Question

Let $f:R\to R,g:R\to R$ be the two given functions, then the function $h\left(x\right)=2\times$ min$\left\{f\right(x)-g(x),0\}$ is same as

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

Answer 1

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

if $f\left(x\right)>g\left(x\right)$, then $h\left(x\right)=2\times min\left(f\right(x)-g(x),0)$ will be equal to 0.
if $f\left(x\right)<g\left(x\right)$ then $h\left(x\right)=2\times min\left(f\right(x)-g(x),0)$ will be same as $2\times \left(f\left(x\right)-g\left(x\right)\right)$
Only option D satisfies both the the conditions. Hence the correct answer is option D.