Q108 / 200 · Manipal (MET) 2011 · Mathematics
The function $f(x)=\max \{(1-x),(1+x), 2\}$, $x \in(-\infty, \infty)$ is equivalent to
- A$f(x)=\left\{\begin{array}{cc}1+x, & x \leq-1 \\ 2, & -1 \lt x \lt 1 \\ 1-x, & x \geq 1\end{array}\right.$
- Correct answerB$f(x)=\left\{\begin{array}{cc}1-x, & x \leq-1 \\ 2, & -1 \lt x \lt 1 \\ 1+x, & x \geq 1\end{array}\right.$
- C$f(x)=\left\{\begin{array}{cc}1-x, & x \leq-1 \\ 1, & -1 \lt x \lt 1 \\ 1+x, & x \geq 1\end{array}\right.$
- DNone of the above
Solution
Given that,
$f(x)=\max \{(1-x), 2,(1+x)\}$
For $x \leq-1$, we find that $1-x \geq 2$ and $1-x \geq 1+x$
$\therefore \max \{(1-x), 2,(1+x)\}=1-x$
For $-1 \lt x \lt 1$, we find that
$0 \lt 1-x \lt 2$ and $0 \lt 1+x \lt 2$
$\therefore \max \{(1-x), 2,(1+x)\}=2$
For $x \geq 1$, we find that
$1+x \geq 2,1+x\gt1-x$
$\therefore \max \{(1-x), 2,(1+x)\}=1+x$
Hence, $f(x)=\left\{\begin{array}{cc}1-x, & x \leq-1 \\ 2, & -1 \lt x \lt 1 \\ 1+x, & x \geq 1\end{array}\right.$
Hence, the correct answer is option: 'B'.