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Question

Quantitative Aptitude Question on Functions

Let r be a real number and f(x)={2xrif xr rif x<rf(x) = \begin{cases} 2x - r & \text{if } x \geq r \\\ r & \text{if } x < r \end{cases}.Then, the equation f(x)=f(f(x))f(x)=f(f(x)) holds for all real values of x where

A

xrx≤r

B

xrx≥r

C

x>rx>r

D

xrx≠r

Answer

xrx≤r

Explanation

Solution

The correct answer is A: xrx ≤ r
Let's analyze the equation f(x)=f(f(x))f(x) = f(f(x)) for different cases:
Case 1: x<rx < r
In this case, the equation f(x)=f(f(x))f(x) = f(f(x)) becomes f(x)=f(2xr)  since  x<rf(x) = f(2x - r) \space{since}\space x < r. Now, from the definition of f(x)f(x), when x<r,f(x)=rx<r, f(x) = r. So, we have r=f(2xr)r = f(2x - r).
Case 2: xrx ≥ r
In this case, the equation f(x)=f(f(x))f(x) = f(f(x)) becomesf(x)=f(x) f(x) = f(x) since xrx ≥ r. This simplifies to f(x)=xf(x) = x, which is true for xrx ≥ r.
Now,combining both cases:
For x<rx < r, we have r=f(2xr)r = f(2x - r).
For xrx ≥ r, we have f(x)=xf(x) = x.
Since the equation r=f(2xr)r=f(2x - r) holds for x<rx < r and f(x)=xf(x) = x holds for xrx ≥ r, the correct answer is: a. x ≤ r