Question

Let f(x) = 2x-5and g(x) = 7-2x. Then |f(x) + g(x)| = |f(x)| + |g(x)| if and only if

• A. 5/2<x<7/2
• B. x≤5/2 or x≥7/2
• C. x<5/2 or x≥7/2
• D. 5/2≤x≤7/2

Answer 1

The Correct Option is D

To solve the problem of determining when \( |f(x) + g(x)| = |f(x)| + |g(x)| \) for the functions \( f(x) = 2x - 5 \) and \( g(x) = 7 - 2x \), first analyze the expressions:

• Calculate \( f(x) + g(x) \): \[ f(x) + g(x) = (2x - 5) + (7 - 2x) = 2 \]
• Since \( f(x) + g(x) = 2 \), its absolute value is constant: \[ |f(x) + g(x)| = |2| = 2 \] 

Now, evaluate \( |f(x)| + |g(x)| \):

• Find critical points by setting each function to zero:
  • \( f(x) = 0 \Rightarrow 2x - 5 = 0 \Rightarrow x = \frac{5}{2} \)
  • \( g(x) = 0 \Rightarrow 7 - 2x = 0 \Rightarrow x = \frac{7}{2} \)
• Analyze the sign changes across intervals:
  • For \( x < \frac{5}{2} \), \( f(x) < 0 \) and \( g(x) > 0 \), hence: \[ |f(x)| + |g(x)| = -(2x-5) + (7-2x) = 12 - 4x \]
  • For \( \frac{5}{2} \le x \le \frac{7}{2} \), both \( f(x) \ge 0 \) and \( g(x) \ge 0 \), so: \[ |f(x)| + |g(x)| = (2x-5) + (7-2x) = 2 \]
  • For \( x > \frac{7}{2} \), \( f(x) > 0 \) and \( g(x) < 0 \), thus: \[ |f(x)| + |g(x)| = (2x-5) - (7-2x) = 4x - 12 \]

Conclusion: The condition \( |f(x) + g(x)| = |f(x)| + |g(x)| \) holds when \( |f(x)| + |g(x)| = 2 \), which occurs only in the interval \( \frac{5}{2} \le x \le \frac{7}{2} \).

Therefore, the answer is \( \frac{5}{2} \le x \le \frac{7}{2} \).