Class 12th

${\overrightarrow{a}}_1=x\widehat{i}-\widehat{j}+\widehat{k}\&{\overrightarrow{a}}_2=\widehat{i}+y\widehat{j}+z\widehat{k}$

given ${\overrightarrow{a}}_1\&{\overrightarrow{a}}_2$ are collinear then ${\overrightarrow{a}}_1=\lambda {\overrightarrow{a}}_2$

$\Rightarrow \left(x\widehat{i}-\widehat{j}+\widehat{k}\right)=\lambda \left(\widehat{i}+y\widehat{j}+z\widehat{k}\right)$       

Since $\widehat{i},\widehat{j}\&\widehat{k}$ are not collinear so

$Sox\widehat{i}+y\widehat{j}+z\widehat{k}=\lambda \widehat{i}-\frac{1}{\lambda }\widehat{j}+\frac{1}{\lambda }\widehat{k}$     

Hence possible unit vector parallel to it be $\frac{1}{\sqrt[]{3}}\left(\widehat{i}-\widehat{j}+\widehat{k}\right)$ for $\lambda$ =

Given f(x) =   _{${\displaystyle {\int }_e^x{e}^{t}f\left(t\right)dt+e^x..........\left(i\right)}$}

using Leibniz rule then

f'(x) = e^xf(x) + e^x

_{$\Rightarrow \frac{dy}{dx}=e^{x}y+e^{x}wherey=f\left(x\right)then\frac{dy}{dx}=f\text{'}\left(x\right)$}            

P = -e^x, Q = e^x

Solution be y. (I.F.) =  ${\displaystyle \int Q\left(I.F.\right)dx+c}$

I. f. =  $e^{{\displaystyle \int -e^{x}dx}=e^{-e^x}}$

$\Rightarrow y.\left(e^{-e^x}\right)={\displaystyle \int e^x.e^{-e^x}dx+c}$   

$y.e^{-e^x}=-{\displaystyle \int dt+c=-t+c=-e^{-e^x}+c..........\left(ii\right)}$

Put x = 0 , in (i) f (0) = 1

$From\left(ii\right),\frac{1}{e}=-\frac{1}{e}+cgivenc=\frac{2}{e}$   

Hence f(x) = 2. $e^{\left(e^x-1\right)}-1$

  $v=\frac{1}{\sqrt[]{\epsilon \mu }}=\frac{1}{\sqrt[]{2{\epsilon }_0.2{\mu }_0}}=\frac{1}{2\sqrt[]{{\epsilon }_0{\mu }_0}}=\frac{c}{2}=15*10^{7}m/s.$

Candidates seeking admission to the BTech programme can enrol for admission with Class 12 marks. Candidates must complete Class 12 to enrol for the BTech course. The college offers only one UG course, i.e., BTech. DPG Institute of Technology and Management admissions are based on the entrance exam.

  $\frac{N{}_p}{N_s}=\frac{V_p}{V_s}\Rightarrow N_p=\frac{V_p}{V_s}*N_s=\frac{220}{12}*24=440$

$l=l_{0}co{s}^2\theta =100*co{s}^2\left(30°\right)=75Lumens$

Note : As the incident light is unpolarized, intensity of emerging light does not depend on the polarization axis of polarizer.

  $A^2=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$A^3=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^3\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]$
$A^4\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^3\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^4\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

Similarly we get A¹⁹ =   $=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^{19}\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]\&A^{20}=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^{20}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

=  $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 4\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$\Rightarrow 1+\alpha +\beta =1gives\alpha +\beta =0........\left(i\right)$

$2^{20}+\left(2^{19}-2\right)\alpha =4from\left(i\right)$

$\Rightarrow \alpha =\frac{4-2^{20}}{2^{19}-2}=\frac{4\left(1-2^{18}\right)}{-2\left(1-2^{18}\right)}=-2$

So, β = 2

Hence β - α = 4

Let the equation of normal is Y – y = - $\frac{1}{m}\left(X-x\right)$  

where m is slope of tangent to the given curve then

  $Y-y=-\frac{dx}{dy}\left(X-x\right)$         

It passes through (a, b) so b – y = $\frac{-dx}{dy}\left(a-x\right)$

=> (a – x) dx = (y – b) dy

On integration     $ax-\frac{x^2}{2}=\frac{y^2}{2}-by+c.........\left(i\right)$  

(ii) passes through (3, -3) &  then

3a – 3b – c = 9       .(ii)

& 4a - $2\sqrt[]{2}b$ - c = 12           .(iii)

also given  $a-2\sqrt[]{2}b=3............(iv)$

Solve (ii), (iii) & (iv) b = 0, a = 3

Hence a² + b² + ab = 9

Quality factor =   $\frac{\omega L}{R}=\frac{2*3.14*10*10^6*2*10^{-4}}{6.28}=2000$

Given $l_{m,n}={\displaystyle {\int }_0^1{x}^{m-1}{\left(1-x\right)}^{n-1}dx............\left(i\right)}$  

put 1 - x = $t\{\begin{array}{l}x=0,t=1\\ x=1,t=0\end{array}$

dx = -dt

From (i) $l_{m,n}={\displaystyle {\int }_1^0{\left(1-t\right)}^{m-1}.t^{n-1}\left(-dt\right)}$

$l_{m,n}={\displaystyle {\int }_0^1{t}^{n-1}{\left(1-t\right)}^{m-1}dt={\displaystyle {\int }_0^1{x}^{n-1}{\left(1-x\right)}^{m-1}dx..........\left(ii\right)}}$    

$Putx=\frac{1}{1+y}in\left(i\right)thendx=-\frac{1}{{\left(1+y\right)}^2}dy$                          

(i)  $l_{m,n}={\displaystyle {\int }_{\infty }^0\frac{1}{{\left(1+y\right)}^{m-1}}}{\left(1-\frac{1}{1+y}\right)}^{n-1}\left(\frac{-dy}{{\left(1+y\right)}^2}\right)={\displaystyle {\int }_0^{\infty }\frac{y^{n-1}}{{\left(1+y\right)}^{m+n}}dy}..........\left(iii\right)$

Similarly by (ii) $l_{m,n}={\displaystyle {\int }_0^{\infty }\frac{y^{m-1}}{{\left(y+1\right)}^{m+n}}dy..........\left(iv\right)}$

Adding (iii) & (iv) $2l_{m,n}={\displaystyle {\int }_0^{\infty }\frac{y^{n-1}+y^{m+1}}{{\left(y+1\right)}^{m+n}}}$

Putting  $\frac{1}{z}\{\begin{array}{l}y=1,z=1\\ y=\infty ,z=0\end{array}$ $\Rightarrow dy=\frac{-1}{z^2}dz$  

Hence $l_{m,n}={\displaystyle {\int }_0^1\frac{x^{m-1}+x^{n-1}}{{\left(1+x\right)}^{m+n}}}$ dx = α lm, n

=> α = 1