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Yes, Arena Animation Jaipur applications are open for 2025 session. Candidates can fill out the admission enquiry form on the institute's website or visit the campus to reach out to the admission team. Aspirants need to fill out the basic details in the form to apply for the preferred course.

The minimum requirement of marks in graduation at National Insurance Academy Pune is 55%. The students belonging to the category, SC/ST/OBC-NC/DA shall be required to be 50%. Provisionally those in final year can apply.

To finish the application and admissions process at MNLU Aurangabad candidates must upload and submit a few required documents. The following are these documents: 

Marksheet for Class 10 
Marksheet for Class 12

Marksheet for graduation
Government-issued Certificate of ID Transfer
Candidates can successfully register for their selected programme by submitting the application form with the aforementioned papers. 

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

$\begin{array}{l}Giventhaty=\sqrt[]{x},x=2y+3,firstquadrantandx-axis.\\ Solvingy=\sqrt[]{x}andx=2y+3,weget\\ y=\sqrt[]{2y+3}\Rightarrow y^2=2y+3\\ \Rightarrow y^2-2y-3=0\Rightarrow y^2-3y+y-3=0\\ \Rightarrow y\left(y-3\right)+1\left(y-3\right)=0\Rightarrow \left(y+1\right)\left(y-3\right)=0\\ \therefore y=-1,3\\ Areaoftheshadedregion={\displaystyle \underset{0}{\overset{3}{\int }}\left(2y+3\right)dy}-{\displaystyle \underset{0}{\overset{3}{\int }}{y}^{2}dy}\\ ={\left[2\frac{y^2}{2}+3y\right]}_0^3-\frac{1}{3}{\left[y^3\right]}_0^3\\ =\left[\left(9+9\right)-\left(0+0\right)\right]-\frac{1}{3}\left[27-0\right]\\ =18-9=9sq.units\\ Hence,therequiredarea=9sq.units.\end{array}$

Admission Parameters National Insurance Academy Pune believes in accepting CAT and CMAT as entrance exams. One needs to sign up to one of these examinations and secure an acceptable score. Through these scores students are shortlisted to appear in GD and PI. The admission process is incomplete without exam.

Arena Animation Jaipur conducts admission in an online mode. Check the steps below to apply for admission:

1: Visit the official website of the college.

2: Click on 'Admission Open' on the Home page and fill out the admission enquiry form.

3: Submit the form.

4: Someone from the admission team then contacts the candidates with further process.

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

$\begin{array}{l}Wearegiventhaty=-x^{2}or{x}^2=-y\\ andthelinex+y+2=0\\ Solvingthetwoequations,weget\\ \therefore x-x^2+2=0\\ \Rightarrow x^2-x-2=0\\ \Rightarrow x^2-2x+x-2=0\\ \Rightarrow x\left(x-2\right)+1\left(x-2\right)=0\\ \Rightarrow \left(x-2\right)\left(x+1\right)=0\therefore x=-1,x=2\\ Areaoftherequiredshadedregion=\left|{\displaystyle \underset{-1}{\overset{2}{\int }}\left(-x-2\right)dx-}{\displaystyle \underset{-1}{\overset{2}{\int }}-x^{2}dx}\right|\\ =\left|-{\left[\frac{x^2}{2}+2x\right]}_{-1}^2+\frac{1}{3}{\left[x^3\right]}_{-1}^2\right|\\ =\left|-\left[\left(\frac{4}{2}+4\right)-\left(\frac{1}{2}-2\right)\right]+\frac{1}{3}\left[8+1\right]\right|\\ =\left|-\left(6+\frac{3}{2}\right)+\frac{1}{3}\left(9\right)\right|=\left|-\frac{15}{2}+3\right|\\ =\left|\frac{-15+6}{2}\right|=\left|\frac{-9}{2}\right|=\frac{9}{2}sq.units\\ Hence,therequiredarea=\frac{9}{2}sq.units.\end{array}$

To apply for any of the courses offered at University College of Engineering, students may either visit the college's campus or the official website to start the process. Before applying, students are advised to check the admission criteria. To submit the form, students need to duly provide basic details, such as personal information, educational qualifications, etc. Once the form is filled, students may submit it to complete the process. It must be noted that this is a general application process and may contain other steps as per the college's guidelines.

Candidates do not require passing an entrance exam for admission. Aspirants are shortlited for admission based on their merit in the last qualifying exam. The scores of the aspirants in Class 12 is considered for UG and Class 10 for certificate courses. Once selected, candidates are required to confirm their seat by paying the course fee. 

The steps to apply for courses at Maharashtra National Law University, Aurangabad are as follows: 

Step 1: Register for an account on the university's official website.

Step 2: Provide the required information on the form. 

Step 3: Upload the necessary documents. 

Step 4: Preview the form prior to the final submission. 

Step 5: Pay the application fee in step five. 

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

$\begin{array}{l}Wearegiventheequationsofcurvey=\sqrt[]{x}andliney=x.\\ Solvingy=\sqrt[]{x}\Rightarrow y^2=xandy=x,weget\\ x^2=x\Rightarrow x^2-x=0\\ \Rightarrow x\left(x-1\right)=0\therefore x=0,1\\ Areaoftherequiredshadedregion={\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{x}dx}-{\displaystyle \underset{0}{\overset{1}{\int }}xdx}\\ =\frac{2}{3}{\left[x^{3/2}\right]}_0^1-\frac{1}{2}{\left[x^2\right]}_0^1\\ =\frac{2}{3}\left[{\left(1\right)}^{3/2}-0\right]-\frac{1}{2}\left[{\left(1\right)}^2-0\right]\\ =\frac{2}{3}-\frac{1}{2}=\frac{4-3}{6}=\frac{1}{6}sq.units\\ Hence,therequiredarea=\frac{1}{6}sq.units.\end{array}$

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

$\begin{array}{l}Here,wehavey=\sqrt[]{a^2-x^2}\\ \Rightarrow y^2=a^2-x^2\\ \Rightarrow x^2+y^2=a^2\\ Areaoftheshadedregion\\ =2\left[{\left(1\right)}^{3/2}-0\right]-\frac{3}{2}\left[{\left(1\right)}^2-0\right]\\ ={\left[\frac{x}{a}\sqrt[]{a^2-x^2}+\frac{a^2}{2}si{n}^{-1}\frac{x}{a}\right]}_0^a\\ =\left[\frac{a}{a}\sqrt[]{a^2-a^2}+\frac{a^2}{2}si{n}^{-1}\frac{a}{a}-0-0\right]\\ =\frac{a^2}{2}si{n}^{-1}\left(1\right)=\frac{a^2}{2}.\frac{\pi }{2}=\frac{\pi a^2}{4}sq.units\\ Hence,therequiredarea=\frac{\pi a^2}{4}sq.units.\end{array}$

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

$\begin{array}{l}Here,wehavey=\sqrt[]{x-1}\\ \Rightarrow y^2=x-1\left(Parabola\right)\\ Areaoftherequiredregion={\displaystyle \underset{1}{\overset{5}{\int }}\sqrt[]{x-1}dx}\\ =\frac{2}{3}{\left[{\left(x-1\right)}^{3/2}\right]}_1^5\\ =\frac{2}{3}\left[{\left(5-1\right)}^{3/2}-0\right]=\frac{2}{3}\times {\left(4\right)}^{3/2}\\ =\frac{2}{3}\times 8=\frac{16}{3}sq.units\\ Hence,therequiredarea=\frac{16}{3}sq.units.\end{array}$

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

$\begin{array}{l}Giventhat2y=5x+7,x-axis,x=2andx=8\\ Letusdrawthegraphof2y=5x+7\Rightarrow y=\frac{5x+7}{2}\\ \begin{array}{cc}\hfill x\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill y\hfill \\ \hfill 6\hfill & \hfill 1\hfill \end{array}\\ Areaoftherequiredshadedregion={\displaystyle \underset{2}{\overset{8}{\int }}\left(\frac{5x+7}{2}\right)dx}\\ =\frac{1}{2}{\left[\frac{5x^2}{2}+7x\right]}_2^8\\ =\frac{1}{2}\left[\frac{5}{2}\left(64-4\right)+7\left(8-2\right)\right]\\ =\frac{1}{2}\left[\frac{5}{2}\times 60+7\times 6\right]=\frac{1}{2}\left[150+42\right]=\frac{1}{2}\times 192=96sq.units\\ Hence,therequiredarea=96sq.units.\end{array}$