Let f(x) and g(x) be two real valued functions then `|f(x) - g(x)| le |f(x)| + |g(x)|`
Let f(x) = x-3 and g(x) = 4-x, then
The number of solution(s) of the above inequality when `x ge 4` is

Options

A. 0
B. 1
C. 2
D. Infinitely many

Answer

To solve the inequality \( |f(x) - g(x)| \leq |f(x)| + |g(x)| \) for the functions \( f(x) = x - 3 \) and \( g(x) = 4 - x \) when \( x \geq 4 \), we will follow these steps:

### Step 1: Substitute the functions into the inequality
We start by substituting \( f(x) \) and \( g(x) \) into the inequality:
\[ |f(x) - g(x)| \leq |f(x)| + |g(x)| \]
This becomes:
\[ | (x - 3) - (4 - x) | \leq |x - 3| + |4 - x| \]

### Step 2: Simplify the left side
Now, simplify the left side:
\[ | (x - 3) - (4 - x) | = | x - 3 - 4 + x | = | 2x - 7 | \]

### Step 3: Simplify the right side
Next, simplify the right side:
\[ |x - 3| + |4 - x| \]
Since \( x \geq 4 \), we have:
- \( |x - 3| = x - 3 \) (since \( x - 3 \geq 0 \))
- \( |4 - x| = 4 - x \) (since \( 4 - x \leq 0 \))

Thus, the right side becomes:
\[ |x - 3| + |4 - x| = (x - 3) + (4 - x) = 1 \]

### Step 4: Set up the inequality
Now, we can set up the inequality:
\[ |2x - 7| \leq 1 \]

### Step 5: Solve the absolute value inequality
To solve \( |2x - 7| \leq 1 \), we can break it down into two cases:
1. \( 2x - 7 \leq 1 \)
2. \( 2x - 7 \geq -1 \)

#### Case 1: \( 2x - 7 \leq 1 \)
\[ 2x \leq 8 \implies x \leq 4 \]

#### Case 2: \( 2x - 7 \geq -1 \)
\[ 2x \geq 6 \implies x \geq 3 \]

### Step 6: Combine the results
From the two cases, we have:
- From Case 1: \( x \leq 4 \)
- From Case 2: \( x \geq 3 \)

Combining these results, we get:
\[ 3 \leq x \leq 4 \]

### Step 7: Determine the number of solutions for \( x \geq 4 \)
Since we are specifically looking for solutions where \( x \geq 4 \), the only solution that satisfies both conditions is:
\[ x = 4 \]

### Conclusion
Thus, the number of solutions for the inequality \( |f(x) - g(x)| \leq |f(x)| + |g(x)| \) when \( x \geq 4 \) is:
\[ \text{Number of solutions} = 1 \]