電子科技大學自主招生數(shù)學試題及答案

1、1.甲、乙、丙3人投籃,投進的概率分別是, , .現(xiàn)3人各投籃1次,求:()3人都投進的概率;()3人中恰有2人投進的概率.2.已知函數(shù)f(x)=sin(2x)+2sin2(x) (xR)()求函數(shù)f(x)的最小正周期 ; (2)求使函數(shù)f(x)取得最大值的x的集合.3.如圖,=l , A, B,點A在直線l 上的射影為A1, 點B在l的射影為B1,已知AB=2,AA1=1, BB1=, 求: () 直線AB分別與平面,所成角的大小; ()二面角A1ABB1的大小. 4已知正項數(shù)列an，其前n項和Sn滿足10Sn=an2+5an+6且a1,a3,a15成等比數(shù)列，求數(shù)列an的通項an .5 如

2、圖,三定點A(2,1),B(0,1),C(2,1); 三動點D,E,M滿足=t, = t , =t , t0,1. () 求動直線DE斜率的變化范圍; ()求動點M的軌跡方程.yxOMDABC11212BE6.已知函數(shù)f(x)=kx33x2+1(k0).()求函數(shù)f(x)的單調(diào)區(qū)間;()若函數(shù)f(x)的極小值大于0, 求k的取值范圍.【參考答案】1解: ()記"甲投進"為事件A1 , "乙投進"為事件A2 , "丙投進"為事件A3,則 P(A1)= , P(A2)= , P(A3)= , P(A1A2A3)=P(A1) ·P

3、(A2) ·P(A3) = × ×= 3人都投進的概率為() 設“3人中恰有2人投進"為事件BP(B)=P(A2A3)+P(A1A3)+P(A1A2) =P()·P(A2)·P(A3)+P(A1)·P()·P(A3)+P(A1)·P(A2)·P() =(1)× × + ×(1)× + × ×(1) = 3人中恰有2人投進的概率為2.解:() f(x)=sin(2x)+1cos2(x) = 2sin2(x) cos2(x)+1 =2sin

4、2(x)+1 = 2sin(2x) +1 T= ()當f(x)取最大值時, sin(2x)=1,有 2x =2k+ 即x=k+ (kZ) 所求x的集合為xR|x= k+ , (kZ).ABA1B1l第2題解法一圖EFABA1B1l第2題解法二圖yxyEF3.解法一: ()如圖, 連接A1B,AB1, , =l ,AA1l, BB1l, AA1, BB1. 則BAB1,ABA1分別是AB與和所成的角.RtBB1A中, BB1= , AB=2, sinBAB1 = = . BAB1=45°.RtAA1B中, AA1=1,AB=2, sinABA1= = , ABA1= 30°.

5、故AB與平面,所成的角分別是45°,30°.() BB1, 平面ABB1.在平面內(nèi)過A1作A1EAB1交AB1于E,則A1E平面AB1B.過E作EFAB交AB于F,連接A1F,則由三垂線定理得A1FAB, A1FE就是所求二面角的平面角.在RtABB1中,BAB1=45°,AB1=B1B=. RtAA1B中,A1B= = . 由AA1·A1B=A1F·AB得 A1F= = ,在RtA1EF中,sinA1FE = = , 二面角A1ABB1的大小為arcsin.解法二: ()同解法一.() 如圖,建立坐標系, 則A1(0,0,0),A(0,0,1

6、),B1(0,1,0),B(,1,0).在AB上取一點F(x,y,z),則存在tR,使得=t , 即(x,y,z1)=t(,1,1), 點F的坐標為(t, t,1t).要使,須·=0, 即(t, t,1t) ·(,1,1)=0, 2t+t(1t)=0,解得t= , 點F的坐標為(, ), =(, ). 設E為AB1的中點,則點E的坐標為(0, ). =(,).又·=(,)·(,1,1)= =0, , A1FE為所求二面角的平面角.又cosA1FE= = = = = ,二面角A1ABB1的大小為arccos.4.解: 10Sn=an2+5an+6, 10a

7、1=a12+5a1+6,解之得a1=2或a1=3. 又10Sn1=an12+5an1+6(n2), 由得 10an=(an2an12)+6(anan1),即(an+an1)(anan15)=0 an+an1>0 , anan1=5 (n2). 當a1=3時,a3=13,a15=73. a1, a3,a15不成等比數(shù)列a13;當a1=2時,a3=12, a15=72, 有a32=a1a15 , a1=2, an=5n3.5.解法一: 如圖, ()設D(x0,y0),E(xE,yE),M(x,y).由=t, = t , 知(xD2,yD1)=t(2,2). 同理 . kDE = = = 12

8、t. t0,1 , kDE1,1.() =t (x+2t2,y+2t1)=t(2t+2t2,2t1+2t1)=t(2,4t2)=(2t,4t22t). , y= , 即x2=4y. t0,1, x=2(12t)2,2.即所求軌跡方程為: x2=4y, x2,2解法二: ()同上.yxOMDABC11212BE第5題解法圖() 如圖, =+ = + t = + t() = (1t) +t, = + = +t = +t() =(1t) +t, = += + t= +t()=(1t) + t = (1t2) + 2(1t)t+t2 .設M點的坐標為(x,y),由=(2,1), =(0,1), =(2,1)得 消去t得x2=4y, t0,1, x2,2.故所求軌跡方程為: x2=4y, x2,26.解: （I）當k=0時, f(x)=3x2+1 f(x)的單調(diào)增區(qū)間為(,0,單調(diào)減區(qū)間0,+).當k>0時 , f '(x)=3kx2