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Yes, Christ University Pune admissions for the academic year 2025-2026 are open and currently ongoing for the various programmes, including BSc, BCom, MBA, and others. Students aiming to take admission must visit the official website and apply for the preferred programme before the last date to apply, which is 13 Aug, 2025.

$\underset{n\to \infty }{lim}{\left(1+\frac{1+\frac{1}{2}+.......+\frac{1}{n}}{n^2}\right)}^n$ limit is in the form of $1^{\infty }$  

$\mathcal{l}=exp\left(\underset{n\to \infty }{lim}\frac{1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{n}}{n^2}\right)$             

$0\le 1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{n}\le 1+\frac{1}{\sqrt[]{2}}+\frac{1}{\sqrt[]{3}}+....+\frac{1}{\sqrt[]{n}}\le 2\sqrt[]{n}-1$        

Taking limit   $\left(n\to \infty \right)$

$\mathcal{l}$ = exp (0) (from sandwich)

  $\mathcal{l}=1$          

Second Method :

$1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{n}\ge ln\left(n+1\right).......\left(i\right)$

$1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{n}\le 1+{\displaystyle {\int }_1^2\frac{1}{2}dx\le 1+lnn..........\left(ii\right)}$

From (i) & (ii)

$ln\left(n+1\right)\le 1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{n}\le 1+Inn,\forall n\in N,n\ge 2$           

As $\underset{n\to \infty }{lim}\frac{ln\left(n+1\right)}{n}=0$  

and $\underset{n\to \infty }{lim}\frac{1+ln\left(n\right)}{n}=0$  

$\therefore$ from sandwich theorem

$\underset{n\to \infty }{lim}\frac{1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{n}}{n}=0$  

$\therefore e^0=1$   

The deadline to fill out the Christ University Pune BCom application form is out on the university's official website. At present, the deadline to fill out the application form for the preferred BCom courses was Aug 13, 2025. Students seeking admission must go to the official website of the university to fill out the application form.

The Union Public Service Commission is expected to release the UPSC CSE (IAS) main exam admit card 2025 soon on its official website - upsc.gov.in. The candidates will be required to download their UPSC admit card from the official website using their UPSC One-Time Registration (OTR) number and password.

The UPSC CSE (IAS) Main exam 2025 will be conducted from August 22 to 31.

The total tuition fees to pursue a BCom course at Christ University Pune is INR 7.8 lakh. Other than this, the fee structure includes other charges such as admission fees, hostel fees, etc. Students who wish to know the entire fee structure can visit the official website of the university.

Note: The above-mentioned fee is as per the official sources. However, it is indicative and subject to change.

2x + y = 1

$m_1=\frac{-2}{1}=-2and{m}_1{m}_2=-1$          

$-2.m_2=-1$           

$m_2=\frac{1}{2}$           

y² = 6x

y² = 4 (3/2) x

$y=mx+\frac{a}{m}$         

$y=\frac{1}{2}x+\frac{\frac{3}{2}}{\frac{1}{2}}$        

$y=\frac{x}{2}+3$           

Candidates who have missed the admissions must not worry. These candidates can apply again the next year. Follow the NIPS Hotel Management Ranchi application process mentioned below:

Step 1: Visit the official website of NIPS Hotel Management Ranchi.

Step 2: Select the required course.

Step 3: Pay the application fee.

Step 4: Fill out and submit the application form with required information.

Yes, candidates can get admission in the BSc course at SIT Siliguri with 60% in Class 12. The institute has not specified the minimum required aggregate for the same, but, since 60% is a good score, candidates are eligible for admission. The institute accepts Class 12 merit for admissions in the BSc course. Candidates must fill out the application forms available online on the official website before the last date. 

Yes, NIPS Hotel Management Ranchi awards various scholarships to students as financial assistance. NIPS Hotel Management students can avail of various Government Scholarship Schemes to support their studies. Some of the scholarships available at NIPS Hotel Management Ranchi are mentioned below:

$A\to \left(B\to A\right)$

$\sim A\vee \left(B\to A\right)$

$\sim A\vee \left(\sim B\vee A\right)$

$\left(\sim A\vee B\right)\vee \left(\sim A\vee A\right)=tA\to \left(A\cup B\right)$

Established in 1993, the NIPS Hotel Management Institute offers undergraduate and Diploma courses to students. These courses are provided across the Management and various other streams. There are five short-term programmes available at the college. The college is ranked among the Top 90 Hotel Management institutions in the World. The college has an approved training partner, which is NSDC. The college placements are also good. Hence, NIPS Hotel Management, Ranchi, is a good college. 

SCIT Pune offers MBA at a PG level in full-time mode on the basis of merit and entrance. Candidates must secure a minimum aggregate of 50% in graduation and valid scores in SNAP or MAH CET. Post entrance qualification, candidates must appear for group exercise and personal interview. After completing all rounds, the shortlisted candidates will be eligible to secure the final admission. 

  $\frac{x^2}{a}+\frac{y^2}{b}=1,\frac{x^2}{c}-\frac{y^2}{\left(-d\right)}=1$

Ellipse and Hyperbola are orthogonal so these will be confocal.

$\sqrt[]{a-b}=\sqrt[]{c+\left(-d\right)}$            

a – b = c – d

Yes, Global Institute of Management and Technology accept Class 12th marks for admission. Candidates seeking admission to the UG and Diploma programmes can enrol for admission with Class 12 marks. Candidates must complete Class 12 to enrol for various courses. The college offers more than 10 certificate courses. Global Institute of Management and Technology admissions are based on entrance-exam scores.

$\overrightarrow{O}P=\left(2,-1,1\right)$

Normal vector to the plane

$=\overrightarrow{A}B\times A\overrightarrow{C}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=\left(3,-1,1\right)=\overrightarrow{n}$

Projection of $\overrightarrow{O}Pon\overrightarrow{n}is\left|\frac{\overrightarrow{O}P.\overrightarrow{n}}{\left|\overrightarrow{n}\right|}\right|$

PN = $\frac{6+1+1}{\sqrt[]{11}}=\frac{8}{\text{exception 25:}}$

$\frac{\text{exception 25:}}{}$Projection of OP on plane = ON =  $\sqrt[]{O{P}^2-P{N}^2}=\sqrt[]{6-\frac{64}{11}}=\sqrt[]{\frac{2}{11}}$

No, Maths is not compulsory for taking admission into Christ Pune BCom courses. As per the university's admission criteria, students who have passed Class 12 examinations from a recognised board of India with a minimum of 60% aggregate in the commerce stream are eligible to apply. Thus, students who did not have Maths as a subject in Class 12 can also apply for the course.

$l={\displaystyle \int \frac{8x^3+20x^2}{x^4+5x^3-7x^2}}dx=4{\displaystyle \int \frac{2x+5}{x^2+5x-7}dx=4ln\left|x^2+5x-7\right|+c}$