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Let f (x) be an even function with period 2 defined on R. when x ∈ [0,1], f (x) = x + 1, then f (3) 2）=（ ） A. 1 B. 2 Three C. 1 Two D. 3 Two
∵ the function f (x) is defined on R with period 2,
f（ 3
2）=f（-1
2+2）=f（-1
2），
And ∵ the function f (x) is an even function defined on R,
∴f（-1
2）=f（1
2），
When x ∈ [0,1], f (x) = x + 1,
∴f（1
2）=1
2+1=3
Two
Therefore, D is selected
Let f (x) be an even function with period 2 defined on R, and the analytic expression on the interval [0,1] is ln (x + 1) Then f (- 6.5), f (- 1) and f (0) are
A:
F (x) is an even function on R: F (- x) = f (x)
The period is 2: F (x) = f (x + 2)
00
f(0)=ln(0+1)=0
So:
f(-1)>f(-6.5)>f(0)
Let f (x) be an even function with period 4. When x ∈ [0,2], f (x) = X-1, then the solution set of inequality XF (x) > 0 on [- 1,3] is______ .
If x ∈ [- 2,0], then - x ∈ [0,2], then f (- x) = - X-1,
∵ f (x) is an even function, ᙽ f (- x) = - X-1 = f (x), that is, f (x) = - X-1, X ∈ [- 2, 0],
If x ∈ [2,4], then x-4 ∈ [- 2,0],
∵ the period of the function is 4, ᙽ f (x) = f (x-4) = - (x-4) - 1 = 3-x,
That is, f (x)=
−x−1， −2≤x≤0
x−1， 0≤x≤2
3 − x, 2 ≤ x ≤ 4, make the function f (x) on [- 1, 3] as shown in the figure,
If 0 < x ≤ 3, then the inequality XF (x) > 0 is equivalent to f (x) > 0, where 1 < x < 3,
If - 1 ≤ x ≤ 0, then the inequality XF (x) > 0 is equivalent to f (x) < 0,
To sum up, the solution set of the inequality XF (x) > 0 on [- 1,3] is (1,3) ∪ - 1,0,
So the answer is: (1,3) ∪ (- 1,0)
The function f (x) = cosx / 2 is an even function with minimum positive period of 4 π
T=2π/(1/2)=4π
f(-x)=cos(-x/2)=cos(x/2)=f(x)
Let f (x) = log4 (4x + 1) + KX (x ∈ R) be even functions (1) Find the value of K; (2) If the equation f (x) - M = 0 has a solution, find the range of M
(1) From the function f (x) = log4 (4x + 1) + KX (x ∈ R) is an even function, we can know that f (x) = f (- x) log4 (4x + 1) + KX = log4 (4-x + 1) - KX ((2 points), namely log44x + 14 − x + 1 = − 2kx ﹥ log44x = - 2kx (4 points) ᚄ x = - 2kx holds for X ∈ R
If f (x) is an even function defined on R with period 3 and f (2) = 0, then the minimum number of solutions of the equation f (x) = 0 in the interval (0, 6) is () A. 5 B. 4 C. 3 D. 2
∵ f (x) is an even function defined on R, and the period is 3, f (2) = 0, ᙽ f (- 2) = 0,
∴f（5）=f（2）=0，f（1）=f（-2）=0，f（4）=f（1）=0．
That is, in the interval (0, 6), the,
f（2）=0，f（5）=0，f（1）=0，f（4）=0，
So answer: B
If f (x) is an even function defined on R with period 3 and f (2) = 0, then the minimum number of solutions of the equation f (x) = 0 in the interval (0, 6) is () A. 5 B. 4 C. 3 D. 2
∵ f (x) is an even function defined on R, and the period is 3, f (2) = 0, ᙽ f (- 2) = 0,
∴f（5）=f（2）=0，f（1）=f（-2）=0，f（4）=f（1）=0．
That is, in the interval (0, 6), the,
f（2）=0，f（5）=0，f（1）=0，f（4）=0，
So answer: B
If f (x) is an even function defined on R with period 3 and f (2) = 0, then the minimum number of solutions of the equation f (x) = 0 in the interval (0, 6) is () A. 5 B. 4 C. 3 D. 2
∵ f (x) is an even function defined on R, and the period is 3, f (2) = 0, ᙽ f (- 2) = 0,
∴f（5）=f（2）=0，f（1）=f（-2）=0，f（4）=f（1）=0．
That is, in the interval (0, 6), the,
f（2）=0，f（5）=0，f（1）=0，f（4）=0，
So answer: B
If f (x) is an odd function defined on R with period 3 as its period, if f (2) = 0, then the number of solutions of the equation f (x) = 0 in the interval (0, 6) () A. It's three B. It's four C. It's five D. More than 5
∵ f (x) is an odd function defined on R with 3 as the period, f (2) = 0. If x ∈ (0, 6), we can get f (5) = f (2) = 0. According to f (x) as odd function, then f (- 2) = - f (2) = 0, and then f (4) = f (1) = f (- 2) = 0
If f (x) is an odd function defined on R with period 3 as its period, if f (2) = 0, then the number of solutions of the equation f (x) = 0 in the interval (0, 6) () A. It's three B. It's four C. It's five D. More than 5
∵ f (x) is an odd function defined on R with 3 as the period, f (2) = 0. If x ∈ (0, 6), we can get f (5) = f (2) = 0. According to f (x) as odd function, then f (- 2) = - f (2) = 0, and then f (4) = f (1) = f (- 2) = 0
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