JEE Advanced 2013 · Paper 2, Mathematics question 44

Simplify. $a+b-|a-b|=2\min(a,b)$, so
$$f(x)=2\min\left(2|x|,\ |x+2|\right).$$

Shape. $2|x|=|x+2|$ at $x=-\frac23$ and $x=2$.

• $x=-2$: $f=0$, a local minimum.
• $x=-\frac23$: the two pieces cross with $f$ switching from rising to falling, a local maximum ($f=\frac83$).
• $x=0$: local minimum (not an option).
• $x=2$: $f$ keeps increasing through it, no extremum.

(A), (B)