CAT 2022 Slot 3 QA Question & Solution

AlgebraMedium

Question

Let $r$ be a real number and

$f(x) = \begin{cases} \ \mathbf{2x - r} \quad \text{if } x \ge r ; \quad \mathbf{r} \quad \text{if } x < r \end{cases} $

Then, the equation $f(x) = f(f(x))$ holds for all real values of $x$ where

Options

$x > r$

$x \leq r$

$x \geq r$

Solution

When $x < r$
$$ f(x) = r $$
We know that $f(x) = f(f(x))$, so:
$$ r = f(r) $$
Substitute $f(r)$ from the second case:
$$ r = 2r - r $$
This simplifies to:
$$ r = r $$

When $x \ge r$
$$ f(x) = 2x - r $$
Now, using $f(x) = f(f(x))$, we get:
$$ 2x - r = f(2x - r) $$
Substitute the expression for $f(2x - r)$:
$$ 2x - r = 2(2x - r) - r $$
Simplify:
$$ 2x - r = 4x - 3r $$
This leads to:
$$ x = r $$

Therefore, we conclude that $x \le r$.