题目内容
给出下列命题:①对于实数u,v,定义一种运算“*“为:u*v=uv+v.若关于x的方程x*(a*x)=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_ST/0.png)
②设直线kx+(k+1)y-1=0(k为正整数)与坐标轴所构成的直角三角形的面积为Sk,则S1+S2+S3+…+S2008=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_ST/1.png)
③函数y=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_ST/2.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_ST/3.png)
④甲、乙、丙3位同学选修课程,从4门课程中,甲选修2门,乙、丙各选修3门,则不同的选修方案共有48种.
其中真命题的个数有( )
A.1个
B.2个
C.3个
D.4个
【答案】分析:①根据新定义整理出一元二次方程,然后根据判别式△<0,方程没有实数根列式得到关于a的不等式,求解不等式即可判断;
②求出直线与坐标轴的交点坐标,再根据直角三角形的面积公式列式得到Sk的表达式,然后利用拆项法整理求解;
③先配方,再根据二次函数的最值问题求解;
④求出每一名同学的可能选修方法的种数,然后相乘即可得解.
解答:解:①根据新定义,x*(a*x)=x*(ax+x),
=x(ax+x)+(ax+x),
=(a+1)x2+(a+1)x,
所以,(a+1)x2+(a+1)x+
=0,
∵方程没有实数根,
∴△=(a+1)2-4(a+1)×
<0,
即a(a+1)<0,
解得-1<a<0,故本小题错误;
②当y=0时,kx-1=0,解得x=
,
当x=0时,(k+1)y-1=0,解得y=
,
所以,与x轴的交点坐标为(
,0),与y轴的交点坐标为(0,
),
∵k为正整数,
∴Sk=
×
×
=![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/9.png)
=
(
-
),
∴S1+S2+S3+…+S2008=
(1-
+
-
+
-
+…+
-
),
=
(1-
),
=
×
,
=
,故本小题正确;
③∵y=-
+
=-(
-
+
)+
=-(
-
)2+
,
∴当
=
,即x=
时,函数有最大值
,故本小题错误;
④设4门课程分别为1,2,3,4,甲选修2门,可有1,2;1,3;1,4;2,3;2,4;3,4共6种情况,
同理乙,丙均可有1,2,3;1,2,4;2,3,4;1,3,4共4种情况,
所以,不同的选修方案共有6×4×4=96种,故本小题错误;
综上所述,真命题有②共1个.
故选A.
点评:本题考查了一元二次方程的根的判别式,一次函数图象上点的坐标特征,二次函数的最值问题,排列组合,综合性较强,难度较大,对同学们的能力要求比较高.
②求出直线与坐标轴的交点坐标,再根据直角三角形的面积公式列式得到Sk的表达式,然后利用拆项法整理求解;
③先配方,再根据二次函数的最值问题求解;
④求出每一名同学的可能选修方法的种数,然后相乘即可得解.
解答:解:①根据新定义,x*(a*x)=x*(ax+x),
=x(ax+x)+(ax+x),
=(a+1)x2+(a+1)x,
所以,(a+1)x2+(a+1)x+
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/0.png)
∵方程没有实数根,
∴△=(a+1)2-4(a+1)×
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/1.png)
即a(a+1)<0,
解得-1<a<0,故本小题错误;
②当y=0时,kx-1=0,解得x=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/2.png)
当x=0时,(k+1)y-1=0,解得y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/3.png)
所以,与x轴的交点坐标为(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/4.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/5.png)
∵k为正整数,
∴Sk=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/6.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/7.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/8.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/9.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/10.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/11.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/12.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/13.png)
∴S1+S2+S3+…+S2008=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/14.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/15.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/16.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/17.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/18.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/19.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/20.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/21.png)
=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/22.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/23.png)
=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/24.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/25.png)
=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/26.png)
③∵y=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/27.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/28.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/29.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/30.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/31.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/32.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/33.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/34.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/35.png)
∴当
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/36.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/37.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/38.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101191316815591968/SYS201311011913168155919005_DA/39.png)
④设4门课程分别为1,2,3,4,甲选修2门,可有1,2;1,3;1,4;2,3;2,4;3,4共6种情况,
同理乙,丙均可有1,2,3;1,2,4;2,3,4;1,3,4共4种情况,
所以,不同的选修方案共有6×4×4=96种,故本小题错误;
综上所述,真命题有②共1个.
故选A.
点评:本题考查了一元二次方程的根的判别式,一次函数图象上点的坐标特征,二次函数的最值问题,排列组合,综合性较强,难度较大,对同学们的能力要求比较高.
![](http://thumb2018.1010pic.com/images/loading.gif)
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