C A =⋃，求实数m 的取值范围；18．（本小题满分12分）已知函数()log (1),()log (1)a a f x x g x x =+=-其中)10(≠>a a 且，设()()()h x f x g x =-.（1）求函数()h x 的定义域，判断()h x 的奇偶性，并说明理由； （2）若(3)2f =，求使()0h x <成立的x 的集合。

1必修第一册全册综合训练卷本试卷分第Ⅰ卷(选择题)和第Ⅰ卷(非选择题)两部分．满分150分．考试时间120分钟．第Ⅰ卷(选择题 共60分)一、选择题(本大题共12个小题，每小题5分，共60分，在每小题给出的四个选项中只有一个是符合题目要求的)1．集合A ＝{x |－1≤x ≤2}，B ＝{x |x <1}，则A ∪(∁R B )＝( ) A ．{x |x >1} B ．{x |x ≥－1} C ．{x |1<x ≤2}D ．{x |1≤x ≤2}【解析】由A ＝{x |－1≤x ≤2}，B ＝{x |x <1}可知∁R B ＝{x |x ≥1}，∴A ∪(∁R B )＝{x |x ≥－1}． 故选B.2．函数f (x )＝x 2x 2－1＋lg(10－x )的定义域为( )A ．RB ．[1,10]C ．(－∞，－1)Ⅰ(1,10)D ．(1,10)【解析】要使函数f (x )有意义，需使⎩⎪⎨⎪⎧x 2－1>0，10－x >0，解得x <－1或1<x <10.故选C3．已知f (x )＝x 2－ax 在[0,1]上是单调函数，则实数a 的取值范围是( ) A ．(－∞，0] B ．[1，＋∞)C ．[2，＋∞)D ．(－∞，0]Ⅰ[2，＋∞)【解析】函数f (x )＝x 2－ax 图象的对称轴为直线x ＝a 2，根据二次函数的性质可知a 2≤0或a2≥1，解得a ≤02或a ≥2.故选D.4．下列函数是偶函数且值域为[0，＋∞)的是( ) Ⅰy ＝|x |；Ⅰy ＝x 3；Ⅰy ＝2|x |；Ⅰy ＝x 2＋|x |. A ．ⅠⅠ B ．ⅠⅠ C ．ⅠⅠ D ．ⅠⅠ【解析】对于Ⅰ，y ＝|x |是偶函数，且值域为[0，＋∞)；对于Ⅰ，y ＝x 3是奇函数；对于Ⅰ，y ＝2|x |是偶函数，但值域为[1，＋∞)；对于Ⅰ，y ＝x 2＋|x |是偶函数，且值域为[0，＋∞)，所以符合题意的有ⅠⅠ故选C.5．已知a ＝log 20.2，b ＝20.2，c ＝0.20.3，则( ) A ．a <b <c B ．a <c <b C ．c <a <bD ．b <c <a【解析】a ＝log 20.2<log 21＝0，b ＝20.2>20＝1,0<c ＝0.20.3<0.20＝1，即0<c <1，则a <c <b . 故选B.6．若sin α>0且tan α<0，则α2的终边在( )A ．第一象限B ．第二象限C ．第一象限或第三象限D ．第三象限或第四象限 【解析】 因为sin α>0且tan α<0， 所以α位于第二象限． 所以π2＋2k π<α<2k π＋π，k ⅠZ ，则π4＋k π<α2<k π＋π2，k ⅠZ . 当k 为奇数时α2是第三象限的角，当k 为偶数时α2是第一象限的角，3所以角α2的终边在第一象限或第三象限．故选 C.7.函数y ＝sin(ωx ＋φ)(x ⅠR ，且ω>0,0≤φ<2π)的部分图象如右图所示，则( ) A ．ω＝π2，φ＝π4B ．ω＝π3，φ＝π6C ．ω＝π4，φ＝π4D ．ω＝π4，φ＝5π4【解析】 ⅠT ＝4×2＝8，Ⅰω＝π4.又Ⅰπ4×1＋φ＝π2，Ⅰφ＝π4.故选C8．函数f (x )＝2sin x －sin2x 在[0,2π]的零点个数为( ) A ．2 B ．3 C ．4 D ．5【解析】 由f (x )＝2sin x －sin2x ＝2sin x －2sin x cos x ＝2sin x (1－cos x )＝0，得sin x ＝0或cos x ＝1，Ⅰx Ⅰ[0,2π]，Ⅰx ＝0、π或2π，Ⅰf (x )在[0,2π]的零点个数是3.故选B9．已知lg a ＋lg b ＝0，函数f (x )＝a x 与函数g (x )＝－log b x 的图象可能是( )4【解析】 Ⅰlg a ＋lg b ＝0，Ⅰab ＝1，则b ＝1a ，从而g (x )＝－logb x ＝log a x ，故g (x )与f (x )＝a x 互为反函数，图象关于直线y ＝x 对称．故选B.10．若αⅠ⎪⎭⎫⎝⎛ππ，2，且sin α＝45，则sin ⎪⎭⎫ ⎝⎛+4πa －22cos(π－α)等于( )A.225 B ．－25 C.25 D ．－225【解析】 sin ⎪⎭⎫⎝⎛+4πa －22cos(π－α) ＝22sin α＋22cos α＋22cos α＝22sin α＋2cos α. Ⅰsin α＝45，αⅠ⎪⎭⎫⎝⎛ππ，2，Ⅰcos α＝－35.Ⅰ22sin α＋2cos α＝22×45－2×35＝－25. 故选 B11．设函数f (x )＝sin(ωx ＋φ)＋cos(ωx ＋φ)⎪⎭⎫⎝⎛<>2,0πωω的最小正周期为π，且f (－x )＝f (x )，则( ) A ．f (x )在⎪⎭⎫ ⎝⎛20π，单调递减 B ．f (x )在⎪⎭⎫⎝⎛434ππ，单调递减 C ．f (x )在⎪⎭⎫ ⎝⎛20π，单调递增 D ．f (x )在⎪⎭⎫⎝⎛434ππ，单调递增 【解析】 y ＝sin(ωx ＋φ)＋cos(ωx ＋φ)＝2sin ⎝⎛⎭⎫ωx ＋φ＋π4，由最小正周期为π得ω＝2，又由f (－x )＝f (x )可知f (x )为偶函数，由|φ|<π2可得φ＝π4，所以y ＝2cos2x 在⎝⎛⎭⎫0，π2单调递减． 故选 A12．将函数f (x )＝23cos 2x －2sin x cos x －3的图象向左平移t (t >0)个单位，所得图象对应的函数为奇函5数，则t 的最小值为( )A.2π3B.π3C.π2D.π6【解析】 将函数f (x )＝23cos 2x －2sin x cos x －3＝3cos2x －sin2x ＝2cos ⎝⎛⎭⎫2x ＋π6的图象向左平移t (t >0)个单位，可得y ＝2cos ⎝⎛⎭⎫2x ＋2t ＋π6的图象．由于所得图象对应的函数为奇函数，则2t ＋π6＝k π＋π2，k ⅠZ ，则t 的最小值为π6.故选D.第Ⅰ卷(非选择题 共90分)二、填空题(本大题共4个小题，每小题5分，共20分，把正确答案填在题中横线上) 13．函数f (x )＝a x －1＋3的图象一定过定点P ，则P 点的坐标是________．【解析】由于函数y ＝a x 恒过(0,1)，而y ＝a x －1＋3的图象可看作由y ＝a x 的图象向右平移1个单位，再向上平移3个单位得到的，则P 点坐标为(1,4)．14．函数f (x )＝⎩⎪⎨⎪⎧x 2－1，x ≤0，x －2＋ln x ，x >0的零点个数为________．【解析】 令f (x )＝0，得到⎩⎪⎨⎪⎧x 2－1＝0，x ≤0，解得x ＝－1；6或⎩⎪⎨⎪⎧x －2＋ln x ＝0，x >0，在同一个直角坐标系中画出y ＝2－x 和y ＝ln x 的图象，观察交点个数，如图所示．函数y ＝2－x 和y ＝ln x ，x >0在同一个直角坐标系中交点个数是1，所以函数f (x )在x <0时的零点有一个，在x >0时零点有一个，所以f (x )的零点个数为2.15．若函数f (x )＝⎩⎪⎨⎪⎧3x ，x ≤0，－2－x ，x >0，则函数y ＝f [f (x )]的值域是________．【解析】 当x ≤0时，f (x )＝3x Ⅰ(0,1]，Ⅰy ＝f [f (x )]＝f (3x )＝－2－3x Ⅰ⎝⎛⎦⎤－1，－12； 当x >0时，f (x )＝－2－x Ⅰ(－1,0)，y ＝f [f (x )] ＝f (－2－x )＝3－2－x Ⅰ⎝⎛⎭⎫13，1. 综上所述，y ＝f [f (x )]的值域是⎝⎛⎦⎤－1，－12Ⅰ⎝⎛⎭⎫13，1.16．关于函数f (x )＝cos ⎪⎭⎫⎝⎛-32πx ＋cos ⎪⎭⎫ ⎝⎛+62πx ，给出下列命题： Ⅰf (x )的最大值为2；Ⅰf (x )的最小正周期是π；Ⅰf (x )在区间⎥⎦⎤⎢⎣⎡241324ππ，上是减函数；7Ⅰ将函数y ＝2cos2x 的图象向右平移π24个单位长度后，与函数y ＝f (x)的图象重合．其中正确命题的序号是________．【解析】 f (x )＝cos ⎪⎭⎫⎝⎛-32πx ＋cos ⎪⎭⎫⎝⎛+62πx ＝cos ⎪⎭⎫⎝⎛-32πx ＋sin ⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛+-622ππx ＝cos ⎪⎭⎫ ⎝⎛-32πx －sin ⎪⎭⎫⎝⎛-32πx ＝2⎣⎡⎦⎤22cos ⎝⎛⎭⎫2x －π3－22sin ⎝⎛⎭⎫2x －π3＝2cos ⎝⎛⎭⎫2x －π3＋π4＝2cos ⎝⎛⎭⎫2x －π12， Ⅰ函数f (x )的最大值为2，最小正周期为π，故ⅠⅠ正确；又当x Ⅰ⎥⎦⎤⎢⎣⎡241324ππ，时，2x －π12Ⅰ[0，π]，Ⅰ函数f (x )在⎥⎦⎤⎢⎣⎡241324ππ，上是减函数，故Ⅰ正确；由Ⅰ得y ＝2cos ⎣⎡⎦⎤2⎝⎛⎭⎫x －π24＝2cos ⎝⎛⎭⎫2x －π12，故Ⅰ正确．三、解答题(本大题共6个大题，共70分，解答应写出文字说明，证明过程或演算步骤) 17．(本小题满分10分)计算（1）21412⎪⎭⎫ ⎝⎛－(－9.6)0－32-833⎪⎭⎫ ⎝⎛＋(1.5)－2； （2）log 2512·log 45－log 133－log 24＋5log 52.【解析】 （1）21412⎪⎭⎫ ⎝⎛－(－9.6)0－32-833⎪⎭⎫ ⎝⎛＋(1.5)－2 ＝2149⎪⎭⎫ ⎝⎛－1－32-827⎪⎭⎫ ⎝⎛＋2-23⎪⎭⎫ ⎝⎛ ＝32－1－2-23⎪⎭⎫⎝⎛＋＝32－1－49＋49＝12. (2)log 2512·log 45－log 133－log 24＋5log 52＝－14＋1－2＋2＝34.818．(本小题满分12分)已知函数f (x )＝2cos x ·sin ⎪⎭⎫⎝⎛+3πx －3sin 2x ＋sin x cos x . （1）当x Ⅰ⎥⎦⎤⎢⎣⎡2,0π时，求f (x )的值域； （2）用“五点法”在下图中作出y ＝f (x )在闭区间⎥⎦⎤⎢⎣⎡-65,6ππ上的简图．【解析】 f (x )＝2cos x ·sin ⎪⎭⎫⎝⎛+3πx －3sin 2x ＋sin x cos x ＝2cos x ⎝⎛⎭⎫sin x cos π3＋cos x sin π3－3sin 2x ＋sin x cos x ＝sin2x ＋3cos2x ＝2sin ⎪⎭⎫ ⎝⎛+32πx . (1)Ⅰx Ⅰ⎥⎦⎤⎢⎣⎡2,0π，Ⅰπ3≤2x ＋π3≤4π3， Ⅰ－32≤sin ⎪⎭⎫ ⎝⎛+32πx ≤1， Ⅰ当x Ⅰ⎥⎦⎤⎢⎣⎡2,0π时，f (x )的值域为[－3，2]． (2)由T ＝2π2，得最小正周期T ＝π，列表：x－π6π12π37π125π692x ＋π3π2π3π22π2sin ⎪⎭⎫⎝⎛+32πx 0 2 0 －2 019．(本小题满分12分) 已知A (cos α，sin α)，B (cos β，sin β)，其中α，β为锐角，且|AB |＝105. （1）求cos(α－β)的值； （2）若cos α＝35，求cos β的值．【解析】 (1)由|AB |＝105， 得(cos α－cos β)2＋(sin α－sin β)2＝105， Ⅰ2－2(cos αcos β＋sin αsin β)＝25，Ⅰcos(α－β)＝45.(2)Ⅰcos α＝35，cos(α－β)＝45，α，β为锐角，10Ⅰsin α＝45，sin(α－β)＝±35.当sin(α－β)＝35时，cos β＝cos[α－(α－β)]＝cos αcos(α－β)＋sin αsin(α－β)＝2425. 当sin(α－β)＝－35时，cos β＝cos[α－(α－β)]＝cos αcos(α－β)＋sin αsin(α－β)＝0. Ⅰβ为锐角，Ⅰcos β＝2425.20．(本小题满分12分)已知函数f (x )是定义在区间[－1,1]上的奇函数，对于任意的m ，n Ⅰ[－1,1]有f (m )＋f (n )m ＋n>0(m ＋n ≠0)．（1）判断函数f (x )的单调性；（2）解不等式f ⎪⎭⎫⎝⎛+21x <f (1－x )． 【解析】 (1)设x 1＝m ，x 2＝－n ，由已知可得f (x 1)－f (x 2)x 1－x 2>0，不妨设x 1<x 2，则f (x 1)<f (x 2)，由函数单调性的定义可得函数f (x )在区间[－1,1]上是增函数．(2)由(1)知函数在区间[－1,1]上是增函数．又由f ⎝⎛⎭⎫x ＋12<f (1－x )，得⎩⎪⎨⎪⎧－1≤x ＋12≤1，－1≤1－x ≤1，x ＋12<1－x ，解得0≤x <14.所以不等式f ⎝⎛⎭⎫x ＋12<f (1－x )的解集为⎩⎨⎧⎭⎬⎫x |0≤x <14.11 21．(本小题满分12分)某村电费收取有以下两种方案供用户选择：方案一：每户每月收管理费2元，月用电不超过30度时，每度0.5元，超过30度时，超过部分按每度0.6元收取．方案二：不收管理费，每度0.58元．（1）求方案一收费L (x )(单位：元)与用电量x (单位：度)间的函数关系；（2）老王家九月份按方案一交费35元，问老王家该月用电多少度？（3）老王家月用电量在什么范围时，选择方案一比选择方案二更好？【解析】 (1)当0≤x ≤30时，L (x )＝2＋0.5x ；当x >30时，L (x )＝2＋30×0.5＋(x －30)×0.6＝0.6x －1，ⅠL (x )＝⎩⎪⎨⎪⎧ 2＋0.5x ，0≤x ≤30，0.6x －1，x >30.(注：x 也可不取0)(2)当0≤x ≤30时，令L (x )＝2＋0.5x ＝35得x ＝66，舍去；当x >30时，由L (x )＝0.6x －1＝35得x ＝60，Ⅰ老王家该月用电60度．(3)设按方案二收费为F (x )元，则F (x )＝0.58x .当0≤x ≤30时，由L (x )<F (x )，得2＋0.5x <0.58x ，解得x >25，Ⅰ25<x ≤30；当x >30时，由L (x )<F (x )，得0.6x －1<0.58x ，解得x <50，Ⅰ30<x <50.综上，25<x <50.故老王家月用电量在25度到50度范围内(不含25度、50度)时，选择方案一比方案二更好．22．(本小题满分12分)已知函数f (x )＝A sin(ωx ＋φ)＋B (A >0，ω>0)的一系列对应值如表： x －π6 π3 5π6 4π3 11π6 7π3 17π612 f (x ) －1 1 3 1 －1 1 3（1（2）根据(1)的结果，若函数y ＝f (kx )(k >0)的周期为2π3，当x Ⅰ⎥⎦⎤⎢⎣⎡30π，时，方程f (kx )＝m 恰有两个不同的解，求实数m 的取值范围．【解析】 (1)设f (x )的最小正周期为T ，则T ＝11π6－⎝⎛⎭⎫－π6＝2π，由T ＝2πω，得ω＝1， 又⎩⎪⎨⎪⎧ B ＋A ＝3，B －A ＝－1，解得⎩⎪⎨⎪⎧A ＝2，B ＝1， 令ω·5π6＋φ＝π2＋2k π，k ⅠZ ， 即5π6＋φ＝π2＋2k π，k ⅠZ ，取φ＝－π3， 所以f (x )＝2sin ⎝⎛⎭⎫x －π3＋1. (2)因为函数y ＝f (kx )＝2sin ⎝⎛⎭⎫kx －π3＋1的周期为2π3，又k >0，所以k ＝3.令t ＝3x －π3， 因为x Ⅰ⎥⎦⎤⎢⎣⎡30π，，所以t Ⅰ⎣⎡⎦⎤－π3，2π3， 如图，sin t ＝s 在⎣⎡⎦⎤－π3，2π3上有两个不同的解，则s Ⅰ⎣⎡⎭⎫32，1，所以方程f (kx )＝m 在x Ⅰ⎣⎡⎦⎤0，π3时恰好有两个不同的解，则m Ⅰ[3＋1,3)，即实数m 的取值范围是[3＋1,3)．13。

最新人教版高中英语必修一综合测试题全套及答案:::£单元检测卷（一）必修1 Unit 1I •阅读理解I had a friend Jordan who had permissive（宽容白勺）parents who permitted him to do whatever he wan ted to do. He seldom atte nded school. His mother stated that since he was a tee nager, he was adult eno ugh to do anything he wan ted to do. He subseque ntly en ded up living in poverty with a series of dead-end （没前途的）jobs and his two daughters did not fare（进展，过日子）any better.Children often complain about having strict parents but strict parents offer clear guidelines for them to adhere to （遵从）.The main goals of strict parents are to have highly successful children who are happy with their lives・ Strict parents want their children's lives to be materially, educationally, and economically better than their lives have been.My beloved father was strict and loving・ Although my mother was strict, she was not as strict as my father was. He argued that socialization and extracurricular （课夕啲）activities should be within strict parameters. He believed that intellectual pursuits （理智H勺追求）were primary. He further claimed that I should read three books weekly and magazines such as Times and Newsweek・He thought that fashion magazines were an utter waste of time.My father let me socialize with other childre n withi n some para mete rs ・He believed that I could not learn from other children and my time would be better spent on more constructive activities ・I spent much of my time reading books, newspapers, and/or other intellectual, educational, and cultural hobbies. I attended parties and other activities within severe limitations while I was attending high school and college. Education was foremost and uppermost in my father' s mind. He holds the strongly belief that parties and other forms of socialization can be freely addicted to after I have completed my education.I was teased by my friends and relatives about having a strict father. However, I am very thankful for this・ As a result of my father's guidanee, I was kept on the straight and narrow path. I was not a bad guy and I was exposed to varied cultural activities such as museums, plays, and dancing school. As a result of my childrearing, I am very appreciative of the better things of life and have no patie nee for the comm on aspects of life ・If a child has strict parents，he/she is surely blessed. Strict parents care and love their children and only want the very best for them・ Furthermore, a child knows why he/she stands with a strict par ent. Strict pare nts are not afraid to be pare nts and the world needs more strict pare nts.:LOn the whole, the first paragraph serves as a ___________ ・A.negative instaneeB. concept definitionC. topic introductionD. detailed explanation2.____________________________________________________________ The meaning ofthe underlined word "parameter" may be closest to " __________________________ "•A.identificationB. considerationC. connection D・ limitation3 ・What is a strict father's attitude towards socialization activities after their kids graduated from college?A.approvingB. negativeC. neutral D・ positive4.Which of the following proverbs may support the main idea of this passage?A.Great hopes make great man.B.Strict masters produce outstanding students.C.To be severe with on eself and kind with others ・D.Your father's honor is to you but a second-hand honor.MEGADIM, Israel （AP） - The worst to rest fire in Israel's history on Thursday destroyed oneof the country's few forested areas, killing at least 36 guards on their way to rescue prisoners there, destroying homes and forcing the evacuation（疏散）of thousands.The fire ran through the Carmel forest in Is rael's Galilee, reaching the coastal city of Haifa, jumping from place to place in the forest. The fire broke out around midday aind quickly spread and was still burning out of control as midnight approached・Investigators（调查者）supposed that the fire could have been set accideintally, or it might have been a criminal act, but pretty much ruled out（列卡除）some sort of attack by a Palestinian group ・"This is a huge disaster/ Prime M inister Benjami n Neta ny ahu said. "The go ver nme nt is using all means to control the fire/ He said some countries such as Cyprus, Italy, Russia and Greece agreed to provide backup.Most of the dead were Prison Service guards racing through the fire toward a prison to evacuate the prisoners, most of them Palestinians. A tree fell across the road, blocking their bus. Some guards were burned alive in side, while others died as they tried to escape ・ Fourtee n bodies were found near the bur nt bus 10 hours after the fire started ・The fire heavily damaged one of Israel's few large forests, made up of natural growth and plan ted areas, a favorite place for camping a nd a home for doze ns of species of wildlife ・ Forestry workers tried to evacuate animals from the fire・The forest recovered slowly from a fire in 1989, but experts said Thursday's big fire was many times worse.5.Some Prison Service guards died when they tried to ______ .A.control the running prisonersB.put out the fierce forest fireC.get away from the burning busD.save the guards trapped in the fire6.According to the investigators, which of the following might be the cause of the fire?A.Prisoners set the fire purposely.B.The fire broke out all by itself.C.A Palestinian group did it.D・ The fire was started by accident・7.We learn from the text that _____ .A.all of the dead were Prison Service guard sB.the forest once suffered a fire in the 1980sC.the animals didn't suffer from the fireD.the fire caused 50 d eaths altogether8.The text is mainly about _____ .A .the worst forest fire in Israel's historyB.the actual cause of the worst ever forest fireC .the damage caused by the forest fireD. the government"s efforts in controlling the fireII •语法填空I was walking with a friend yesterday 1 ________ I found an iPhone 5s in a pile of snow. I picked 2 _____ up and it still worked! I started looking through the name list and dialing numbersto see whether I could figure out whose phone it was・ 3 ________ , only one person answered and she didn't recognize the number. Since that lady didn't know whose phone it was, my friend suggested that we go back to 4 _______________ house in front of which we found the iPhone 5s. I knoekedon the door and two college-age persons answered・ They 5 ______ (lose) theiir phones, but as Itold them the story and mentioned some 6________ on the name list, the young man said that healso had those names and it 7 _______ be the phone of their friends. I really didn't have anythingto say, because in my mind I was just doing the right thing. I 8 _______ (immediate) tried to findthe owner, because that is exactly 9 ______ I would want some one to do if either I or any of my family members lost phones. The purpose of helping others is not for return. 10 __________ (help) others is more than enough for me and it is the most endless joy of life.1.______2. _________3. __________4. _________5. _________6. _______7. _________8. __________9. _________ 10. ________III.短文改错导Conflicts with others are common in everyday life. During the basketball game yesterday after no on, Su Hua and Li Jiang bumped into each other, tried to catch the ball ・ The n they started shouting and yelling, but it turned into a horrible quarrel.To be honest, it was Sifs fault, but Li was also to blame for—they were ruot calmly enough and both said some really mean things. They cared too much winning and losing・As a matter of fact, blocking, pushing and bumping are just part of a toughly game・To avoid such conflicts, we should be kind to one another; that is essential to enjoy a harmonious life. It is also a virtue to forgive and forget, especial in such a competitive and stressful society ・ In stead of blami ng each other, we should comm un icate more and put us in others' place. Don't be self-ce nter and try to be con siderate ・ We must lear n to handle conf licts calmly and wisely. 答案：I •阅读理解A语篇导读：本文作者认为严格的父母更容易造就成功的子女。

高二语文选择性必修一第一章综合练习
一、阅读理解
1. A
2. B
3. C
二、词语填空
1. 应用
2. 丰富
3. 熟悉
三、改错题
1. 将“是用”的“是”删除
2. 将“变傻”的“傻”改为“痴”
3. 将“讲究好的属”的“讲究”改为“讲靠”，将“属”改为“树”
四、写作题
请根据自己的经历或观点，写出不少于200字的作文。

主题：对于电子书的看法
随着科技的发展，电子书逐渐取代了传统纸质书成为主流。

对于这一现象，我的看法是：
首先，电子书的出现提供了更多方便快捷的阅读方式。

只需要一台电子设备，就可以随时随地获取各种图书，无需携带厚重的纸质书。

这方便了我们的阅读体验。

其次，电子书的存储量远远超过了纸质书。

传统纸质书在一定空间限制下，只能容纳有限的书籍。

而电子书可以通过云存储等技术，储存大量的图书资源，满足读者不同的阅读需求。

然而，我也认识到电子书带来的一些问题。

首先，长时间阅读电子书对眼睛造成一定的伤害。

其次，与纸质书相比，电子书的版权保护较为脆弱，容易出现盗版等问题。

综上所述，电子书既带来了便利又存在一些问题。

我们应该合理应用电子书，享受其便利的同时也要关注其问题，并努力寻找解决方案。

五、完形填空
1. A
2. C
3. D
以上是对高二语文选择性必修一第一章综合练习的答案整理。