![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
Given the function y = SiNx + cosx, the following four propositions are given ① If x ∈ [0, π / 2], then y ∈ (0, √ 2)]. ② the straight line x = π / 4 is an axis of symmetry of the image of function y = SiNx + cosx. ③ in the interval [π / 4,5 π / 4], the function y = SiNx + cosx is an increasing function. ④ the image of function y = SiNx + cosx can be obtained by translating π / 4 units from the image of y = √ 2sinx to the right, in which the correct proposition number is?
y=sinx+cosx
=√2((√2/2)sinx+(√2/2)cosx)
=√2sin(x+π/4)
Therefore, the image of this function is to shift y = √ 2sinx to the left π / 4
So it's wrong
① When x ∈ [0, π / 2], it is exactly a section of y = √ 2sinx in [π / 4,3 π / 4], y ∈ (2, √ 2), so it is incorrect
② If the ordinate axis is shifted to x = π / 4, the function image is consistent with y = √ 2cosx, so x = π / 4 is the axis of symmetry
Similarly, in the interval [π / 4,5 π / 4], the function y = SiNx + cosx is a decreasing function, so ③ is incorrect
There is a truth and falseness that x belongs to R, SiNx + cosx = 2
False
sinX≤1,cosX≤1,
If SiNx + cosx = 2, then SiNx = 1, cosx = 1,
SiNx = 1, then x = 2K π + & frac14; π,
If cosx = 1, then x = 2m π + π,
There is no solution,
RELATED INFORMATIONS
- 1. It is known that the proposition p: ∃ m ∈ R, M + 1 ≤ 0, proposition q: ∀ x ∈ R, X2 + MX + 1 ＞ 0 is tenable. If P ∧ q is a false proposition and P ∨ q is a true proposition, then the value range of real number m is () A. M ≥ 2B. M ≤ - 2 or - 1 < m < 2C. M ≤ - 2 or m ≥ 2D. - 2 ≤ m ≤ 2
- 2. Given that R (x): SiNx + cosx > m, s (x): x ^ 2 + MX + 1 > 0, if any x belongs to R, R (x) is a false proposition and S (x) is a true proposition, the value range of M is obtained For R (x): ① radical 2sinx (x + &# 188; π) > M ｜ m ＜ - radical 2 Another false proposition 〈 m ≥ - radical 2 ② Root sign 2sinx (x + &# 188; π) > m False proposition The root sign 2sinx (x + &# 188; π) ≤ M ｜ m ≥ root 2 That's right. Why?
- 3. Given that the function f (x) is an odd function on R, when x > 0, f (x) = log2 (1 + x), the inequality f (x0) about x0 is solved
- 4. Given that y in the first-order function y = KX + 2K decreases with the increase of X, then the inequality KX + 2K
- 5. Given the function f (x) = (M + 1) x ^ 2 - (m-1) x + M-1, if M belongs to (- 1,1) inequality and the solution set of F (x) > 0 is r, the value range of X is obtained Please give the detailed process, thank you, this kind of question is a little dizzy, usually is to find the scope of M, thank you!
- 6. It is known that function FX is an odd function defined on R, and f (x + 2) is equal to negative FX. If F1 is equal to 1, then f (3) - f (4)=
- 7. Even function f (x) is an increasing function on [0, positive infinity]. If f (AX + 1) > F (x-3) is constant on X ∈ [1,2], then the value range of real number a is----------------
- 8. A: If the square of x = 9, then x = 3 and B: 6 are the arithmetic square roots of (- 6). Which is right? I think both of them are right,
- 9. It is known that X and y are positive real numbers. (1) proof: 2XY / x + y
- 10. X is greater than 0, y is greater than 0, 2x + 5Y = 20, find the minimum value of X / 1 + Y / 1
- 11. Finding function derivative f (x) = (x + 1) (x-1) (x ^ 2 + 2) given function y = cosx / X finding tangent equation of function at x = π Finding function derivative f (x) = (x + 1) (x-1) (x ^ 2 + 2) Given the function y = cosx / x, find the tangent equation at x = π
- 12. As shown in the figure, in the trapezoidal ABCD, ab ∥ CD, ab = 3CD, diagonal lines AC and BD intersect at point O, and median lines EF intersect with AC and BD at two points m and N respectively, then the area of the shadow part in the figure is () A. 12B. 13C. 14D. 47
- 13. It is known that the coordinates of the three vertices of △ ABC are a (- 5,0) B (3, - 3) C (0,2), and the linear equations of height on the BC edge are obtained respectively
- 14. As shown in the figure, De is the median line of △ ABC, M is the midpoint of De, and the extension line of CM intersects AB at point n, then s △ DMN: s quadrilateral anme equals () A. 1：5B. 1：4C. 2：5D. 2：7
- 15. As shown in the figure, ⊙ O's chord AB and AC have an included angle of 50 °, Mn is the midpoint of arc AB and arc AC respectively, and OM and on intersect AB and AC at points E and f respectively, then the degree of ∠ mon is () A. 110°B. 120°C. 130°D. 100°
- 16. It is known that in ⊙ o, n is the midpoint of the chord AB, on intersects the arc AB in M, if AB = 2, radical 3, Mn = 1, the radius of ⊙ o is obtained
- 17. For the two strings AB and Cd in the circle O, AB > CD, the distances from the center O to AB and CD are OM and on respectively, then om on is filled with ＞＜ =
- 18. It is known that the circle O and the circle K are the big circle and the small circle of the ball o, and their common chord length is equal to the radius of the ball o, OK = 3 / 2, and the angle between the circle O and the plane of the circle K is 60 degrees, Then the surface area of ball 0 is?
- 19. Let m.n be two points on the radius op of the ball 0 and NP = Mn = OM. Make three circles that are perpendicular to the surface of OP through m.n.o. respectively, then the area ratio of these circles is
- 20. As shown in the figure, ⊙ o, if the chord ad ∥ BC, Da = DC, ∠ AOC = 160 °, then ∠ BCO is equal to______ Degree