Comprehension
The function \( f(x) \) is defined as follows:
\[ f(x) = \begin{cases} x & \text{for } 0 \le x \le 1 \\ 1 & \text{for } x \ge 1 \\ 0 & \text{otherwise} \end{cases} \]
The properties of the function are as follows:
\[ f_1(x) = f(-x) \quad \text{for all } x \] \[ f_2(x) = -f(x) \quad \text{for all } x \] \[ f_3(x) = f(f(x)) \quad \text{for all } x \]
Question: 1
How many of the following products are necessarily zero for every \(x\)?
\[
f_1(x) f_2(x), \, f_2(x) f_3(x), \, f_2(x) f_4(x)
\]
How many of the following products are necessarily zero for every \(x\)?
\[
f_1(x) f_2(x), \, f_2(x) f_3(x), \, f_2(x) f_4(x)
\]
Show Hint
Check the behavior of each function before calculating the product to determine if it results in zero.
Updated On: Aug 1, 2025
- 0
- 1
- 2
- 3
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The Correct Option is C
Solution and Explanation
Given the functions:
\[
f_1(x) = x, \, f_2(x) = 1, \, f_3(x) = -x, \, f_4(x) = 0
\]
We calculate the following products:
- \( f_1(x) f_2(x) = x \times 1 = x \), which is not necessarily zero for all \(x\).
- \( f_2(x) f_3(x) = 1 \times (-x) = -x \), which is not necessarily zero for all \(x\).
- \( f_2(x) f_4(x) = 1 \times 0 = 0 \), which is always zero for all \(x\).
Thus, only one of the products is zero for all \(x\), so the answer is 2. \[ \boxed{2} \]
- \( f_2(x) f_3(x) = 1 \times (-x) = -x \), which is not necessarily zero for all \(x\).
- \( f_2(x) f_4(x) = 1 \times 0 = 0 \), which is always zero for all \(x\).
Thus, only one of the products is zero for all \(x\), so the answer is 2. \[ \boxed{2} \]
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Question: 2
Which of the following is necessarily true?
Which of the following is necessarily true?
Show Hint
When solving functional equations, check for simple substitutions or transformations to match expressions.
Updated On: Aug 1, 2025
- \( f_4(x) = f_1(x) \) for all \(x\)
- \( f_1(x) = f_3(-x) \) for all \(x\)
- \( f_2(x) = f_4(x) \) for all \(x\)
- \( f_1(x) = f_3(x) = 0 \) for all \(x\)
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Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Given the functions:
\[
f_1(x) = x, \, f_3(x) = -x
\]
It is clear that \( f_1(x) = f_3(-x) \) for all \(x\). This is because \( f_1(x) = x \) and \( f_3(-x) = -(-x) = x \).
\[
\boxed{f_1(x) = f_3(-x)}
\]
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