The correct option is
A Surjective but not injective
f(x)=x3+6x2+11x+6=0.
One one roots as
x=−1Hence
f(x)=(x+1)(x2+5x+6)
=(x+1)(x+3)(x+2)
Hence
f(x)=0 has three roots −1,−2,−3
Hence y meets the x axis at three points.
Hence
The above function is surjective but it is not injective.