CAT 2022 Slot 3 · Quant · Question 6

The function is defined piecewise, so test each piece separately.

Take x<r first. Then f\left( x \right)=r by the second rule.

To find f\left( f\left( x \right) \right), evaluate f at r itself.

Since r\ge r, the first rule applies: f\left( r \right)=2r-r=r.

So both sides equal r, and the equation holds for every x<r.

Now take x\ge r. Then f\left( x \right)=2x-r.

Check which rule applies next: 2x-r\ge r exactly when 2x\ge 2r, that is x\ge r — which is true here.

So f\left( f\left( x \right) \right)=2\left( 2x-r \right)-r=4x-3r.

Setting the two equal, 2x-r=4x-3r, which gives 2r=2x and so x=r.

So on this branch only the single point x=r works.

Combining, the equation holds for all x<r together with x=r, that is for x\le r — the first option.

Shortcut. The function flattens everything below r to the constant r, which is a fixed point since f\left( r \right)=r. So the equation is automatic on the flat side, and the rising side can only agree where it meets the flat side.