Ask & Answer: India's Largest Education Community

Candidates must know their K J Somaiya College of Arts and Commerce eligibility criteria and apply accordingly. Follow the K J Somaiya College of Arts and Commerce application process mentioned below:

Step 1: Visit the official website of the KJSAC.

Step 2: Select the required course.

Step 3: Pay the application fee.

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

Average salary offer for BTech is mostly around 3 to 4 LPA, same like other courses. Some IT students may get higher package depends on skills and company. So learning extra tech and projects really help boost pay.

KJSAC was established in 1959. K J Somaiya College of Arts and Commerce is affiliated with the University of Mumbai.  KJ Somaiya College of Arts and Commerce is accredited with A Grade by NAAC (National Assessment and Accreditation Council). About 6,500 students of both the Senior and Junior Colleges and about 150 PG and Research students take part in the academic and other activities of the college each and every year.

Placement can’t be 100% guarantee but yes, university do their best for connecting students with industries. Final selection depends on student skills, interviews and what company wants. Training and internships also helps lot.

$\left(ta{n}^{-1}y\right)-x)dy=\left(1+y^2\right)dx$

$\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{ta{n}^{-1}y}{1+y^2}$

$l.F=e^{{\displaystyle \int \frac{1}{1+y^2}dy}}=e^{ta{n}^{-1}y}$

$x.e^{ta{n}^{-1}y}{\displaystyle \int \frac{e^{ta{n}^{-1}y}ta{n}^{-1}y}{1+y^2}dy}$

Let

$e^{ta{n}^{-1}y}=t$

$=x{e}^{ta{n}^{-1}y}=e^{ta{n}^{-1}y}y-e^{ta{n}^{-1}y}+c...(i)$

$?$ It passes through (1, 0) = c = 2

Now put y = tan 1, then

ex = e – e + 2

$\Rightarrow x=\frac{2}{6}$

Candidates must know their K J Somaiya College of Arts and Commerce eligibility criteria and apply accordingly. Follow the K J Somaiya College of Arts and Commerce application process mentioned below:

Step 1: Visit the official website of the KJSAC.

Step 2: Select the required course.

Step 3: Pay the application fee.

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

There is quite many companies that visits campus such as Wonder Cement, TCS, Crest Infosystems, Macleods Pharmaceutical, Adani Port, NPCIL, DB Engineers, Logicwind and more. IT streams got more software companies too this year. For other specialisations too, the university is bringing more companies.

${\displaystyle {\int }_0^1\frac{1}{7^{\left(\frac{1}{x}\right)}}dx,let\frac{1}{x}=t}$

$\frac{-1}{x^2}dx=dt$

$={\displaystyle {\int }_{\infty }^1\frac{1}{-t^2{7}^{\left|t\right|}}dt={\displaystyle {\int }_1^{\infty }\frac{1}{t^2{7}^{\left|t\right|}}dt}}$

$=\frac{1}{7}{\left[-\frac{1}{t}\right]}_1^2+\frac{1}{7^2}{\left[\frac{-1}{t}\right]}_2^3+\frac{1}{7^3}{\left[-\frac{1}{t}\right]}_2^3+...$

$={\displaystyle {\sum }_{n=1}^{\infty }\frac{1}{7_n}\left(\frac{1}{n}-\frac{1}{n+1}\right)}$

$=1+6log\frac{6}{7}$

TAPMI Manipal is liked by majority of students. The Institute is considered in the highest regards for placement facilities, faculty, curriculum, infrastructure, and various other opportunities offered. The Institute is accredited and recognised by various government regulated bodies. Furthermore, during TAPMI Placements 2024, the the highest package offered to the MBA – General Programme students stood at INR 32.02 LPA. Further, a total of 90 companies visited the campus during TA Pai Management Institute MBA General Program placements 2025. Overall, it is worth pursuing academia from TAPMI Manipal. 

K J Somaiya College of Arts and Commerce is a big name and admissions to this college is tough due to the high competition. To apply for various courses offered at K J Somaiya College of Arts and Commerce, students must first visit the official website. Then, click on the Apply Now tab. Then, select the required course. Then, pay the application fee. Then, fill out and submit the application form with required information. 

${\displaystyle {\int }_{cosx}^1{t}^{2}f\left(t\right)dt=si{n}^{3}x+cosx}$

$\Rightarrow sinxco{s}^{2}xf\left(cosx\right)=3si{n}^{2}xcosx-sinx$

$\Rightarrow f\text{'}\left(cosx\right)\left(-sinx\right)=3se{c}^{2}x-2se{c}^{2}tanx$

$cosx=\frac{1}{\sqrt[]{3}}$

$\therefore f\text{'}\left(\frac{1}{\sqrt[]{3}}\right)\left(-\frac{\sqrt[]{2}}{\sqrt[]{3}}\right)=9-6\sqrt[]{2}$

$\frac{1}{\text{exception 25:}}f\text{'}\left(\frac{1}{\text{exception 25:}}\right)=6-$$\frac{9}{}$

$f\left(x\right)={\displaystyle {\int }_0^{x^2}\frac{t^2-5t+4}{2+e^t}dt}$

$f\text{'}\left(x\right)=2x\left(\frac{x^4-5x^2+4}{2+e^{x^2}}\right)=0$

$x=0,or\left(x^2-4\right)\left(x^2-1\right)=0$

$x=0,x=\pm 2,\pm 1$

Now,

$f\text{'}\left(x\right)=\frac{2x\left(x+1\right)\left(x-1\right)\left(x+2\right)\left(x-2\right)}{e^{x^2}+2}$

Changes sign from positive to negative at x = 1, 1 So, number of local maximum points = 2

Changes sign from negative to positive at

x = 2, 0, 2 So, number of local minimum points = 3

$\therefore m=2,n=3$

MDU Rohtak offers variety of specialisations for the students enrolled in their BTech programme. The specialisations include:

• CSE
• Mechanical
• ECE 
• Civil  
• AI 
• AI and ML
• Biotechnology.

This variety allows students to choose a specialisation according to their interests and career plans.

No, N.L. Dalmia does not offers direct admission. The selection is strictly based on merit of the aspirants. The institute considers scores of the aspirants in national-level entrance exams such as CAT, XAT, CMAT, GMAT, etc. for admission. Once selected, candidates have to further pass the personal interview round for final selection. The institute also considers academic performance, Work Experience and Extra-curricular of the aspirants for admission in the PGDM programme.

Candidates need to have a score in CAT, MAT, XAT, CMAT, etc. for admission to PGDM programme. Management entrance exams are competitive and need strong preparation to pass the test. Once the aspirants are selected based on their entrance exam scores, they need to pass the GDPI round. Hence, right preparation and strategy is required to get a seat in the institute.

Yes, you can get into Chennai Institute of Technology after Class 12, given that you satisfy the admission criteria. The institute offers BTech at the undergraduate level, admission to which is entrance-based. The minimum eligibility requires students to complete Class 12 with 45%. However, as admissions are done through TNEA, students need to participate in TNEA counselling for seat allotment. For admission to the PG courses, Class 12 is not enough.

Yes, GATE scores are accepted by Chennai Institute of Technology to admission into its M.Tech programs. CEETA PG scores are also another option through which one can apply. The single window counselling process requires all the applicants to go through the process. The admission is only based on suiting the qualification and availing of seats.

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

$\begin{array}{l}Given\text{\hspace{0.17em}points}areA\left(2,3,4\right),B\left(-1,2,-3\right)andC\left(-4,1,-10\right)\\ AB=\sqrt[]{{\left(2+1\right)}^2+{\left(3-2\right)}^2+{\left(4+3\right)}^2}=\sqrt[]{9+1+49}=\sqrt[]{59}\\ BC=\sqrt[]{{\left(-1+4\right)}^2+{\left(2-1\right)}^2+{\left(-3+10\right)}^2}=\sqrt[]{9+1+49}=\sqrt[]{59}\\ AC=\sqrt[]{{\left(2+4\right)}^2+{\left(3-1\right)}^2+{\left(4+10\right)}^2}=\sqrt[]{36+4+196}=\sqrt[]{236}=2\sqrt[]{59}\\ \therefore AB+BC=AC\\ \sqrt[]{59}+\sqrt[]{59}=2\sqrt[]{59}\\ Hence,A,BandCarecollinearandAC:BC=2\sqrt[]{59}:\sqrt[]{59}=2:1\\ Hence,CdividesABis2:1externally.\end{array}$

The final Chennai Institute of Technology 2025 admission date will be announced by DOTE Chennai. The applicants are required to apply TNEA or CEETA PG within the timeline set by the institutions. Applications forms will be out in either May or June of 2025.