​The intensity of competition can be inferred from knowing the carrying capacity (K) and the population size (N) in the equation below:

​$$\frac {dn}{dt}=rN \frac{(K-N)}K$$

Assume that populations have the same intrinsic growth rates (r) and carrying capacities (K). Then, at which one of the following values of the second term (K−N)/K in the equation, is the intraspecific competition likely to be the highest?​

Correct option is A

The Term ( $$\frac{K-N}{K}$$) indicates the remaining portion of carrying capacity . As the population N increases and nears K , this term becomes very small, approaching 0.

-  When $$\frac{K-N}{K}$$ = 0.001 , it means that Nis 99.99% of K, i.e. the population is extremely close to the carrying capacity.

- At this point , resources are almost exhausted, leading to maximum intraspecific competition (within the same species) , since every individual is surviving for limited remaining resources. 

- Hence, intraspecific competition is highest  when $$\frac{K-N}{K}$$ is the  smallest , i.e. 0.001

Therefore, the correct answer is (a) 0.001.