Maths Case study (4 Marks)

1. (c) 3

Solution:

We have, $2\frac{\text{d}^2\text{y}}{\text{dx}^2}+3\sqrt{1-\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\text{y}=0}.$

$\therefore\ \ 2\frac{\text{d}^2\text{y}}{\text{dx}^2}=-3\sqrt{1-\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\text{y}}$

Squaring both sides, we get

$4\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)^2=9\Bigg[1-\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\text{y}\Bigg]$

Here, highest order derivative is $\frac{\text{d}^2\text{y}}{\text{dx}^2}$ and its power is 2. So, its degree is 2.

2. (d) 1, 4

Solution:

We have, $\text{y}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\frac{\text{dy}}{\text{dx}}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^3}$

$\Rightarrow\ \ \text{y}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+\text{y}\Big(\frac{\text{dy}}{\text{dx}}\Big)^4=\text{x}$

⇒ Here, highest order derivative is $\frac{\text{d}\text{y}}{\text{dx}}.$ So, its order is I and degree is 4.

3. (a) Order = 3, degree = undefined.

Solution:

We have, y'" + y² + e^y' = 0.

$\frac{\text{d}^3\text{y}}{\text{dy}^3}+\text{y}^2+\text{e}^\frac{\text{dy}}{\text{dx}}=0$

Highest order derivative is $\frac{\text{d}^3\text{y}}{\text{dy}^3}.$ So, its order is 3. Also, the given differential cannot be expressed as a polynomial. So, its degree is not defined.

4. (c) 1

Solution:

The given differential equation is,

$\sqrt{\text{a+x}}\times\Big(\frac{\text{dy}}{\text{dx}}\Big)+\text{x}=0$

$\Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=\frac{-\text{x}}{\sqrt{\text{a+x}}}$

Clearly, degree = 1.

5. (b) 2, 3

Solution:

We have, $\Bigg(1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^3\Bigg)^\frac{7}{3}=7\frac{\text{d}^2\text{y}}{\text{dx}^2}$

$\Rightarrow\ \ \Bigg(1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^3\Bigg)^7=\Bigg(7\frac{\text{d}^2\text{y}}{\text{dx}^2}\Bigg)^3$

$\therefore$ Order is 2 and degree is 3.

1. (b) $\frac{\cos\text{x}}{1+\sin\text{x}},\ \frac{\text{-x}}{1+\sin\text{x}}$

Solution:

he given differential equation can be written as $\frac{\text{dy}}{\text{dx}}+\frac{\cos\text{x}}{1+\sin\text{x}}\text{y}=\frac{\text{-x}}{1+\sin\text{x}}$

Compare it with $\frac{\text{dy}}{\text{dx}}+\text{py}=\text{Q},$ we get $\text{P}=\frac{\cos\text{x}}{1+\sin\text{x}}$ and $\text{Q}=\frac{\text{-x}}{1+\sin\text{x}}$

2. (c) $1+\sin\text{x}$

Solution:

$\text{I.F.}=\text{e}^{\int\text{pdx}}=\text{e}^{\int\frac{\cos\text{x}}{1+\sin\text{x}}\text{dx}}$

Put $1+\sin\text{x}=\text{t}\Rightarrow\cos\text{x}\ \text{dx}=\text{dt}$

$\therefore\ \text{I.F.}=\text{e}^{\int\frac{1}{\text{t}}\text{dt}}=\text{e}^{\log\text{t}}=\text{t}=1+\sin\text{x}$

3. (d) $\text{y}(1+\sin\text{x})=\frac{\text{-x}^2}{2}\text{+c}$

Solution:

Solution of given differential equation is given by $\text{y}\times\text{(I.F.)}=\int\text{Q}\text{(I.F.)}\ \text{dx}+\text{c}$

$\Rightarrow\ \text{y}(1+\sin\text{x})=\int\frac{\text{-x}}{1+\sin\text{x}}\times(1+\sin\text{x})\ \text{dx + c}$

$\Rightarrow\ \text{y}(1+\sin\text{x})=\frac{\text{-x}^2}{2}+\text{ c}$

4. (a) $\frac{2-\text{x}^2}{2(1+\sin\text{x})}$

Solution:

We have, y(0) = 1 i.e., x = 0, y = 1

$\therefore\ 1(1+\sin0)=\text{c}\Rightarrow\text{c}=1$

$\therefore\ \ \text{y}(1+\sin\text{x})=\frac{\text{-x}^2}{2}+1=\frac{2-\text{x}^2}{2}$

$\therefore\ \ \ \text{y}=\frac{2-\text{x}^2}{2(1+\sin\text{x})}$

5. (b) $\frac{8-\pi^2}{16}$

Solution:

We have, $\text{y}=\frac{2-\text{x}^2}{2(1+\sin\text{x})}$

$\therefore\ \text{y}\Big(\frac{\pi}{2}\Big)=\frac{2-\Big(\frac{\pi}{2}\Big)^2}{2\Bigg(1+\sin\frac{\pi}{2}\Bigg)}=\frac{2-\frac{\pi^2}{4}}{4}=\frac{8-\pi^2}{16}$

1. (c) 2

Solution:

The given differential equation can be written as $\frac{\text{dy}}{\text{dx}}+2\text{y}\cot\text{x}=\text{cosec x}$

$\therefore\ \text{I.F}=\text{e}^{\int2\cot\text{xdx}}=\text{e}^{2\log|\sin\text{x}|}=(\sin\text{x})^2$

$\therefore\ \lambda=2$

2. (c) $\sqrt{1-\text{x}^2}$

Solution:

We have, $(1-\text{x}^2)\frac{\text{dy}}{\text{dx}}-\text{xy}=1$

$\Rightarrow\ \frac{\text{dy}}{\text{dx}}-\frac{\text{x}}{1-\text{x}^2}\cdot\text{y}=\frac{1}{1-\text{x}^2}$

$\therefore\ \text{I.F.}=\text{e}^{-\int\frac{\text{x}}{1-\text{x}^2}\text{dx}}=\text{e}^{\frac{1}{2}\int\frac{-2\text{x}}{1-\text{x}^2}\text{dx}}$

$=\text{e}^{\frac{1}{2}\log(1-\text{x}^2)}=\text{e}^{\log(1-\text{x}^2)^\frac{1}{2}}=\sqrt{1-\text{x}^2}$

3. (b) $\text{y}=\text{xe}^{-\text{x}}$

Solution:

We have, $\frac{\text{dy}}{\text{dx}}+\text{y}=\text{e}^{-\text{x}}$

It is a linear differential equation with $\text{I.F.}=\text{e}^{\int\text{dx}}=\text{e}^\text{x}$

Now, solution is $\text{y}\cdot\text{e}^\text{x}=\int\text{e}^{-\text{x}}\text{dx}+\text{c}$

$\Rightarrow\text{ye}^\text{x}=\int\text{dx}+\text{c}$

$\Rightarrow\text{ye}^\text{x}=\text{x}+\text{c}$

$\Rightarrow\text{y}=\text{xe}^{-\text{x}}+\text{ce}^{-\text{x}}$

$\because\ \text{y}(0)=0\Rightarrow\text{c}=0$

$\therefore\ \text{y}=\text{xe}^{-\text{x}}$

4. (a) $\text{y}\sec\text{y}=\tan\text{x}+\text{c}$

Solution:

We have, $\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}=\sec\text{x}$

It is a linear differential equation with,

$\text{I.F.}=\text{e}^{\int\tan\text{xdx}}=\text{e}^{\log|\sec\text{x}|}=\sec\text{x}$

Now, solution is $\text{y}\sec\text{x}=\int\sec^2\text{x dx}+\text{c}$

$\Rightarrow\text{y}\sec\text{x}=\tan\text{x}+\text{c}$

5. (c) $\text{e}^{-3\text{x}}$

Solution:

We have, $\frac{\text{dy}}{\text{dx}}3\text{y}=\sin2\text{x}$

It is a linear differential equation with,

$\text{I.F.}=\text{e}^{\int-3\text{dx}}=\text{e}^{-3\text{x}}$

1. (d) 1000

Solution:

Since, maximum number of students in hostel is 1000

$\therefore$ Maximum value of n(t) is 1000.

2. (a) n(1000 - n)

Solution:

Clearly, according to given information,

$\frac{\text{dn}}{\text{dt}}=\lambda\text{n}(1000-\text{n})$ where $\lambda$ is constant of proportionality.

3. (b) 50

Solution:

Since, 50 students are infected after 4 days.

$\therefore$ n(4) = 50.

4. (c) $\frac{1}{1000}\log\Big(\frac{\text{n}}{1000-\text{n}}\Big)=\lambda\text{t}+\text{c}$

Solution:

We have, $\frac{\text{dn}}{\text{dt}}=\lambda\text{n}(1000-\text{n})$

$\Rightarrow\ \ \int\frac{\text{dn}}{\text{n}(1000-\text{n})}=\lambda\int\text{dt}$

$\Rightarrow\frac{1}{1000}\int\Big(\frac{1}{1000-\text{n}}+\frac{1}{\text{n}}\Big)\text{dn}=\lambda\int\text{dt}$

$\Rightarrow\frac{1}{1000}\Big[\frac{\log(1000-\text{n})}{-1}+\log\text{n}\Big]=\lambda\text{t+c}$

$\Rightarrow\frac{1}{1000}\log\Big(\frac{\text{n}}{1000-\text{n}}\Big)=\lambda\text{t}+\text{c}$

5. (a) $\text{n(t)}=\frac{1000}{1+999\text{e}^{-0.9906\text{t}}}$

Solution:

When, t = 0, n = 1

This condition is satisfied by option (a) only.

1. (b) T - 50

Solution:

Given, T is the temperature of the body at any time t. Then, by Newton's law of cooling, we get $\frac{\text{dT}}{\text{dt}}=\text{k(T} - \text{50)},$ where k is the constant of proportionatity.

2. (c) 70ºF

Solution:

From given information, we have

At 8 p.m. temperature is 70ºF

$\therefore$ At t = 0, T = 70ºF

3. (b) 60ºF

Solution:

From given information, we have

At 10 p.m., temperature is 60ºF

$\therefore$ At t = 2, T = 60ºF

4. (c) $50+20\Big(\frac{1}{2}\Big)^\frac{\text{t}}{2}$

Solution:

$\frac{\text{dT}}{\text{dt}}=\text{k}(\text{T}-50)$

$\Rightarrow\frac{\text{dT}}{\text{T}-50}=\text{k}\ \text{dt}$

On integrating both sides, we get

$\log|\text{T}-50|=\text{kt}+\log\text{C}$

⇒ T - 50 = Ce^kt

Clearly, for t = 0, T = 70º

⇒ C = 20

Thus, T - 50 = 20e^kt

For t = 2, T = 60º

⇒ 10 = 20e^2k

$\Rightarrow2\text{k}=\log\Big(\frac{1}{2}\Big)$

$\Rightarrow\text{k}=\frac{1}{2}\log\Big(\frac{1}{2}\Big)$

Hence, $\text{T}=50+20\Big(\frac{1}{2}\Big)^\frac{\text{t}}{2}$

5. (b) 5:30 p.m.

Solution:

We have, $\text{T}=50+20\Big(\frac{1}{2}\Big)^\frac{\text{t}}{2}$

Now, $\text{98.6}=50+20\Big(\frac{1}{2}\Big)^\frac{\text{t}}{2}$

$\Rightarrow\frac{48.6}{20}=\Big(\frac{1}{2}\Big)^\frac{\text{t}}{2}$

$\Rightarrow\log\Big(\frac{48.6}{20}\Big)=\frac{\text{t}}{2}\log\Big(\frac{1}{2}\Big)$

$\Rightarrow\frac{\text{t}}{2}=\frac{\log\Big(\frac{48.6}{20}\Big)}{\log\Big(\frac{1}{2}\Big)}$

$\Rightarrow{\text{t}}={2}=\begin{pmatrix}\frac{\log\Big(\frac{48.6}{20}\Big)}{\log\Big(\frac{1}{2}\Big)}\end{pmatrix}=-2.56$

So, it appears that the person was murdered 2.5 hours before 8 p.m. i.e., about 5 : 30 p.m.

1. (c) 5000

Solution:

Since, size of population is 5000.

$\therefore$ Maximum value of y(t) is 5000.

2. (d) y(5000 - y)

Solution:

Clearly, according to given information, $\frac{\text{dy}}{\text{dt}}=\text{ky}(5000-\text{y}),$ where k is the constant of proportionality.

3. (a) 100

Solution:

Since, rumour is initiated with 100 people.

$\therefore$ When t = 0, then y = 100

Thus y(0) = 100

4. (b) 500

Solution:

Since, rumour is spread in 500 people, after 2 days.

$\therefore$ When t = 2, then y = 500.

Thus, y(2) = 500

5. (c) $\text{y}=\frac{5000}{_\text{49e}-5000\text{kt}_{+1}}$

Solution:

We know that, when t = 0, then y = 100 This condition is satisfied by option (c) only.

1. (b) -2

Solution:

We have, $\frac{\text{dy}}{\text{dx}}=\frac{\text{ax+3}}{\text{2y+f}}$

$\Rightarrow\ \ (\text{ax+3})\text{dx}=(2\text{y}+\text{f})\text{dy}$

$\Rightarrow\text{a}\frac{\text{x}^2}{2}+\text{3x}=\text{y}^2+\text{fy}+\text{c}$ (Integrating)

$\Rightarrow-\frac{\text{a}}{2}\text{x}^2+\text{y}^2-\text{3x}+\text{fy}+\text{C}=0$

This will represent a circle, if $\frac{-\text{a}}{2}=1\Rightarrow\text{a}=-2$

[ $\therefore$ In circle, coefficient of x² = coefficient of y²)

2. (c) Fixed radius 1 and variable centre on x-axis

Solution:

We have, $\frac{\text{ydy}}{\sqrt{1-\text{y}^2}}=\text{dx}$

On integration, we get $-\sqrt{1-\text{y}^2}=\text{x+c}$

⇒ 1 - y² = (x + c)² ⇒ (x + c)² + y² = 1 which represents a circle with radius I and centre on the x-axis.

3. (c) 3

Solution:

$\text{y}'=\text{y}+1\Rightarrow\frac{\text{dy}}{\text{y}+1}=\text{dx}$

⇒ In (y + 1) = x + c (integrating)

Now, y(0) = 1 ⇒ c = In 2

$\therefore \ \text{In}\Bigg(\frac{\text{y}+1}{2}\Bigg)=\text{x}$

⇒ y + 1 = 2e^x

So, y (In 2) = -1 + 2e^{In 2} = -1 + 4 = 3

4. (c) $\text{e}^\text{y}=\frac{\text{x}^3}{3}+\text{e}^\text{x}+\text{c}$

Solution:

From the given differential equation, we have

$\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^\text{x}+\text{x}^\text{2}}{\text{e}^\text{y}}$

$\Rightarrow\ \text{e}^\text{y}\text{dy}=(\text{e}^\text{x}+\text{x}^2)\text{dx}$

Integrating, we get $\text{e}^\text{y}=\text{e}^\text{x}+\frac{\text{x}^3}{3}+\text{c}$

5. (a) $\text{y}=\text{e}^{\sin^2}\text{x}$

Solution:

We have, $\frac{\text{dy}}{\text{dx}}=\text{y}\sin2\text{x}$

$\Rightarrow\ \frac{\text{dy}}{\text{y}}=\sin2\text{x}\ \text{dx}$

$\Rightarrow\ \log\text{y}=-\frac{\cos2\text{x}}{2}+\text{c}$

Since x = 0, y = 1

therefore $\text{C}=\frac{1}{2}$

Now, $\log\text{y}=\frac{1}{2}(1-\cos2\text{x})$

$\Rightarrow\ \log\text{y}=\sin^2\text{x}\Rightarrow\text{y}=\text{e}^{\sin^2}\text{x}$

1. (b) $\tan^{-1}\frac{\text{y}}{\text{x}}=\log|\text{x}|+\text{c}$

Solution:

We have, $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2+\text{xy}+\text{y}^2}{\text{x}^2}$

Put y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v+x}\frac{\text{dv}}{\text{dx}}$

$\therefore\ \text{v+x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2+\text{x}\times\text{vx}+\text{v}^2\text{x}^2}{\text{x}^2}=1+\text{v}+\text{v}^2$

$\Rightarrow \text{x}\frac{\text{dv}}{\text{dx}}=1+\text{v}^2\Rightarrow\int\frac{\text{dv}}{1+\text{v}^2}=\int\frac{\text{dx}}{\text{x}}+\text{c}$

$\Rightarrow\tan^{-1}\text{v}=\log|\text{x}|+\text{c}$

$\Rightarrow\tan^{-1}\frac{\text{y}}{\text{x}}=\log|\text{x}|+\text{c}$

2. (d) x² + y² = cx³

Solution:

We have,

$2\text{xy}\frac{\text{dy}}{\text{dx}}=\text{x}^2+3\text{y}^2$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2+3\text{y}^2}{2\text{xy}}$

Put y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v+x}\frac{\text{dv}}{\text{dx}}$

$\therefore\ \text{v+x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2+3\text{v}^2\text{x}^2}{2\text{vx}^2}$

$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+3\text{v}^2}{2\text{v}}-\text{v}$

$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}^2}{2\text{v}}$

$\Rightarrow\int\frac{2\text{v}}{1+\text{v}^2}\text{dv}=\int\frac{\text{dx}}{\text{x}}+\log\text{c}$

$\Rightarrow\log|1+\text{v}^2|=\log|\text{x}|+\log|\text{c}|$

$\Rightarrow\log|\text{v}^2+1|=\log|\text{xc}|$

$\Rightarrow\text{v}^2+1=\text{xc}\Rightarrow\frac{\text{y}^2}{\text{x}^2}+1=\text{xc}$

$\Rightarrow\text{x}^2+\text{y}^2=\text{x}^3\text{c}$

3. (d) $\frac{\text{x}}{\text{x+y}}+\log\text{x = c}$

Solution:

We have, 

(x² + 3xy + y²) dx - x² dy = 0

$\Rightarrow\frac{\text{x}^2+3\text{xy}+\text{y}^2}{\text{x}^2}=\frac{\text{dy}}{\text{dx}}$

Put y= vx and $\frac{\text{dy}}{\text{dx}}=\text{v+x}\frac{\text{dv}}{\text{dx}}$

$\therefore\ \frac{\text{x}^2+3\text{x}^2\text{v}+\text{x}^2\text{v}^2}{\text{x}^2}=\Big(\text{v+x}\frac{\text{dv}}{\text{dx}}\Big)$

$\Rightarrow1+3\text{v}+\text{v}^2=\text{v+x}\frac{\text{dv}}{\text{dx}}$

$\Rightarrow1+2\text{v}+\text{v}^2=\text{x}\frac{\text{dv}}{\text{dx}}$

$\Rightarrow\int\frac{\text{dx}}{\text{x}}-\int(\text{v+1})^{-2}=\text{dv}=\text{c}$

$\log{\text{x}}+\frac{1}{\text{v+1}}=\text{c}$

$\Rightarrow\log\text{x}+\frac{\text{x}}{\text{x+y}}=\text{c}$

4. (c) $\log\frac{\text{y}}{\text{x}}=\text{cx}$

Solution:

We have, $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}\bigg\{\log\Big(\frac{\text{y}}{\text{x}}\Big)+1\bigg\}$

Put y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v+x}\frac{\text{dv}}{\text{dx}}$

$\therefore\ \text{v+x}\frac{\text{dv}}{\text{dx}}=\text{v}\{\log(\text{v}+1\}$

$\Rightarrow\text{x}\frac{\text{dv}}{\text{dx}}=\text{v}\log\text{v}$

$\Rightarrow\int\frac{\text{dv}}{\text{v}\log\text{v}}=\int\frac{\text{dx}}{\text{x}}\Rightarrow\log|\log\text{v}|=\log|\text{x}|+\log|\text{c}|$

$\Rightarrow\log\Big(\frac{\text{y}}{\text{x}}\Big)=\text{cx}$

5. (a) $\text{e}^\frac{\text{y}}{\text{x}}-\sin\text{x = c}$

Solution:

We have, $\Big(\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}\Big)\text{e}^\frac{\text{y}}{\text{x}}=\text{x}^2\ \cos\text{x}$

$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}}\Big)\text{e}^\frac{\text{y}}{\text{x}}=\text{x}\ \cos\text{x}$

Put y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v+x}\frac{\text{dv}}{\text{dx}}$

$\Rightarrow\Big(\text{v+x}\frac{\text{dv}}{\text{dx}}-\text{v}\Big)\text{e}^\text{v}=\text{x}\cos\text{x}$

$\Rightarrow\text{xe}^\text{v}\frac{\text{dv}}{\text{dx}}=\text{x}\cos\text{x}$

$\Rightarrow\int\text{e}^\text{v}\text{dv}=\int\cos\text{x}\ \text{dx}$

$\Rightarrow\text{e}^\text{v}=\sin\text{x + c}$

$\Rightarrow\ \text{e}^\frac{\text{y}}{\text{x}}-\sin\text{x = c}$

1. (b) $\frac{\text{Pr}}{100}$

Solution:

Here, P denotes the principal at any time t and the rate of interest be r% per annum compounded continuously, then according to the law given in the problem, we get

$\frac{\text{dP}}{\text{dt}}=\frac{\text{Pr}}{100}$

2. (a) $\log\Big(\frac{\text{P}}{\text{P}_0}\Big)=\frac{\text{rt}}{100}$

Solution:

We have, $\frac{\text{dP}}{\text{dt}}=\frac{\text{Pr}}{100}$

$\Rightarrow\frac{\text{dP}}{\text{P}}=\frac{\text{r}}{100}\text{dt}$

$\Rightarrow\int\frac{\text{1}}{\text{P}}\text{dP}=\frac{\text{r}}{100}\int\text{dt}$

$\Rightarrow\log\text{P}=\frac{\text{rt}}{100}+\text{C}...\text{(i)}$

$\text{At t}=0,\ \ \text{P}=\text{P}_0.$

$\therefore\ \ \text{C}=\log\text{P}_0$

So, $\log\text{P}=\frac{\text{rt}}{100}+\log\text{p}_0$

$\Rightarrow\log\Big(\frac{\text{P}}{\text{P}_0}\Big)=\frac{\text{rt}}{100}...\text{(ii)}$

3. (c) 13.862 years

Solution:

We have, r = 5, P₀ = ₹ 100 and P = ₹ 200 = 2P₀ Substituting these values in (2), we get

$\log2=\frac{5}{100}\text{t}$

$\Rightarrow\text{t}=20\log_\text{e}$

$2 = 20 × 0.6931\ \text{years} = 13.862\ \text{years}$

4. (d) 6.931%

Solution:

We have,

P₀ = ₹ 100, P = ₹ 200 = 2P₀ and t = 10 years Substituting these values in (2), we get

$\log2\frac{10\text{r}}{100}\Rightarrow\text{r}=10\log2=10\times0.6931=6.931$

5. (a) ₹ 1648

Solution:

We have P₀ = ₹ 1000, r = 5 and t = 10 Substituting these values in (2), we get

$\log\Big(\frac{\text{P}}{1000}\Big)=\frac{5\times10}{100}=\frac{1}{2}=0.5$

$\Rightarrow\frac{\text{P}}{1000}=\text{e}^{0.5}$

$\Rightarrow\text{ P} = 1000 × 1.648 = ₹ \ 1648$

1. (d) $-\lambda(\text{T - S})$

Solution:

Given, at any time t the thermometer reading be T°F and the outside temperature be S°F. Then, by Newton's law of cooling, we have

$\frac{\text{dT}}{\text{dt}}\propto(\text{T -S })\Rightarrow\frac{\text{dT}}{\text{dt}}=-\lambda(\text{T - S})$

2. (d) 60°F

Solution:

Since, after 5 minutes, thermometer reads 60"F

$\therefore$ Value of T(S) = 60°F

3. (a) 50°F

Solution:

Clearly from given information, value of T(10) is 5O°F.

4. (b) $\log(\text{T - S})=\lambda{\text{t + c}}$

Solution:

We have, $\frac{\text{dT}}{\text{dt}}=-\lambda({\text{T - S}})$

$\Rightarrow\frac{\text{dT}}{\text{T - S}}=-\lambda\text{dt}\int\frac{1}{\text{T - S}}\text{dt}=-\lambda\int\text{dt}$

$\Rightarrow\log(\text{T - S})=-\lambda\text{t + c}$

5. (c) $\log(80\ -\ \text{S})$

Solution:

Since, at t = 0, T = 80°F

$\therefore\ \log(80\ -\ \text{S})=0\ +\ \text{c}$

$\Rightarrow\text{c}=\log(80\ -\ \text{S})$