Ask & Answer: India's Largest Education Community

Maharaja Ganga Singh University Bikaner has not released the 2024 and 2023 placement reports. However, the university has established a career counseling and placement cell that provides exposure to the range of job opportunities available for the students in their respective sectors. As per the AQAR report, the highest package offered during MGSU Bikaner placements 2022 was INR 5.71 LPA. The top recruiters during the 2022 placement drive were, Govt AYUSH Wellness Center, Dunes Milk Processing Pvt. Ltd., ASKIN Softech, among others.

Yes, UCSD provides both on-campus and off-campus housing options to its students. The on-campus housing rates for 2025-26 are below:

1 USD = INR 86.67

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Here,slopeoftheofthecurve=\frac{dy}{dx}\\ andthedifferencebetweentheabscissaandordinate=x-y.\\ \therefore Asperthecondition,\frac{dy}{dx}={\left(x-y\right)}^2\\ Putx-y=v\\ 1-\frac{dy}{dx}=\frac{dv}{dx}\\ \therefore \frac{dy}{dx}=1-\frac{dv}{dx}\\ \therefore theequationbecomes\\ 1-\frac{dv}{dx}=v^2\Rightarrow \frac{dv}{dx}=1-v^2\Rightarrow \frac{dv}{1-v^2}=dx\\ Integratingbothsides,weget\\ {\displaystyle \int \frac{dv}{1-v^2}={\displaystyle \int dx}}\\ \Rightarrow \frac{1}{2}log\left|\frac{1+v}{1-v}\right|=x+c\Rightarrow \frac{1}{2}log\left|\frac{1+x-y}{1-x+y}\right|=x+c\dots \left(1\right)\\ Since,thecurveisthrough\left(0,0\right),then\\ \Rightarrow \frac{1}{2}log\left|\frac{1+0-0}{1-0+0}\right|=0+c\Rightarrow c=0\\ \therefore Onputtingc=0ineq.\left(1\right)weget\\ \frac{1}{2}log\left|\frac{1+x-y}{1-x+y}\right|=x\Rightarrow log\left|\frac{1+x-y}{1-x+y}\right|=2x\\ \therefore \frac{1+x-y}{1-x+y}=e^{2x}\\ \Rightarrow \left(1+x-y\right)=e^{2x}\left(1-x+y\right)\\ Hence,therequiredequationis\left(1+x-y\right)=e^{2x}\left(1-x+y\right).\end{array}$

According to NIRF ranking 2024, NLSIU Bangalore emerged as the No. 1 ranked Law college in Karnataka secured 1st position followed by other colleges like Christ University and Alliance School of Law.

Hi, the table below shows the top government Engineering colleges accepting Andhra Pradesh EAMCET scores in Andhra Pradesh along with their total tuition fees:

Disclaimer: This information is sourced from the official website/ sanctioning body and may vary.

The table below compares the IIT Kanpur placement statistics 2023 with other closely ranked IITs:

*The placement data is fetched from multiple media reports.

Candidates wanting to compare SLS, PDEU BBA (H) with other preferred colleges/ universities can do it using certain parameters. It helps the students in comparing colleges in a fair manner. Some of the commonly used parameters are:

• Course fees
• Placement package
• Faculty
• Location, etc. 

There are about 3 top law colleges in Karnataka of which 2 colleges are privately owned and rest 1 college is owned by the government organisations. NLSIU Bangalore, School of Law, Christ University and Alliance School of Law are the best Law colleges in Karnataka. CLAT,  AUSAT and CUET are the popular entrance exams for admission to top Law colleges in Karnataka.

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhattheslopeoftangenttoacurveat\left(x,y\right)is\\ \frac{dy}{dx}=\frac{y-1}{x^2+x}\Rightarrow \frac{dy}{y-1}=\frac{dx}{x^2+x}\\ Integratingbothsides,wehave\\ {\displaystyle \int \frac{dy}{y-1}={\displaystyle \int \frac{dx}{x^2+x}}}\\ \Rightarrow {\displaystyle \int \frac{dy}{y-1}={\displaystyle \int \frac{dx}{x^2+x+\frac{1}{4}-\frac{1}{4}}}}\left[Makingperfectsquare\right]\\ \Rightarrow {\displaystyle \int \frac{dy}{y-1}={\displaystyle \int \frac{dx}{{\left(x+\frac{1}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2}}}\\ \Rightarrow log\left|y-1\right|=\frac{1}{2\times \frac{1}{2}}log\left|\frac{x+\frac{1}{2}-\frac{1}{2}}{x+\frac{1}{2}+\frac{1}{2}}\right|+logc\\ \Rightarrow log\left|y-1\right|=log\left|\frac{x}{x+1}\right|+logc\\ \Rightarrow log\left|y-1\right|=log\left|c\left(\frac{x}{x+1}\right)\right|\\ \therefore y-1=\frac{cx}{x+1}\Rightarrow \left(y-1\right)\left(x+1\right)=cx\\ Since,thelineispassingthroughthepoint\left(1,0\right),then\\ \left(0-1\right)\left(1+1\right)=c\left(1\right)\Rightarrow c=2\\ Hence,therequiredsolutionis\left(y-1\right)\left(x+1\right)=2x.\end{array}$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhattheslopeoftangenttoacurveat\left(x,y\right)is\\ \frac{dy}{dx}=\frac{x^2+y^2}{2xy}\\ Itisahomogeneousdifferentialequation\\ So,puty=vx\Rightarrow \frac{dy}{dx}=v+x.\frac{dv}{dx}\\ \Rightarrow v+x.\frac{dv}{dx}=\frac{x^2+v^2{x}^2}{2x.vx}\\ \Rightarrow v+x.\frac{dv}{dx}=\frac{1+v^2}{2v}\\ \Rightarrow x.\frac{dv}{dx}=\frac{1+v^2}{2v}-v\Rightarrow x.\frac{dv}{dx}=\frac{1+v^2-2v^2}{2v}\\ \Rightarrow x.\frac{dv}{dx}=\frac{1-v^2}{2v}\Rightarrow \frac{2v}{1-v^2}dv=\frac{dx}{x}\\ Integratingbothsides,wehave\\ {\displaystyle \int \frac{2v}{1-v^2}dv={\displaystyle \int \frac{dx}{x}}}\\ \Rightarrow -log\left|1-v^2\right|=logx+logc\\ \Rightarrow -log\left|1-\frac{y^2}{x^2}\right|=logx+logc\Rightarrow -log\left|\frac{x^2-y^2}{x^2}\right|=logx+logc\\ \Rightarrow log\left|\frac{x^2}{x^2-y^2}\right|=log\left|xc\right|\Rightarrow \frac{x^2}{x^2-y^2}=xc\\ Since,thecurveispassingthroughthepoint\left(2,1\right)\\ \therefore \frac{{\left(2\right)}^2}{{\left(2\right)}^2-{\left(1\right)}^2}=2c\Rightarrow \frac{4}{3}=2c\Rightarrow c=\frac{2}{3}\\ Hence,therequiredequationis\frac{x^2}{x^2-y^2}=\frac{2}{3}x\Rightarrow 2\left(x^2-y^2\right)=3x\end{array}$

BMS has excellent scope in India especially because companies today are actively hiring management graduates with strong communication, leadership and analytical skills. Here is what makes BMS valuable in India:

• Rising number of startups and MNCs
• Growing need for marketing, sales, and business development professionals
• Demand in BPO, retail, finance, logistics, and digital services sectors
• Popular route to MBA, PGDM, and competitive exams

So whether you want a job or want to go for higher education, BMS opens up a variety of career pathways.

The registration process for GS College of Commerce is as follows: 

Step 1: Visit the college application portal

Step 2: Get the RTMNU registration form

Step 3: Fill out the form with accurate details 

Step 4: Pay the form fee and download the form to keep for future reference. 

No. Every college has different questions. So, it is better to change your MBA essay for each college. Answer what they ask.

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Givenequationis\frac{dy}{dx}-3y=sin2x\\ Here,P=-3andQ=sin2x\\ IntegratingfactorI.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int -3.dx}}=e^{-3x}\\ \therefore Solutionoftheequationis\\ y\times I.F.={\displaystyle \int Q\times I.F.dx+c}\\ \Rightarrow y.e^{-3x}={\displaystyle \int sin2x.e^{-3x}dx+c}\\ LetI={\displaystyle \int \underset{I}{sin2x}.\underset{II}{e^{-3x}}dx}\\ \Rightarrow I=sin2x.{\displaystyle \int e^{-3x}dx-{\displaystyle \int \left(D\left(sin2x\right).{\displaystyle \int e^{-3x}dx}\right)dx}}\\ \Rightarrow I=sin2x.\frac{e^{-3x}}{-3}-{\displaystyle \int 2cos2x.}\frac{e^{-3x}}{-3}dx\\ \Rightarrow I=\frac{e^{-3x}}{-3}sin2x+\frac{2}{3}{\displaystyle \int \underset{I}{cos2x}.\underset{II}{e^{-3x}}dx}\\ \Rightarrow I=\frac{e^{-3x}}{-3}sin2x+\frac{2}{3}\left[cos2x.{\displaystyle \int e^{-3x}dx-{\displaystyle \int \left(D\left(cos2x\right).{\displaystyle \int e^{-3x}dx}\right)dx}}\right]\\ \Rightarrow I=\frac{e^{-3x}}{-3}sin2x+\frac{2}{3}\left[cos2x.\frac{e^{-3x}}{-3}-{\displaystyle \int 2sin2x.\frac{e^{-3x}}{-3}dx}\right]\\ \Rightarrow I=\frac{e^{-3x}}{-3}sin2x-\frac{2}{9}cos2x.e^{-3x}-\frac{4}{9}{\displaystyle \int sin2x.e^{-3x}dx}\\ \Rightarrow I=\frac{e^{-3x}}{-3}sin2x-\frac{2}{9}cos2x.e^{-3x}-\frac{4}{9}I\\ \Rightarrow I+\frac{4}{9}I=\frac{e^{-3x}}{-3}sin2x-\frac{2}{9}{e}^{-3x}cos2x\\ \Rightarrow \frac{13}{9}I=-\frac{1}{9}\left[3e^{-3x}sin2x+2e^{-3x}cos2x\right]\\ \Rightarrow I=-\frac{1}{13}{e}^{-3x}\left[3sin2x+2cos2x\right]\\ \therefore Theequationbecomes\\ \Rightarrow y.e^{-3x}=-\frac{1}{13}{e}^{-3x}\left[3sin2x+2cos2x\right]+c\\ \therefore y=-\frac{1}{13}\left[3sin2x+2cos2x\right]+c.e^{3x}\\ Hence,therequiredsolutionisy=-\left[\frac{3sin2x+2cos2x}{13}\right]+c.e^{3x}\end{array}$