The correct option is B surjective but not injective
f(x)=x1+x2
Let f(x1)=f(x2), x1,x2∈R
⇒x11+x21=x21+x22
⇒x11+x21−x21+x22=0
⇒(x1−x2)(1−x1x2)(1+x21)(1+x21)=0
⇒x1=x2, x1=1x2
Thus, f is not injective.
Let y=x1+x2
⇒yx2−x+y=0
⇒x=1±√1−4y22y
∵x∈R, 1−4y2≥0 & y≠0
⇒y∈[−12,12]−{0}
Also, for x=0, y=0
∴Rf=[−12,12]=Co-domain of f
Alternatively, put x=tanθ, θ∈(−π2,π2)
Then y=12sin2θ
−1≤sin2θ≤1
⇒−12≤y≤12
Thus, f is surjective.