Key Concepts and Formulas

• Greatest Integer Function: The greatest integer function, denoted by $[x]$, gives the greatest integer less than or equal to $x$. It has jump discontinuities at every integer value of $x$.
• Continuity of a Product of Functions: If two functions, $g(x)$ and $h(x)$, are continuous at a point $x=a$, then their product $(g \cdot h)(x) = g(x)h(x)$ is also continuous at $x=a$.
• Continuity of Trigonometric Functions: Trigonometric functions like $\cos(x)$ are continuous for all real numbers.

Step-by-Step Solution

The function is given by $f(x) = [x]\cos\left(\frac{2x-1}{2}\pi\right)$. We need to determine the continuity of this function.

Step 1: Analyze the continuity of each factor of the function.

The function $f(x)$ is a product of two functions: $g(x) = [x]$ and $h(x) = \cos\left(\frac{2x-1}{2}\pi\right)$.

• The function $g(x) = [x]$ is the greatest integer function. We know that $[x]$ has jump discontinuities at every integer value of $x$. For any non-integer value of $x$, $[x]$ is constant in an interval around $x$, and hence is continuous at non-integer values.

• The function $h(x) = \cos\left(\frac{2x-1}{2}\pi\right)$ is a composition of the cosine function and a linear function. The cosine function $\cos(u)$ is continuous for all real $u$. The argument of the cosine function, $u(x) = \frac{2x-1}{2}\pi = x\pi - \frac{\pi}{2}$, is a linear function of $x$, which is continuous for all real $x$. Therefore, the composition $h(x) = \cos(u(x))$ is continuous for all real values of $x$.

Step 2: Examine the continuity of $f(x)$ at non-integer values of $x$.

Let $a$ be any non-integer real number. At $x=a$, $g(x) = [x]$ is continuous. At $x=a$, $h(x) = \cos\left(\frac{2x-1}{2}\pi\right)$ is continuous.

Since both $g(x)$ and $h(x)$ are continuous at $x=a$, their product $f(x) = g(x)h(x)$ is also continuous at $x=a$. This means $f(x)$ is continuous for all non-integer real numbers.

Step 3: Examine the continuity of $f(x)$ at integer values of $x$.

Let $n$ be an integer. We need to check the limit of $f(x)$ as $x \to n$ and compare it with $f(n)$.

First, let's evaluate $f(n)$: $f(n) = [n]\cos\left(\frac{2n-1}{2}\pi\right)$ Since $n$ is an integer, $[n] = n$. $f(n) = n\cos\left(\frac{2n-1}{2}\pi\right)$ We can rewrite the argument of cosine: $\frac{2n-1}{2}\pi = \left(n - \frac{1}{2}\right)\pi = n\pi - \frac{\pi}{2}$. Using the identity $\cos(A - B) = \cos A \cos B + \sin A \sin B$: $\cos\left(n\pi - \frac{\pi}{2}\right) = \cos(n\pi)\cos\left(\frac{\pi}{2}\right) + \sin(n\pi)\sin\left(\frac{\pi}{2}\right)$ Since $\cos\left(\frac{\pi}{2}\right) = 0$ and $\sin(n\pi) = 0$ for any integer $n$, we have: $\cos\left(n\pi - \frac{\pi}{2}\right) = (\pm 1)(0) + (0)(1) = 0$. So, $f(n) = n \cdot 0 = 0$ for any integer $n$.

Now let's consider the limit as $x \to n$. We need to evaluate the left-hand limit and the right-hand limit.

Step 3a: Evaluate the left-hand limit as $x \to n^-$. As $x \to n^-$, $[x]$ approaches $n-1$. The argument of cosine approaches $\frac{2n-1}{2}\pi$. So, $\lim_{x \to n^-} f(x) = \lim_{x \to n^-} [x]\cos\left(\frac{2x-1}{2}\pi\right) = (n-1) \cos\left(\frac{2n-1}{2}\pi\right)$. From Step 3, we know that $\cos\left(\frac{2n-1}{2}\pi\right) = 0$. Therefore, $\lim_{x \to n^-} f(x) = (n-1) \cdot 0 = 0$.

Step 3b: Evaluate the right-hand limit as $x \to n^+$. As $x \to n^+$, $[x]$ approaches $n$. The argument of cosine approaches $\frac{2n-1}{2}\pi$. So, $\lim_{x \to n^+} f(x) = \lim_{x \to n^+} [x]\cos\left(\frac{2x-1}{2}\pi\right) = n \cos\left(\frac{2n-1}{2}\pi\right)$. From Step 3, we know that $\cos\left(\frac{2n-1}{2}\pi\right) = 0$. Therefore, $\lim_{x \to n^+} f(x) = n \cdot 0 = 0$.

Step 4: Compare the limits with the function value at integer points.

For any integer $n$: Left-hand limit: $\lim_{x \to n^-} f(x) = 0$. Right-hand limit: $\lim_{x \to n^+} f(x) = 0$. Function value: $f(n) = 0$.

Since the left-hand limit, the right-hand limit, and the function value are all equal at every integer $n$, the function $f(x)$ is continuous at every integer point.

Step 5: Conclude the overall continuity of the function.

From Step 2, we concluded that $f(x)$ is continuous at all non-integer real numbers. From Step 4, we concluded that $f(x)$ is continuous at all integer real numbers. Therefore, $f(x)$ is continuous for every real number $x$.

Common Mistakes & Tips

• Misinterpreting the Greatest Integer Function: Be careful about the behavior of $[x]$ at integer points. It has jump discontinuities. However, when multiplied by another function, the product might become continuous.
• Simplifying Trigonometric Expressions: Always simplify trigonometric expressions involving integer multiples of $\pi$ or $\frac{\pi}{2}$ carefully. In this case, $\cos\left(\frac{2n-1}{2}\pi\right)$ simplifies to 0 for all integers $n$.
• Checking Both Factors Separately: Before concluding about the product, analyze the continuity of each component function. The continuity of the product depends on the continuity of its factors.

Summary

The function $f(x) = [x]\cos\left(\frac{2x-1}{2}\pi\right)$ is a product of the greatest integer function $[x]$ and a continuous trigonometric function $\cos\left(\frac{2x-1}{2}\pi\right)$. While $[x]$ is discontinuous at integers, the trigonometric part simplifies to zero at all integer values of $x$. This causes the product $f(x)$ to evaluate to zero at every integer. Consequently, when we check the limits at integers, both left-hand and right-hand limits match the function value, proving continuity at integers. Since the function is also continuous at non-integers (as it's a product of continuous functions there), $f(x)$ is continuous for every real number $x$.

The final answer is \boxed{A}.