CAT 2022 Slot 1 · Quant · Question 10

Read each modulus as a distance: \left| x+a \right| is the distance from x to -a, and \left| x-1 \right| is the distance from x to 1.

So the left side is the total distance from x to the two fixed points -a and 1.

For any x lying between those two points, that total is exactly the gap between them, \left| 1-\left( -a \right) \right|=\left| 1+a \right|.

For any x outside the gap, the total is strictly larger.

So the smallest possible value of the left side is \left| 1+a \right|, and it is achieved across the whole stretch between the two points.

Infinitely many solutions therefore occur exactly when that minimum equals 2, so every point of the stretch is a solution.

Setting \left| 1+a \right|=2 gives 1+a=2 or 1+a=-2.

So a=1 or a=-3.

The larger of the two is a=1 — the third option.

A check at a=1: the equation reads \left| x+1 \right|+\left| x-1 \right|=2, which holds for every x from -1 to 1.

Shortcut. If the minimum were below 2 the equation would have just two solutions, and if it were above 2 it would have none — infinitely many happens only at the knife edge \left| 1+a \right|=2. Since the question asks for the largest a, take the positive branch and stop.