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The key highlights of Kurukshetra University MBA placements 2024 are presented below:

Note: The above-mentioned data are gathered from the KUK Management NIRF report 2025.

  $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}=r$

For ‘B’, x = 2r + 1, y = 3r – 1, z = -2r + 1

As AB is perpendicular to the line,

$\left\{\left(2r+1\right)-0\right\}2+\left\{\left(3r-1\right)-1\right\}3+\left\{\left(-2r+1\right)-2\right\}$

$r=\frac{2}{17}\Rightarrow B\left(\frac{2}{17},\frac{-11}{7},\frac{13}{17}\right)$ ->

direction ratios of AB

(2r + 1, 3r – 2, -2r – 1)

Equation of AB

$\frac{x}{-3}=\frac{y-1}{4}=\frac{z-2}{3}$  

ax² + bx + c = 0

D = b² – 4ac

D = 0

b² – 4ac = 0

b² = 4ac

(i) AC = 1, b = 2 (1, 2, 1) is one way

(ii) AC = 4, b = 4

$\begin{array}{l}a=4c=1\\ a=2c=2\\ a=1c=4\end{array}\}3ways$  

(iii) AC = 9, b = 6, a = 3, c = 3 is one way

1 + 3 + 1 = 5 way

Required probability =  $\frac{5}{216}$   

Let sinθ= t ${\displaystyle \int \frac{sin\theta \left(2sin\theta .cos\theta \right)\left(si{n}^6\theta +si{n}^4\theta +si{n}^2\theta \right)\sqrt[]{2si{n}^4\theta +3si{n}^2\theta +6}}{2si{n}^2\theta }}$ dθ

sinθ = t

cos θ. dθ = dt

${\displaystyle \int \frac{u^{1/2}}{12}}du=\frac{u^{3/2}}{18}+C=\frac{{\left(2t^6+3t^4+6t^2\right)}^{3/2}}{18}+C=\frac{{\left(2si{n}^6\theta +3si{n}^4\theta +6si{n}^2\theta \right)}^{3/2}}{18}+C$

WSU offers more than 300 clubs and organizations offer social and leadership possibilities, student government, a vibrant Greek community, and more.

At Altera Institute, students are selected based on several parameters. This includes fulfilling the admission criteria of completing a Bachelor's degree in any discipline from a recognised university. Additionally, the final selection is made based on scores secured in the accepted entrance examination. The institute accepts several national-level exams, such as MAT/ CAT, etc., in addition to conducting its own test.

l + m – n = 0

l + m = n . (i)

l² + m² = n²

Now from (i)

l² + m² = (l + m)²

->2lm = 0

->lm = 0

l = 0 or m = 0

->m = n Þ l = n

if we take direction consine of line

$0,\frac{1}{\sqrt[]{2}},\frac{1}{\sqrt[]{2}}and\frac{1}{\sqrt[]{2}},0,\frac{1}{\sqrt[]{2}}$  

cos α =   $\frac{1}{2}$

$si{n}^4\alpha +co{s}^4\alpha ={\left(\frac{\sqrt[]{3}}{2}\right)}^4={\left(\frac{1}{2}\right)}^4=\frac{9}{16}+\frac{1}{16}=\frac{10}{16}=\frac{5}{8}$           

xyz = 24

24 = 2³ × 3

Let’s distribute 2, 3 among 3 variables. No. of positive integral solution =

No. of ways to distribute = 

To be eligible for the program, a candidate must meet the following criteria:

• Academic Requirement: Class 12 with a 50% or an equivalent score in UG from a recognised institution.
• Qualifying Entrance Exams: The institute accepts scores from CAT, NMAT, XAT & GMAT. Aspirants who do not have any of these exams need to take the Altera Institute National Aptitude Test (ALNAT) after submitting the application.

To enrol in a BCA course at Global Institute of Management and Technology, candidates must meet the eligibility criteria set by the institute. The institute offers full-time BCA courses lasting four years. Students are selected based on merit. In addition to the selection criteria, students must pass Class 12 (common eligibility).

To get into Altera Institute, students need to go through several steps. The institute offers entrance-based admission to PGP in Applied marketing. The college accepts various entrance exam scores such as CAT/ GMAT/ XAT/ NMAT  or equivalent examinations scores for admission to PGP in Applied marketing. Further, candidates must fulfil the eligibility criteria to secure admission i.e. undergraduate degree with a score of 60% or equivalent from a recognised university/institute.

Yes, BCA course is available at the Global Institute of Management and Technology. The institute offers BCA courses at the UG level. Candidate must meet the eligibility criteria to enrol for BCA course admission set by the college. Aspirants must pass Class 12 with a 40% aggregate. Global Institute of Management and Technology's admission process is merit-based.

$\therefore In\Delta AQP$         

$tan30°=\frac{PQ}{AQ}$ $\frac{1}{\sqrt[]{3}}=\frac{h}{x+y}$        

x + y = $\sqrt[]{3}h.....\left(i\right)$  

$\therefore ln\Delta BQP$          

$tan45°=\frac{h}{y}$          

h = y .(ii)

(i) & (ii) x + y = $\sqrt[]{3}y$  

$\Rightarrow x=\left(\sqrt[]{3}-1\right)y.......\left(iii\right)$     

Let the speed be S

$\frac{x}{S}=20$

x = 20.S

from (iii)

$\frac{y}{S}=10\left(\sqrt[]{3}+1\right)$          

The UPSC CSE (IAS) exam result 2024-25 was released on June 11. The Commission released the final UPSC CSE (IAS) 2024 cutoff for all the categories and stages of exam. The Commission released the UPSC Cutoff 2025 for all the stages of exam as given below:

Yes, to get admission for the BTech, BBA, BCA, and After 10 Diploma courses at Global Institute of Management and Technology, candidates must meet the eligibility criteria set by the college. Candidates must pass Class 12 for the UG courses and Class 10 for the After 10 Diploma course. The selection criteria are based on entrance-exam scores.

Yes, applications are open at the Global Institute of Management and Technology for various courses. The mode of application is online. Aspirants need to apply at the official website for admission to the various programme. Candidates can fill out the application form on the institute's official website.

sin 2qθ+ tan 2θ> 0

$\frac{2tan\theta }{1+ta{n}^2\theta }+\frac{2tan\theta }{1-ta{n}^2\theta }>0$           

Let tan q = x

  $\frac{2x}{1+x^2}+\frac{2x}{1-x^2}>0$          

$tan\theta <-1or0<tan\theta <1$          

  $\theta \in \left(0,\frac{\pi }{4}\right)\cup \left(\frac{\pi }{2},\frac{3\pi }{4}\right)\cup \left(\pi ,\frac{5\pi }{4}\right)\cup \left(\frac{3\pi }{2},\frac{7\pi }{4}\right)$          

Global Institute of Management and Technology application window opens online on the official website of the institute. Below are the steps to apply for the Global Institute of Management and Technology admission:

Step 1: Visit the official website of GIMT

Step 2: Click on 'Apply Now' on the Home page.

Step 3: Register with the basic details to generate login credentials.

Step 4: Login to fill out the application form.

Step 5: Upload the documents and pay the application fee

Step 6: Submit the application form.