Let |x| be the greatest integer less than or equal to x, Then f(x)=`x cos (pi (x+[x])` is continous at

Options

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

Explanation

To determine the continuity of the function \( f(x) = x \cos(\pi (x + [x])) \), where \([x]\) is the greatest integer less than or equal to \(x\), we will check the continuity at various points, specifically at \(x = -1\), \(x = 0\), \(x = 1\), and \(x = 2\).

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

**Left-hand limit as \(x\) approaches \(-1\):**
\[ \lim_{x \to -1^-} f(x) = \lim_{h \to 0} (-1 - h) \cos(\pi(-1 - h) + \pi[-1 - h]) \]
Here, \([-1 - h] = -2\) since \(-1 - h\) is less than \(-1\).
\[ = \lim_{h \to 0} (-1 - h) \cos(\pi(-1 - h) - 2\pi) \]
\[ = \lim_{h \to 0} (-1 - h) \cos(-\pi + 3\pi) = \lim_{h \to 0} (-1 - h) \cos(3\pi) \]
Since \(\cos(3\pi) = -1\):
\[ = \lim_{h \to 0} (-1 - h)(-1) = \lim_{h \to 0} (1 + h) = 1 \]

**Right-hand limit as \(x\) approaches \(-1\):**
\[ \lim_{x \to -1^+} f(x) = \lim_{h \to 0} (-1 + h) \cos(\pi(-1 + h) + \pi[-1 + h]) \]
Here, \([-1 + h] = -1\) since \(-1 + h\) is greater than \(-1\).
\[ = \lim_{h \to 0} (-1 + h) \cos(\pi(-1 + h) - \pi) \]
\[ = \lim_{h \to 0} (-1 + h) \cos(-\pi) = \lim_{h \to 0} (-1 + h)(-1) \]
\[ = \lim_{h \to 0} (1 - h) = 1 \]

Since the left-hand limit and right-hand limit are equal, but not equal to \(f(-1)\), we conclude that \(f(x)\) is discontinuous at \(x = -1\).

### Step 2: Check continuity at \(x = 0\)

**Left-hand limit as \(x\) approaches \(0\):**
\[ \lim_{x \to 0^-} f(x) = \lim_{h \to 0} (0 - h) \cos(\pi(0 - h) + \pi[0 - h]) \]
Here, \([0 - h] = -1\):
\[ = \lim_{h \to 0} (-h) \cos(-\pi h - \pi) = \lim_{h \to 0} (-h) \cos(-\pi(h + 1)) \]
Since \(\cos(-\pi(h + 1)) = -\cos(\pi(h + 1))\):
\[ = \lim_{h \to 0} (-h)(-\cos(\pi(h + 1))) = \lim_{h \to 0} h \cos(\pi(h + 1)) = 0 \]

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

Since both limits are equal and equal to \(f(0) = 0\), \(f(x)\) is continuous at \(x = 0\).

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

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

**Right-hand limit as \(x\) approaches \(1\):**
\[ \lim_{x \to 1^+} f(x) = \lim_{h \to 0} (1 + h) \cos(\pi(1 + h) + \pi[1 + h]) \]
Here, \([1 + h] = 1\):
\[ = \lim_{h \to 0} (1 + h) \cos(\pi(1 + h) + \pi) = \lim_{h \to 0} (1 + h)(-\cos(\pi h)) = -1 \]

Since both limits are equal but not equal to \(f(1)\), \(f(x)\) is discontinuous at \(x = 1\).

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

**Left-hand limit as \(x\) approaches \(2\):**
\[ \lim_{x \to 2^-} f(x) = \lim_{h \to 0} (2 - h) \cos(\pi(2 - h) + \pi[2 - h]) \]
Here, \([2 - h] = 1\):
\[ = \lim_{h \to 0} (2 - h) \cos(\pi(2 - h) + \pi) = \lim_{h \to 0} (2 - h)(-\cos(\pi h)) = -2 \]

**Right-hand limit as \(x\) approaches \(2\):**
\[ \lim_{x \to 2^+} f(x) = \lim_{h \to 0} (2 + h) \cos(\pi(2 + h) + \pi[2 + h]) \]
Here, \([2 + h] = 2\):
\[ = \lim_{h \to 0} (2 + h) \cos(\pi(2 + h) + 2\pi) = \lim_{h \to 0} (2 + h)(\cos(\pi h)) = 2 \]

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

### Summary of Continuity
- \(f(x)\) is discontinuous at \(x = -1\), \(x = 1\), and \(x = 2\).
- \(f(x)\) is continuous at \(x = 0\).