The function `f(x)=lim_(nrarroo)cos^(2n)(pix)+[x]` is (where, `[.]` denotes the greatest integer function and `n in N`)

Options

A. continuous at `x=1` but discontinuous at `x=(3)/(2)`
B. continuous at `x=1` and `x=(3)/(2)`
C. discontinuous at `x=1` and `x=(3)/(2)`
D. discontinuous at `x=1` but continuous at `x=(3)/(2)`

Answer

To solve the problem, we need to analyze the function defined as:

\[ f(x) = \lim_{n \to \infty} \cos^{2n}(\pi x) + [x] \]

where \([x]\) denotes the greatest integer function.

### Step 1: Analyze the limit part of the function

The term \(\cos^{2n}(\pi x)\) behaves differently depending on the value of \(x\):

- If \(x\) is an integer, \(\cos(\pi x) = \cos(k\pi) = (-1)^k\) where \(k\) is an integer. Thus, \(\cos^{2n}(\pi x) = 1\).
- If \(x\) is not an integer, then \(\cos(\pi x)\) will be a value between -1 and 1. As \(n\) approaches infinity, \(\cos^{2n}(\pi x)\) approaches 0.

Therefore, we can summarize this behavior as follows:
- For integer \(x\), \(\lim_{n \to \infty} \cos^{2n}(\pi x) = 1\).
- For non-integer \(x\), \(\lim_{n \to \infty} \cos^{2n}(\pi x) = 0\).

### Step 2: Define \(f(x)\) based on \(x\)

From the above analysis, we can define \(f(x)\) as:

\[ f(x) = \begin{cases} 1 + [x] & \text{if } x \text{ is an integer} \\ 0 + [x] & \text{if } x \text{ is not an integer} \end{cases} \]

This simplifies to:

\[ f(x) = \begin{cases} 1 + [x] & \text{if } x \in \mathbb{Z} \\ [x] & \text{if } x \notin \mathbb{Z} \end{cases} \]

### Step 3: Check continuity at \(x = 1\)

Now, we need to check the continuity of \(f(x)\) at \(x = 1\):

- **Left-hand limit** as \(x\) approaches 1 from the left (\(1 - h\)):
\[ f(1 - h) = [1 - h] = 0 \quad \text{(since \(1 - h\) is not an integer)} \]

- **Right-hand limit** as \(x\) approaches 1 from the right (\(1 + h\)):
\[ f(1 + h) = 1 + [1 + h] = 1 + 1 = 2 \quad \text{(since \(1 + h\) is not an integer)} \]

- **Value at \(x = 1\)**:
\[ f(1) = 1 + [1] = 1 + 1 = 2 \]

Since the left-hand limit (0) does not equal the right-hand limit (2), \(f(x)\) is discontinuous at \(x = 1\).

### Step 4: Check continuity at \(x = \frac{3}{2}\)

Now, we check the continuity of \(f(x)\) at \(x = \frac{3}{2}\):

- **Left-hand limit** as \(x\) approaches \(\frac{3}{2}\) from the left (\(\frac{3}{2} - h\)):
\[ f\left(\frac{3}{2} - h\right) = [\frac{3}{2} - h] = 1 \quad \text{(since \(\frac{3}{2} - h\) is not an integer)} \]

- **Right-hand limit** as \(x\) approaches \(\frac{3}{2}\) from the right (\(\frac{3}{2} + h\)):
\[ f\left(\frac{3}{2} + h\right) = [\frac{3}{2} + h] = 1 \quad \text{(since \(\frac{3}{2} + h\) is not an integer)} \]

- **Value at \(x = \frac{3}{2}\)**:
\[ f\left(\frac{3}{2}\right) = [\frac{3}{2}] = 1 \]

Since the left-hand limit (1), right-hand limit (1), and the function value (1) are all equal, \(f(x)\) is continuous at \(x = \frac{3}{2}\).

### Conclusion

Thus, the function \(f(x)\) is discontinuous at \(x = 1\) and continuous at \(x = \frac{3}{2}\).

### Final Answer
The function \(f(x)\) is discontinuous at \(x = 1\) and continuous at \(x = \frac{3}{2}\).

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