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Candidates applying to Indian institute of Science education and research thiruvananthapuram (IISER TVM) should submit their Class 10th and 12th marksheets, candidate scorecards of the entrance exam, which in this case can be JEE Advanced or GATE and a valid photo ID. They are also required to carry category certificates, scanned photograph and signature, OCI/PIO certificates as the case may be. Any PwD candidate has to provide a good medical certificate.

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Giventhaty-axis,y=cosx,y=sinx,0\le x\le \frac{\pi }{2}\\ Requiredarea={\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}cosx}dx-{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}sinx}dx\\ ={\left[sinx\right]}_0^{\frac{\pi }{4}}-{\left[-cosx\right]}_0^{\frac{\pi }{4}}\\ =\left[sin\frac{\pi }{4}-sin0\right]+\left[cos\frac{\pi }{4}-cos0\right]\\ =\left[\frac{1}{\sqrt[]{2}}-0+\frac{1}{\sqrt[]{2}}-1\right]=\frac{2}{\sqrt[]{2}}-1=\left(\sqrt[]{2}-1\right)sq.units\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l}Givenequationsarey=1+\left|x+1\right|,x=-3andx=3,y=0\\ Takingy=1+\left|x+1\right|\\ \Rightarrow y=1+x+1\\ \Rightarrow y=x+2andy=1-x-1\Rightarrow y=-x\\ Onsolvingwegetx=-1\\ \therefore Areaofrequiredregions\\ ={\displaystyle \underset{-3}{\overset{-1}{\int }}-x}dx+{\displaystyle \underset{-1}{\overset{3}{\int }}\left(x+2\right)}dx\\ =-{\left[\frac{x^2}{2}\right]}_{-3}^{-1}+{\left[\frac{x^2}{2}+2x\right]}_{-1}^3=-\left[\frac{1}{2}-\frac{9}{2}\right]+\left[\left(\frac{9}{2}+6\right)-\left(\frac{1}{2}-2\right)\right]\\ =-\left(-4\right)+\left[\frac{21}{2}+\frac{3}{2}\right]=4+12=16sq.units\\ Hence,therequiredarea=16sq.units.\end{array}$

The seat intake for the courses at Maharashtra National Law University, Aurangabad are as follows: 

Candidates are requried to participate in the admission process to stand a chance to get the final admission offer. Candidates must pay the course fee to secure their seat in their applied to course. 

There are about 15+ top Low-cost Pharmacy colleges in India and given below are some of them with thier tuition fees:

The Maharashtra CET Cell releases the free mock test papers for candidates a few days before the exam, which can be accessed online through the official website. Meanwhile, candidates can practice the previous year MH CET Law mock test papers, you can also go through the memory based questions compiled up here to get an idea about the exam structure and level of exam.

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l}Givenequationofthecurveisy=2cosx\\ \therefore Areaofshadedregion\\ ={\displaystyle \underset{0}{\overset{2\pi }{\int }}2cosx}dx={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}2cosx}dx+{\displaystyle \underset{\frac{\pi }{2}}{\overset{\frac{3\pi }{2}}{\int }}\left|2cosx\right|}dx+{\displaystyle \underset{\frac{3\pi }{2}}{\overset{2\pi }{\int }}2cosx}dx\\ =2{\left[sinx\right]}_0^{\frac{\pi }{2}}+\left|{\left[sinx\right]}_{\frac{\pi }{2}}^{\frac{3\pi }{2}}\right|+2{\left[sinx\right]}_{\frac{3\pi }{2}}^{2\pi }\\ =2\left[sin\frac{\pi }{2}-sin0\right]+\left|2\left(sin\frac{3\pi }{2}-sin\frac{\pi }{2}\right)\right|+2\left[sin2\pi -sin\frac{3\pi }{2}\right]\\ =2\left(1\right)+\left|2\left(-1-1\right)\right|+2\left(0+1\right)=2+4+2=8sq.units\\ Hence,therequiredarea=8sq.units.\end{array}$

Candidates can get the previous year questions through the Shiksha's website. The MH CET Law examination cell does not releases the question paper online. The candidates can find the memory based questions along with the answers here.

It really depends on your goals and the kind of campus experience you're dreaming of. If you're aiming for the prestige, exposure, and opportunities that Delhi University offers, a focused drop year with CUET prep could be worth it. But if you prefer not to lose a year, direct admission in a reputed private BBA college can also give you a great start. Reflect on your priorities campus life, placements, fees and talk to current students if possible. Both paths can lead to success it's about where you'll grow best, not just where you begin.

The NATA 2025 exam, required for BArch admission at RNS School of Architecture, is being conducted in multiple phases—typically in April, June, and July. The JEE Main Paper -2 2025 was held on Apr 9. Candidates must meet the eligibility criteria set by the Council of Architecture and the institute. Registration and exam details for NATA are available on the official NATA website.

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l}Giventhaty=4x+5\dots \left(i\right)\\ y=5-x\dots \left(ii\right)\\ and4y=x+5\dots \left(iii\right)\\ \begin{array}{cc}\hfill x\hfill & \hfill 0\hfill \\ \hfill -5/4\hfill & \hfill y\hfill \\ \hfill 5\hfill & \hfill 0\hfill \end{array}\begin{array}{cc}\hfill x\hfill & \hfill 0\hfill \\ \hfill 5\hfill & \hfill y\hfill \\ \hfill 5\hfill & \hfill 0\hfill \end{array}\begin{array}{cc}\hfill x\hfill & \hfill 0\hfill \\ \hfill -5\hfill & \hfill y\hfill \\ \hfill 5/4\hfill & \hfill 0\hfill \end{array}\\ Solvingeqn\left(i\right)and\left(ii\right)weget\\ 4x+5=5-x\\ \therefore x=0andy=5\\ CoordinatesofA=\left(0,5\right)\\ Solvingeqn\left(ii\right)and\left(iii\right)weget\\ y=5-x\\ \therefore 4y=x+5\\ 5y=10\\ \therefore y=2andx=3\\ \therefore CoordinatesofB=\left(3,2\right)\\ Solvingeqn\left(i\right)and\left(iii\right)weget\\ y=4x+5and4y=x+5\\ \Rightarrow 4\left(4x+5\right)=x+5\\ \Rightarrow 16x+20=x+5\Rightarrow 15x=-15\\ \therefore x=-1andy=1\\ \therefore CoordinatesofC=\left(-1,1\right)\\ \therefore Areaofrequiredregions\\ ={\displaystyle \underset{-1}{\overset{0}{\int }}{y}_{AC}}dx+{\displaystyle \underset{0}{\overset{3}{\int }}{y}_{AB}}dx-{\displaystyle \underset{-1}{\overset{3}{\int }}{y}_{CB}}dx\\ ={\displaystyle \underset{-1}{\overset{0}{\int }}\left(4x+5\right)}dx+{\displaystyle \underset{0}{\overset{3}{\int }}\left(5-x\right)}dx-{\displaystyle \underset{-1}{\overset{3}{\int }}\left(\frac{x+5}{4}\right)}dx\\ ={\left[4\frac{x^2}{2}+5x\right]}_{-1}^0+{\left[5x-\frac{x^2}{2}\right]}_0^3-\frac{1}{4}{\left[\frac{x^2}{2}+5x\right]}_{-1}^3\\ =\left[\left(0+0\right)-\left(2-5\right)\right]+\left[\left(15-\frac{9}{2}\right)-\left(0-0\right)\right]-\frac{1}{4}\left[\left(\frac{9}{2}+15\right)-\left(\frac{1}{2}-5\right)\right]\\ =3+\frac{21}{2}-\frac{1}{4}\left[\frac{39}{2}+\frac{9}{2}\right]=3+\frac{21}{2}-\frac{1}{4}\times 24\\ =3+\frac{21}{2}-6=\frac{21}{2}sq.units\\ Hence,therequiredarea=\frac{21}{2}sq.units.\end{array}$

For getting admission in RNS School of Architecture the candidates need to appear for JEE Main Paper-2 which is and NATA which is National Aptitude Test in Architecture. Additionally, candidates who seek admission to this course must ensure that they fulfil the below eligibility criteria.

NMIMS Mukesh Patel School of Technology Shirpur accepts multiple entrance exams for BTech admission, including JEE Main, MHT CET, NMIMS CET, and NMIMS-NPAT. This allows students from different academic backgrounds to compete for admission through a range of national and institutional-level tests.

Yes, BTech admission at NMIMS Mukesh Patel School of Technology Shirpur is currently open for 2025. The institute is actively conducting Phase 2 of NMIMS CET counselling, and MHT CET CAP registrations are ongoing. Interested candidates should apply and complete the necessary steps before the deadlines.

The eligibility criteria for the courses at Maharashtra National Law University, Aurangabad  is as follows: 

LLD: In order to be admitted to the University's LLD program, a candidate must have completed exceptional research in the relevant field and met the minimal requirements listed below. In addition to having at least 10 years of teaching experience or outstanding academic accomplishments in the field of law, the candidate must: (a) have earned a PhD or an equivalent degree in the relevant discipline from any accredited university or from a foreign university of standing.

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l}Giventhatx+2y=2\dots \left(i\right)\\ y-x=1\dots \left(ii\right)\\ and2x+y=7\dots \left(iii\right)\\ \begin{array}{cc}\hfill x\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill y\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\begin{array}{cc}\hfill x\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill y\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\begin{array}{cc}\hfill x\hfill & \hfill 0\hfill \\ \hfill 7/2\hfill & \hfill y\hfill \\ \hfill 7\hfill & \hfill 0\hfill \end{array}\\ Solvingeqn\left(ii\right)and\left(iii\right)weget\\ y=1+x\\ \therefore 2x+1+x=7\\ 3x=6\Rightarrow x=2\\ \therefore y=1+2=3\\ CoordinatesofB=\left(2,3\right)\\ Solvingeqn\left(i\right)and\left(iii\right)weget\\ x+2y=2\\ \therefore x=2-2y\\ 2x+y=7\\ 2\left(2-2y\right)+y=7\\ \Rightarrow 4-4y+y=7\Rightarrow -3y=3\\ \therefore y=-1andx=4\\ \therefore CoordinatesofC=\left(4,-1\right)andCoordinatesofA=\left(0,1\right)\\ Takingthelimitsony-axis,weget\\ {\displaystyle \underset{-1}{\overset{3}{\int }}{x}_{BC}}dy-{\displaystyle \underset{-1}{\overset{1}{\int }}{x}_{AC}}dy-{\displaystyle \underset{1}{\overset{3}{\int }}{x}_{AB}}dy\\ ={\displaystyle \underset{-1}{\overset{3}{\int }}\frac{7-y}{2}}dy-{\displaystyle \underset{-1}{\overset{1}{\int }}\left(2-2y\right)}dy-{\displaystyle \underset{1}{\overset{3}{\int }}\left(y-1\right)}dy\\ =\frac{1}{2}{\left[7y-\frac{y^2}{2}\right]}_{-1}^3-2{\left[y-\frac{y^2}{2}\right]}_{-1}^1-{\left[\frac{y^2}{2}-y\right]}_1^3\\ =\frac{1}{2}\left[\left(21-\frac{9}{2}\right)-\left(-7-\frac{1}{2}\right)\right]-2\left[\left(1-\frac{1}{2}\right)-\left(-1-\frac{1}{2}\right)\right]-\left[\left(\frac{9}{2}-3\right)-\left(\frac{1}{2}-1\right)\right]\\ =\frac{1}{2}\left[\frac{33}{2}+\frac{15}{2}\right]-2\left[\frac{1}{2}+\frac{3}{2}\right]-\left[\frac{3}{2}+\frac{1}{2}\right]\\ =\frac{1}{2}\times 24-2\times 2-2=12-4-2=6sq.units\\ Hence,therequiredarea=6sq.units.\end{array}$