Question

If $f:R\to R,g:R\to R$ be two giveen functions, then $f\left(x\right)=2min\{|f\left(x\right)-g\left(x\right)|,0\}$ equals

• A. $f\left(x\right)+g\left(x\right)-|g\left(x\right)-f\left(x\right)|$
• B. $f\left(x\right)+g\left(x\right)+|g\left(x\right)-f\left(x\right)|$
• C. $f\left(x\right)-g\left(x\right)+|g\left(x\right)-f\left(x\right)|$
• D. $f\left(x\right)-g\left(x\right)-|g\left(x\right)-f\left(x\right)|$

Answer 1

The correct option is D $f\left(x\right)-g\left(x\right)-|g\left(x\right)-f\left(x\right)|$

$h\left(x\right)=2min\left\{f\right(x)-g(x),0\}$

Case 1: When $f\left(x\right)>g\left(x\right)$

$\Rightarrow f\left(x\right)>g\left(x\right)\Rightarrow f\left(x\right)-g\left(x\right)>0$

$\Rightarrow 2min\left\{f\right(x)-g(x),0\}=0$

Therefore $h\left(x\right)=0$

Case 2: When $f\left(x\right)<g\left(x\right)$

$\Rightarrow f\left(x\right)<g\left(x\right)\Rightarrow f\left(x\right)-g\left(x\right)<0$

$\Rightarrow 2min\left\{f\right(x)-g(x),0\}=2\left(f\right(x)-g(x\left)\right)$

$\Rightarrow h\left(x\right)=2\left(f\right(x)-g(x\left)\right)$

Therefore, $h\left(x\right)=f\left(x\right)-g\left(x\right)-|g\left(x\right)-f\left(x\right)|$