Class 12th

$\underset{x\to a}{lim}\frac{xf\left(a\right)-af\left(x\right)}{x-a}\left(\frac{0}{0}\right)$

By L'hospital Rule

$=\underset{x\to a}{lim}\frac{f\left(a\right)-af\text{'}\left(x\right)}{1}=f\left(a\right)-af\text{'}\left(a\right)$

$=4-2a$

Now equation of line OA be

3, 4, 5, 5

In remaining six places you have to arrange

3, 4, 5,5

So no. of ways $=\frac{6!}{2!2!2!}$

Total no. of seven digits nos. = $\frac{7!}{2!3!2!}*1$

Hence Req. prob. $\frac{\frac{6!}{2!2!2!}}{\frac{7!}{2!3!2!}}=\frac{6!3!}{2!7!}=\frac{3}{7}$

Electrophilic addition of bromine to an alkene is anti-addition, in which cis-alkene gives two enantiomers and trans – alkene gives meso form

Here; trans-but-2-ene will give meso products

Calamine is the ore of zinc i.e. ZnCO₃.

Malachite is the ore of copper i.e. CuCO₃.Cu (OH)₂.

  $g\left(2\right)=\underset{x\to 2}{lim}g\left(x\right)=\underset{x\to 2}{lim}\frac{x^2-x-2}{2x^2-x-6}=\underset{x\to 2}{lim}\frac{\left(x-2\right)\left(x+1\right)}{2x\left(x-2\right)+3\left(x-2\right)}$         

$Nowfog=f\left(g\left(x\right)\right)=si{n}^{-1}g\left(x\right)=si{n}^{-1}\left(\frac{x^2-x-2}{2x^2-x-6}\right)$

$\frac{3x^2-2x-8}{2x^2-x-6}\ge 0\&\frac{-x^2+4}{2x^2-x-6}\le 0$

On solving we get  $x\in \left(-\infty ,-2\right)\cup \left[\frac{-4}{3},\infty \right)$

As x = 2 also lies in domain since g(2) $=\underset{x\to 2}{lim}g\left(x\right)$

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

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.

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