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CAT 2022 Slot 2 — QA Question 3

Composite FunctionsEasy

Suppose for all integers x, there are two functions f and g such that f(x) + f(x - 1) - 1 = 0 and g(x) = x2. If f(x2 - x) = 5, then the value of the sum f(g(5)) + g(f(5)) is

Answer & solution

Answer: 12

Solution

Easy

The recurrence f(x)+f(x−1)=1f(x)+f(x-1)=1 forces ff to alternate between just two values. Pin down f(0)f(0) from the clue f(x2−x)=5f(x^2-x)=5, then ride the recurrence to get ff at any integer. Finally evaluate the composite using g(x)=x2g(x)=x^2.

1

Find f(0)f(0). Put x=0x=0 in f(x2−x)=5f(x^2-x)=5: since 02−0=00^2-0=0,

f(0)=5.f(0)=5.
2

Use the recurrence f(x)=1−f(x−1)f(x)=1-f(x-1) to step along:

f(1)=1−f(0)=1−5=−4f(2)=1−f(1)=1−(−4)=5\begin{aligned} &f(1)=1-f(0)=1-5=-4\\ &f(2)=1-f(1)=1-(-4)=5 \end{aligned}

So f(even)=5f(\text{even})=5 and f(odd)=−4f(\text{odd})=-4.

3

Evaluate the two pieces. Here g(5)=25g(5)=25 and f(5)=−4f(5)=-4:

f(g(5))=f(25)=−4(25 is odd)g(f(5))=g(−4)=(−4)2=16\begin{aligned} &f(g(5))=f(25)=-4 \quad(\text{$25$ is odd})\\ &g(f(5))=g(-4)=(-4)^2=16 \end{aligned}

f(g(5))+g(f(5))=−4+16=12f(g(5))+g(f(5))=-4+16=\mathbf{12}.

CAT 2022 Slot 2 QA Q3: Suppose for all integers x, there are two functions f and g such that f(x) + f(x - 1) - 1 = 0 and g(x) = x 2 . — Solution | TheCATExam