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Some the top recruiters for Marwari College, Ranchi are listed below:

Chettinad Academy of Research and Education BSc in Medical Biotechnology curriculum is divided into six semesters. The curriculum includes various core and elective subjects. A list of some of these subjects is presented in the table below:

The given D.E.is

$\frac{dy}{dx}+\left(secx\right)y=tanx$ Which is in the form $\frac{dy}{dx}+Py=Q$

So, $P=secx\&Q=tanx$

$\therefore$ $\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int secxdx}}=e^{log\left|secx+tanx\right|}=secx+tanx$

Thus, the general solution is ,

$\begin{array}{l}y\times I.F={\displaystyle \int Q\times I.Fdx+c}\\ =y\times \left(secx+tanx\right)={\displaystyle \int tanx}\left(secx+tanx\right)dx+c\end{array}$

$\begin{array}{l}={\displaystyle \int \left(tanxsecx+ta{n}^{2}x\right)dx+c}\\ ={\displaystyle \int \left(tanxsecx+se{c}^2-1\right)}dx+c\\ =sec+tanx-x+c\end{array}$

$=\left(secx+tanx\right)y=secx+tanx-x+c\left\{?se{c}^{2}x=ta{n}^{2}x+1\right\}$

The average package offered at Marwari College, Ranchi placement has not released the average package. Check out the table for the highest package offered in 2023:

8. Yellow colour in NaCl is due to the defect called as metal excess defect. In this defect anionic vacancies get created due to the diffusion of Cl-ions to the surface of the crystal and there after unpaired electrons occupy anionic sites. These sites are known as F-centres. The electrons at F-centres then absorb energy from the visible region and undergo excitation which makes the crystal appear yellow.

The first step in the procedure is to check eligibility criteria. Aspirants who meet the entry requirements can register and apply in both online and offline mode. The applications are further shortlisted by Motherhood University for BPharm course on merit-basis. Hence, the applicants performance in Class 12 or equivalent is evaluated. Selected candidates can further get their documents verified and pay the fees. On payment of first semester fee or admission fee amount, the seat is confirmed. 

136. Given, $sin(A+B)=sinAcosB+cosAsinB$

Differentiating w r t. ‘x’ we get,

$\frac{d}{dx}sin(A+B)=\frac{d}{dx}(sinAcosB+cosAsinB)$

$\to cos(A+B)-\frac{d}{dx}(A+B)=sinA\frac{d}{dx}cosB+cosB\frac{d}{dx}sinA+cosA\frac{d}{dx}sinB+sinB\frac{d}{dx}cosA$

$\to cos(A+B)\left(\frac{dA}{dx}+\frac{dB}{dx}\right)=-sinAsinB\frac{dB}{dx}+cosBcosA\frac{dA}{dx}+cosAcosB\frac{dB}{dx}-sinAsinB\frac{dA}{dx}$

$\to cos(A+B)\left(\frac{dA}{dx}+\frac{dB}{dt}\right)=cosAcosB\left(\frac{dA}{dx}+\frac{dB}{dx}\right)-sinAsinB\left(\frac{dA}{dx}+\frac{dB}{dx}\right)$

$=(cosAcosB-sinAsinB)\left(\frac{dA}{dx}+\frac{dB}{dx}\right).$

$\to cos(A+B)=cosAcosB-sinAsinB.$

The given D.E. is

$\frac{dy}{dx}+\frac{y}{x}=x^2$ Which is in the form $\frac{dy}{dx}+Py=Q$

So,  $P=\frac{1}{x}=\&Q=x^2$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int \frac{1}{2}dx}}=e^{logx}=x\left\{?e^{logx}=x\right\}$

Thus, the general solution is

$\begin{array}{l}y\times I.F={\displaystyle \int Q\times I.Fdx+c}\\ y.x={\displaystyle \int x^2.xdx+c}\\ =xy={\displaystyle \int x^{3}dx+c}\\ =xy=\frac{x^4}{4}+c\end{array}$

The MBA course offered by SRM University Andhra Pradesh is a management course that runs for two years. During this course duration, students study different facets of Management, such as Finance, Marketing, Operation, SCM, and more. Those who wish to enrol in this course must meet the eligibility requirements set by the university and apply for it on the university's official website.

BA Public Administration courses are ideal for those with a deep interest in the field of public services. Students who pursue BA Pubic Administration courses learn some core skills such as Leadership, Budget Analysis, Policy Enhancement, etc. They get a chance to work with both the government and private sectors.

Read more:

BA Public Administration

Marwari College, Ranchi recorded a total of 450 students securing placements across various sectors in 2023. Check out the table for clarity:

Note- The institute has not released the 2024 placement report, once released it will be updated here. 

7. Crystals have long range ordered arrangement of  their constituent particles but usually these crystals are not perfect as during the process of crystallisation some deviations as compared to such ideal arrangement set in depending upon the rate of cooling or presence of impurities in solution  also this process occurs at such a rate that the constituent particles may not get the sufficient time to arrange themselves in a perfect order  and hence these deviations or irregularities in arrangement is being termed as defects or imperfections. Therefore, crystals are usually not perfect. 

Given, D.E. is

$\frac{dy}{dx}+3y=e^{-2x}$ which is of the form

$\frac{dy}{dx}+Py=Q$

Where $\begin{array}{l}P=3\\ \&Q=e^{-2x}\end{array}$

So, I.F $=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int 3dx}}=e^{3x}$

So, the solution is $=y\times I.F={\displaystyle \int e^{-2x}\left(I.F\right).dx+c}$

$\begin{array}{l}=y\times e^{3x}={\displaystyle \int e^{-2x}.}{e}^{3x}dx+c\\ =e^{3x}y={\displaystyle \int e^{x}dx+c}\\ =e^{3x}y=e^x+c\\ =y=\frac{e^x}{e^{3x}}+\frac{c}{e^{3x}}\\ =y=e^{-2x}+c{e}^{-3x}\end{array}$

Is the required general solution.

135. Given, $f(x)=|x{|}^3=\{\begin{array}{cc}\hfill x^3\hfill & \hfill if x\ge 0\hfill \\ \hfill -x^3\hfill & \hfill if x<0\hfill \end{array}$

For $x\ge 0,f(x)=|x{|}^3=x^3$

and $\begin{array}{cc}\hfill f^{\prime }(x)=3x^2\hfill & \hfill f^{\prime \prime }(x)=6x\hfill \end{array}$

For $x<0,f(x)=|x{|}^3={(-x)}^3=-x^3.$

so, $\begin{array}{cc}\hfill f^{\prime }(x)=-3x^2\hfill & \hfill f^{\prime \prime }(x)=-6x\hfill \end{array}$

Hence, $f^{\prime \prime }(x)=\{\begin{array}{cc}\hfill 6x,\hfill & \hfill if x\ge 0\hfill \\ \hfill -6x,\hfill & \hfill if x<0\hfill \end{array}$

Each semester of MBA at SRM University Andhra Pradesh comprises various core subjects. These subjects are compulsory for all students to attend. Listed below are some of the core subjects included in the curriculum:

• Business Statistics
• Accounting for Managers
• Operations Research 
• Marketing Management
• Organizational Behavior
• Business Research Methods
• Human Resource Management
• Decision Science and Analytics
• Operations Management
• Financial Management 

6. In solids the constituent particles are found to possess fixed positions and can only oscillate about their mean positions and hence they are said to be incompressible and rigid. 

Marwari College, Ranchi provides good placements to its graduating students. Check out the table below to know the key highlights of the Marwari College, Ranchi placement in 2023:

Note- The institute has not released the 2024 placement report, once released it will be updated here. 

The given D.E. is

$\frac{dy}{dx}+2y=2sinx$ which is of form $\frac{dy}{dx}+Px=Q$

We have, P = 2

$Q=sinx$

So, I.F. $=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int 2dx}}=e^{2x}$

The solution is $y\times I.F={\displaystyle \int Q\times \left(I.F\right)dx+c}$

$\begin{array}{l}y{e}^{2x}={\displaystyle \int sinx{e}^{2x}dx+c}\\ =y{e}^{2x}=I+c---------(1)\\ Where,I={\displaystyle \int sinx{e}^{2x}}\\ =sinx{\displaystyle \int e^{2x}-{\displaystyle \int \left(\frac{d}{dx}sinx{\displaystyle \int e^{2x}dx}\right)dx}}\\ =sinx\frac{e^{2x}}{2}-\frac{1}{2}\left[{\displaystyle \int cosx{e}^{2x}dx}\right]\\ =\frac{e^{2x}sinx}{2}-\frac{1}{2}\left[cos{\displaystyle \int e^{2x}dx-{\displaystyle \int \left(\frac{d}{dx}cosx{\displaystyle \int e^{2x}dx}\right)dx}}\right]\\ =\frac{e^{2x}sinx}{2}-\frac{1}{2}\left[cosx\frac{e^{2x}}{2}\frac{1}{2}{\displaystyle \int \left(-cosx\right)e^{2x}dx}\right]\\ =\frac{e^{2x}sinx}{2}-\frac{e^{2x}cosx}{4}-\frac{1}{4}{\displaystyle \int sinx{e}^{2x}dx}\end{array}$

$\begin{array}{l}=I=e^{2x}\frac{sinx}{2}-e^{2x}\frac{cosx}{4}-\frac{1}{4}1\\ =I+\frac{1}{4}I=2e^{2x}\frac{sinx-e^{2x}cosx}{4}\\ =\frac{5}{4}I=e^{2x}\frac{\left(2sinx-cosx\right)}{4}\\ =I=e^{2x}\frac{\left(2sinx-cosx\right)}{5}\end{array}$

Hence, equation, (1) becomes,

$\begin{array}{l}y{e}^{2x}=\frac{e^{2x}}{5}\left(2sinx-cosx\right)+c\\ =y=\frac{\left(2sinx-cosx\right)}{5}+c{e}^{2x}\end{array}$

Is the required solution.

5. The liquids and gases both are categorised as fluids as they show the ability to flow and also their molecules can move past one another freely. 

134. Given, $x=a(cost+tsint)$ and $y=a(sint-tcost).$

Differentiating w r t. ‘t’ we get,

$\frac{dx}{dt}=a\frac{d}{dt}(cost+tsint).$

$=a\left(-sint+t\frac{d}{dt}sint+sint\frac{dt}{dt}\right).$

$=a(-sint+tcost+sint)=atcost$

$\frac{dy}{dt}=\frac{ad}{dt}(csint-tcost)$

$=a\left(cost-t\frac{d}{dt}cost-cot\frac{dt}{dt}\right)$

$=a(cost+tsint-cost)=atsint$

$\frac{dy}{dx}=\frac{dy/dt}{dx/at}=\frac{atsint}{atcost}=tant$

So, $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}(tant)=\frac{d}{dt}(tant)\cdot \frac{dt}{dx}.$

$=se{c}^{2}t\times \frac{dt}{dx}.$

$=se{c}^{2}t\times \frac{1}{(dx/dt)}$

$=se{c}^{2}t\times \frac{1}{ at cost}$

$=\frac{se{c}^{3}t}{at}$