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NM College of Commerce & Economics, Mumbai offers the flagship BCom course to students who wish to pursue a career in the Commercial field. Students can choose from among the following NM College BCom specializations:

1. BCom (Regular)
2. B.Com. in Accounts and Finance
3. B.Com. in Financial Markets
4. B.Com in Economics and Analytics
5. Bachelor of Commerce (B.Com. Hons.)
6. B.Com in Economics

Symbiosis College of Arts & Commerce, Pune provides the Bachelor of Commerce (BCom) programme to students who wish to pursue a career in Commerce. Students can choose from 3 distinct specializations for Symbiosis College BCom program:

1. BCom (Regular)
2. B.Com. in Accounting and Finance (Integrated with ACCA)
3. B.Com. in Management Accounting (Integrated with CMA)

As soon as the Manipur HSLC routine is announced, the students must check as it gives them a clear idea about how much time is left in exam. It also help the students plan their preparation. While making the end time preparation plan, ansure that the entire syllabus is covered and revised. Students should also keep the time to practice the official Manipur HSLC sample papers and previous year's question papers.

K J Somaiya College of Arts and Commerce provides the Master of Arts (MA) programme in various specializations. The course is sought-after by students who wish to pusue further studies in Humanities. Below are the various K J Somaiya College of Arts and Commerce MA specializations:

1. MA in Psychology
2. MA in English
3. MA in Sociology
4. MA in History
5. MA in Philosophy
6. MA in Marathi

Amity University Online, Noida BCA programme is offered to students who wish to make a solid career in Computer Application. Students can choose from the following diverse specializations in AUO Noida BCA:

1. BCA (Regular)
2. BCA in Data Analytics
3. BCA in Cloud and Security
4. BCA in Software Engineering
5. BCA in Data Engineering

IDOL Mumbai MSc programme is a popular selection for students who wish to pursue PG programs in the Scientific realm. The MSc course is offered in 3 distinct specializations:

1. MSc Computer Science
2. MSc Information Technology
3. MSc Mathematics

Students can choose their relevant IDOL Mumbai MSc specialisation according to the specific field they are seeking to be enrolled in.

Vignan's Foundation for Science, Technology & Research, also known as VFSTR, is a private college and is located in Hyderabad, Telangana. Vignan's Foundation for Science, Technology & Research courses are AICTE-approved and are offered to students in full-time mode. Further, the college was established in 1977 and has been accredited by NAAC with an A+ grade. 

  $\left(1,1\right)\left(1,4\right)\left(4,1\right)\left(2,4\right)\left(4,2\right)\left(3,4\right)\left(4,3\right)\left(4,4\right)$  all have only one image.

(2, 1) (1, 2), (2, 2) each element has 3 choice.

(3, 2) (2, 3) (3, 1) (1, 3) (3, 3) each element has two choices.

$\therefore$ total function = 3 × 3 × 2 × 2 × 2 = 72

Case I

None of the pre image have 3 as image, total functions = 2 × 2 × 1 × 1 × 1 = 4

Case II

None of the pre images have 2 as image then number of function = 2⁵ = 32

Case III

None of the pre image have either 3 or 2 as image

Total function = 1⁵ = 1

$\therefore$ Total number of onto function

= 72 – 4 – 32 + 1 = 37

$\tilde{z}=i{z}^2+z^2-z$

$z+\overline{Z}=z^2\left(i+1\right)$

$z+\overline{Z}=z^2\left(i+1\right)$ Let z be equal to (x + iy)

(x + iy) + (x – iy) = (x + iy)² (i + 1)

$2x=\left(x^2-y^2+2ixy\right)\left(i+1\right)$               

Equating the real & in eg part.

$\left(x^2-y^2+2ixy\right)=0.........\left(i\right)$

$\left(x^2-y^2-2xy\right)=\left(2x\right)..............\left(ii\right)$               

(i) & (ii)

 4xy = -2x Þ x = 0 or y = $\left(\frac{-1}{2}\right)$  

(for x = 0, y = 0)

For y = $\frac{-1}{2}$  

x²   $-\frac{1}{4}+2\left(\frac{-1}{2}\right)x=0$

x =   $\frac{4\pm \sqrt[]{16+16}}{2.4}$

=  $\left(\frac{1+\sqrt[]{2}}{2}\right)or\left(\frac{1-\sqrt[]{2}}{2}\right)$  

$\therefore sum$ of   ${\left|z\right|}^2={\left(\frac{1+\sqrt[]{2}}{2}\right)}^2+\frac{1}{4}+{\left(\frac{1-\sqrt[]{2}}{2}\right)}^2+\frac{1}{4}+0^2+O^2$

=   $\frac{3}{4}+\frac{\sqrt[]{2}}{2}+\frac{1}{4}+\frac{3}{4}-\frac{\sqrt[]{2}}{2}+\frac{1}{4}=\frac{3}{2}+\frac{1}{2}=2$

Jai Hind College offers the flagship Bachelor of Science (BSc) programme to students who wish to pursue a career in the Scientific field. The Jai Hind College BSc course is offered in 3 distinct specializations of BSc (Regular), BSc (Biotechnology), and BSc (Information Technology). Students can choose from their relevant specialization. For instance, students interested in the Medical-related fields can choose to pursue BSc Biotechnology.

$\left(\begin{array}{cc}\hfill 1+i\hfill & \hfill 1\hfill \\ \hfill -i\hfill & \hfill 0\hfill \end{array}\right)$

$A^2=\left(\begin{array}{cc}\hfill 1+i\hfill & \hfill 1\hfill \\ \hfill -i\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{cc}\hfill 1+i\hfill & \hfill 1\hfill \\ \hfill -i\hfill & \hfill 0\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill i\hfill & \hfill 1+i\hfill \\ \hfill 1-i\hfill & \hfill -i\hfill \end{array}\right)$

$A^4=\left(\begin{array}{cc}\hfill i\hfill & \hfill 1+i\hfill \\ \hfill 1-i\hfill & \hfill -i\hfill \end{array}\right)\left(\begin{array}{cc}\hfill i\hfill & \hfill 1+i\hfill \\ \hfill 1-i\hfill & \hfill -i\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)$

$\therefore A^5=A$

A⁹ = A

$\therefore forn=1,5,9,.....,97$

total possible values of n = 25

BTech programme is a popular course of University of Mumbai. It is in demand among UG and PG students. It offers solid Engineering foundation to students. There are many specialisations to choose from like CSE, Civil Engineering, Mechanical Eng, Electronics Eng, Textile Technology, IT etc.

Yes, the TNPSC Group 4 Answer Key PDF is available here for downloading purposes. The Tamil Nadu Public Service Commission will release the TNPSC Group 4 answer key 2026 by December 26, 2026. The Commission released two answer keys, one each for the two papers. The Commission released the TNPSC Group 4 answer key separately for both the papers.

$\begin{array}{cc}\hfill 2x-3y=\gamma +5\hfill & \hfill \alpha x+5y=\left(\beta +1\right)\hfill \end{array}\}infinitelymanysolution$

$\therefore \frac{2}{\alpha }=\frac{-3}{5}=\left(\frac{\gamma +5}{\beta +1}\right)$

(i) $\frac{2}{\alpha }=\frac{-3}{5}=\frac{\gamma +5}{\beta +1}$

$\alpha =\frac{5\times 2}{-3}$ 5x + 25 = -3β - 3

^{$5\gamma +3\beta =-28$}

^{$\left|9\alpha +5\gamma +3\beta \right|=\left|9\times \frac{-10}{3}-28\right|$}

= ^{$\left|-30-28\right|=58$}

BTech from Woxsen University without JEE Main is possible. The other accepted entrance exams are VITEEE, SAT India, MHT CET, TS EAMCET, Andhra Pradesh EAMCET, CUET, and others. Candidates willing to get admission must visit the official website. Then, click on the Apply Now tab. Then, select the required course and pay the application fee, and fill out and submit the application form with required information. 

$S_n=\frac{n^2}{1-\frac{1}{{\left(n+1\right)}^2}}=\frac{n{\left(n+1\right)}^2}{\left(n+2\right)}={\left(n+1\right)}^2-2\left(n+1\right)+2-\frac{2}{\left(n+2\right)}$

$\frac{1}{26}+{\displaystyle \sum _{n=1}^{50}\left(S_n+\frac{2}{\left(n+1\right)}-\left(n+1\right)\right)=\frac{1}{26}+{\displaystyle \sum _{n=1}^{50}{\left(n+1\right)}^2}-3\left(n+1\right)+2+2\left(\frac{1}{\left(n+1\right)}-\frac{1}{\left(n+2\right)}\right)}$

$=\frac{1}{26}+45525-3\times 1325+2\times 50+2\left(\frac{1}{2}-\frac{1}{52}\right)$

$=\frac{1}{26}+41550+100+1-\frac{1}{26}=41651$