Given `f(x)={3-[cot^(-1)((2x^3-3)/(x^2))] for x >0 and {x^2}cos(e^(1/x)) for x<0` (where {} and [] denotes the fractional part and the integral part functions respectively). Then which of the following statements do/does not hold good?

Options

A. `f(0^(0-)) = 0`
B. `f(0^(+)) = 0`
C. `f(0) = 0 rArr` Continuous at x = 0
D. Irremovable discontinuity at x = 0

Answer

To solve the problem, we need to analyze the given piecewise function and determine the continuity at \( x = 0 \).

### Step 1: Define the function
The function is defined as:
\[ f(x) = \begin{cases} 3 - \lfloor \cot^{-1}\left(\frac{2x^3 - 3}{x^2}\right) \rfloor & \text{for } x > 0 \\ x^2 \cos\left(e^{\frac{1}{x}}\right) & \text{for } x < 0 \end{cases} \]

### Step 2: Find the right-hand limit as \( x \to 0^+ \)
For \( x > 0 \):
\[ f(x) = 3 - \lfloor \cot^{-1}\left(\frac{2x^3 - 3}{x^2}\right) \rfloor \]
As \( x \to 0^+ \):
\[ \frac{2x^3 - 3}{x^2} \to \frac{-3}{0^+} \to -\infty \]
Thus, \( \cot^{-1}(-\infty) = \pi \). Therefore:
\[ \lfloor \cot^{-1}\left(\frac{2x^3 - 3}{x^2}\right) \rfloor \to \lfloor \pi \rfloor = 3 \]
So,
\[ f(0^+) = 3 - 3 = 0 \]

### Step 3: Find the left-hand limit as \( x \to 0^- \)
For \( x < 0 \):
\[ f(x) = x^2 \cos\left(e^{\frac{1}{x}}\right) \]
As \( x \to 0^- \):
\[ x^2 \to 0 \quad \text{and} \quad \cos\left(e^{\frac{1}{x}}\right) \text{ oscillates between } -1 \text{ and } 1 \]
Thus, the left-hand limit is:
\[ f(0^-) = 0 \cdot \text{(bounded value)} = 0 \]

### Step 4: Compare the limits
We have:
\[ \lim_{x \to 0^+} f(x) = 0 \quad \text{and} \quad \lim_{x \to 0^-} f(x) = 0 \]
Since both limits are equal, we can conclude that:
\[ \lim_{x \to 0} f(x) = 0 \]

### Step 5: Check continuity at \( x = 0 \)
For continuity at \( x = 0 \), we need:
\[ \lim_{x \to 0} f(x) = f(0) \]
However, \( f(0) \) is not defined in the piecewise function. Therefore, the function is not continuous at \( x = 0 \).

### Conclusion
The function \( f(x) \) has a removable discontinuity at \( x = 0 \) since the limits from both sides exist and are equal, but the function is not defined at that point.

### Final Answer
The statement that does not hold good is that there is an irreparable discontinuity at \( x = 0 \).

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