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Vidyavardhini College of Engineering and Technology offers several opportunities outside the four walls of the College to provide industry exposure. The College has various clubs and societies, moderated by the student body and faculty members. Furthermore, several collaborations and partnerships eases the way into the transformations within the years of study for a student. These clubs and societies helps the students enhance the skills and develop appropriate skill-set for the real world. 

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

Currently, the M.Tech cutoff of the latest closing rank for admission to the 2025 batch at NIT Raipur has been released. Moreover, the closing ranks for all the specialisations is yet to be released. For a better overview, check out the below table:

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

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

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{x^{1/2}}{1+x^{3/4}}}dx\\ Putx=t^4\Rightarrow dx=4t^{3}dt\\ ={\displaystyle \int \frac{t^2.4t^3}{1+t^3}}dt=4{\displaystyle \int \frac{t^5}{1+t^3}}dt\\ =4{\displaystyle \int \left(t^2-\frac{t^2}{t^3+1}\right)}dt=4{\displaystyle \int t^2}dt-4{\displaystyle \int \frac{t^2}{t^3+1}}dt\\ LetI=I_1-I_2\\ Now{I}_1=4{\displaystyle \int t^2}dt=4.\frac{t^3}{3}+C_1=\frac{4}{3}{x}^{3/4}+C_1\\ and{I}_2=4{\displaystyle \int \frac{t^2}{t^3+1}}dt\\ Put{t}^3+1=z\Rightarrow 3t^{2}dt=dz\Rightarrow t^{2}dt=\frac{1}{3}dz\\ \therefore I_2=\frac{4}{3}{\displaystyle \int \frac{dz}{z}}=\frac{4}{3}log\left|z\right|+C_2=\frac{4}{3}log\left|t^3+1\right|+C_2\\ =\frac{4}{3}log\left|{\left(x\right)}^{3/4}+1\right|+C_2\\ \therefore I=I_1-I_2\\ =\frac{4}{3}{x}^{3/4}+C_1-\frac{4}{3}log\left|{\left(x\right)}^{3/4}+1\right|-C_2\\ \therefore I=\frac{4}{3}{x}^{3/4}-\frac{4}{3}log\left|{\left(x\right)}^{3/4}+1\right|+C_1-C_2\\ Hence,I=\frac{4}{3}\left[x^{3/4}-log\left|{\left(x\right)}^{3/4}+1\right|\right]+C.\left[?C_1-C_2=C\right]\end{array}$

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

ILAM Dehradun - Institute of Logistics and Aviation Management Dehradun admissions are based on entrance for the MBA course, specifically through the MAT/XAT scores. Candidates must qualify in the required exam to be eligible for admission. Candidates applying to other courses are accepted based on their merit . Candidates can learn more through the official website of ILAM Dehradun - Institute of Logistics and Aviation Management Dehradun. 

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

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

IIIT Dharwad offers admission to BTech on the basis of valid scores of accepted entrance exams such as JEE Main and KCET. Shortlisted candidates must qualify the counselling round and must pay the fees to confirm a seat. In addition to this, candidates must secure a minimum aggregate of 75% in Class 12 with PCM. Candidates must apply for the respective entrance exam to register the candidature. 

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

The following factors determine the M.Tech cut-off (admission/qualifying):

• Number of aspirants who appeared for entrance exam
• The difficulty level of the exam
• Total number of seats available
• Total number of applications received for a particular programme
• The category under which admission is sought

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

Sol:

$\begin{array}{l}LetI={\displaystyle \int ta{n}^{2}x}.se{c}^{4}xdx\\ ={\displaystyle \int ta{n}^{2}x.se{c}^{2}x.se{c}^{2}xdx=}{\displaystyle \int ta{n}^{2}x.\left(1+ta{n}^{2}x\right).se{c}^{2}xdx}\\ Puttanx=t,\therefore se{c}^{2}xdx=dt\\ \therefore I={\displaystyle \int t^2\left(1+t^2\right)dt}={\displaystyle \int \left(t^2+t^4\right)dt}={\displaystyle \int t^{2}dt+{\displaystyle \int t^4}}dt\\ =\frac{1}{3}{t}^3+\frac{1}{5}{t}^5=\frac{1}{3}ta{n}^{3}x+\frac{1}{5}ta{n}^{5}x+C\\ Hence,therequiredsolutionis\frac{1}{3}ta{n}^{3}x+\frac{1}{5}ta{n}^{5}x+C.\end{array}$

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

Sol:

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