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

Options

A. continuous at x = 1 but discontinuous at x = 0
B. continuous at x = 1 and x = 0
C. discontinuous at x = 1 and x = 0
D. discontinuous at x = 1 but continuous at x = 0

Answer

To solve the problem, we need to analyze the function given:

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

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

### Step 1: Analyze the limit as \(n\) approaches infinity

The term \(\cos^{2n}(\pi x^2)\) will behave differently based on the value of \(\cos(\pi x^2)\):
- If \(|\cos(\pi x^2)| < 1\), then \(\cos^{2n}(\pi x^2) \to 0\) as \(n \to \infty\).
- If \(|\cos(\pi x^2)| = 1\), then \(\cos^{2n}(\pi x^2) = 1\).

### Step 2: Determine when \(\cos(\pi x^2) = 1\)

The cosine function equals 1 when its argument is an integer multiple of \(2\pi\):
\[ \pi x^2 = 2k\pi \quad \text{for } k \in \mathbb{Z} \]
This simplifies to:
\[ x^2 = 2k \quad \Rightarrow \quad x = \sqrt{2k} \quad \text{or} \quad x = -\sqrt{2k} \]

### Step 3: Evaluate the limit based on the value of \(x\)

1. **For \(x = 0\)**:
- \(\cos(0) = 1\), thus \(\cos^{2n}(0) = 1\).
- \([0] = 0\).
- Therefore, \(f(0) = 1 + 0 = 1\).

2. **For \(x = 1\)**:
- \(\cos(\pi) = -1\), thus \(\cos^{2n}(\pi) = 1\).
- \([1] = 1\).
- Therefore, \(f(1) = 1 + 1 = 2\).

3. **For \(0 < x < 1\)**:
- Here, \(\cos(\pi x^2)\) will be positive and less than 1, so \(\cos^{2n}(\pi x^2) \to 0\).
- \([x] = 0\).
- Therefore, \(f(x) = 0 + 0 = 0\).

4. **For \(x > 1\)**:
- For \(x > 1\), \(\cos(\pi x^2)\) will oscillate and be less than 1, thus \(\cos^{2n}(\pi x^2) \to 0\).
- \([x] = 1\) if \(1 < x < 2\) and \([x] = 2\) if \(2 \leq x < 3\), etc.
- Therefore, \(f(x) = 0 + [x]\).

### Step 4: Check continuity at \(x = 0\) and \(x = 1\)

1. **At \(x = 0\)**:
- Left-hand limit (LHL) as \(x \to 0^{-}\): \(f(x) \to 0\).
- Right-hand limit (RHL) as \(x \to 0^{+}\): \(f(x) \to 0\).
- Thus, \(f(0) = 1\) and LHL = RHL = 0, so \(f(x)\) is discontinuous at \(x = 0\).

2. **At \(x = 1\)**:
- LHL as \(x \to 1^{-}\): \(f(x) \to 0\).
- RHL as \(x \to 1^{+}\): \(f(x) \to 1\).
- Thus, \(f(1) = 2\) and LHL ≠ RHL, so \(f(x)\) is discontinuous at \(x = 1\).

### Conclusion

The function \(f(x)\) is discontinuous at both \(x = 0\) and \(x = 1\).

### Final Answer

The correct option is that the function is discontinuous at \(x = 0\) and \(x = 1\).

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