Discussion

Explanation:

Here, the critical points are x = -3 and 2

Case 1: x > 2
∴ x – 2 + x + 3 < 2
⇒ 2x < 1
⇒ x < ½ (this contradicts our assumption that x > 2)
∴ No solution

Case 2: -3 < x < 2
∴ -(x - 2) + x + 3 < 2
⇒ 5 < 2 (not true)
∴ No solution

Case 3: x < -3
∴ -(x - 2) – (x + 3) < 2
⇒ -2x < 3
⇒ x > -3/2 (this contradicts our assumption that x < -3/2)
∴ No solution

∴ There is no real value of x for which |x − 2| + |x + 3| < 2.

Hence, option (d).

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