The correct option is B (2,2)
y=12x4−5x3+18x2−19x
dydx=2x3−15x2+36x−19
Let f(x)=2x3−15x2+36x−19
f′(x)=6x2−30x+36=0
For extrema, x2−5x+6=0
⇒x=2,3
f′′(x)=12x−30
f′′(x)<0 for x=2
So, at x=2, slope is maximum.
y=8−40+72−38
=72−70=2
Hence, maximum slope occurs at (2,2).