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Topic source: homework help

The function f(x)=|2x+a|+2a, a∈R is known.

(i) If f (x) satisfies f(x)=f(3? X), find the value of a;

(ii) if x∈R exists, f(x)|2x? 1|+a holds, and the range of real number a is realistic.

Test center/site

Solution of absolute triangle inequality absolute inequality

analyse

(i) Find the symmetry axis of the function, get the equation about a, and solve it;

(2) Equivalent to | 2x+a |+2x-1|+a ≤ 0, let g(x) = | 2x+A |+2x- 1 |+A, find the minimum value of g(x), get the inequality about a, and then solve it.

explain

(i) Because f(x)=f(3? X), x∈R, so the image of f(x) is symmetrical about x=32.

What about the image with f(x)=2|x+a2|+2a? A2 is symmetrical, so? A2=32, so a=? 3.

(ⅱ)f(x)| 2x？ 1|+a is equivalent to |2x+a|+|2x? 1|+a？ 0.

Let g(x)=|2x+a|+|2x? 1|+a，

So g(x)min=|(2x+a)? (2x？ 1)|+a=|a+ 1|+a。

G(x)min？ 0, that is |a+ 1|+a? 0.

When a 1, a+ 1+a? 0, a 12, so? 1? a 12；

When a<? In 1,? (a+ 1)+a？ 0,? 1? 0, so a

To sum up, a 12.