(备战2013中考15分钟精华题)考点34新概念及阅读理解型问题

新概念及阅读理解型问题

一、选择题

1.古希腊著名的毕达哥拉斯派1、3、6、10、…这样的数称为“三角形数”，而把1、4、9、16…这样的数称为“正方形数”.从图中可以发现，任何一个大于1的“正方形数”都可以看作两个相邻的“三角形数”之和，下列等式中，符合这一规律的是（） A.13=3+10 B.25=9+16 C.36=15+21 D.49=18+31

2.已知222

22112

11,c x b x a y c x b x a y ++=++=且满足

)1,0(2

2121≠===k k c c b b a a .则称抛物线21,y y 互为“友好抛物线”，则下列关于“友好抛物线”的说法不正确的是（）

A 、y 1,y 2开口方向，开口大小不一定相同

B 、因为y 1,y 2的对称轴相同

C 、如果y 2的最值为m ，则y 1的最值为km

D 、如果y 2与x 轴的两交点间距离为d ，则y 1与x 轴的两交点间距离为d k

3.科技馆为某机器人编制一段程序，如果机器人在平地上按照图

中所示的步骤行走，那么该机器人所走的总路程为（）．

A ．6米

B ．8米

C ．12米

D ．不能确定

二、填空题 1.【原创】刘谦的魔术表演风靡全国，小明也学起了刘谦发明了

一个魔术盒，当任意实数对(a ，b)进入其中时，会得到一个新的实数：a 2＋b －1，例如把(3，－2)放入其中，就会得到：32＋（－2）－1＝6．现将实数对(－1，3)放入其中，得到实数m ，再将实数对(m ，1)放入其中后，得到的实数是_______.

2.中国已经进入一个老龄化社会，“老人”是一个模糊概念，?有人想用“老人系数”来表

示一个人的老年化程度，其中一个人的“老人系数”与年龄的关系如图所示，按照这

第3题

样的规定，一个年龄为70岁的人，他的“老人系数”为 ▲ ．

3.我们常用的数是十进制数，而计算机程序处理数据使用的只有数码0和1的二进制数，这

二者可以相互换算，如将二进制数110换算成十进制数应为：1×22+1×21+0×20

=6 按此方式，则将十进制数11换算成二进制数应为___________

4.规定一种新运算a ※b=a 2

－2b,如1※2=-3，则2※（－2）= .

5.刘谦的魔术表演风靡全国，小明也学起了刘谦发明了一个魔术盒，当任意实数对(a ，b)进入其中时，会得到一个新的实数：a 2

＋b －1，例如把(3，－2)放入其中，就会得到：32

＋（－2）－1＝6．现将实数对 (－1，3)放入其中，得到实数m ，再将实数对(m ，1)放入其中后，得到的实数是．

6.定义新运算“*”，规则：()

()a a b a b b a b ≥?*=?

，如122*=

，(

=。若

210x x +-=的两根为12,x x ，则12x x *＝ ▲ ．

7.如图是某种计算机的程序示意图, 初始端输入x 后经“运算中心式子”

c bx ax ++2( 、

a b c 是常数, 且0>a , 0≠bc )处理后得到一个结果. 若这个结果大于0, 则输出此结果; 否则就将第一次得到的结果作为输入的x 再次运行程序……直到输出结

果为止. 若该程序满足条件：“存在实数，当输入x 的值等于时, 该程序的运算无法停止(即会一直循环运行) ”，请写出一个符合条件的运算中心式子以及相应的能使它一直循环运行的x 的值__ ▲____，____ ▲_____.

8.在数学中，为了简便，记1n

k k =∑＝1+2+3+…+(n －1)+ n ．1！＝1，2！＝2×1，3！＝3×2

×1，…，n ！＝n ×(n －1)×(n －2)×…×3×2×1．则k k ∑

=2009

－∑=2010

k k +

2009!

2010= ．

（第9题图）

9.如图是瑞典人科赫（Koch ）在1906年构造的能够描述雪花形状的科赫雪花图案．图形的作法是，从一个正三角形开始，把每条边分成三等份，然后以各边的中间长度为底边．分别向外作正三角形，再把“底边”线段抹掉．反复进行这一过程，就会得到一个“雪花”样子的曲线．这是一个极有特色的图形：在图形不断变换的过程中，它的周长趋于无穷大，而其面积却趋于定值．如果假定原正三角形边长为a ，则可算出下图每步变换后科赫雪花的周长：1C =3a ，

2C = ，3C = ，…，则n C = ．

答案：选择题 1、C 2、D 3、C

填空题 1、【答案】9 2、答案：0.5 3、答案：1011 4、答案6 5、答案：9 6、答案：

5- 7、答案：说明：只需使方程x c bx ax =++2有一个负数根即可；如32--x x ，1- 8、答案：0

9、答案：2C =433a ；3C =24()33a ；n C =14()33

n a - ，（1+1+2分）

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