The correct option is C (43,2e)
Given curve, y=xex2
And its slope of tangent =dydx
dydx=x⋅2x⋅ex2+ex2
=2x2ex2+ex2
=ex2(1+2x2)
At point (1,e), dydx=3e
For given point (1,e) equation of the tangent is
(y−e)=3e(x−1)
Hence, the point (43,2e) lies on the above line.