Let f(x)=cos(pi*x/3)

g(x)=1^x+(−1)^x
h(x)=f(x)⋅g(x)
where x is a positive integer.

Given that h(x) has a range of exactly 3 different values: a, b and c,
what is the value of a^2+b^2+c^2?

for successive values of

g(x) = 0 if x is odd
g(x) = 2 if x is even

f(x) = 1/2 if x = 1,5,9,...
f(x) = -1/2 if x = 2,4,8,10,...

x f g
1 1/2 0
2 -1/2 2
3 -1 0
4 -1/2 2
5 1/2 0
6 1 2
This repeats, so fg has the values 0,1,-1

a^2+b^2+c^2 = 2

sory sir, but it says incorrect

To find the range of h(x), we need to evaluate h(x) for different values of x and observe the resulting values. Let's begin by analyzing each function separately:

1. f(x) = cos(pi*x/3)
The function f(x) is a periodic function with a period of 6. It oscillates between -1 and 1. When x is a multiple of 6, f(x) takes the value 1. When x is 3 more than a multiple of 6, f(x) takes the value -1. For other values of x, f(x) takes a value between -1 and 1.

2. g(x) = 1^x + (-1)^x
The function g(x) involves two components: 1^x and (-1)^x.
- For positive even values of x, both components are equal to 1, so g(x) = 1 + 1 = 2.
- For positive odd values of x, 1^x remains 1, but (-1)^x alternates between 1 and -1. Thus, g(x) will alternate between 2 and 0.

Now, let's analyze the composition of f(x) and g(x) to find h(x):

h(x) = f(x) * g(x)
= (cos(pi*x/3)) * (1^x + (-1)^x)

To determine the range of h(x), we need to determine all possible combinations of f(x) and g(x).

Case 1: When f(x) = 1 and g(x) = 2.
If f(x) = 1 and g(x) = 2, then h(x) = 1 * 2 = 2.

Case 2: When f(x) = -1 and g(x) = 2.
If f(x) = -1 and g(x) = 2, then h(x) = -1 * 2 = -2.

Case 3: When f(x) takes any value between -1 and 1, and g(x) = 0.
In this case, h(x) will be equal to 0.

Therefore, the range of h(x) is {2, -2, 0}.

To find a^2 + b^2 + c^2, we need to square each value and sum the squares:
a^2 + b^2 + c^2 = 2^2 + (-2)^2 + 0^2 = 4 + 4 + 0 = 8.

Therefore, a^2 + b^2 + c^2 = 8.