CAT 2022 Slot 2 · Quant · Question 10

The first relation says f\left( x \right)+f\left( x-1 \right)=1 for every integer x.

Applying it at x and at x-1 gives f\left( x \right)+f\left( x-1 \right)=1 and f\left( x-1 \right)+f\left( x-2 \right)=1.

Subtracting, f\left( x \right)=f\left( x-2 \right), so f takes one value on all even integers and another on all odd integers.

Call those values p for even inputs and q for odd, with p+q=1 from the original relation.

Now look at the given condition: {x}^{2}-x=x\left( x-1 \right) is a product of consecutive integers, so it is always even.

So f\left({x}^{2}-x \right) is always p, and the condition says p=5.

Then q=1-p=1-5=-4.

For the first term, g\left( 5 \right)=25, which is odd, so f\left( g\left( 5 \right) \right)=f\left( 25 \right)=q=-4.

For the second term, f\left( 5 \right)=q=-4 since 5 is odd, so g\left( f\left( 5 \right) \right)=g\left( -4 \right)={\left( -4 \right)}^{2}=16.

The sum is -4+16=12.

Shortcut. The relation f\left( x \right)+f\left( x-1 \right)=1 makes f alternate between two values summing to 1, so only the parity of the input ever matters. Spotting that {x}^{2}-x is always even pins down the even value at once, and the odd value follows by subtraction.