Question

$f\left(x\right)=$ max $\left\{2-x,2+x,4\right\}x\in R$

• A. $f\left(x\right)=\left\{\begin{array}{ll}2-x& x\ge 2\\ 4& -2<x<2\\ 2+x& x\le -2\end{array}\right.$
• B. $f\left(x\right)=\left\{\begin{array}{ll}2-x& -2<x<2\\ 4& x\ge 2\\ 2+x& x\le -2\end{array}\right.$
• C. $f\left(x\right)=\left\{\begin{array}{ll}2-x& x\le -2\\ 4& x\ge 2\\ 2+x& -2<x<2\end{array}\right.$
• D. $f\left(x\right)=\left\{\begin{array}{ll}2-x& x\le -2\\ 4& -2<x<2\\ 2+x& x\ge 2\end{array}\right.$

Answer 1

The correct option is D

$f\left(x\right)=\left\{\begin{array}{ll}2-x& x\le -2\\ 4& -2<x<2\\ 2+x& x\ge 2\end{array}\right.$

Explanation for the correct option:

There will be three cases:

Case 1. If $\mathit{x}\mathbf{}\mathbf{\le }\mathbf{}\mathbf{-}\mathbf{2}$, then

$-x\ge 2$

$\Rightarrow$$2-x\ge 2+2$

$\Rightarrow 2-x\ge 4$

Also $2-x\ge 2+x$

$\therefore$Max $(2–x,2+x,4)$ is $2–x$

Case 2. If $\mathbf{-}\mathbf{2}\mathbf{}\mathbf{<}\mathbf{}\mathit{x}\mathbf{}\mathbf{<}\mathbf{}\mathbf{2}$, then

$2>-x>-2$

$\Rightarrow$$2+2>2-x>2-2$

$\Rightarrow$ $4>2-x>0$

Now, $-2<x<2$

$\Rightarrow$$2-2<2+x<2+2$

$\Rightarrow$ $0<2+x<4$

$\therefore$Max $(2–x,2+x,4)$ is $4$

Case 3. If $\mathit{x}\mathbf{}\mathbf{\ge }\mathbf{}\mathbf{2}$, then

$\Rightarrow$$x+2\ge 2+2$

$\Rightarrow$ $x+2\ge 4$

Now, $x\ge 2$

$\Rightarrow$ $-x\le -2$

$\Rightarrow$$2-x\le 2-2$

$\Rightarrow$$2-x\le 0$

$\therefore$Max $(2–x,2+x,4)$ is $2+x$

Thus, $f\left(x\right)=$max $\left\{2-x,2+x,4\right\}x\in R$ is $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\{\begin{array}{ll}2-x& x\le -2\\ 4& -2<x<2\\ 2+x& x\ge 2\end{array}$

Hence, Option ‘D’ is Correct.