I. Multiple choice questions
1. If the function f(x) is odd function and there are three zeros x 1, x2, x3, then the value of x 1+x2+x3 is ().
A.- 1
C.3 D. Not sure
,
x0=- 1+02=-0.5
f(x0)= 3.375 & gt； 0 [- 1,-0.5]
x 1 =- 1+(-0.5)2 =-0.75 f(x 1)≈ 1.578 & gt； 0 [- 1,-0.75]
x2 =- 1+(-0.75)2 =-0.875 f(x2)≈0.393 & gt； 0 [- 1,-0.875]
x3 =- 1-0.8752 =-0.9375 f(x3)≈- 0.277 & lt； 0 [-0.9375,-0.875]
∫|-0.875-(-0.9375)| = 0.0625 & lt； 0. 1,
The original equation is accurate to 0. 1 inch (-1, 0), and the approximate solution is -0.9.
17. If the function f(x)=log3(ax2-x+a) has zero, find the value range of a. 。
[resolution] ∫f(x)= log3(ax2-x+a) has zero.
∴log3(ax2-x+a)=0 has a solution. ∴ax2-x+a= 1 There is a solution.
When a=0, x=- 1.
When a≠0, if ax2-x+a- 1=0 has a solution,
Then δ = 1-4a (a- 1) ≥ 0, that is, 4a2-4a- 1≤0,
The solution is 1-22≤a≤ 1+22 and a≠0.
To sum up, 1-22≤a≤ 1+22.
18. Judge whether the equation x3-x- 1=0 has a real number solution in the interval [1, 1.5]; If yes, find the approximate solution (accurate to 0. 1).
[Analysis] Let the function f(x)=x3-x- 1, because f (1) =- 1
Take the midpoint of the interval (1, 1.5) as x 1= 1.25, and use a calculator to calculate F (1.25) =-0.30.
Then take the midpoint of (1.25, 1.5) x2= 1.375, and use a calculator to calculate f (1.375) ≈ 0.22 >; 0. Because of f( 1.25)? f( 1.375)& lt； 0, so x0∈( 1.25, 1.375).
Similarly, we can get x0∈( 1.3 125, 1.375), x0 ∈ (1.31.
Because | 1.34375-1.3 125 | < 0. 1, when the interval (1.3125,1.345)
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