Before applying the medicine: bacterial growth

It is given that bacterial growth follows first-order kinetics.

For first-order growth, $\frac{dN}{dt}=kN$ where $N$ is the number of bacteria.

Integrating, $N=N_0 e^{kt}$

So, before medicine is applied:

• $N$ increases exponentially with time.
• Hence the $N$ vs $t$ graph should be an increasing, concave upward curve.

After applying the medicine: bacterial decay

It is given that the rate of bacterial decay is proportional to the square of the existing number of bacteria.

Therefore, $-\frac{dN}{dt}=kN^2$ or $\frac{dN}{dt}=-kN^2$

Rearranging, $\frac{dN}{N^2}=-k\,dt$

Integrating, $-\frac{1}{N}=-kt+C$ $\frac{1}{N}=kt+C'$

If at $t=0$, $N=N_0$, then $\frac{1}{N}=kt+\frac{1}{N_0}$

Thus, $N=\frac{1}{kt+\frac{1}{N_0}}=\frac{N_0}{1+kN_0 t}$

So after medicine is applied:

• $N$ decreases with time.
• The decay is not exponential.
• The $N$ vs $t$ graph is a decreasing hyperbola-like curve, falling rapidly at first and then flattening out.

What kind of graphs are expected?

Therefore, the correct set must show:

• Before treatment: exponential increase in $N$ vs $t$
• After treatment: second-order decay, i.e. a decreasing curve of the form $N=\frac{N_0}{1+kN_0 t}$

Matching with options

Among the given options, the graph set that corresponds to:

• first-order bacterial growth before medicine, and
• second-order decay after medicine

is Option C.

Final answer

$\boxed{\text{C}}$