Modulus Function

Equations with sums of absolute values are analyzed by splitting the real line at the points where each modulus changes sign, a standard JEE Advanced casework method. The critical points are x = 1 and x = 3. For x < 1, both expressions are negative inside, giving (1 - x) + (3 - x) = 4 - 2x = 4, so x = 0, which satisfies x < 1 and is valid. For 1 ≤ x ≤ 3, the sum equals (x - 1) + (3 - x) = 2, a constant never equal to 4, so no solutions here. For x > 3, the sum is (x - 1) + (x - 3) = 2x - 4 = 4, giving x = 4, which satisfies x > 3 and is valid. Thus exactly two solutions, x = 0 and x = 4, exist. Option 1 misses one branch. Option infinitely many wrongly treats the middle plateau as equal to 4. Option 0 ignores both outer branches. Hence the count is 2. Plausibility check: the expression represents total distance from 1 and 3, whose minimum is the gap 2, so the value 4 exceeding the minimum is attained symmetrically at two points equidistant outside the interval, confirming two solutions.