For a parameter a, suppose u is the smallest positive real number such that 100cos^2u−100sin^2u=a+25 and v is the largest negative real number such that 25√3sinv−75cosv=a. For how many integer values of a is u+v non-negative?

16

wrong

36

wrong

13

To find the number of integer values of a for which u + v is non-negative, we need to determine the ranges of u and v individually and then find the possible overlapping values.

1. Range of u:
We are given the equation 100cos²u - 100sin²u = a + 25.

The left side of the equation can be simplified using the trigonometric identity: cos²u - sin²u = cos(2u).

So, the equation becomes 100cos(2u) = a + 25.

Taking the inverse cosine, we have 2u = arccos((a + 25)/100).

Since u is the smallest positive real number, the range of u can be written as u ∈ [0, 0.5arccos((a + 25)/100)].

2. Range of v:
We are given the equation 25√3sin(v) - 75cos(v) = a.

Dividing both sides by 25, we get √3sin(v) - 3cos(v) = a/25.

Squaring both sides of the equation, we have 3sin²(v) - 6√3sin(v)cos(v) + 9cos²(v) = (a/25)².

Using the trigonometric identity sin²(v) + cos²(v) = 1, the equation becomes 2cos²(v) - 6√3sin(v)cos(v) + 8cos²(v) = (a/25)².

Simplifying further, we have 10cos²(v) - 6√3sin(v)cos(v) = (a/25)².

Again, using the trigonometric identity cos²(v) - sin²(v) = cos(2v), the equation becomes 10cos(2v) = (a/25)² + 6√3sin(v)cos(v).

Taking the inverse cosine, we have 2v = arccos((a/25)² + 6√3sin(v)cos(v))/10.

Since v is the largest negative real number, the range of v can be written as v ∈ [-0.5arccos((a/25)² + 6√3sin(v)cos(v))/10, 0].

Now, we need to find the overlapping range of u and v to ensure that u + v is non-negative. To do that, we need to determine the values of a for which the lower bound of u is less than or equal to the upper bound of v.

0 ≤ 0.5arccos((a + 25)/100) ≤ -0.5arccos((a/25)² + 6√3sin(v)cos(v))/10

Simplifying the above inequalities, we get:

0 ≤ arccos((a + 25)/100) ≤ -arccos((a/25)² + 6√3sin(v)cos(v))/5

Since the range of the arccosine function is between 0 and π, and the second inequality implies a negative angle, we only need to consider the first inequality.

Let's solve the first inequality:

0 ≤ arccos((a + 25)/100)

Taking the cosine of both sides, we have:

cos(0) ≤ (a + 25)/100

Simplifying, we get:

1 ≤ (a + 25)/100

Multiplying both sides by 100, we have:

100 ≤ a + 25

Subtracting 25 from both sides, we get:

75 ≤ a

Therefore, there are 75 non-negative integer values for a.