1. Beijing middle school students' mathematics competition has a long history. In the past ten years, the mathematics competition for middle school students in Beijing has been held in Grade Two and Grade One. Starting from 1990, it is divided into initial test and second test. The initial test is mainly based on popularization, and the second test is moderately improved. The proposition is closely combined with the reality of middle school mathematics teaching, vivid but not difficult, interesting but not strange, clever but not biased, and strives to embody the comprehensive unity of science, knowledge, application, inspiration and interest. Math contest is a kind of extracurricular activity deeply loved by teenagers. Develop intelligence and guide students to improve their mathematics quality through interesting, innovative and level topics. Mathematical competition is a good form to implement mathematical quality. After more than ten years of mathematics competitions in Beijing, a number of good topics with shining mathematical ideas and wisdom have been accumulated. It is a happy thing to guide students to learn and appreciate. Below, the author tries to share the brilliant sunshine of mathematical wisdom with readers by solving the first question of the first grade mathematics competition in Beijing.

Second, the topic

Beijing 1992 Mathematics Competition Question 4 "II. The fill-in-the-blank questions for Grade One are as follows:

4. If sin2x+cosx+a=0 has real roots, what is the range of real number A?

The topic is short and dry, with a full score of 8.

Third, try to solve it

The number of knowledge in the equation is X, and X has two trigonometric functions, Sinx and Cosx. And Sin2x= 1-cos2x, well, change it, and the original equation becomes.

cos2x-cosx- 1-a=0 ①

If x in the original equation has a real root, then cosx will have a corresponding real number, so t= cosx, so equation ① becomes

t2-t- 1-a=0 ②

So equation ② should have real roots, so its discriminant △ = (-1) 2-4 (-1-a) = 4a+5 ≥ 0, so a≥-(5/4).

So the range of the real number A is a≥-(5/4).

Is this the right answer?

When a≥-(5/4), there must be △≥0, and equation ② must have a real number root. The question is, does cosx=t have a real number root and x has a real number root? Note that the range of cosine function is cosx ∈ [- 1, 1], so ② having real roots does not guarantee that cosx=t must be within [- 1, 1], indicating that the above solution is not rigorous, and students who are not careful in thinking may make mistakes here. This is a hidden trap set by the test questions.

Fourth, reflection.

What should we do?

If the real number solution T of equation ② can be guaranteed to be in the interval [- 1, 1], then the simplest triangular equation cosx=t must have a real number solution x = 2kπ arccost. Good, so the problem becomes when equation ② has real roots in [- 1, 1].

According to equation ②:

Therefore, when a ∈ [-(5/4), 1]∨[-(5/4),-1] = [-(5/4), 1], the original equation has a real number root about X.

The above method uses the root formula of quadratic equation of one variable, the inequality group composed of two unreasonable inequalities, and the intersection and union of sets. I feel practical in my heart, but the calculation is complicated. Is there a better way?

Improvement of verb (abbreviation of verb)

If the left end of equation ② is f(t), i.e.

f(t)=t2-t- 1-a

Then equation ② has a real number solution in [- 1, 1], which is equivalent to the image parabola of quadratic function f(t)=t2-t- 1-a intersecting with the t axis in [- 1, 1]. Number is transformed into shape and shape help number. All right, give it a try.

When there is only one intersection point between the parabola and the T axis in [- 1, 1], if and only if.

F(- 1)f( 1)≤0。

(1-a)(- 1-a)≤0, and the solution is-1≤ a ≤1; ③

When the parabola and the T-axis have two intersections in [- 1, 1], if and only if.

It can be seen from ③ ④ that when a ∈ [- 1] ∨ [-(5/4), 1] = [-(5/4),-1], y = f (.

Because of the calculation of f( 1), f(- 1), δ, etc. Simple. Is the above solution simpler?

Sixth, look at the problem from another angle.

The poem says: "Seen from one side of the ridge, the peaks are different in height." . I don't know the true face of Lushan Mountain, but I am only on this mountain. "Our previous problem-solving thinking focused on' the equation has a real root', and we couldn't jump out of the palm of the Tathagata's hand. The solution in "Five" is an ingenious solution, which permeates the transformation of numbers and shapes. If we look at the problem from another angle, we will convert the term of equation ① into.

a=cos2x-cosx- 1

Think of A as a function of X, and think in reverse: If X has a real number solution, there is COSX ∈ [- 1, 1], and a = [COSX-( 1/2)] 2-(5/4) When COSX = (6544). There is a maximum value of A =(9/4)-(5/4)= 1, so on the other hand, when a takes a value in [-(5/4), 1], cosx must be in [- 1,/kloc-0. You see, isn't the range of a found out?

VII. Variants

1989 the Monkey King in the Journey to the West is miraculous and changeable. Good math problems also have some variations. What else can you think of from the above solution? Can you adapt a corresponding topic? Give it a try.

Coincidentally, nine years later, in the first year of the new millennium, 200 1, the last question was "Two. The fill-in-the-blank question in the first year of Beijing middle school students' mathematics competition is like this: "8. If the equation sin2x+sinx+a=0 has a real number solution, find the sum of the maximum and minimum values of the number A. "

Readers appreciate this, will they "smile"

Eight. enlighten

Looking back at the above problem-solving process, we used the idea of equation, equivalent transformation, combination of numbers and shapes, changing perspective and reverse thinking. Thought endows wisdom, wisdom produces ingenious solutions, and ingenious solutions are intoxicating.