Observe the graph and choose the correct option from the following:

Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A = \(\frac{dN}{dt}\) = rN; B = \(\frac{dN}{dt}\) = rN(\(\frac{K-N}{K}\))
A population grows in one of two ways. When resources are unlimited it grows in an exponential (geometric) way: \(\displaystyle\frac{dN}{dt} = rN\), and the plot of N against time is a J-shaped curve. When resources are limited the growth slows and stops at the carrying capacity (K). This is logistic growth, \(\displaystyle\frac{dN}{dt} = rN\left(\frac{K-N}{K}\right)\), and the plot is a sigmoid curve. In the graph, curve A keeps rising (curve a of the book’s Figure 11.3) and curve B levels off at K (curve b).
Option (1) A = \(\frac{dN}{dt}\) = rN; B = \(\frac{dN}{dt}\) = rN(\(\frac{K-N}{K}\)) — Correct. A is J-shaped, so it is exponential growth, rN. B flattens at K, so it is logistic growth. As N comes close to K, the term (K−N)/K becomes nearly zero, and growth almost stops.
Option (2) A = \(\frac{dN}{dt}\) = rN; B = \(\frac{dN}{dt}\) = rN(\(\frac{K}{K-N}\)) — Incorrect. A is right, but B is wrong. In K/(K−N) the value gets larger as N comes close to K, so growth would speed up near the carrying capacity instead of stopping.
Option (3) A = \(\frac{dN}{dt}\) = rN(\(\frac{K-N}{K}\)); B = \(\frac{dN}{dt}\) = rN. — Incorrect. This swaps the two curves. A J-shaped curve never levels off, so it cannot carry the (K−N)/K term, and the sigmoid curve B cannot be the unlimited growth rN.
Option (4) A = \(\frac{dN}{dt}\) = rN(\(\frac{K-N}{K}\)); B = \(\frac{dN}{dt}\) = rN(\(\frac{K}{K-N}\)) — Incorrect. Both parts are wrong. A is not logistic, and K/(K−N) does not describe any growth that levels off at K.
Note: Swap the a and b of Figure 11.3 for A and B, and the book’s graph turns into a question. Learn each curve with its own equation, and a relabelled figure will not slow you down.
NCERT Extract:
Population Growth – “… The above equation describes the exponential or geometric growth pattern of a population (Figure 11.3) and results in a J-shaped curve when we plot N in relation to time.”
Population Growth – “A population growing in a habitat with limited resources show initially a lag phase, followed by phases of acceleration and deceleration and finally an asymptote, when the population density reaches the carrying capacity. A plot of N in relation to time (t) results in a sigmoid curve. This type of population growth is called Verhulst-Pearl Logistic Growth (Figure 11.3) and is described by the following equation:
\(\displaystyle\frac{dN}{dt} = rN\left(\frac{K-N}{K}\right)\)
Where N = Population density at time t
r = Intrinsic rate of natural increase
K = Carrying capacity”

Figure 11.3 Population growth curve a when responses are not limiting the growth, plot is exponential, b when responses are limiting the growth, plot is logistic, K is carrying capacity