Hi, I'm in need of some math help. I'll award points for people with answers.

1) the range of f(x)=-1/4sin4x is

a) -1/4≤y≤0
b) -1/4≤y≤1/4
c) 0≤y≤1/4
d) -1≤y≤1
e) -4≤y≤4

2) if 0≤x≤2π and 4sin^2 x +4cosx-1=0, which of the following sets contains all values of x?

a) {π/3, π/6}
b) {π/3, 2π/3"
c) {2π/3, -2π/3}
d) {π/3, -π/3}
e) {2π/3, 4π/3}

3) which term in the expansion of (1/x + x^2) ^9 contains no power of x?

a) no term
b) sixth
c)fifth
d) third
e) fourth

4) if f(a) =2^a, then log2f(a)=

a)2
b) f(a)
c)a
d) 1/2^a
e) a^2

5) if log10(x+4)+log10(x-4)=1, then x=

a) negative radical 6
b) radical 26
c) 5
d) radical 6
e) 5/4

6) a rectangular box is inscribed in a cylinder of height 5 and with a circular base of radius 2.5 and AB=4. The volume of the box is

a) 20
b) 45
c) 15
d) 30
e) 60

1) f(x) = -1/4 sin 4x
On the graph, the amplitude is 1/4, meaning the sine function oscillates. The highest point reached vertically is +1/4 and the lowest point is -1/4. Thus the range is b) -1/4 ≤ y ≤ 1/4.

2) 4sin²x + 4cosx - 1 = 0
First you want to rewrite everything in terms of cosines. Recall the Pythagorean identity sin²x + cos²x = 1. Subtracting cos²x to both sides you get sin²x = 1 - cos²x. Therefore, you can substitute (1 - cos²x) in for sin²x into this equation:
4(1 - cos²x) + 4cosx - 1 = 0
Distribute.
4 - 4cos²x + 4cosx - 1 = 0
Simplify.
-4cos²x + 4cosx + 3 = 0
Divide by -1 (to make factoring easier).
4cos²x - 4cosx - 3 = 0
Factor.
(2cosx - 3)(2cosx + 1) = 0
Set each factor equal to 0. First factor:
2cosx - 3 = 0
Add 3.
2cosx = 3
Divide by 2.
cosx = 3/2
The cosine of any multiple cannot be greater than 1; this solution is invalid. The other factor:
2cosx + 1 = 0
Subtract 1.
2cosx = -1
Divide by 2.
cosx = -1/2
In the interval [0, 2π], the cosine is -1/2 twice: once when the angle is 2π/3 and once when it is 4π/3. Therefore, the answer is e) {2π/3, 4π/3}.