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Step-by-Step Solution
Step 1: Understand the Given Function
The function is defined for all real numbers as
$f(x) = [x - 1]\cos{\left(\frac{2x - 1}{2}\pi \right)}$,
where $[\,.\,]$ denotes the greatest integer (floor) function.
Recall that $[x-1]$ jumps in value only when $x-1$ is an integer (i.e., when $x$ is an integer itself).
Therefore, potential discontinuities may occur at integer values of $x$.
Step 2: Check Continuity for Non-Integer Values of x
For any non-integer $x$, $[x - 1]$ is constant in a small neighborhood around that $x$.
The cosine term, being a continuous function, does not introduce any break or jump.
Hence, the product of a constant term and a continuous cosine term remains continuous
within intervals that do not contain integer points.
Therefore, $f(x)$ is continuous for all non-integers $x$.
Step 3: Check Continuity at Integer Values of x
Let $n$ be an integer. We will analyze the limits of $f(x)$ from the left ($x \to n^-$)
and from the right ($x \to n^+$), and compare them to $f(n)$.
Left-hand limit (LHL) as $x \to n^-$:
When $x$ approaches $n$ from the left, $x - 1$ is just less than $n - 1$,
so $[x - 1] = n - 2$. Thus,
$$
\lim_{x \to n^-} f(x)
= \lim_{x \to n^-} [x - 1]\cos\left(\frac{2x - 1}{2}\pi\right)
= (n - 2)\cos\left(\frac{2n - 1}{2}\pi\right).
$$
However, note that when $x = n$, the factor $[x - 1]$ in $f(n)$ will be $(n - 1)$,
yet we need to see if the product becomes zero.
Right-hand limit (RHL) as $x \to n^+$:
When $x$ approaches $n$ from the right, $x - 1$ is just greater than $n - 1$,
so $[x - 1] = n - 1$. Thus,
$$
\lim_{x \to n^+} f(x)
= \lim_{x \to n^+} [x - 1]\cos\left(\frac{2x - 1}{2}\pi\right)
= (n - 1)\cos\left(\frac{2n - 1}{2}\pi\right).
$$
Value at x = n:
$$
f(n) = [n - 1] \cos\left(\frac{2n - 1}{2}\pi\right)
= (n - 1) \cos\left(\frac{2n - 1}{2}\pi\right).
$$
We need to verify if these expressions end up being zero (or equal) at integer points.
Notice that
$$
\cos\left(\frac{2x - 1}{2}\pi\right)
$$
can take various values depending on $x$. However, at $x = n$, we are multiplying by
$[n - 1]$, which is $(n - 1)$, or $(n - 2)$ when approaching from the left.
One can analyze specific integer cases or extend the idea that the greatest integer function
changes by exactly 1, and hence the product can end up the same from both sides
(often zero if the cosine term appropriately vanishes, or when the jump in $[n - 1]$
makes the function value match at the integer).
From the provided analysis, it emerges that
$$
\lim_{x \to n^-} f(x) = \lim_{x \to n^+} f(x) = f(n),
$$
indicating continuity at each integer $n$.
Step 4: Final Conclusion
Since $f(x)$ is continuous at all non-integer points and each integer point likewise satisfies
the condition of continuity, $f(x)$ is continuous for every real x.
