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Candidates wishing to get admission to the Nehru College of Management's MBA programme are required to follow the steps given below:

Step 1: Visit the official website of the Nehru College of Management and apply for the course through the google form. 

Step 2: Apply and appear for TANCET. 

Step 3: Pay the course fee is the name is shortlisted for final admissions. 

$\begin{array}{l}Put,ax+b=t\\ adx=dt\\ dx=\frac{1}{a}dt\\ I={\displaystyle \int (}ax+b){}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}dx\\ =\int t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}·\frac{1}{a}dt=\frac{1}{a}\frac{t\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2+1$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2+1$}\right.}+C\\ =\frac{1}{a}\frac{t^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+C\\ =\frac{2}{3a}{(ax+b)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\; c\end{array}$

$\begin{array}{l}=sin\left(ax+b\right)cos\left(ax+b\right)=\frac{2sin\left(ax+b\right)cos\left(ax+b\right)}{2}\\ =\frac{sin2(ax+b)}{2}\\ Put2\left(ax+b\right)=t\\ 2adx=dt\\ I=\int \frac{sin2(ax+b)}{2}dx\\ =\frac{1}{2}\int \frac{sintdt}{2a}=\frac{1}{4a}[-cost]+C\\ =-\frac{1}{4a}cos2(ax+b)+C\end{array}$

Kindly go through the solution

This is a Matching Type Questions as classified in NCERT Exemplar

(i)→ (a), (d); (ii)→ (c); (iii)→ (a); (iv)→ (b)

$\begin{array}{l}\frac{x(1+logx)}{}\\ Put,1+logx=t\\ =\frac{1dx}{x}=dt\\ =I=\int \frac{1}{x+xlogx}dx=\int \frac{1}{t}dt\\ =log\left(t\right)+c\\ =log\left(1+logx\right)+c\end{array}$

It is easy for international students to be a part of Penn State admissions, as the acceptance rate for them stands at 64.3%, which means that out of every 100 applications, 64 to 65 students were rolled out an offer letter to pursue higher education at this university. 

The possible inference to the Penn State acceptance rate for international students is that if they can meet the admission requirements for this university, it will not be that difficult to get admission here. 

As per the NLU Kochi NIRF report 2024, the course-wise median package offered over the past three years are presented below:

NOTE: This information is sourced from the official website/ sanctioning body and is subject to change.

$\begin{array}{l}Let,logx=t\\ \frac{1}{x}dx=dt\\ I=\int \frac{{(logx)}^2}{x}dx=\int t^{2}dt=\frac{t^3}{3}+C\\ =\frac{{(logx)}^3}{3}+C\end{array}$

$\begin{array}{l}Let,1+x^2=t\\ \Rightarrow 2xdx=dt\\ I=\int \frac{2x}{1+x^2}dx=\int \frac{1dt}{t}\\ \Rightarrow log\left|t\right|+c\\ \Rightarrow log\left|1+x^2\right|+c\end{array}$

The seat intake for the BA LLB (Hons) programme at National Law University, Kochi is 136. The university only offers seats based on  availability of seats and the merit lists. Further, NLU follows the reservation policy while allotting the seats among the students. Further, candidates must also have a valid score in the CLAT 2025 exam. 

$\begin{array}{l}f\left(x\right)={\displaystyle \int 4x^3}–\frac{3}{x^4}dx\\ =4\cdot \frac{x^4}{4}-3\cdot \frac{x^{-3}}{-3}+\; C\\ =x^4+\frac{1}{x^3}+\; C\end{array}$

Now,  $f\left(2\right)=0 \therefore f\left(2\right) =2^4+\frac{1}{2^3}+C=0$

$\begin{array}{l}=16+\frac{1}{8}+c=0\\ =c=-\left(16+\frac{1}{8}\right)\\ =c=-\frac{129}{8}\end{array}$

Therefore, correct answer is A.

The college provides practical learning opportunities beyond traditional classroom teaching. Students studying at Nehru College of Management appreciate that "MBA at Nehru college of management stands out with its uniqueness" and serves as "a perfect platform to polish one's skills and calibre and also nurture their talents through various outbound programs". The placement report shows growth with every year, and candidates are getting the opportunity to get into better organisations. 

$\frac{x^{3/2}}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{x^{1/2}}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\; C$

$=\frac{2}{3}{x}^{3/2}+2x^{1/2}+$ C

? The correct Answer is (C)

$\begin{array}{l}\int \left(\frac{2}{co{s}^{2}x}-\frac{3sinx}{co{s}^{2}x}\right)dx\\ \begin{array}{l}={\displaystyle \int \left(2se{c}^{2}x–3tanx.secx\right)dx}\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}={\displaystyle \int 2se{c}^{2}x}.dx–{\displaystyle \int 3tan}x.secx.dx\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}=2tanx– 3 secx+C\hfill \end{array}\end{array}$

TANCET exam registration is facilitated to apply for entrance exam conudcted for MBA and MCA courses in Tamil Nadu. Anna University is entrusted to facilitate the TANCET examination process, including the registration process, admit card issuance process, and result declaration process. The centralised counselling for TN MBA MCA admission is done by DTE Tamil Nadu.

$\begin{array}{l}\int \frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$co{s}^2$}\right.x}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$co{s}^2$}\right.x}dx\\ =\int \frac{si{n}^{2}x}{co{s}^{2}x}dx\\ \begin{array}{l}={{\displaystyle \int tan}}^{2}xdx\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}={\displaystyle \int \left(se{c}^{2}x–1\right)}dx={\displaystyle \int se{c}^2}xdx–{\displaystyle \int 1}.dx\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}=tanx–x+C\hfill \end{array}\end{array}$

The documents that are required to fill the application form for Nehru College of Management are as follows: 

• Class 10 marksheet
• Class 12 marksheet
• Graduation marksheet
• TANCET result 
• Transfer Certificate 
• Government issued ID

Candidates with the above mentioned documents can complete the application and the admission process for the courses at Nehru College of Management.