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【中考调研室押题】2014中考特训卷专题四：阅读理解型问题

专题四阅读理解型问题

⊙热点一：阅读试题所提供新定义、新定理，解决新问题

1．(2013年上海)当三角形中一个内角α是另一个内角β的2倍时，我们称此三角形为“特征三角形”，其中α称为“特征角”．如果一个“特征三角形”的“特征角”为100°，那么这个“特征三角形”的最小内角的度数为__________．

2．(2012年湖南张家界)阅读材料：对于任何实数，我们规定符号??????a c b d 的意义是????

??a c b d ＝ad －bc .例如：??????1 23 4＝1×4－2×3＝－2，????

??－2 43 5＝(－2)×5－4×3＝－22. (1)按照这个规定，请你计算????

??5 67 8的值； (2)按照这个规定，请你计算：当x 2－4x ＋4＝0时，????

??x ＋1 2x x －1 2x －3的值．

⊙热点二：阅读试题信息，归纳总结提炼数学思想方法 1．(2013年湖北黄石)在计数制中，通常我们使用的是“十进位制”，即“逢十进一”，而计数制方法很多，如60进位制：60秒化为1分，60分化为1小时；24进位制：24小时化为一天；7进位制：7天化为1周等…而二进位制是计算机处理数据的依据．已知二进位十进位制 0 1 2 3 4 5 6 …

二进位制 0 1 10 11 100 101 110 …

(二)2．(2013年四川凉山州)先阅读以下材料，然后解答问题：

材料：将二次函数y ＝－x 2＋2x ＋3的图象向左平移1个单位，再向下平移2个单位，求平移后的抛物线的解析式(平移后抛物线的形状不变)．

解：在抛物线y ＝－x 2＋2x ＋3图象上任取两点A (0,3)，B (1,4)，由题意知：点A 向左平移1个单位得到A ′(－1,3)，再向下平移2个单位得到A ″(－1,1)；点B 向左平移1个单位得到B ′(0,4)，再向下平移2个单位得到B ″(0,2)．

设平移后的抛物线的解析式为y ＝－x 2＋bx ＋c .则点A ″(－1,1)，B ″(0,2)在抛物线上．

可得：????? －1－b ＋c ＝1，c ＝2.解得?

????

b ＝0，

c ＝2. 所以平移后的抛物线的解析式为y ＝－x 2＋2.

根据以上信息解答下列问题：

将直线y ＝2x －3向右平移3个单位，再向上平移1个单位，求平移后的直线的解析式．

⊙热点三：阅读试题信息，借助已有方法或通过归纳探索解决新问题

1．(2012年湖北十堰)阅读材料：

例：说明代数式x 2＋1＋(x －3)2＋4的几何意义，并求它的最小值．

解：x 2＋1＋(x －3)2＋4＝(x －0)2＋12＋(x －3)2＋22，如图Z4-5，建立平面直角坐标系，点P (x,0)是x 轴上一点，则(x －0)2＋12可以看成点P 与点A (0,1)的距离，(x －3)2＋22可以看成点P 与点B (3,2)的距离，所以原代数式的值可以看成线段P A 与PB 的长度之和，它的最小值就是P A ＋PB 的最小值．

设点A 关于x 轴的对称点为A ′，则P A ＝P A ′，因此，求P A ＋PB 的最小值，只需求P A ′＋PB 的最小值，而点A ′，B 间的直线段距离最短，所以P A ′＋PB 的最小值为线段A ′B 的长度．为此，构造直角三角形A ′CB ，因为A ′C ＝3，CB ＝3，所以A ′B ＝3 2，即原式的最小值为3 2.

图Z4-5

根据以上阅读材料，解答下列问题：

(1)代数式(x －1)2＋12＋(x －2)2＋9的值可以看成平面直角坐标系中点P (x,0)与点

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