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在求自变量的取值范围时,要注意题中隐含的条件,否则将给解题带来失误。现举例如下。 例1 已知关于x的方程(x-1)~2=2-t有两个实根x_1、x_2。设y=(x_1-x_2)~2。求y与t之间的函数关系式,并求自变量的取值范围。 错解:原方程可化为 x~2-2x+t-1=0。 则由根与系数的关系可得 x_1+x_2=2,x_1x_2=t-1。 ∴y=(x_1-x_2)~2=(x_1+x_2)~2-4x_1x_2 =2~2-4(t-1)=-4t+8。 t的取值范围是一切实数。 评析:上面的解法没有注意到题中隐含的条件,方程有实根,则△=4-4(t-1)≥0。 解得t≤2。因此,正确答案应为t≤2。 例2 如图1,ABCD是半径为1的圆内接等腰梯形,其下底是圆的直径。试求周长y与腰长x之间的函数关系式,并求自变量x的取值范围。 错解:过C作CE⊥AB于E,连AC。易证 Rt△ABC∽Rt△CBE。 ∴(BC)/(BE)=(AB)/(BC), 即BC~2=AB·BE。 于是,x~2=2BE。 ∴BE=1/2x~2,DC=AB-2BE=2-x~2。 ∴y=2+(2-x~2)+2x=-x~2+2x+4。
When seeking the range of the value of the independent variable, we must pay attention to the conditions implied in the question, otherwise it will cause mistakes in solving the problem. Examples are as follows. Example 1 It is known that the equation (x-1)~2=2-t for x has two real roots x_1, x_2. Let y=(x_1-x_2)~2. Find the function relation between y and t, and find the range of the value of the argument. Misunderstanding: The original equation can be changed to x~2-2x+t-1=0. Then the relation between the root and the coefficient is x_1+x_2=2 and x_1x_2=t-1. ∴y=(x_1-x_2)~2=(x_1+x_2)~2-4x_1x_2=2~2-4(t-1)=-4t+8. The range of t is all real numbers. Comment: The above solution did not notice the implied condition in the question. If the equation has a real root, △=4-4(t-1)≥0. Solution t ≤ 2. Therefore, the correct answer should be t ≤ 2. Example 2 As shown in Figure 1, ABCD is a circle with a radius of 1 inscribed in an isosceles trapezoid and its bottom is the diameter of the circle. Try to find the functional relationship between the perimeter y and the waist length x, and find the range of the value of the independent variable x. Misunderstanding: Over C for CE, AB for E, and AC. Easy to prove Rt△ABC∽Rt△CBE. ∴(BC)/(BE)=(AB)/(BC), ie BC~2=AB·BE. Thus, x~2=2BE. ∴ BE=1/2x~2, DC=AB-2BE=2-x~2. ∴y=2+(2-x~2)+2x=-x~2+2x+4.