Ask & Answer: India's Largest Education Community

While practicing only Maharashtra HSC previous year question papers can help you significantly in clearing the exam, it's not recommended to solely rely on them as a complete study method. You should also learn the syllabus and concepts from the prescribed textbooks to fully understand the topics.

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

Sol:

$\begin{array}{l}Givenequationis\left(\frac{2+sinx}{1+y}\right)\frac{dy}{dx}=-cosx\\ \Rightarrow \left(\frac{2+sinx}{cosx}\right)\frac{dy}{dx}=-\left(1+y\right)\Rightarrow \frac{dy}{\left(1+y\right)}=-\left(\frac{cosx}{2+sinx}\right)dx\\ Integratingbothsides,weget\\ {\displaystyle \int \frac{dy}{\left(1+y\right)}=-{\displaystyle \int \left(\frac{cosx}{2+sinx}\right)dx}}\\ \Rightarrow log\left|1+y\right|=-log\left|2+sinx\right|+logc\\ \Rightarrow log\left|1+y\right|+log\left|2+sinx\right|=logc\\ \Rightarrow log\left(1+y\right)\left(2+sinx\right)=logc\\ \Rightarrow \left(1+y\right)\left(2+sinx\right)=c\\ Putx=0andy=1,weget\\ \Rightarrow \left(1+1\right)\left(2+sin0\right)=c\Rightarrow c=4\\ \therefore equationis\left(1+y\right)\left(2+sinx\right)=4\\ Nowputx=\frac{\pi }{2}\\ \therefore \left(1+y\right)\left(2+sin\frac{\pi }{2}\right)=4\Rightarrow \left(1+y\right)\left(2+1\right)=4\\ \Rightarrow 1+y=\frac{4}{3}\Rightarrow y=\frac{4}{3}-1=\frac{1}{3}\\ So,y\left(\frac{\pi }{2}\right)=\frac{1}{3}\\ Hence,therequiredsolutionisy\left(\frac{\pi }{2}\right)=\frac{1}{3}.\end{array}$

SLS, Pandit Deendayal Energy University offers four-year BBA (H) across three specialisations. The UG-level programme is available in Marketing, Finance, and Human Resource Management. Admission criteria is entrance-based at the school. Before proceeding with application formalities, the candidates are advised to check out the prescribed eligibility requirements. 

Maharaja Ganga Singh University provides hostel facilities for both, boys and girls. The campus is equipped with CCTV cameras to provide a safe and secure surrounding. The girls hostel is called 'Gargi Mahila Chhatrawas'. The hostels are equipped with basic amenities like mess, water purifiers, electricity, etc. and monitored by wardens to ensure safety and discipline. Rental and paying guest facilities are also available in the vicinity of the campus. 

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

Sol:

$\begin{array}{l}Givendifferentialequationis\left(x+2y^3\right)\frac{dy}{dx}=y\\ \Rightarrow \frac{dy}{dx}=\frac{y}{x+2y^3}\Rightarrow \frac{dx}{dy}=\frac{x+2y^3}{y}\\ \Rightarrow \frac{dx}{dy}=\frac{x}{y}+\frac{2y^3}{y}\Rightarrow \frac{dx}{dy}-\frac{x}{y}=2y^2\\ \therefore P=-\frac{1}{y}andQ=2y^2\\ IntegratingfactorI.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int -\frac{1}{y}.dx}}=e^{-logy}=e^{log\frac{1}{y}}=\frac{1}{y}\\ \therefore Solutionoftheequationis\\ x\times I.F.={\displaystyle \int Q.I.F.dy+c}\\ \Rightarrow x.\frac{1}{y}={\displaystyle \int 2y^2.\frac{1}{y}dy+c}\Rightarrow \frac{x}{y}=2{\displaystyle \int ydy+c}\\ \Rightarrow \frac{x}{y}=2.\frac{y^2}{2}+c\Rightarrow \frac{x}{y}=y^2+c\\ \therefore x=y^3+cy=y\left(y^2+c\right)\\ Hence,therequiredvalueisx=y\left(y^2+c\right).\end{array}$

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

Sol:

 $\begin{array}{l}Givenequationis\frac{dy}{dx}=1+x+y^2+x{y}^2\\ \Rightarrow \frac{dy}{dx}=1\left(1+x\right)+y^2\left(1+x\right)\\ \Rightarrow \frac{dy}{dx}=\left(1+x\right)\left(1+y^2\right)\Rightarrow \frac{dy}{1+y^2}\Rightarrow \left(1+x\right)dx\\ Integratingbothsides,weget\\ {\displaystyle \int \frac{dy}{1+y^2}}={\displaystyle \int \left(1+x\right)dx}\Rightarrow ta{n}^{-1}y=x+\frac{x^2}{2}+c\\ Putx=0andy=0,wegetta{n}^{-1}\left(0\right)=0+0+c\Rightarrow c=0\\ \therefore ta{n}^{-1}y=x+\frac{x^2}{2}y=tan\left(x+\frac{x^2}{2}\right)\\ Hence,therequiredsolutionisy=tan\left(x+\frac{x^2}{2}\right).\end{array}$

Maharaja Ganga Singh University offers scholarships to encourage meritorious students and academic excellence. It also offers scholarships to students belonging to the reserved category as per the government guidelines. Scholarships offered at MGSU Bikaner are:

• Post-matric Scholarship is a centrally sponsored scheme, implemented by State Government. It provides financial assistance to students belonging to ST, SC, OBC/SEBC/EBC, pursuing higher education at post-secondary stage.

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

Sol:

$\begin{array}{l}Givenequationisydx-xdy=x^{2}ydx\\ \Rightarrow ydx-x^{2}ydx=xdy\\ \Rightarrow y\left(1-x^2\right)dx=xdy\\ \Rightarrow \left(\frac{1-x^2}{x}\right)dx=\frac{dy}{y}\Rightarrow \left(\frac{1}{x}-x\right)dx=\frac{dy}{y}\\ Integratingbothsides,wehave\\ {\displaystyle \int \left(\frac{1}{x}-x\right)dx}={\displaystyle \int \frac{dy}{y}}\Rightarrow logx-\frac{x^2}{2}=logy+logc\\ \Rightarrow logx-\frac{x^2}{2}=logyc\Rightarrow logx-logyc=\frac{x^2}{2}\\ \Rightarrow log\frac{x}{yc}=\frac{x^2}{2}\Rightarrow \frac{x}{yc}=e^{\frac{x^2}{2}}\Rightarrow \frac{yc}{x}=e^{\frac{-x^2}{2}}\\ \Rightarrow yc=x{e}^{\frac{-x^2}{2}}\Rightarrow y=\frac{1}{c}.x{e}^{\frac{-x^2}{2}}\Rightarrow y=kx{e}^{\frac{-x^2}{2}}\left[?k=\frac{1}{c}\right]\\ Hence,therequiredsolutioniskx{e}^{\frac{-x^2}{2}}.\end{array}$

No, students interested in taking admission to the NLU Kolkata BA LLB (H) course cannot do so without having a valid CLAT score. Thus, all students interested in taking admission are required to appear for the CLAT exam. It is a national-level entrance exam conducted in pen and paper mode. The application forms for the exam can be filled out by Consortium of NLUs.

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

Sol:

$\begin{array}{l}Givenequationis\frac{dy}{dx}+1=e^{x+y}\\ Putx+y=t\\ \therefore 1+\frac{dy}{dx}=\frac{dt}{dx}\\ \therefore \frac{dt}{dx}=e^t\Rightarrow \frac{dt}{e^t}=dx\Rightarrow e^{-t}dt=dx\\ Integratingbothsides,wehave\\ {\displaystyle \int e^{-t}dt}={\displaystyle \int dx}\Rightarrow -e^{-t}=x+c\\ -e^{-\left(x+y\right)}=x+c\Rightarrow \frac{-1}{e^{\left(x+y\right)}}x+c\Rightarrow \left(x+c\right).e^{\left(x+y\right)}=-1\\ Hence,therequiredsolutionis\left(x+c\right).e^{\left(x+y\right)}+1=0.\end{array}$

Hi, the table below shows the top specialisations offered by Engineering colleges in Andhra Pradesh Accepting Andhra Pradesh EAMCET score and their no. of colleges:

Yes, SLS PDEU offer scholarship schemes for the UG course aspirants. The eligible students can apply for its merit-cum-means and merit scholarships. However, to avail the benefits of a scheme, it is important that the prescribed criteria is met. Moreover, the SLS PDEU BBA applicants who get scholarship shall receive the scheme in the form of tuition fee waiver.

Nehru Arts and Science College appoints well-trained and expert faculty members. The faclty focuses on therotical and practical knowledge of the studnets under the guidance of experts. The faculty focuses on research and quality education for the students. The faculty have years of experince in the related field. 

Hi, It is estimated that joining Engineering colleges in Andhra Pradesh that accept Andhra Pradesh EAMCET scores can be affordable, as most colleges have tuition fees between INR 1-2 lakh. Listed below are some of the colleges along with their fees:

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

Sol:

$\begin{array}{l}Givendifferentialequationis\frac{dy}{dx}+ay=e^{mx}\\ \therefore P=aandQ=e^{mx}\\ IntegratingfactorI.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int a.dx}}=e^{ax}\\ \therefore Solutionoftheequationis\\ y\times I.F.={\displaystyle \int Q.I.F.dx+c}\\ \Rightarrow y.e^{ax}={\displaystyle \int e^{mx}.e^{ax}dx+c}\Rightarrow y.e^{ax}={\displaystyle \int e^{\left(m+a\right)x}dx+c}\\ \Rightarrow y.e^{ax}=\frac{e^{\left(m+a\right)x}}{\left(m+a\right)}+c\Rightarrow y=\frac{e^{\left(m+a\right)x}}{\left(m+a\right)}.e^{-ax}+c.e^{-ax}\\ \therefore y=\frac{e^{mx}}{\left(m+a\right)}+c.e^{-ax}\\ Hence,therequiredvalueisy=\frac{e^{mx}}{\left(m+a\right)}+c.e^{-ax}.\end{array}$