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Both are prominent B-Schools in Bangalore. We have compared ABBS and Dayananda Sagar Business School below:

1. Ranking: Business Today ranked ABBS 107th in 2022. On the other hand, DSBS Bangalore was ranked #45 by the Times Survey 2024.
2. Tuition Fees: ABBS Bangalore MBA tuition fee ranges between INR 9.5 Lakhs and 11.5 Lakhs, and at DSBS, it is INR 9.5 Lakhs.
3. Median Salary: The median salary at ABBS is INR 6.12 LPA, while at Dayananda Sagar Business School, it is INR 6.8 LPA.

No, students cannot request changes in the official SEBA HSLC routine. The schedule is decided after careful planning by SEBA and is the same for all students. However, if there are any clashes due to technical issues or other emergencies, SEBA may issue revised dates. Such updates are published on the official website. Students should stay informed and prepare accordingly.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{dx}{1+cosx}}\\ ={\displaystyle \int \frac{dx}{2co{s}^{2}x/2}\left[?1+cosx=2co{s}^{2}x/2\right]}\\ =\frac{1}{2}{\displaystyle \int se{c}^2\frac{x}{2}dx}=\frac{1}{2}.2tan\frac{x}{2}+C=tan\frac{x}{2}+C\\ Hence,therequiredsolutionistan\frac{x}{2}+C.\end{array}$

NEHU Shillong offers a lively and academia centric environment for the students. the University conducts various activities, and events to provide industry-oriented exposure for students within the walls of the University. Furthermore, the students receive various opportunities through the extracurricular activities conducted by the University proactively. Apart from this, NEHU Shillong organises events and fests to celebrate the culture and social diversity present at the university.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{1+cosx}{x+sinx}}dx\\ Putx+sinx=t\Rightarrow \left(1+cosx\right)dx=dt\\ \therefore I={\displaystyle \int \frac{dt}{t}}=log\left|t\right|=log\left|x+sinx\right|+C\\ Hence,therequiredsolutionislog\left|x+sinx\right|+C.\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{e^{6logx}-e^{5logx}}{e^{4logx}-e^{3logx}}}dx\\ \therefore I={\displaystyle \int \frac{e^{log{x}^6}-e^{log{x}^5}}{e^{log{x}^4}-e^{log{x}^3}}}dx\\ ={\displaystyle \int \frac{x^6-x^5}{x^4-x^3}dx}={\displaystyle \int \frac{x^2\left(x^4-x^3\right)}{x^4-x^3}dx}={\displaystyle \int x^{2}dx}\\ =\frac{1}{3}{x}^3+C\\ Hence,therequiredsolutionis\frac{1}{3}{x}^3+C.\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{\left(x^2+2\right)}{x+1}}dx\\ \therefore I={\displaystyle \int \left[\left(x-1\right)+\frac{3}{x+1}\right]}dx\\ ={\displaystyle \int \left(x-1\right)dx}+3{\displaystyle \int \frac{1}{x+1}dx}\\ =\frac{x^2}{2}-x+3log\left|x+1\right|+C\\ Hence,therequiredsolutionis\frac{x^2}{2}-x+3log\left|x+1\right|+C.\end{array}$

Both City College and St Claret College are some of the popular institutes offering BCA in the Bangalore region. Below are some of the prominent comparison points between the 2 colleges:

1. Ranking: While City College has not received any official ranking from agencies, St Claret College is ranked #61 by Outlook 2024.
2. Tuition Fees: The total tuition fees at City College Bangalore is INR 2.6 lakhs, while for SCC Bangalore it is INR 2.31 Lakhs - 2.91 Lakhs.
3. Placements: The average salary at City College is INR 4 LPA, on the other hand the median package at St Claret College is INR 428 LPA.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}L.H.S.={\displaystyle \int \frac{2x+3}{x^2+3x}}dx\\ Put{x}^2+3x=t\\ \therefore \left(2x+3\right)dx=dt\\ \Rightarrow {\displaystyle \int \frac{dt}{t}}=log\left|t\right|\Rightarrow log\left|x^2+3x\right|+C=R.H.S.\\ L.H.S.=R.H.S.\\ Hence,proved.\end{array}$

The admission procedure of SCMS Cochin includes multiple steps, such as application, selection, and admission confirmation. Students must satisfy the admission criteria prior to applying for the programme. Students can check the course-wise eligibility from the table presented below:

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}L.H.S.={\displaystyle \int \frac{2x-1}{2x+3}}dx\\ \Rightarrow {\displaystyle \int \left(1-\frac{4}{2x+3}\right)}dx\left[Dividingthenumeratorbythe\text{\hspace{0.17em}denominator}\right]\\ \Rightarrow {\displaystyle \int 1.}dx-4{\displaystyle \int \frac{1}{2x+3}dx}\Rightarrow {\displaystyle \int 1.}dx-\frac{4}{2}{\displaystyle \int \frac{1}{x+\frac{3}{2}}dx}\\ \Rightarrow {\displaystyle \int 1.}dx-2{\displaystyle \int \frac{1}{x+\frac{3}{2}}dx}\Rightarrow x-2log\left|x+\frac{3}{2}\right|+C\\ \Rightarrow x-2log\left|\frac{2x+3}{2}\right|+C\Rightarrow x-log\left|{\left(\frac{2x+3}{2}\right)}^2\right|+C\\ \left[?nlogm=log{m}^n\right]\\ \Rightarrow x-log\left|{\left(2x+3\right)}^2\right|-log{2}^2+C\\ \Rightarrow x-log\left|{\left(2x+3\right)}^2\right|+C_1=R.H.S.\left[where{C}_1=C-log{2}^2\right]\\ L.H.S.=R.H.S.\\ Hence,proved.\end{array}$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}log\left|sinx+cosx\right|dx}\dots \left(i\right)\\ ={\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}log\left|sin\left(\frac{\pi }{4}-\frac{\pi }{4}-x\right)+cos\left(\frac{\pi }{4}-\frac{\pi }{4}-x\right)\right|dx}\left[Using{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx={\displaystyle \underset{a}{\overset{b}{\int }}f\left(a+b-x\right)dx}}\right]\\ ={\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}log\left|sin\left(-x\right)+cosx\right|dx}\\ ={\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}log\left|cosx-sinx\right|dx}\dots \left(ii\right)\\ Adding\left(i\right)and\left(ii\right)\\ 2I={\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}log\left|cosx+sinx\right|dx}+{\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}log\left|cosx-sinx\right|dx}\\ 2I={\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}log\left|\left(cosx+sinx\right)\left(cosx-sinx\right)\right|dx}\\ 2I={\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}log\left|co{s}^{2}x-si{n}^{2}x\right|dx}\\ \therefore 2I={\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}logcos2xdx}\\ 2I=2{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}logcos2xdx}\left[?{\displaystyle \underset{-a}{\overset{a}{\int }}f\left(x\right)dx=2{\displaystyle \underset{0}{\overset{a}{\int }}f\left(x\right)dxiff\left(-x\right)=f\left(x\right)}}\right]\\ \therefore I=\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}logcos2xdx}\\ Put2x=tdx=\frac{dt}{2}\\ Whenx=0\therefore t=0;whenx=\frac{\pi }{4}\therefore t=\frac{\pi }{2}\\ I=\frac{1}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logcostdt}\text{\&thi}\end{array}$

$\begin{array}{l}Onadding\left(iii\right)and\left(iv\right),weget\\ 2I=\frac{1}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(logcost+logsint\right)dt}\\ 2I=\frac{1}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logsintcostdt}\\ 2I=\frac{1}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{log2sintcostdt}{2}}\\ 2I=\frac{1}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(logsin2t-log2\right)dt}\\ 4I={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logsin2tdt}-{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}log2dt}\\ Put2t=u\Rightarrow 2dt=du\Rightarrow dt=\frac{du}{2}\\ \therefore 4I=\frac{1}{2}{\displaystyle \underset{0}{\overset{\pi }{\int }}logsinudu}-{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}log2dt}\left[Changingthelimit\right]\\ 4I=\frac{1}{2}\times 2{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logsinudu}-log2{\left[t\right]}_0^{\frac{\pi }{2}}\\ 4I={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logsinudu}-log2.\frac{\pi }{2}\\ 4I=2I-\frac{\pi }{2}.log2\end{array}$