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BITS Pilani Goa cutoff 2024 has been released based on BITSAT exam. The cutoff was released for different categories of candidates belonging to the All India quota. The cutoff scores among the General AI category candidates varied between 234 and 301. The most in-demand specialisation as per the cutoff scores was BE in Computer Science due to its the highest cutoff score.

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{tanx}{1+m^{2}ta{n}^{2}x}dx}\\ I={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\frac{sinx}{cosx}}{1+m^2\frac{si{n}^{2}x}{co{s}^{2}x}}dx}={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\frac{sinx}{cosx}}{\frac{co{s}^{2}x+m^{2}si{n}^{2}x}{co{s}^{2}x}}dx}\\ ={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sinxcosx}{co{s}^{2}x+m^{2}si{n}^{2}x}dx}={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sinxcosx}{1-si{n}^{2}x+m^{2}si{n}^{2}x}dx}\\ ={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sinxcosx}{1-si{n}^{2}x+\left(1-m^2\right)}dx}\\ Putsi{n}^{2}x=t\\ 2sinxcosxdx=dt\\ sinxcosxdx=\frac{dt}{2}\\ Changingthelimitsweget,\\ Whenx=0\therefore t=si{n}^{2}0=0;whenx=\frac{\pi }{2}\therefore t=si{n}^2\frac{\pi }{2}=1\\ \therefore I=\frac{1}{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{1+\left(m^2-1\right)t}}=\frac{1}{2}{\left[\frac{log\left[1+\left(m^2-1\right)t\right]}{m^2-1}\right]}_0^1\\ =\frac{1}{2\left(m^2-1\right)}\left[log\left(1+m^2-1\right)-log\left(1\right)\right]=\frac{log\left|m^2\right|}{2\left(m^2-1\right)}\\ Hence,I=\frac{log\left|m^2\right|}{2\left(m^2-1\right)}=\frac{log\left|m\right|}{m^2-1}.\end{array}$

Ganpat University offers a comprehensive BPharma course. Thus, the total fee to pursue this programme amounts to INR 5.92 lakh. This fee comprises two components: tuition fees and hostel fees. Moreover, students must note that this fee is applicable for a duration of four years. To know more, students can check out the below table:

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

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \underset{0}{\overset{1}{\int }}\frac{dx}{e^x+e^{-x}}}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}\frac{dx}{e^x+\frac{1}{e^x}}}={\displaystyle \underset{0}{\overset{1}{\int }}\frac{dx}{\frac{e^{2x}+1}{e^x}}}={\displaystyle \underset{0}{\overset{1}{\int }}\frac{e^{x}dx}{e^{2x}+1}}\\ Put{e}^x=t\Rightarrow e^{x}dx=dt\\ Changingthelimits,wehave\\ Whenx=0\therefore t=e^0=1\\ Whenx=1\therefore t=e^1=e\\ \therefore I={\displaystyle \underset{1}{\overset{e}{\int }}\frac{dt}{t^2+1}}={\left[ta{n}^{-1}t\right]}_1^e=\left[ta{n}^{-1}e-ta{n}^{-1}\left(1\right)\right]=ta{n}^{-1}e-\frac{\pi }{4}\\ Hence,I=ta{n}^{-1}e-\frac{\pi }{4}.\end{array}$

After the completion of a BA (Hons) from JNU, candidates can apply for different government jobs. Graduates tend to seek employment in public administration, teaching, and social services. Candidates can apply for competitive exams such as UPSC, SSC, and State Public Service Commissions for the positions of IAS, IPS, or other government departments.

Besides, there are opportunities in banking, railways, and other PSUs, where a BA degree is a qualifying criterion. With appropriate preparation and focus, graduates with a BA can get employment in the government sector. 

Kochi Business School (KBS) offers one postgraduate program, namely MBA. Accepted entrance exams for the MBA programme include CAT, CMAT, KMAT, and MAT. Candidates must have a valid score in at least one of these exams to be selected for admission. Interested candidates can apply online on the official website for the MBA course.

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \underset{0}{\overset{2}{\int }}{e}^{x}dx}\\ Usingtheformula,\\ {\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}=\underset{h\to 0}{lim}h\left[f\left(a\right)+f\left(a+h\right)+f\left(a+2h\right)+\dots +f\left(a+\overline{n-1}h\right)\right]\\ whereh=\frac{b-a}{n}Here,a=0andb=2\\ \therefore h=\frac{2-0}{n}\therefore nh=2\\ Here,f\left(x\right)=e^x\\ f\left(0\right)=e^0=1\\ f\left(0+h\right)=e^{0+h}=e^h\\ f\left(0+2h\right)=e^{0+2h}=e^{2h}\\ \dots \dots \dots \dots \dots \dots \\ \dots \dots \dots \dots \dots \dots \\ f\left(0+\overline{n-1}h\right)=e^{0+\left(n-1\right)h}=e^{\left(n-1\right)h}\\ Now,{\displaystyle \underset{0}{\overset{2}{\int }}{e}^{x}dx}=\underset{h\to 0}{lim}h\left[1+e^h+e^{2h}+\dots +e^{\left(n-1\right)h}\right]\\ =\underset{h\to 0}{lim}h\left[\frac{1\left(e^{nh}-1\right)}{e^h-1}\right]\left[?a+ar+a{r}^2+\dots +a{r}^{n-1}=\frac{a\left(r^n-1\right)}{r-1}\right]\\ =\underset{h\to 0}{lim}\frac{e^{nh}-1}{\frac{e^h-1}{h}}=\frac{e^2-1}{1}=e^2-1\left[?\underset{x\to 0}{lim}\frac{e^x-1}{x}=1\right]\\ Hence,{\displaystyle \underset{0}{\overset{2}{\int }}{e}^{2}dx}=e^2-1.\end{array}$

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \underset{0}{\overset{2}{\int }}\left(x^2+3\right)dx}\\ \text{\hspace{0.17em}Using}theformula,\\ {\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}=\underset{h\to 0}{lim}h\left[f\left(a\right)+f\left(a+h\right)+f\left(a+2h\right)+\dots +f\left(a+\overline{n-1}h\right)\right]\\ whereh=\frac{b-a}{n}Here,a=0andb=2\\ \therefore h=\frac{2-0}{n}\therefore nh=2\\ Here,f\left(x\right)=x^2+3\\ f\left(0\right)=0^2+3=0+3=3\\ f\left(0+h\right)={\left(0+h\right)}^2+3=h^2+3\\ f\left(0+2h\right)={\left(0+2h\right)}^2+3=4h^2+3=3\\ \dots \dots \dots \dots \dots \dots \\ \dots \dots \dots \dots \dots \dots \\ f\left(0+\overline{n-1}h\right)={\left(0+\overline{n-1}h\right)}^2+3{\left(n-1\right)}^2{h}^2+3\\ Now,{\displaystyle \underset{0}{\overset{2}{\int }}\left(x^2+3\right)dx}=\underset{h\to 0}{lim}h\left[3+h^2+3+4h^2+3+\dots +{\left(n-1\right)}^2{h}^2+3\right]\\ =\underset{h\to 0}{lim}h\left[\left(3+3+3+\dots +n\right)+\left\{h^2+4h^2+\dots +{\left(n-1\right)}^2{h}^2\right\}\right]\\ =\underset{h\to 0}{lim}h\left[3n+h^2\left\{1+4+\dots +{\left(n-1\right)}^2\right\}\right]\\ =\underset{h\to 0}{lim}h\left[3n+h^2\frac{n\left(n-1\right)\left(2n-1\right)}{6}\right]\\ \left[?1+4+9+\dots +{\left(n-1\right)}^2=\frac{n\left(n-1\right)\left(2n-1\right)}{6}\right]\\ =\underset{h\to 0}{lim}\left[3nh+\frac{h^{3}n\left(n-1\right)\left(2n-1\right)}{6}\right]\\ =\underset{h\to 0}{lim}\left[3nh+\frac{nh\left(nh-h\right)\left(2nh-h\right)}{6}\right]\\ =\left[3\times 2+\frac{2\left(2-0\right)\left(2\times 2-0\right)}{6}\right]\left[?nh=2andh=0\right]\\ =\left[6+\frac{2\times 2\times 4}{6}\right]=6+\frac{8}{3}=\frac{26}{3}\\ Hence,{\displaystyle \underset{0}{\overset{2}{\int }}\left(x^2+3\right)dx}=\frac{26}{}\end{array}$

Admissions to the BTech programme offered at Dr. D.Y. Patil Institute of Technology is entrance-based. Students seeking to apply for the course must fulfil both the eligibility and selection criteria for the university. The university accepts JEE Main or MHT CET score for admission to its BTech programme. Hence, a student can still get into the Dr. D.Y. Patil Institute of Technology BTech course without appearing for JEE Main, given they have a valid MHT CET score. For further admission related queries, students can contact the admission help desk of the university.

Ganpat University BPharma admissions commence with the application stage. During this stage, students are required to fill out the application form available at the university's official website. Once the application window is closed, the admissions committee ranks applicants based on scores obtained by them in the CUET UG scores/qualifying examination marks. Next, the university shortlists students based on their rank and the number of available seats. At last, those who get selected for admissions must report to the university for verification of documents and payment of fees.

This is a Short answer type Questions as classified in NCERT Exemplar

Yes, it is possible to get admission to BITS Pilani with a score of 250 in BITSAT. BITS Pilani Goa cutoff 2024 has been released for the last round. The cutoff was released as opening and closing cutoff ranks for differnet categories belonging to the All India quota. The cutoff ranks among the General AI category candidate varied between 234 and 301. Candidates can check the branch wise cutoff for BITS Pilani on the basis of BITSAT exam in the table below:

After graduation from BTech at Dr. D.Y. Patil Institute of Technology, students have a range of career opportunities awaiting them. Both government and private sectors provide numerous opportunities. Some of the most desired government career opportunities for BTech graduates from Dr. D.Y. Patil Institute of Technology are as follows:

Central Government Jobs: