Question

The solution set of the inequality \(|x-2|+|x+3|\ge7\) is:

• A. \((-\infty,-5]\cup[4,\infty)\)
• B. \([-5,4]\)
• C. \((-\infty,\infty)\)
• D. \([-4,5]\)

Hint:

Break modulus equations into intervals.

Answer 1

The Correct Option is A

We are given the inequality:
\(|x-2| + |x+3| \ge 7\) 

Consider the critical points where expressions inside absolute values change sign: 
\(x = -3\) and \(x = 2\) 

Case 1: \(x \ge 2\)
\(|x-2| = x-2,\quad |x+3| = x+3\)
So, \((x-2) + (x+3) \ge 7 \Rightarrow 2x + 1 \ge 7 \Rightarrow x \ge 3\)

Case 2: \(-3 \le x < 2\)
\(|x-2| = 2-x,\quad |x+3| = x+3\)
So, \((2-x) + (x+3) = 5\)
Since \(5 \not\ge 7\), no solution in this interval. 

Case 3: \(x < -3\)
\(|x-2| = 2-x,\quad |x+3| = -(x+3)\)
So, \((2-x) + (-(x+3)) \ge 7 \Rightarrow -2x -1 \ge 7 \Rightarrow x \le -4\)

Combining all cases:
\(x \le -5 \quad \text{or} \quad x \ge 4\)

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Final Solution Set:
\(\boxed{(-\infty,-5]\cup[4,\infty)}\)