Key Concepts and Formulas

• Continuity of a Function: A function $f(x)$ is continuous at a point $x=c$ if the following three conditions are met:
  1. $f(c)$ is defined.
  2. $\lim_{x \to c} f(x)$ exists.
  3. $\lim_{x \to c} f(x) = f(c)$.
• Greatest Integer Function: The greatest integer function, denoted by $[x]$, gives the largest integer less than or equal to $x$. It is discontinuous at every integer value of $x$.
• Trigonometric Functions: The cosine function, $\cos(\theta)$, is continuous for all real values of $\theta$.
• Continuity of Product of Functions: If two functions $g(x)$ and $h(x)$ are continuous at a point $x=c$, then their product $p(x) = g(x)h(x)$ is also continuous at $x=c$.

Step-by-Step Solution

Step 1: Analyze the function and identify potential points of discontinuity. The given function is $f(x) = [x - 1]\cos \left( \frac{2x - 1}{2} \right)\pi$. The function is a product of two functions: $g(x) = [x - 1]$ and $h(x) = \cos \left( \frac{2x - 1}{2} \right)\pi$. The greatest integer function, $[x-1]$, is discontinuous at values where $x-1$ is an integer. This occurs when $x$ is an integer. The cosine function, $\cos(\theta)$, is continuous everywhere. The argument of the cosine function, $\frac{2x-1}{2}\pi$, is a linear function of $x$ multiplied by $\pi$, which is also continuous for all real $x$. Therefore, $h(x) = \cos \left( \frac{2x - 1}{2} \right)\pi$ is continuous for all real $x$. The continuity of $f(x)$ will depend on the continuity of $g(x)$ and the behavior of $h(x)$ at the points of discontinuity of $g(x)$, i.e., at integer values of $x$.

Step 2: Check the continuity of $f(x)$ at integer values of $x$. Let $x = n$, where $n$ is an integer. We need to check if $\lim_{x \to n} f(x) = f(n)$. First, let's evaluate $f(n)$: $f(n) = [n - 1]\cos \left( \frac{2n - 1}{2} \right)\pi$ Since $n$ is an integer, $n-1$ is also an integer. Thus, $[n-1] = n-1$. $f(n) = (n - 1)\cos \left( \frac{2n - 1}{2} \right)\pi$

Now, let's evaluate the cosine term: $\cos \left( \frac{2n - 1}{2} \right)\pi = \cos \left( n\pi - \frac{\pi}{2} \right)$ Using the identity $\cos(A - B) = \cos A \cos B + \sin A \sin B$: $\cos \left( n\pi - \frac{\pi}{2} \right) = \cos(n\pi)\cos(\frac{\pi}{2}) + \sin(n\pi)\sin(\frac{\pi}{2})$ We know that $\cos(\frac{\pi}{2}) = 0$ and $\sin(\frac{\pi}{2}) = 1$. Also, $\sin(n\pi) = 0$ for any integer $n$. So, $\cos \left( n\pi - \frac{\pi}{2} \right) = \cos(n\pi) \cdot 0 + 0 \cdot 1 = 0$.

Therefore, $f(n) = (n - 1) \cdot 0 = 0$ for any integer $n$.

Step 3: Evaluate the Left-Hand Limit (LHL) at integer values of $x$. Let $x \to n^-$. This means $x$ approaches $n$ from values slightly less than $n$. So, $x = n - \epsilon$, where $\epsilon > 0$ and $\epsilon \to 0$. Then, $x - 1 = n - \epsilon - 1 = (n - 1) - \epsilon$. Since $n-1$ is an integer, for sufficiently small $\epsilon$, $[x - 1] = [(n - 1) - \epsilon] = n - 2$. Now consider the cosine term: $\cos \left( \frac{2x - 1}{2} \right)\pi = \cos \left( \frac{2(n - \epsilon) - 1}{2} \right)\pi = \cos \left( \frac{2n - 2\epsilon - 1}{2} \right)\pi$ $= \cos \left( n\pi - \epsilon - \frac{\pi}{2} \right)$ This can be written as $\cos \left( (n - \frac{1}{2})\pi - \epsilon \right)$. As $\epsilon \to 0$, this approaches $\cos \left( (n - \frac{1}{2})\pi \right) = \cos \left( n\pi - \frac{\pi}{2} \right) = 0$.

Therefore, LHL = $\mathop {\lim }\limits_{x \to {n^ - }} [x - 1]\cos \left( {{{2x - 1} \over 2}} \right)\pi$ $= \mathop {\lim }\limits_{x \to {n^ - }} (n - 2) \cos \left( {{{2x - 1} \over 2}} \right)\pi$ Since $\cos \left( {{{2x - 1} \over 2}} \right)\pi \to 0$ as $x \to n^-$, and $(n-2)$ is a constant, the limit is: LHL = $(n - 2) \cdot 0 = 0$.

Step 4: Evaluate the Right-Hand Limit (RHL) at integer values of $x$. Let $x \to n^+$. This means $x$ approaches $n$ from values slightly greater than $n$. So, $x = n + \epsilon$, where $\epsilon > 0$ and $\epsilon \to 0$. Then, $x - 1 = n + \epsilon - 1 = (n - 1) + \epsilon$. Since $n-1$ is an integer, for sufficiently small $\epsilon$, $[x - 1] = [(n - 1) + \epsilon] = n - 1$. Now consider the cosine term: $\cos \left( \frac{2x - 1}{2} \right)\pi = \cos \left( \frac{2(n + \epsilon) - 1}{2} \right)\pi = \cos \left( \frac{2n + 2\epsilon - 1}{2} \right)\pi$ $= \cos \left( n\pi + \epsilon - \frac{\pi}{2} \right)$ This can be written as $\cos \left( (n - \frac{1}{2})\pi + \epsilon \right)$. As $\epsilon \to 0$, this approaches $\cos \left( (n - \frac{1}{2})\pi \right) = \cos \left( n\pi - \frac{\pi}{2} \right) = 0$.

Therefore, RHL = $\mathop {\lim }\limits_{x \to {n^ + }} [x - 1]\cos \left( {{{2x - 1} \over 2}} \right)\pi$ $= \mathop {\lim }\limits_{x \to {n^ + }} (n - 1) \cos \left( {{{2x - 1} \over 2}} \right)\pi$ Since $\cos \left( {{{2x - 1} \over 2}} \right)\pi \to 0$ as $x \to n^+$, and $(n-1)$ is a constant, the limit is: RHL = $(n - 1) \cdot 0 = 0$.

Step 5: Compare LHL, RHL, and $f(n)$ at integer values. From Step 2, $f(n) = 0$. From Step 3, LHL = 0. From Step 4, RHL = 0. Since LHL = RHL = $f(n)$ for every integer $n$, the function $f(x)$ is continuous at all integer values of $x$.

Step 6: Check continuity at non-integer values of $x$. Let $x_0$ be a non-integer real number. Then $x_0 - 1$ is also a non-integer. For any $x$ in a small neighborhood around $x_0$ that does not contain an integer, $[x - 1]$ will be a constant integer value, say $k$. So, in such a neighborhood, $f(x) = k \cdot \cos \left( \frac{2x - 1}{2} \right)\pi$. Since $k$ is a constant and $\cos \left( \frac{2x - 1}{2} \right)\pi$ is continuous for all real $x$, their product $f(x)$ is continuous in this neighborhood. Specifically, at $x = x_0$ (a non-integer), $f(x)$ is continuous because:

1. $f(x_0) = [x_0 - 1]\cos \left( \frac{2x_0 - 1}{2} \right)\pi$ is defined.
2. $\lim_{x \to x_0} f(x) = \lim_{x \to x_0} [x - 1]\cos \left( \frac{2x - 1}{2} \right)\pi$. Since $[x-1]$ is constant in a neighborhood of $x_0$, this limit is $[x_0 - 1]\cos \left( \frac{2x_0 - 1}{2} \right)\pi$.
3. Thus, $\lim_{x \to x_0} f(x) = f(x_0)$.

Therefore, the function $f(x)$ is continuous at all non-integer values of $x$.

Step 7: Conclude the continuity of the function. From Step 5, $f(x)$ is continuous at all integer values of $x$. From Step 6, $f(x)$ is continuous at all non-integer values of $x$. Combining these results, $f(x)$ is continuous for every real $x$.

Common Mistakes & Tips

• Misinterpreting the Greatest Integer Function: A common mistake is to assume the greatest integer function is discontinuous at all real numbers. Remember it's only discontinuous at integer values.
• Incorrectly Evaluating Limits of $[x-1]$: When taking limits, ensure you correctly determine the value of $[x-1]$ as $x$ approaches an integer from the left or right. For $x \to n^-$, $[x-1]$ becomes $n-2$, and for $x \to n^+$, $[x-1]$ becomes $n-1$.
• Forgetting to Evaluate $f(n)$: Always ensure all three conditions for continuity are met: the function value must exist, the limit must exist, and they must be equal.

Summary

The function $f(x) = [x - 1]\cos \left( \frac{2x - 1}{2} \right)\pi$ is a product of the greatest integer function $[x-1]$ and a continuous cosine function. The greatest integer function is discontinuous at integers. However, at integer values $x=n$, the term $\cos \left( \frac{2n - 1}{2} \right)\pi$ evaluates to 0. This makes $f(n) = 0$. The left-hand and right-hand limits at integer $x=n$ also evaluate to 0 due to the behavior of $[x-1]$ near $n$ and the fact that the cosine term approaches 0. Thus, the function is continuous at all integer points. For non-integer points, $[x-1]$ is constant in a small neighborhood, and since the cosine term is continuous, the product is also continuous. Therefore, the function is continuous for every real number.

The final answer is \boxed{A}.