Let `f(x)=|x|+|x-1|,` then
(a)`f(x)` is continuous at `x=0,` as well at `x=1`
(b)`f(x)` is continuous at `x=0,` but not at `x=1`
(c)`f(x)` is continuous at `x=1,` but not at `x=0`
(d)none of these

Explanation

To determine the continuity of the function \( f(x) = |x| + |x - 1| \) at the points \( x = 0 \) and \( x = 1 \), we will analyze the function piecewise.

### Step 1: Break down the function into intervals
The function \( f(x) \) involves absolute values, which means we need to consider different cases based on the values of \( x \).

1. For \( x < 0 \):
\[ f(x) = -x + (1 - x) = -2x + 1 \]

2. For \( 0 \leq x < 1 \):
\[ f(x) = x + (1 - x) = 1 \]

3. For \( x \geq 1 \):
\[ f(x) = x + (x - 1) = 2x - 1 \]

### Step 2: Evaluate continuity at \( x = 0 \)
To check continuity at \( x = 0 \), we need to find \( f(0) \), \( \lim_{x \to 0^-} f(x) \), and \( \lim_{x \to 0^+} f(x) \).

- \( f(0) = |0| + |0 - 1| = 0 + 1 = 1 \)

- For \( x \to 0^- \):
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-2x + 1) = -2(0) + 1 = 1 \]

- For \( x \to 0^+ \):
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 1 = 1 \]

Since \( f(0) = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = 1 \), the function is continuous at \( x = 0 \).

### Step 3: Evaluate continuity at \( x = 1 \)
To check continuity at \( x = 1 \), we need to find \( f(1) \), \( \lim_{x \to 1^-} f(x) \), and \( \lim_{x \to 1^+} f(x) \).

- \( f(1) = |1| + |1 - 1| = 1 + 0 = 1 \)

- For \( x \to 1^- \):
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} 1 = 1 \]

- For \( x \to 1^+ \):
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x - 1) = 2(1) - 1 = 1 \]

Since \( f(1) = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = 1 \), the function is continuous at \( x = 1 \).

### Conclusion
Since \( f(x) \) is continuous at both \( x = 0 \) and \( x = 1 \), the correct answer is:
**(a) \( f(x) \) is continuous at \( x = 0 \), as well at \( x = 1 \)**.