For a concave lens of focal length $$f$$, the relation between object and image distance $$u$$ and $$v$$, respectively, from its pole can best be represented by ($$u = v$$ is the reference line):

$$\frac{1}{v} = \frac{1}{u} + \frac{1}{f} \implies v = \frac{uf}{u+f}$$

$$u \to 0 \implies v \to 0$$ and $$u \to \infty \implies v \to f$$

$$\frac{dv}{du} = \frac{f(u+f) - uf}{(u+f)^2} = \frac{f^2}{(u+f)^2}$$

$$\left.\frac{dv}{du}\right|_{u=0} = 1$$

Since the initial slope is $$1$$, the curve starts tangentially to the line $$u=v$$ and lies below it as $$u$$ increases ($$v < u$$).

Answer: Option (A): curve starts along $$u=v$$ and approaches $$v=f$$ asymptotically from below.