Question

A real valued function \( f \) is defined as \[ f(x) = \begin{cases} -1 & \text{if} \, -2 \leq x \leq 0 \\ x - 1 & \text{if} \, 0 \leq x \leq 2 \end{cases} \] \(\text{Which of the following statements is FALSE?}\)

• A. \( f(|x|) = |x| - 1, \text{ if } 0 \leq x \leq 1 \)
• B. \( |f(x)| = x - 1, \text{ if } 1 \leq x \leq 2 \)
• C. \( f(|x|) + f(x) = 1, \text{ if } 0 \leq x \leq 1 \)
• D. \( f(|x|) - |f(x)| = 0, \text{ if } 1 \leq x \leq 2 \)

Hint:

When working with piecewise functions, always check the behavior of the function over the defined intervals and ensure all conditions are satisfied for each case.

Answer 1

The Correct Option is C

We are given the piecewise function: \[ f(x) = \begin{cases} -1 & \text{if} \, -2 \leq x \leq 0 \\ x - 1 & \text{if} \, 0 \leq x \leq 2 \end{cases} \] Now, let's check each of the given statements to identify the FALSE one. 

Option (a): \( f(|x|) = |x| - 1, \text{ if } 0 \leq x \leq 1 \) For \( x \in [0, 1] \), we have \( f(x) = x - 1 \), so \( f(|x|) = f(x) = x - 1 \). Thus, this statement is correct. 

Option (b): \( |f(x)| = x - 1, \text{ if } 1 \leq x \leq 2 \) For \( x \in [1, 2] \), \( f(x) = x - 1 \), so \( |f(x)| = |x - 1| \). 

Since \( x - 1 \geq 0 \) for \( x \in [1, 2] \), we get: \[ |f(x)| = x - 1 \] Thus, this statement is also correct. 

Option (c): \( f(|x|) + f(x) = 1, \text{ if } 0 \leq x \leq 1 \) For \( x \in [0, 1] \), \( f(x) = x - 1 \) and \( f(|x|) = f(x) = x - 1 \). 

So, \[ f(|x|) + f(x) = (x - 1) + (x - 1) = 2x - 2 \] For \( 0 \leq x \leq 1 \), \( 2x - 2 \neq 1 \). Hence, this statement is false. 

Option (d): \( f(|x|) - |f(x)| = 0, \text{ if } 1 \leq x \leq 2 \) For \( x \in [1, 2] \), \( f(x) = x - 1 \) and \( f(|x|) = f(x) = x - 1 \). Also, \( |f(x)| = x - 1 \). So, \[ f(|x|) - |f(x)| = (x - 1) - (x - 1) = 0 \] Thus, this statement is correct. 

Conclusion: The FALSE statement is option (c). Thus, the correct answer is \( \boxed{(c)} \).