The correct option is D 𝑓 is bijective function
For one-one :
Let x,y∈R−{−b} such that f(x)=f(y)
⇒x+ax+b=y+ay+b
⇒xy+ay+bx+ab=xy+xa+yb+ab
⇒x(b−a)=y(b−a)⇒x=y (∵a≠b)
∴f is one-one function.
For onto :
Let y∈R such that f(x)=y
⇒x+ax+b=y
⇒x+a=xy+yb
⇒x=a−byy−1
⇒y∈R−{1}
∴f is onto function.
Hence, f is one-one and onto function, it is bijective function.