![](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>)
Find the range. ① y = 2sinxcos ^ 2x / 1 + SiNx; ② y = Tan ^ 2x - TaNx + 1 / Tan ^ 2x + TaNx + 1
① Y = 2sinxcos ^ 2x (1-sinx) / (1 + SiNx) (1-sinx) = 2sinx (1-sinx), then the formula is y = - 2 [SiNx + (1 / 2)] ^ 2 + (1 / 2) from the image, y ∈ [- 4,1 / 2] ② y = 1 - [2tanx / (Tan ^ 2x + TaNx + 1)] = 1 - {2 / [TaNx + (1 / TaNx) + 1]} ∵ TaNx + (1 / TaNx) ≥ 2 ∩ y ∈ [1 / 3,1)
RELATED INFORMATIONS
- 1. The range of function y = (tan-1) / (TaNx + 1) is
- 2. The range of y = log2 (the square of X - 4x + 6)
- 3. Find the range y = log2 (square of X + 4x + 5)
- 4. Monotone interval of y = log2 {4x-x square}
- 5. Find the range of y = Log1 / 2 (x ^ 2 + 2x + 5)
- 6. Find the range of function y = Log & nbsp; 12 (- x2-2x + 3)
- 7. Find the range of function FX = - x2 + 2x + 3 (- 2 ≤ x ≤ 3)
- 8. Given the function FX = cos ^ 2x + asinx. When a = 2, find the range of function FX
- 9. The range of function - X & # 178; + 2x + 4 is
- 10. Given that the value range of function FX = ㏑ (X & # 178; + 1) is (0, 1, 2), then the number of functions satisfying this condition?
- 11. Find the range: y = x square - 2x - 2 / 5 (- 2)
- 12. The range of y = x squared + 2x / 3 squared - 4 is
- 13. Find the range y = - 2cos ^ 2x + 2sinx + 3 / 2, X ∈ [- π / 4, π / 4]
- 14. The range of y = 2cos (2x + π / 3) + 1, X ∈ [0, π / 2]
- 15. Find the range of y = (x2-2x-3) / (2x2 + 2x + 1) The range of y = (x2-2x-3) / (2x2 + 2x + 1) (2 after X means square)
- 16. The range of x2-2x-3 / 2x2 + 2x + 1
- 17. Let x1, X2 be the two real roots of the equation x2 + PX + q = 0, and X1 + 1, X2 + 1 be the two real roots of the equation x2 + QX + P = 0 about X, then P=______ ，q=______ ．
- 18. The root ratio of x ^ 2 + PX + q = 0 is 1:2, and the discriminant value is 1. Find the value of PQ and solve the equation
- 19. The solution of the equation PX + 8q = 333 of X is 1, and P, q are prime numbers, P
- 20. p. If q is prime and the equation x ^ 2-px + q = 0 has a positive integer root, then p =, Q=