Prove that the function f (x) = cos(πx/2) + cos(πx) has a root in eachinterval [4k − 1, 4k] and [4k, 4k + 1], for every k ∈ Z. Relevant theoremsand properties of the cosine function may be freely used, if stated clearly

To prove that the function f(x) = cos(πx/2) + cos(πx) has a root in each interval [4k - 1, 4k] and [4k, 4k + 1], for every k ∈ Z, we can use the Intermediate Value Theorem (IVT) and properties of the cosine function.

The Intermediate Value Theorem states that if a function f is continuous on a closed interval [a, b], and k is any number between f(a) and f(b), then there exists at least one number c in the interval (a, b) such that f(c) = k.

The cosine function is continuous for all real numbers, so f(x) = cos(πx/2) + cos(πx) is also continuous for all real numbers.

Now, let's consider the interval [4k - 1, 4k].

At x = 4k - 1, we have f(4k - 1) = cos(2πk - π/2) + cos(2πk - π) = 0 - (-1) = 1.

At x = 4k, we have f(4k) = cos(2πk) + cos(2πk) = 1 + 1 = 2.

Since 0 is between 1 and 2, by the IVT, there exists a number c in the interval (4k - 1, 4k) such that f(c) = 0. Therefore, f(x) has a root in the interval [4k - 1, 4k].

Similarly, for the interval [4k, 4k + 1],

At x = 4k, we have f(4k) = cos(2πk) + cos(2πk) = 1 + 1 = 2.

At x = 4k + 1, we have f(4k + 1) = cos(2πk + π/2) + cos(2πk + π) = 0 - 1 = -1.

Since 0 is between -1 and 2, by the IVT, there exists a number c in the interval (4k, 4k + 1) such that f(c) = 0. Therefore, f(x) has a root in the interval [4k, 4k + 1].

So, for every k ∈ Z, the function f(x) = cos(πx/2) + cos(πx) has a root in each interval [4k - 1, 4k] and [4k, 4k + 1].