Ncert Solutions Maths class 12th

113. Solution:

By Rolle’s Theorem, for a function $f\left[a,b\right]\to R,$ if

f is continuous on $\left[a,b\right]$

f is differentiable on $\left(a,b\right)$

f(a)= f(b)

then, there exists some $c\in \left(a,b\right)$ such that $f^{\text{'}}\left(c\right)=0$

therefore, Rolle’s Theorem is not applicable to those functions that do not satisfy any of the three conditions of the hy

The given D.E. is

$\begin{array}{l}{y}^1=\frac{x+y}{x}\\ =\frac{dy}{dx}=1+\frac{y}{x}=f\left(\frac{y}{x}\right)\end{array}$

Hence, the given D.E is homogenous.

Let,  $y=vx\to \frac{y}{x}=v,sothat,\frac{dy}{dx}=v+\frac{xdv}{dx}$

So, the D.E. becomes

$\begin{array}{l}v+\frac{xdv}{dx}=1+v\\ \Rightarrow \frac{xdv}{dx}=1\\ \Rightarrow dv=\frac{dx}{x}\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int dv={\displaystyle \int \frac{dx}{x}}}\\ v=log\left|x\right|+c\end{array}$

Putting $v=\frac{y}{x}$ back we get,

$\frac{y}{x}=log\left|x\right|+c$

112.  Given, $f\left(x\right)=x^2+2x-8$ , being polynomial function is continuous in $\left[-4,2\right]$ and also differentiable in $\left(-4,2\right)$ .

$\begin{array}{l}f\left(-4\right)={\left(-4\right)}^2+2\left(-4\right)-8\\ =16-8-8\\ =0\\ f\left(2\right)={\left(2\right)}^2+2\times 2-8\\ =4+4-8\\ =0\end{array}$

Therefore, $f\left(-4\right)=f\left(2\right)=0$

The value of $f\left(x\right)$ at -4 and 2 coincides.

Rolle’s Theorem states that there is a point $c\in \left(-4,2\right)$ such that $f^{\text{'}}\left(c\right)=0$ $f\left(x\right)=x^2+2x-8$

Therefore, $f^{\text{'}}\left(x\right)=2x+2$

Hence,

$\begin{array}{l}{f}^{\text{'}}\left(c\right)=0\\ 2c+2=0\\ c=-1\end{array}$

Thus, $c=-1\in \left(-4,2\right)$

Henc

The given D.E. is

$\begin{array}{l}\frac{dy}{dx}=\frac{x^2+y^2}{x^2+x+y}=\frac{x^2\left(1+\frac{y^2}{x^2}\right)}{x^2\left(1+\frac{y}{x}\right)}=F\left(x,y\right)\\ Then,F\left(\lambda x,\lambda y\right)=\frac{{\lambda }^2{x}^2}{{\lambda }^2{x}^2}\left[\frac{1+\frac{{\lambda }^2{y}^2}{{\lambda }^2{x}^2}}{1+\frac{\lambda y}{\lambda x}}\right]={\lambda }^{0}F\left(x,y\right)\\ ={\lambda }^{2}F\left(x,y\right)\end{array}$

Hence, $F\left(x,y\right)$ is a homogenous $f{x}^n$ of degree 2.

To solve it, let

$\begin{array}{l}y=vx,sothat,\frac{dy}{dx}=v.\frac{dx}{dx}+x\frac{dv}{dx}\\ v=\frac{y}{x}\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\end{array}$

The D.E. now becomes,

$\begin{array}{l}v+x\frac{dv}{dx}=\frac{1+v^2}{1+v}\\ \Rightarrow x\frac{dv}{dx}=\frac{1+v^2}{1+v}-v=\frac{1+v^2-v-v^2}{1+v}=\frac{1-v}{1+v}\\ \Rightarrow \left(\frac{1+v}{1-v}\right)dv=\frac{dx}{x}\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{1+v}{1-v}dv={\displaystyle \int \frac{dx}{x}}}\\ \Rightarrow {\displaystyle \int \frac{1+v}{1-v}dv=logx+c}\\ \Rightarrow {\displaystyle \int \frac{2-\left(1-v\right)}{1-v}dv=logx+c}\\ \Rightarrow {\displaystyle \int \frac{2}{1-v}dv-{\displaystyle \int 1dv=logx+c}}\\ \Rightarrow \frac{2log\left|1-v\right|}{-1}-v=logx+c\\ \Rightarrow log{\left(1-v\right)}^2+v=-logx-c\end{array}$

Put $v=\frac{y}{x},$

$\begin{array}{l}log{\left(1-\frac{y}{x}\right)}^2+\frac{y}{2}=-logx+c\\ \Rightarrow log{\left(\frac{x-y}{x}\right)}^2+logx=-\frac{y}{x}-c\end{array}$

$\begin{array}{l}\Rightarrow log\left[{\left(\frac{x-y}{x}\right)}^2\times x\right]=-\frac{y}{x}-c\\ \Rightarrow \frac{x-y}{x}=e^{-\frac{y}{x}-c}=c_1{e}^{-\frac{y}{x}-c},where,c_{1}e\end{array}$

$\Rightarrow {\left(x-y\right)}^2=c_1.x{e}^{-\frac{y}{x}}$ is the required solution of the D.E.

The given D.E. is

$\begin{array}{l}\frac{dy}{dx}=e^{x+y}\\ \Rightarrow \frac{dy}{dx}=e^x.e^y\\ \Rightarrow \frac{dy}{e^y}=e^{x}dx\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{dy}{e^y}={\displaystyle \int e^{x}dx}}\\ \Rightarrow \frac{e^{-y}}{-1}=e^x+c_1\\ \Rightarrow e^{-y}=-e^x-c_1\\ \Rightarrow e^{-y}+e^x=c,where,c=-c_1\end{array}$

$\therefore$ Option (A) is correct.

Let ‘x’ be the number of bacteria present in instantaneous time t.

Then, $\frac{dx}{dt}\propto x$

$\Rightarrow \frac{dx}{dt}=kx,where,k=$ constant of proportionality.

$\Rightarrow \frac{dx}{x}=kdt$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{dx}{x}={\displaystyle \int kdt}}\\ \Rightarrow logx=kt+c\end{array}$

Given, at $t=0,x=x_0(say)then,$

$log{x}_0=c\left(Initial,x_0=100000\right)$

So, the differential equation is

$\begin{array}{l}logx=kt+log{x}_0\\ \Rightarrow logx-log{x}_0=kt\\ \Rightarrow log\frac{x}{x_0}=kt\end{array}$

As the bacteria number increased by 10% in 2 hours.

The number of bacteria increased in 2hours $=10\mathrm{\%}\times 100000=10000$

Hence, at t=2,

$x=100000+10000=110000$

So,&nb

Let P and t the principal and time respectively.

Then, increase in principal $\frac{dP}{dt}=P\times 5\mathrm{\%}$

$\Rightarrow \frac{dP}{P}=\frac{5}{100}dt$

Integrating both sides,

$\begin{array}{l}\Rightarrow {\displaystyle \int \frac{dP}{P}={\displaystyle \int \frac{1}{20}dt}}\\ \Rightarrow logP=\frac{t}{20}+c\\ \Rightarrow P=e^{\frac{t}{20}+c}\end{array}$

At, t=0, P=1000

So,  $\begin{array}{l}1000=e^{\frac{0}{20}+c}\\ \Rightarrow e^c=1000\end{array}$

And at t=10,

$\begin{array}{l}P=e^{\frac{10}{100}+c}=e^{0.5}.e^c\\ \Rightarrow P=1.648\times 1000=1648\end{array}$

P = ?1648

Let P, r and t be the principal rate and time respectively.

Then, increase in principal $\frac{dP}{dt}=P\times r\mathrm{\%}$

$\begin{array}{l}\Rightarrow \frac{dP}{dt}=P.\frac{r}{100}\\ \Rightarrow \frac{dP}{P}=\frac{r}{100}dt\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{dP}{P}={\displaystyle \int \frac{r}{100}dt}}\\ \Rightarrow logP=\frac{rt}{100}+c\\ \Rightarrow P=e^{\frac{rt}{100}+c}\end{array}$

Given at t=0,P=100

So, $100=e^{\frac{r\times 0}{100}+c}$

$\begin{array}{l}\Rightarrow 100=e^0\times e^c\\ \Rightarrow e^c=100\left(?e^0=1\right)\end{array}$

And at $t=10,P=2\times 100=200$

So, $200=e^{\frac{r10}{100}+c}$

$\begin{array}{l}\Rightarrow 200=e^{\frac{r}{10}}.e^e\\ \Rightarrow e^{\frac{r}{10}}=\frac{200}{100}=2\end{array}$

$\begin{array}{l}\Rightarrow \frac{r}{10}=log2\\ \Rightarrow \frac{r}{10}=0.6931\\ \Rightarrow r=6.931\end{array}$

Hence, the rate is 6.931%

111. Given, $y={\left(ta{n}^{-1}x\right)}^2$

So, $y_1=\frac{dy}{dx}=2\left(ta{n}^{-1}x\right)\frac{d}{dx}ta{n}^{-1}x$

$\Rightarrow y_1=2ta{n}^{-1}x\times \frac{1}{1+x^2}$

$\left(x^2+1\right)y_1=2ta{n}^{-1}x$

Differentiating again w r t ‘x’ we get,

$\left(x^2+1\right)\frac{d{y}_1}{dx}+y_1\frac{d}{dx}\left(x^2+1\right)=2\frac{d}{dx}ta{n}^{-1}x$

$\Rightarrow \left(x^2+1\right)y_2+y_1\left(2x\right)=\frac{2}{1+x^2}$

$\Rightarrow {\left(x^2+1\right)}^2{y}_2+2x\left(1+x^2\right)y_1=2$

Hence proved.

Let ‘r’ and U be the radius and volume of the spherical balloon.

Then, $\frac{dU}{dt}=k,$ k = constant

$\begin{array}{l}\frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)=k\\ \Rightarrow 4\pi r^2\frac{dr}{dt}=k\\ \Rightarrow 4\pi r^{2}dr=kdt\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int 4\pi r^{2}dr={\displaystyle \int kdt}}\\ \Rightarrow \frac{4}{3}\pi r^3=kt+c\end{array}$

Given at t = 0, r = 3

So, 4π(3)³ = c

C = 36π

And, at t=3, r=6

So, $\frac{4}{3}\pi {\left(6\right)}^3=3k+36\pi \left(c=36\pi \right)$

$\begin{array}{l}\Rightarrow 288\pi -36\pi =3k\\ \Rightarrow k=\frac{252\pi }{3}=84\pi \end{array}$

Hence, putting value of c and k in,

$\frac{4}{3}\pi r^3=kt+c$ , we get,

$\begin{array}{l}\frac{4}{3}\pi r^3=84\pi .t+36\pi \\ \Rightarrow r^3=\frac{3}{4\pi }\left(84\pi .t+36\pi \right)\\ \Rightarrow r^3=63t+27\\ \Rightarrow r={\left[63t+27\right]}^{\frac{1}{3}}\end{array}$