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CAT 2022 Slot 3 — QA Question 16

Composite FunctionsEasy

Let r be a real number and f(x) = {2x-rif x≥rrif x<r.Then, the equation f(x) = f(f(x)) holds for all real values of x where

Answer & solution

  • A

    x ≠ r

  • B

    x ≥ r

  • C

    x > r

  • x ≤ r

Solution

Easy

ff is piecewise. Test the identity f(x)=f(f(x))f(x)=f(f(x)) region by region. Wherever f(x)=rf(x)=r (a fixed point of ff), the identity holds; wherever ff grows, it fails.

1

Case $x here f(x)=rf(x)=r, and f(r)=2r−r=rf(r)=2r-r=r, so:

f(f(x))=f(r)=r=f(x) ✓f(f(x))=f(r)=r=f(x)\ \checkmark
2

Case x=rx=r: f(r)=2r−r=rf(r)=2r-r=r, and applying ff again gives rr:

f(f(r))=f(r)=r=f(r) ✓f(f(r))=f(r)=r=f(r)\ \checkmark
3

Case x>rx>r: f(x)=2x−r>rf(x)=2x-r>r, so applying ff again uses the same branch:

f(f(x))=2(2x−r)−r=4x−3r4x−3r=2x−r ⇒ x=r, contradicting x>r\begin{aligned} &f(f(x))=2(2x-r)-r=4x-3r\\ &4x-3r=2x-r\ \Rightarrow\ x=r,\ \text{contradicting }x>r \end{aligned}

So the identity fails for x>rx>r.

f(x)=f(f(x))f(x)=f(f(x)) holds for all xx with x≤r\mathbf{x\le r}.

For x≤rx\le r, ff always outputs rr, which is a fixed point (f(r)=rf(r)=r). Applying ff again can't change it — so the identity is automatic there.

CAT 2022 Slot 3 QA Q16: Let r be a real number and f(x) = 2 x - r if x &ge; r r if x < r . Then, the equation f(x) = f(f(x)) holds for — Solution | TheCATExam