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Given the function f (x) = x ^ 2-cosx, for any x1, X2 on [- π / 2, π / 2], we have the following conditions: A.x1>x2 B.x1^2>x2^2 C.|x1|>x2 The condition that f (x1) > F (x2) is constant is B. why
F (x) = x ^ 2-cosx is an even function on [- π / 2, π / 2] and decreases on [- π / 2,0] and increases on [0, π / 2]. So f (x) = x ^ 2-cosx decreases on [0, π / 2] by the symmetry of the even function
The range of function f (x) = LG (x2 + 8) is
The decreasing interval of the function y = Log1 / 2 ^ (x ^ 2-3x + 2) is
Y = log2x / 4log2x, X belongs to the range of [1 / 2,4]
The decreasing interval of y = log3 (x2 + 6x + 10) is
What are the answers
1. The range of y = LG (x ^ 2 + 8) is [3lg2, + ∞)
2. The decreasing interval of y = Log1 / 2 (x ^ 2-3x + 2) is (1, + ∞)
3. (the function is not very clear)
4. The decreasing interval of y = log3 (x ^ 2 + 6x + 10) is (- ∞, - 3]
If f (x) = [LG (10 ^ x + 1) + ax] (1 + 2 / 2 ^ x-1) is an odd function, then a is equal to
If f (x) = [LG (10 ^ x + 1) + ax] [1 + 2 / (2 ^ x-1)] is an odd function, then a is equal to?
A.0 B.1 C.1/2 D.-1/2
Let g (x) = LG (10 ^ x + 1) + AXH (x) = 1 + 2 / (2 ^ x-1) = (2 ^ X-1 + 2) / (2 ^ x-1) = (2 ^ x + 1) / (2 ^ x-1) H (- x) = [2 ^ (- x) + 1] / [2 ^ (- x) - 1] = (1 + 2 ^ x) / (1-2 ^ x) = - H (x) know h (x) odd function f (x) = g (x) * H (x) f (- x) = g (- x) * H (- x) = - G (- x) * H (x) get g (x) = g (- x) from F (x) = - f (- x)
RELATED INFORMATIONS
- 1. When x belongs to R, the function y = f (x) satisfies f (2002 + x) + F (2004 + x) = f (2003 + x), and f (1) = Lg3 / 2, f (2) = LG15, then f (2007) =?
- 2. The function f (x) is an even function on R, and is a decreasing function on (- 00,0). Compare the size of F (- 7 / 8) and f (2a ^ 2-A + 1)
- 3. It is known that the function f (x) defined on [0,1] satisfies the following three conditions simultaneously It is known that the function f (x) defined on [0,1] satisfies the following condition: when x1 ≥ 0, X2 ≥ 0, X1 + x2 ≤ 1, f (x1 + x2) ≥ f (x1) + F (x2) always holds, and whether the function g (x) = 2 ^ X - 1 satisfies this condition in the interval [0.1]
- 4. The range of the image of the function y = LG (2 + x) + LG (2-x) with respect to the symmetric function f (x) = LG (x2-2x + 1)
- 5. It is known that the odd function f (x) = - X's square + 2x (x > 0) 0 (x = 0) x's Square = MX (x)
- 6. Draw the image of function y = 1 / 2x + 1 in the given rectangular coordinate system (fill in the following table first, and then draw the line of points) x=-3 y=?…… x=3 y=?
- 7. Given the function f (x) = 2Sin (x + π / 3) - 2sinx, X belongs to - π / 2,0 (1) If cosx = √ 3 / 3, find the value of function f (x) (2) Finding the range of function f (x) I'm in a hurry
- 8. It is proved that the periods of F (x + a) = - f (x) and f (x + a) = 1 / F (x) are 2A The approximate step is f (x + 2a) = f [(x + a) + a] = - f (x + a) = f (x), so f (x + 2a) = f (x), so the period is 2A F (x + 2a) = f [(x + a) + a] = 1 / F (x + a) = f (x), so f (x + 2a) = f (x), so the period is 2A These are the teacher's steps, but I can't understand them now,
- 9. It is known that f (x) is an even function defined on R, and the image of y = f (x) is symmetric with respect to the line x = 2. It is proved that f (x) is a periodic function
- 10. We know that vector a = (cosx, SiNx), vector b = (cosy, siny), | A-B | = 2 * & 5 / 5, 1. Find the value of COS (X-Y) 2. If 0 Please write the specific steps (O & # 180; &;'O) ぉ (O & # 180; Д'o) は (O & # 180; ω'o) ょ
- 11. Known sequence {an}, {BN} satisfy: A1 = 1 / 4, an + BN = 1, BN + 1 = BN / 1-an ^ 2. Find {BN} general term formula
- 12. Given function, f (x) = 2x + 1, G (x0 = x), (x belongs to R, sequence, {an}, {BN} satisfies A1 = 1, an = f (BN) = g (BN + 1) to find the general term formula of sequence {an}
- 13. It is proved that the sequence LG (1 + an) is an equal ratio sequence It's known
- 14. Given that the image of the function y = MX - (4m-3) (M is a constant) passes through the origin, what is the value of M
- 15. We know that the image of the first-order function passes through the point P (2,3), and the image and the y-axis intersect at the point m, and the distance from the point m to the origin is equal to 5. Find the expression of the first-order function and draw the qualified image
- 16. If the image of the linear function y = MX - (M-3) passes through the origin, then the value of M is
- 17. If the image of the function y = MX - (4m-4) passes through the origin, then M=______ Y increases with the increase of X______ ．
- 18. (1) It is known that the image of the first-order function is parallel to the image of the positive scale function y = - 2 / 3x, and the value of the function is positive when it passes through points m (0,4), (2) x? And (3) if y is greater than or equal to - 1 and less than 2, then the value range of X is?
- 19. What is the vertex coordinate of the image of quadratic function y = (x - |) & # - 2?
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