The smallest integral value of a such that `|x+a-3|+|x-2a|=|2x-a-3|` is true `AA x in R` is

Options

A. `0`
B. `1`
C. `2
D. `3`

Answer

To solve the equation \( |x + a - 3| + |x - 2a| = |2x - a - 3| \) for the smallest integral value of \( a \) such that it holds true for all \( x \in \mathbb{R} \), we can follow these steps:

### Step 1: Analyze the equation
We start with the equation:
\[ |x + a - 3| + |x - 2a| = |2x - a - 3| \]
This equation involves absolute values, which means we need to consider different cases based on the expressions inside the absolute values.

### Step 2: Identify critical points
The critical points where the expressions inside the absolute values change are:
1. \( x + a - 3 = 0 \) → \( x = 3 - a \)
2. \( x - 2a = 0 \) → \( x = 2a \)
3. \( 2x - a - 3 = 0 \) → \( x = \frac{a + 3}{2} \)

### Step 3: Set up cases based on critical points
We will analyze the equation by breaking it into cases based on the values of \( x \):
1. Case 1: \( x < 3 - a \)
2. Case 2: \( 3 - a \leq x < 2a \)
3. Case 3: \( 2a \leq x < \frac{a + 3}{2} \)
4. Case 4: \( x \geq \frac{a + 3}{2} \)

### Step 4: Solve for each case
For each case, we will express the absolute values without the modulus and solve the resulting equations.

#### Case 1: \( x < 3 - a \)
Here, all expressions are negative:
\[ -(x + a - 3) - (x - 2a) = -(2x - a - 3) \]
This simplifies to:
\[ -2x + a + 3 = -2x + a + 3 \]
This case holds true for all \( x < 3 - a \).

#### Case 2: \( 3 - a \leq x < 2a \)
Here, we have:
\[ (x + a - 3) - (x - 2a) = -(2x - a - 3) \]
This simplifies to:
\[ 3a - 3 = -x + a + 3 \]
Rearranging gives:
\[ x = 2a - 3 \]

#### Case 3: \( 2a \leq x < \frac{a + 3}{2} \)
Here, we have:
\[ (x + a - 3) + (x - 2a) = -(2x - a - 3) \]
This simplifies to:
\[ 2x - a - 3 = -2x + a + 3 \]
Rearranging gives:
\[ 4x = 2a + 6 \quad \Rightarrow \quad x = \frac{a + 3}{2} \]

#### Case 4: \( x \geq \frac{a + 3}{2} \)
Here, we have:
\[ (x + a - 3) + (x - 2a) = (2x - a - 3) \]
This simplifies to:
\[ 2x - a - 3 = 2x - a - 3 \]
This case holds true for all \( x \geq \frac{a + 3}{2} \).

### Step 5: Find conditions for all cases to hold
To ensure that the equation holds for all \( x \), we need to check the boundaries of the cases:
1. From Case 1, \( 3 - a \) must be less than or equal to \( 2a \).
2. From Case 2, \( 2a - 3 \) must be less than or equal to \( \frac{a + 3}{2} \).

### Step 6: Solve inequalities
1. From \( 3 - a \leq 2a \):
\[ 3 \leq 3a \quad \Rightarrow \quad a \geq 1 \]
2. From \( 2a - 3 \leq \frac{a + 3}{2} \):
\[ 4a - 6 \leq a + 3 \quad \Rightarrow \quad 3a \leq 9 \quad \Rightarrow \quad a \leq 3 \]

### Step 7: Conclusion
The smallest integral value of \( a \) that satisfies both inequalities \( 1 \leq a \leq 3 \) is:
\[ \boxed{1} \]