CAT 2022 Question Paper | Quant Slot 3

CAT Previous Year Paper | CAT Quant Questions | Question 6

CAT 2022 Quant was dominated by Arithmetic followed by Algebra. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.

Question 6 : Let \(r\) be a real number and \(f(x)=\left\{\begin{array}{cl}2 x-r & \text { if } x \geq r \\ r & \text { if } x\lt r\end{array}\right.\). Then, the equation \(f(x)=f(f(x))\) holds for all real values of \(x\) where

  1. \(x \leq r\)
  2. \(x \geq r\)
  3. \(x \gt r\)
  4. \(x \neq r\)

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Explanatory Answer

Observe that f(r) = 2r - r = r
Therefore, f(f(r)) = f(r) = r
For any x < r,
f(x) = r
f(f(x)) = f(f(r)) = r
For any x > r,
(let’s assume that x = kr, where k is some number greater than 1)
f(x) = f(kr) = 2kr - r = (2k - 1)r
Since k is greater than 1, (2k - 1) is greater than k.
This means that every time we put an input that is greater than r, the output is greater than the input.
So, for any x > r,
f(f(x)) \( \neq \) f(f(r))
Therefore, for f(f(x)) = f(x) to be satisfied, the necessary condition is, \( x \leq r \).

The answer is '\(x \leq r\)'

Choice A is the correct answer.

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