Let `[x]` be the greatest integer less than or equal to `xdot` Then, at which of the following point (s) function `f(x)=xcos(pi(x+[x]))` is discontinuous? `x=1` (b) `x=-1` (c) `x=0` (d) `x=2`

Options

A. x=-1
B. x=0
C. x=2
D. x=1

Answer

To determine the points of discontinuity for the function \( f(x) = x \cos(\pi (x + [x])) \), we will check the left-hand limit and right-hand limit at the given points: \( x = 1 \), \( x = -1 \), \( x = 0 \), and \( x = 2 \).

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

**Left-hand limit:**
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x \cos(\pi (x + [x])) \]
For \( x < 1 \), \( [x] = 0 \). Thus:
\[ = \lim_{x \to 1^-} x \cos(\pi (x + 0)) = \lim_{x \to 1^-} x \cos(\pi x) \]
Substituting \( x = 1 \):
\[ = 1 \cdot \cos(\pi) = 1 \cdot (-1) = -1 \]

**Right-hand limit:**
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} x \cos(\pi (x + [x])) \]
For \( x > 1 \), \( [x] = 1 \). Thus:
\[ = \lim_{x \to 1^+} x \cos(\pi (x + 1)) = \lim_{x \to 1^+} x \cos(\pi x + \pi) \]
Using the property \( \cos(\theta + \pi) = -\cos(\theta) \):
\[ = \lim_{x \to 1^+} x (-\cos(\pi x)) = -1 \cdot \cos(2\pi) = -1 \cdot 1 = 1 \]

**Conclusion at \( x = 1 \):**
The left-hand limit is \(-1\) and the right-hand limit is \(1\). Since they are not equal, \( f(x) \) is discontinuous at \( x = 1 \).

### Step 2: Check at \( x = -1 \)

**Left-hand limit:**
\[ \lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} x \cos(\pi (x + [x])) \]
For \( x < -1 \), \( [x] = -2 \). Thus:
\[ = \lim_{x \to -1^-} x \cos(\pi (x - 2)) = \lim_{x \to -1^-} x \cos(\pi x - 2\pi) \]
\[ = \lim_{x \to -1^-} x \cos(\pi x) = -1 \cdot \cos(-\pi) = -1 \cdot (-1) = 1 \]

**Right-hand limit:**
\[ \lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} x \cos(\pi (x + [x])) \]
For \( x > -1 \), \( [x] = -1 \). Thus:
\[ = \lim_{x \to -1^+} x \cos(\pi (x - 1)) = \lim_{x \to -1^+} x \cos(\pi x - \pi) \]
\[ = \lim_{x \to -1^+} x (-\cos(\pi x)) = -1 \cdot (-1) = 1 \]

**Conclusion at \( x = -1 \):**
Both limits are equal to \( 1 \), so \( f(x) \) is continuous at \( x = -1 \).

### Step 3: Check at \( x = 0 \)

**Left-hand limit:**
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x \cos(\pi (x + [x])) \]
For \( x < 0 \), \( [x] = -1 \):
\[ = \lim_{x \to 0^-} x \cos(\pi (x - 1)) = \lim_{x \to 0^-} x \cos(\pi x + \pi) \]
\[ = \lim_{x \to 0^-} x (-\cos(\pi x)) = 0 \]

**Right-hand limit:**
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x \cos(\pi (x + [x])) \]
For \( x > 0 \), \( [x] = 0 \):
\[ = \lim_{x \to 0^+} x \cos(\pi x) = 0 \]

**Conclusion at \( x = 0 \):**
Both limits are equal to \( 0 \), so \( f(x) \) is continuous at \( x = 0 \).

### Step 4: Check at \( x = 2 \)

**Left-hand limit:**
\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} x \cos(\pi (x + [x])) \]
For \( x < 2 \), \( [x] = 1 \):
\[ = \lim_{x \to 2^-} x \cos(\pi (x + 1)) = \lim_{x \to 2^-} x \cos(\pi x + \pi) \]
\[ = \lim_{x \to 2^-} x (-\cos(\pi x)) = 2 \cdot (-\cos(2\pi)) = 2 \cdot (-1) = -2 \]

**Right-hand limit:**
\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} x \cos(\pi (x + [x])) \]
For \( x > 2 \), \( [x] = 2 \):
\[ = \lim_{x \to 2^+} x \cos(\pi (x + 2)) = \lim_{x \to 2^+} x \cos(\pi x + 2\pi) \]
\[ = \lim_{x \to 2^+} x \cos(\pi x) = 2 \cdot \cos(2\pi) = 2 \cdot 1 = 2 \]

**Conclusion at \( x = 2 \):**
The left-hand limit is \(-2\) and the right-hand limit is \(2\). Since they are not equal, \( f(x) \) is discontinuous at \( x = 2 \).

### Final Conclusion
The function \( f(x) \) is discontinuous at the points \( x = 1 \), \( x = 2 \), and continuous at \( x = -1 \) and \( x = 0 \).

### Points of Discontinuity:
- \( x = 1 \)
- \( x = 2 \)

### Summary of Discontinuities:
The function \( f(x) \) is discontinuous at:
- \( x = 1 \)
- \( x = 2 \)