Question

Previous year Question of jee mains

mathamatics shift :1
24 feb 2021 ​

Answer 1

EXPLANATION.

Population of a time at time t,

$\implies \dfrac{dp(t)}{dt} = (0.5)p(t) - 450$

$\implies p(0) = 850$

As we know that,

We can write equation as,

$\implies \dfrac{dp(t)}{dt} = \dfrac{1}{2} p(t) - 450.$

$\implies \dfrac{dp(t)}{dt} - \dfrac{1}{2} p(t) = - 450.$

As we know that,

This equation is in the form of linear differential equations.

Now, first we find integrating factor.

$\implies I.F = e^{\int -dt/2} = e^{(-t/2)}$

$\implies p(t) \times e^{(-t/2)} = \displaystyle \int -450 \times e^{(-t/2)} dt$

$\implies p(t) \times e^{(-t/2)} = 900 \times e^{(-t/2)} + K.$

$\implies p(t) = 900 + Ke^{(-1/2)}$

$\implies p(0) = 850$

$\implies 850 = 900 + K.$

$\implies K = - 50$

$\implies p(t) = 900 - 50e^{(-1/2)}$

$\implies 900 - 50e^{(-1/2)} = 0$

$\implies 900 = 50e^{(-1/2)}$

$\implies 18 = e^{(-1/2)}$

$\implies 2 ln(18)$

Option [C] is correct answer.

Answer 2

Step-by-step explanation:

Given :- population of a town at time 't' is given by differential equation dP(t)/dt = 0.5 (t) - 450

• P(0) = 850

we have to find time when population of town becomes "zero "

Let at time 't₁' population of town becomes zero

• Given equation ,dP(t) / dt = p(t) - 900 / 2

• ₀∫¹dP(t) / P(t) - 900 = ₀∫¹dt/2

• {ln |P(t) - 900 | }₀¹ = {1/2}₀ᵗ

• |ln|P(t) - 900 | - ln|P(0) - 900| = t/2

• [∵P(0) = 850]

• ln|P(t) - 900 | - ln|850 - 900| = t/2

• ln|P(t) - 900| - ln50 = t/2

∵ population of town becomes 0 at time t₁

• so, Put t = t₁ and P(t) = 0

• Then, ln|0 - 900| - ln50 = t₁ /2

• ∵ lna - lnb = lna/b {property}

• ∴ ln 900 /50 = t₁/2

• ln18 = t₁/2

• 2ln18 = t₁

So at time 2ln18 population of town becomes zero