The correct option is B 39;38
y=2x3−15x2+36x+11
dydx=6x2−30x+36
dydx=0=6x2−30x+36=0
x2−5x+6=0
x2−2x−3x+6=0
(x−3)(x−2)=0
Roots of the quadratic equation are
x=2, 3
Now double differentiation of given equation
d2ydx2=12x−30
At x=2, d2ydx2=−6
Hence at x=2 there is maxima of value ymax=39
At x=3, d2ydx2=+6
Hence at x=3, there is minima of value ymin=38