Let <span class="math-inline">f(x) = \begin{cases} x^2 & ; 0 < x < 2 \\ 2x - 3 & ; 2 \leq x < 3 \\ x + 2 & ; x \geq 3 \end{cases}</span>. Then :
(A) <span class="math-inline">f(f(f(\frac{2}{3}))) = f(\frac{3}{2})</span>
(B) <span class="math-inline">1 + f(f(f(\frac{5}{2}))) = f(\frac{5}{2})</span>
(C) <span class="math-inline">f(f(1)) = f(1) = 1</span>
(D) none of these
Step-by-Step Solution
Key Concept: The function f(x) has two different definitions based on the domain: f(x) = x² for 0 ≤ x ≤ 1, and f(x) = 2 - x for x > 1. We must carefully track which formula applies at each composition step and ensure intermediate values fall in the correct domain.
Step 1: Determine the piecewise function definition
The function $f(x)$ is defined as:
$$ f(x) = \begin{cases} x^2 & ; 0 \le x \le 1 \\ 2-x & ; x > 1 \end{cases} $$
The domain of $f(x)$ is $[0, \infty)$.
Step 2: Check Option (A): $f(f(f(\frac{2}{3}))) = f(\frac{3}{2})$
First, evaluate the left-hand side (LHS):
* For $x = \frac{2}{3}$: Since $0 \le \frac{2}{3} \le 1$, we use $f(x) = x^2$.
$$ f\left(\frac{2}{3}\right) = \left(\frac{2}{3}\right)^2 = \frac{4}{9} $$
* For $x = \frac{4}{9}$: Since $0 \le \frac{4}{9} \le 1$, we use $f(x) = x^2$.
$$ f\left(f\left(\frac{2}{3}\right)\right) = f\left(\frac{4}{9}\right) = \left(\frac{4}{9}\right)^2 = \frac{16}{81} $$
* For $x = \frac{16}{81}$: Since $0 \le \frac{16}{81} \le 1$, we use $f(x) = x^2$.
$$ f\left(f\left(f\left(\frac{2}{3}\right)\right)\right) = f\left(\frac{16}{81}\right) = \left(\frac{16}{81}\right)^2 = \frac{256}{6561} $$
Now, evaluate the right-hand side (RHS):
* For $x = \frac{3}{2}$: Since $\frac{3}{2} > 1$, we use $f(x) = 2-x$.
$$ f\left(\frac{3}{2}\right) = 2 - \frac{3}{2} = \frac{1}{2} $$
Comparing LHS and RHS:
$$ \frac{256}{6561} \ne \frac{1}{2} $$
Therefore, Option (A) is false.
Step 3: Check Option (B): $1 + f(f(f(\frac{5}{2}))) = f(\frac{5}{2})$
First, evaluate $f(\frac{5}{2})$:
* For $x = \frac{5}{2}$: Since $\frac{5}{2} > 1$, we use $f(x) = 2-x$.
$$ f\left(\frac{5}{2}\right) = 2 - \frac{5}{2} = -\frac{1}{2} $$
Next, we need to evaluate $f(f(\frac{5}{2})) = f(-\frac{1}{2})$.
* For $x = -\frac{1}{2}$: Since $-\frac{1}{2} < 0$, this value is outside the defined domain of $f(x)$, which is $[0, \infty)$.
* Therefore, $f(-\frac{1}{2})$ is undefined.
* Consequently, $f(f(f(\frac{5}{2})))$ is undefined.
An equality involving an undefined term cannot be true. Therefore, Option (B) is false.
Step 4: Check Option (C): $f(f(1)) = f(1) = 1$
First, evaluate $f(1)$:
* For $x = 1$: Since $0 \le 1 \le 1$, we use $f(x) = x^2$.
$$ f(1) = 1^2 = 1 $$
Next, evaluate $f(f(1))$:
* $f(f(1)) = f(1) = 1$.
Thus, $f(f(1)) = 1$ and $f(1) = 1$.
Therefore, Option (C) is true.
Step 5: Conclusion
Based on the evaluation of each option, only Option (C) is true.
Correct Answer: A,B,C