The correct option is D f(x) is strictly decreasing in (1,2)
Given : f(x)=x2−4|x|
As, f(−x)=f(x); function is even(symmetric about Y−axis)
For x>0,f(x)=x2−4x
⇒f′(x)=2x−4=0, gives x=2
⇒f′(x)<0, when x∈(0,2)→ decreasing.
f′(x)>0, when x∈(2,∞)→ increasing.
Now for x<0;f(x)=x2+4x
⇒f′(x)=2x+4=0, gives x=−2
⇒f′(x)<0, when x∈(−∞,−2)→ decreasing.
f(x)>0, when x∈(−2,0)→ increasing.