HSC Commerce (English Medium) 12th Standard Board Exam - Maharashtra State Board Question Bank Solutions for Mathematics and Statistics

Choose the correct alternative:

The solution of `("d"y)/("d"x) + x^2/y^2` = 0 is

The integrating factor of the differential equation `("d"y)/("d"x) - y` = x is ______

The solution of `("d"y)/("d"x) + y` = 3 is  ______

Integrating factor of `("d"y)/("d"x) + y/x` = x³ – 3 is ______

The rate of growth of population is proportional to the number present. If the population doubled in the last 25 years and present population is 1 lac., when will the city have population 4,00,000?

Solution: Let p be the population at time t. 

Then the rate of increase of p is `"dp"/"dt"` which is proportional to p.

∴ `"dp"/"dt" ∝ "p"`

∴ `"dp"/"dt"` = kp, where k is a constant

∴ `"dp"/"p"` = kdt

On integrating, we get

`int "dp"/"p" = "k"int "dt"`

∴ log p = kt + c

Initially, i.e., when t = 0, let p = 100000

∴ log 100000 = k × 0 + c

∴ c = `square`

∴ log p = kt + log 100000

∴ log p – log 100000 = kt

∴ `log ("P"/100000)` = kt  ......(i)

Since the number doubled in 25 years, i.e., when t = 25, p = 200000

∴ `log (200000/100000)` = 25k

∴ k = `square`

∴ equation (i) becomes, `log("p"/100000) = square`

When p = 400000, then find t.

∴ `log(400000/100000) = "t"/25 log 2`

∴ `log 4 = "t"/25 log 2`

∴ t = `25 (log 4)/(log 2)`

∴ t = `square` years

In a certain culture of bacteria, the rate of increase is proportional to the number present. If it is found that the number doubles in 4 hours, find the number of times the bacteria are increased in 12 hours.

Solution: Let x be the number of bacteria in the culture at time t.

Then the rate of increase of x is `"dx"/"dt"` which is proportional to x.

∴ `"dx"/"dt" ∝  "x"`

∴ `"dx"/"dt"` = kx, where k is a constant

∴ `square`

On integrating, we get

`int "dx"/"x" = "k" int "dt"`

∴ log x = kt + c

Initially, i.e. when t = 0, let x = x₀

∴ log x₀ = k × 0 + c

∴ c = `square`

∴ log x = kt + log x₀ 

∴ log x - log x₀ = kt

∴ `log ("x"/"x"_0)`= kt    ......(1)

Since the number doubles in 4 hours, i.e. when t = 4,

x = 2x₀ 

∴ `log ((2"x"_0)/"x"_0)` = 4k

∴ k = `square`

∴ equation (1) becomes, `log ("x"/"x"_0) = "t"/4` log 2

When t = 12, we get

`log ("x"/"x"_0) = 12/4` log 2 = 3 log 2

∴ `log ("x"/"x"_0)` = log 2³

∴ `"x"/"x"_0 = 8`

∴ x = `square`

∴ number of bacteria will be 8 times the original number in 12 hours.

Find the population of city at any time t given that rate of increase of population is proportional to the population at that instant and that in a period of 40 years the population increased from 30000 to 40000.

Solution: Let p be the population at time t.

Then the rate of increase of p is `"dp"/"dt"` which is proportional to p.

∴ `"dp"/"dt" prop "p"`

∴ `"dp"/"dt"` = kp, where k is a constant.

∴ `"dp"/"p"` = k dt

On integrating, we get

`int "dp"/"p" = "k" int "dt"`

∴ log p = kt + c

Initially, i.e. when t = 0, let p = 30000

∴ log 30000 = k × 0 + c       

∴ c = `square`

∴ log p = kt + log 30000

∴ log p - log 30000 = kt

∴ `log("p"/30000)` = kt          .....(1)     

when t = 40, p = 40000

∴ `log (40000/30000) = 40"k"`

∴ k = `square`

∴ equation (1) becomes, `log ("p"/30000)` = `square`

∴ `log ("p"/30000) = "t"/40 log (4/3)`

∴ p = `square`

Bacteria increases at the rate proportional to the number of bacteria present. If the original number N doubles in 4 hours, find in how many hours the number of bacteria will be 16N.

Solution: Let x be the number of bacteria in the culture at time t.

Then the rate of increase of x is `("d"x)/"dt"` which is proportional to x.

∴ `("d"x)/"dt" ∝ x`

∴ `("d"x)/"dt"` = kx, where k is a constant

∴ `("d"x)/x` = kdt

On integrating, we get

`int ("d"x)/x = "k" int "dt"`

∴ log x = kt + c    .....(1)

∴ x = ae^kt where a = e^c 

Initially, i.e.,when t = 0, let x = N

∴ N = ae^k(0)

∴ a = `square`

∴ a = N, x = Ne^kt    ......(2)

When t = 4, x = 2N

From equation (2), 2N = Ne^4k

∴ e^4k = 2

∴  e^k = `square`

Now we have to find out t, when x = 16N

From equation (2),

16N = Ne^kt 

∴ 16 = e^kt 

∴ `"t"/4 = square` hours

Hence, number of bacteria will be 16N in `square` hours

The population of city doubles in 80 years, in how many years will it be triple when the rate of increase is proportional to the number of inhabitants. `("Given" log3/log2 = 1.5894)`

Solution: Let p be the population at time t.

Then the rate of increase of p is `"dp"/"dt"` which is proportional to p.

∴ `"dp"/"dt"  ∝  "p"`

∴ `"dp"/"dt"` = kp, where k is a constant

∴ `"dp"/"p"` = kdt

On integrating, we get

`int "dp"/"p" = "k" int "dt"`

∴ log p = kt + c

Initially, i.e., when t = 0, let p = N

∴ log N = k × 0 + c

∴ c = `square`

When t = 80, p = 2N

∴ log 2N = 80k + log N

∴ log 2N – log N = 80k

∴ `log ((2"N")/"N")` = 80k

∴ log (2) = 80k

∴ k = `square`

∴ p = 3N, then t = ?

∴ log p = `log2/80  "t" + log "N"`

∴ log 3N – log N = `square`

∴ t = `square` = `square` years

In a certain culture of bacteria, the rate of increase is proportional to the number present. If it is found that the number doubles in 4 hours, complete the following activity to find the number of times the bacteria are increased in 12 hours.

The rate of disintegration of a radioactive element at time t is proportional to its mass at that time. The original mass of 800 gm will disintegrate into its mass of 400 gm after 5 days. Find the mass remaining after 30 days.

Solution: If x is the amount of material present at time t then `dx/dt = square`, where k is constant of proportionality.

`int dx/x = square + c` 

∴ logx = `square`

x = `square` = `square`.e^c

∴ x = `square`.a where a = e^c

At t = 0, x = 800

∴ a = `square`

At t = 5, x = 400

∴ e^–5k = `square`

Now when t = 30 

x = `square` × `square` = 800 × (e^–5k)⁶ = 800 × `square` = `square`.

The mass remaining after 30 days will be `square` mg.

In a certain culture of bacteria, the rate of increase is proportional to the number present. If it is found that the number doubles in 4 hours, find the number of times the bacteria are increased in 12 hours.

Solution:

Let N be the number of bacteria present at time ‘t’.

Since the rate of increase of N is proportional to N, the differential equation can be written as –

`(dN)/dt αN`

∴ `(dN)/dt` = KN, where K is constant of proportionality

∴ `(dN)/N` = k . dt

∴ `int 1/N dN = K int 1 . dt`

∴ log N = `square` + C   ...(1)

When t = 0, N = N₀ where N₀ is initial number of bacteria.

∴ log N₀ = K × 0 + C

∴ C = log N₀

Also when t = 4, N = 2N₀

∴ log (2 N₀) = K . 4 + `square`   ...[From (1)]

∴ `log((2N_0)/N_0)` = 4K,

∴ log 2 = 4K

∴ K = `square`   ...(2)

Now N = ? when t = 12

From (1) and (2)

log N = `1/4 log 2  . (12) + log N_0`

log N – log N₀ = 3 log 2

∴ `log(N_0/N_0)` = `square`

∴ N = 8 N₀

∴ Bacteria are increased 8 times in 12 hours.

The rate of growth of population is proportional to the number present. If the population doubled in the last 25 years and the present population is 1,00,000, when will the city have population 4,00,000?

Let ‘p’ be the population at time ‘t’ years.

∴ `("dp")/"dt" prop "p"`

∴ Differential equation can be written as  `("dp")/"dt" = "kp"`

where k is constant of proportionality.

∴ `("dp")/"p" = "k.dt"`

On integrating we get

`square` = kt + c   ...(i)

(i) Where t = 0, p = 1,00,000

∴ from (i)

log 1,00,000 = k(0) + c

∴  c = `square`

∴  log `("p"/(1,00,000)) = "kt"`       ...(ii)

(ii) When t = 25, p = 2,00,000

as population doubles in 25 years

∴ from (ii) log2 = 25k

∴  k = `square`

∴  log`("p"/(1,00,000)) = (1/25log2).t`

(iii) ∴ when p = 4,00,000

`log ((4,00,000)/(1,00,000)) = (1/25log2).t`

∴ `log 4 = (1/25 log2).t`

∴ t = `square ` years