CAT 2024 Slot 3 · Quant · Question 18

For any two numbers, the larger minus the smaller is just the distance between them.

So \max \{x,2\}-\min \{x,2\}=\left| x-2 \right|, and the equation becomes \left| x-2 \right|=\left| x+2 \right|-\left| x-2 \right|.

Bringing the like terms together, 2\left| x-2 \right|=\left| x+2 \right|.

The critical points are x=2 and x=-2, so examine the three stretches they create.

For x\ge 2, both x-2 and x+2 are non-negative, so 2\left( x-2 \right)=x+2.

That gives 2x-4=x+2, so x=6, which does lie in x\ge 2 and is accepted.

For -2\le x<2, x-2 is negative while x+2 is non-negative, so 2\left( 2-x \right)=x+2.

That gives 4-2x=x+2, so 3x=2 and x=\frac{2}{3}, which lies in the range and is accepted.

For x<-2, both expressions are negative, so 2\left( 2-x \right)=-\left( x+2 \right).

That gives 4-2x=-x-2, so x=6, which contradicts x<-2 and is rejected.

Two values survive, x=\frac{2}{3} and x=6, so the count is 2.

Shortcut. Recognising \max -\min as \left| x-2 \right| collapses the whole question to 2\left| x-2 \right|=\left| x+2 \right|. Squaring both sides gives 4{\left( x-2 \right)}^{2}={\left( x+2 \right)}^{2}, a quadratic whose two roots are \frac{2}{3} and 6 — and both satisfy the original equation, so neither is extraneous.