MHT-CET Mathematics Full Test-10 QUE-35 Solution

Question-35 A population p(t) of 1000 bacteria introduced into a nutrient medium grows according to the relation p(t)=1000+\frac{1000t}{100+t^2}. The maximum size of this bacterial population is

(A) 1100

(B) 1250

(C) 1050

(D) 950

Answer: (C)

Explanation:

p(t)=1000+\frac{1000t}{100+t^2} \Rightarrow p'(t)=\frac{(100+t^2)(1000)-1000t(2t)}{(100+t^2)^2} =\frac{1000(100-t^2)}{(100+t^2)^2}

For maximum

p'(t)=0 \Rightarrow \frac{1000(100-t^2)}{(100+t^2)^2}=0 \Rightarrow t=\pm10

Also,

p''(t)=1000\left[\frac{(100+t^2)^2(-2t)-(100-t^2)2(100+t^2)(2t)}{(100+t^2)^2}\right] p''(10)=\frac{1000[(40000)(-20)-0]}{40000} =-2000<0

∴ P(max) at (t=10)

=1000+\frac{1000(10)}{100+(10)^2} =1000+\frac{10000}{200} =1050