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 for `x lt 3` 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 \( g(x) = 4 - x \), 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:

\[ |f(x) - g(x)| = |(x - 3) - (4 - x)| = |x - 3 - 4 + x| = |2x - 7| \]

Next, we calculate \( |f(x)| \) and \( |g(x)| \):

\[ |f(x)| = |x - 3| \quad \text{and} \quad |g(x)| = |4 - x| \]

### Step 2: Rewrite the inequality

Now, we rewrite the original inequality using our substitutions:

\[ |2x - 7| \leq |x - 3| + |4 - x| \]

### Step 3: Simplify the right-hand side

We can simplify the right-hand side:

\[ |4 - x| = |-(x - 4)| = |x - 4| \]

Thus, we have:

\[ |x - 3| + |x - 4| \]

### Step 4: Analyze the cases for the absolute values

Since we are interested in the solutions for \( x < 3 \), we need to consider the signs of the expressions in the absolute values.

1. **For \( x < 3 \)**:
- \( |x - 3| = 3 - x \) (since \( x - 3 < 0 \))
- \( |x - 4| = 4 - x \) (since \( x - 4 < 0 \))

So, we can rewrite the inequality as:

\[ |2x - 7| \leq (3 - x) + (4 - x) = 7 - 2x \]

### Step 5: Solve the inequality \( |2x - 7| \leq 7 - 2x \)

This absolute value inequality can be split into two cases:

**Case 1**: \( 2x - 7 \leq 7 - 2x \)

\[ 2x - 7 \leq 7 - 2x \implies 4x \leq 14 \implies x \leq 3.5 \]

**Case 2**: \( -(2x - 7) \leq 7 - 2x \)

\[ -2x + 7 \leq 7 - 2x \implies 7 \leq 7 \quad \text{(always true)} \]

### Step 6: Combine the results

From Case 1, we have \( x \leq 3.5 \). However, since we are only interested in solutions where \( x < 3 \), we take the intersection of \( x < 3 \) and \( x \leq 3.5 \).

Thus, the solution set is:

\[ x < 3 \]

### Conclusion

Since there are infinitely many values of \( x \) that satisfy \( x < 3 \), the number of solutions for the inequality \( |f(x) - g(x)| \leq |f(x)| + |g(x)| \) for \( x < 3 \) is **infinitely many**.

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