11.
Consider the following “logistic” equation that describes the growth of a population of
organisms: 𝑑𝑥/𝑑𝑡 = 𝑥(1 − 𝑥). The stabilities of the fixed (equilibrium) points are:
a. 𝑥 = 0 (stable) and 𝑥 = 1 (stable)
b. 𝑥 = 0 (stable) and 𝑥 = 1 (unstable)
c. 𝑥 = 0 (unstable) and 𝑥 = 1 (stable)
d. 𝑥 = 0 (unstable) and 𝑥 = 1 (unstable)
The logistic equation dx/dt = x(1-x) models population growth with a carrying capacity of 1. Fixed points occur where dx/dt = 0, so x(1-x) = 0, giving x=0 and x=1. Stability is determined by the sign of f'(x) = 1-2x at these points: if f'(x^*) > 0, unstable; if f'(x^*) < 0, stable.
Fixed Points Analysis
At x=0, f'(0)=1 > 0, so perturbations grow, making it unstable—populations near zero increase away from extinction.
At x=1, f'(1)=-1 < 0, so perturbations decay, making it stable—populations approach the carrying capacity.
Option Evaluation
Correct answer: c.
Understanding Fixed Points
Fixed points solve x(1-x)=0, yielding x^*=0 (extinction) and x^*=1 (carrying capacity). Linear stability uses f'(x^*)=1-2x^*: positive eigenvalue means unstable, negative means stable.
Stability Determination
For x^*=0, f'(0)=1 > 0 (unstable)—small populations explode.
For x^*=1, f'(1)=-1 < 0 (stable)—populations converge regardless of starting size (except exactly 0).
CSIR NET Exam Relevance
In ecology/evolution sections, this tests phase line analysis: arrows point right of 0 (growth), left of 1 (decline). Solution x(t) = 1 / (1 + (1/x_0 – 1)e^{-t}) confirms asymptotic approach to 1. Practice identifies correct option c for such MCQs.
