Don't really need to measure any slopes. The reason I thought it was (A) was that f''(1) looked like 0 (inflexion), so f''(1) had to be smallest. (This narrows it down to between (A) and (B).) Assuming this then, it is also clear that f"(x) >= 0 for 0 < x < 1. So the slope at the right endpoint of the interval [0,1], namely f'(1), must be greater than the slope of the secant line joining the points on the graph at the endpoints (this is visually clear, and also follows from the Mean Value Theorem, which iirc was proved in this year's 3U BOS Trial). But the slope of this secant line is just (f(1) - 0)/(1 - 0) = f(1), and the only option between (A) and (B) that has f'(1) > f(1) is (A). So the answer is (A).
I don't think we can actually just deduce from that graph that it is necessarily the case that f'(1) > 1 (unless they meant for you to interpret it as to scale and measure it, which is not usually what HSC 3U/4U does). The Q. should have said which is a possible correct answer (only (A) is).