Verified Solution • 2025–26 Session moderate

QJEEM21Ma6a52: JEE Main 2021 Mathematics - Class 12 JEEM21Ma6a52 Solution: The rate of growth of bacteria in a culture is proportional to…

The rate of growth of bacteria in a culture is proportional to the number of bacteria present and the bacteria count is 1000 at initial time t = 0. The number of bacteria is increased by 20% in 2 hours. If the population of bacteria is 2000 after klog⁡e(65){k \over {{{\log }_e}\left( {{6 \over 5}} \right)}} hours, then (klog⁡e2)2{\left( {{k \over {{{\log }_e}2}}} \right)^2} is equal to :

Multiple Choice Options

  • A 16
  • B 8
  • C 2
  • D 4

Step-by-Step Verified Solution

dxdt∝x{{dx} \over {dt}} \propto x

dxdt=λx{{dx} \over {dt}} = \lambda x

∫1000xdxx=∫0tλdt\int\limits_{1000}^x {{{dx} \over x} = \int\limits_0^t {\lambda dt} }

ln⁡x−ln⁡1000=λt\ln x - \ln 1000 = \lambda t

ln⁡(x1000)=λt\ln \left( {{x \over {1000}}} \right) = \lambda t

Put t = 2, x = 1200

ln⁡(1210)=2λ⇒λ=12ln⁡65\ln \left( {{{12} \over {10}}} \right) = 2\lambda \Rightarrow \lambda = {1 \over 2}\ln {6 \over 5}

Now, ln⁡(x1000)=t2ln⁡(65)\ln \left( {{x \over {1000}}} \right) = {t \over 2}\ln \left( {{6 \over 5}} \right)

x=1000et2ln⁡(65)x = 1000{e^{{t \over 2}\ln \left( {{6 \over 5}} \right)}}

x=2000x = 2000 at t=kln⁡(65)t = {k \over {\ln \left( {{6 \over 5}} \right)}}

⇒2000=1000ek2ln⁡(6/5)×ln⁡(6/5)\Rightarrow 2000 = 1000{e^{{k \over {2\ln (6/5)}} \times \ln (6/5)}}

⇒2=ek/2\Rightarrow 2 = {e^{k/2}}

⇒ln⁡2=k2\Rightarrow \ln 2 = {k \over 2}

⇒kln⁡2=2\Rightarrow {k \over {\ln 2}} = 2

⇒(kln⁡2)2=4\Rightarrow {\left( {{k \over {\ln 2}}} \right)^2} = 4

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring