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This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Giventhat:y=cosx,x=0,x=\pi \\ Requiredarea={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}cosx}dx+\left|{\displaystyle \underset{\frac{\pi }{2}}{\overset{\pi }{\int }}cosx}dx\right|dx\\ ={\left[sinx\right]}_0^{\frac{\pi }{2}}+\left|{\left[sinx\right]}_{\frac{\pi }{2}}^{\pi }\right|\\ =\left[sin\frac{\pi }{2}-sin0\right]+\left|\left[sin\pi -sin\frac{\pi }{2}\right]\right|\\ =\left(1-0\right)+\left|0-1\right|=1+1=2sq.units\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

For the General AI category students, the IIT Kanpur JAM cutoff 2025 was 50 for M.Sc. in Statistics admission. The cutoff ranks vary by rounds and categories. To view the category-wise last round cutoff ranks, check out the table below: 

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Givenequationofcircleis{x}^2+y^2=32\Rightarrow x^2+y^2={\left(4\sqrt[]{2}\right)}^2\\ andthelineisy=xandthex-axis.\\ Solvingthetwoequationswehave\\ x^2+x^2=32\Rightarrow 2x^2=32\\ \Rightarrow x^2=16\therefore x=\pm 4\\ Requiredarea={\displaystyle \underset{0}{\overset{4}{\int }}x}dx+{\displaystyle \underset{4}{\overset{4\sqrt[]{2}}{\int }}\sqrt[]{{\left(4\sqrt[]{2}\right)}^2-x^2}}dx\\ =\frac{1}{2}{\left[x^2\right]}_0^4+{\left[\frac{x}{2}\sqrt[]{{\left(4\sqrt[]{2}\right)}^2-x^2}+\frac{32}{2}si{n}^{-1}\frac{x}{4\sqrt[]{2}}\right]}_4^{4\sqrt[]{2}}\\ =\frac{1}{2}\left[16-0\right]+\left[0+16si{n}^{-1}\left(\frac{4\sqrt[]{2}}{4\sqrt[]{2}}\right)-2\sqrt[]{32-16}-16si{n}^{-1}\frac{x}{4\sqrt[]{2}}\right]\\ =8+\left[16si{n}^{-1}\left(1\right)-8-16si{n}^{-1}\frac{1}{\sqrt[]{2}}\right]\\ =8+16.\frac{\pi }{2}-8-16.\frac{\pi }{4}=8\pi -4\pi =4\pi sq.units\\ Hence,thecorrectoptionis\left(b\right).\end{array}$

Documents typically required include Class 10 and 12 mark sheets, entrance exam scorecard, government-issued ID proof, passport-size photographs, and any category/reservation certificates if applicable. All documents must be uploaded during the NMIMS Mukesh Patel School of Technology BTech application and later verified during admission.

The final step in the NMIMS Mukesh Patel School of Technology Shirpur BTech admission process is fee payment. Selected candidates must pay the tuition fee, which ranges between INR 6.4 Lacs and INR 13.8 lakh, to confirm their admission and secure their seat in the chosen specialisation.

Selection for the BTech course at NMIMS Mukesh Patel School of Technology Shirpur is based on merit in entrance exams such as JEE Main, MHT CET, NMIMS CET, or NMIMS-NPAT. Only candidates meeting the eligibility criteria and securing a valid score in any of these exams are considered for admission.

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Here,equationofcurveisy=\sqrt[]{16-x^2}\\ Requiredarea=2\left[{\displaystyle \underset{0}{\overset{4}{\int }}\sqrt[]{16-x^2}}dx\right]\\ =2{\left[\frac{x}{2}\sqrt[]{16-x^2}+\frac{16}{2}si{n}^{-1}\frac{x}{4}\right]}_0^4\\ =2\left[\left(0+8si{n}^{-1}\frac{4}{4}\right)-\left(0+0\right)\right]\\ =2\left[8si{n}^{-1}\left(1\right)\right]=16.\frac{\pi }{2}=8\pi sq.units\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

The authority will release the IBPS PO 2026 notification at ibps.in. The application form online link will also be activated along with the notification. The IBPS PO 2026 application fee payment dates will be provided in the notification PDF. 

Rungta International Skills University admissions are currently open for all courses. Candidates are selected for courses based on entrance exam scores as per course requirements. Aspirants can apply online on the official website of the university. The applications are open for various courses. Interested candidates meeting the eligibility criteria can fill out the form.

Rungta International Skills University offers various courses such as BTech, BCA, MBA, etc. at the UG and PG level. The selection criteria are based on entrance. The application process at Rungta International Skills University conducted online. To secure a seat, students can check the following steps presented below:

1. Visit the official website and complete the online application process. 
2. Ensure eligibility and selection criteria for all programme.
3. The selection criteria is based on entrance-exam.
4. Upon selected, pay the specified fee to confirm and secure your seat.

The Indian Institute of Science Education and Research Thiruvananthapuram runs its MSc course in four specialisations. The specialisations have a total of 20 students which results in the degree being small and competitive.

There is no fee associated with the CAS letter or the letter for the "Confirmation of Acceptance for Studies". The fees that is often linked with the CAS letter is the tuition fees deposit that the universities abroad ask international students to pay alongside accepting the offer awarded to them.  

Thiruvananthapuram MSc programme at IISER is interdisciplinary and the syllabus is flexible, with electives being introduced at the second semester and onwards. The student is able to select courses in a variety of schools and is able to tailor the time spent on the MSc project to their field of research interests. The course work consists of 1-week modular programs and learning activities with BS -MS and Integrated PhD students.

Both Administrative Management College and T. John College are popular institutes in the Bangalore region for the MBA course. Given below are some comparison points between the two:

1. Rankings: Outlook 2024 has ranked Administrative Management College at the 15th spot for the "Hotel/ Hospitality Management" category.
2. Tuition Fees: The total tuition fees at Administrative Management College is INR 5.14 L, while at T. John Group of Institutions, it is INR 4.92 L - 6.08 L.
3. Placements: The median salary at Administrative Management College is INR 5.5 LPA. On the other hand, the highest salary recorded at T. John College is INR 10.60 LPA.

The IBPO PO exam application process is expected to start in June 2026. The last date to submit the application form will be mentioned in the notification PDF once released. The exam is conducted every year. Candidates will be able to fill the online form at ibps.in.

The candidate is required to have a three to four year degree in sciences, engineering, mathematics or equivalent to qualify as the applicant to join the MSc programme at the Indian Institute of Science Education and Research Thiruvananthapuram. They must have an aggregate or CGPA score of 60 percent or CGPA 6.5 on a scale of 10 or SC/ST/PwD must have 50 percent or CGPA 5.5 schooling.

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Giventhat:Theequationofparabolais{x}^2=4y\dots \left(i\right)\\ andequationofstraightlinex=4y-2\dots \left(ii\right)\\ Solvingeqn\left(i\right)and\left(ii\right)weget\\ y=\frac{x^2}{4}\Rightarrow x=4\left(\frac{x^2}{4}\right)-2\\ \Rightarrow x=x^2-2\Rightarrow x^2-x-2=0\\ \Rightarrow x^2-2x+x-2=0\Rightarrow x\left(x-2\right)+1\left(x-2\right)=0\\ \Rightarrow \left(x-2\right)\left(x+1\right)=0\therefore x=-1,x=2\\ Requiredarea={\displaystyle \underset{-1}{\overset{2}{\int }}\frac{x+2}{4}}dx-{\displaystyle \underset{-1}{\overset{2}{\int }}\frac{x^2}{4}}dx\\ =\frac{1}{4}{\left[\frac{x^2}{2}+2x\right]}_{-1}^2-\frac{1}{4}.\frac{1}{3}{\left[x^3\right]}_{-1}^2\\ =\frac{1}{4}\left[\left(\frac{4}{2}+2\right)-\left(\frac{1}{2}-2\right)\right]-\frac{1}{12}\left[8+1\right]\\ =\frac{1}{4}\left[6+\frac{3}{2}\right]-\frac{1}{12}\left[9\right]=\frac{1}{4}\times \frac{15}{2}-\frac{3}{4}=\frac{15}{8}-\frac{3}{4}=\frac{9}{8}sq.units\\ Hence,thecorrectoptionis\left(d\right).\end{array}$