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The highest package offered during the 2025 placement drive at RGIPT stood at INR 25 LPA. The same is tabulated below: 

Rajiv Gandhi Institute of Petroleum Technology offers decent placements annually to its BTech and MBA students. A total of 55 recruiters participted during the placement session 2025 and presented 148 jobs to batch 2025. The overall highest and average package offered during RGIPT placements 2025 was INR 25 LPA and INR 10.79 LPA.

The application form for BVP CET 2025 has been released. Candidates can apply online from the university website. The last date to fill out the application form was on May 30, 2025. Interested candidates were required to complete the application process by filling out the online form, uploading the necessary documents, and paying the prescribed application fee within the stipulated timeline.

Yes, during the final stage of M.Tech admissions at Takshashila University, students who are offered admission must pay the first installment of the first-year fees. In general, at this stage, students are required to pay admission fees and some part of the tuition fees. For more information, students can visit the official website of the university.

Here are some low-cost Online Photoshop colleges in India you can consider:

Note: The information is sourced from the official/counselling body of the institute/exam.

The deadlines to fill out a BPharm application form at Poona College of Pharmacy for the academic batch 2025 have been released. The last date to apply for BVP CET 2025 was May 30, 2025. Moreover, the counselling process was held in June. Prospective candidates should look out for the upcoming 2026 batch.

Draw g(t) = t³ – 3t

g’(t) = 3(t² – 1)

g(1) is maximum in (-2, 2)

So, maximum (t³ – 3t) = $\{\begin{array}{cc}\hfill t^3-3t\hfill & \hfill ;-2<t<-1\hfill \\ \hfill 2\hfill & \hfill ;-1<t<2\hfill \end{array}$

$I={\displaystyle \underset{-2}{\overset{2}{\int }}f}\left(x\right)dx$

= ${\displaystyle \underset{-2}{\overset{-1}{\int }}\left(t^3-3t\right)dt}+{\displaystyle \underset{-1}{\overset{2}{\int }}2dt}$

I = $\frac{27}{4}$

again rewrite the f(x)

$f\left(x\right)=\left\{\begin{array}{ccc}\hfill \begin{array}{cc}\hfill x^3-3x\hfill & \hfill 2\hfill \end{array}\hfill & \hfill ;\hfill & \hfill \begin{array}{cc}\hfill x\le -1\hfill & \hfill -1<x\le 2\hfill \end{array}\hfill \\ \hfill \begin{array}{cc}\hfill x^2+2x-6\hfill & \hfill 9\hfill \end{array}\hfill & \hfill ;\hfill & \hfill \begin{array}{cc}\hfill 2<x<3\hfill & \hfill 3\le x<4\hfill \end{array}\hfill \\ \hfill \begin{array}{ccc}\hfill 10\hfill & \hfill 11\hfill & \hfill 2x+1\hfill \end{array}\hfill & \hfill ;\hfill & \hfill \begin{array}{ccc}\hfill 4\le x<5\hfill & \hfill x=5\hfill & \hfill x>5\hfill \end{array}\hfill \end{array}\right\}$

$f\text{'}\left(x\right)=\left\{\begin{array}{cccccc}\hfill 3x^2-3\hfill & \hfill ;\hfill & \hfill x<-1\hfill & \hfill 0\hfill & \hfill ;\hfill & \hfill -1<x<2\hfill \\ \hfill 2x+2\hfill & \hfill ;\hfill & \hfill 2<x<3\hfill & \hfill 0\hfill & \hfill ;\hfill & \hfill 3<x<4\hfill \\ \hfill 0\hfill & \hfill ;\hfill & \hfill 4<x<5\hfill & \hfill 2\hfill & \hfill ;\hfill & \hfill x>5\hfill \end{array}\right\}$

So f(x) is not differentiable at x = 2, 3, 4, 5

so m = 4

Because this is a creative sector, candidates must have a creative mindset and the capacity to visualise. The following are some other abilities needed in the field:

Note: The information is sourced from external sites and may vary.

Image of pt (2,4,7) in the plane 3x – y + 4z = 2 is

$\frac{x-2}{3}=\frac{y-4}{-1}=\frac{z-7}{4}=\frac{-2\left(6-4+28-2\right)}{3^2+{\left(-1\right)}^2+4^2}$

Let $\frac{x-2}{3}=\frac{y-4}{-1}=\frac{z-7}{4}=\frac{-28}{13}=\lambda$

$x=3\lambda +2y=-\lambda +4z=4\lambda +7$

Now according to the question

(a, b, c) = $\left(3\lambda +2,-\lambda +4,4\lambda +7\right)$

Now $2a+b+2c=6\lambda +4-\lambda +4+8\lambda +14$

= $13\lambda +22$

= 28 + 22 [Use $\lambda =\frac{-28}{13}$ ]

= 6

Takshashila University offers various specialisations as part of the M.Tech course. Candidates willing to take admission in any of the courses must fulfil the course-specific eligibility criteria. As per the criteria, students with a BTech/BE in Computer Science and Engineering, Information Technology, or related disciplines with a minimum of 50% marks are eligible to apply. Hence, studying maths is not compulsory.

The average fee structure for a Bachelor of Business Administration (BBA) programme may significantly vary based on several factors which includes the country, type of institution (public or private) and the specific program's prestige. Here's a general overview of the fee structure in various contexts:

If you take admission into a Government Colleges, the fees can range between INR 2,000 to INR 1 Lakh. If you want to pursue BBA from a Private University it will be a little bit Costlier which can range from INR 1 Lacs to 7 Lakh.

Once students have filled out the Takshashila University M.Tech application form, they must stay updated with the release of the merit list. The list is generally published on the university's official website by the admission committee. Thus, students must keep checking the official website to stay updated. Moreover, the university's official website states that the university may require some students to appear for an interview process.

Power gain = ${\left(\frac{\Delta i_c}{\Delta i_b}\right)}^2\times \frac{R_0}{R_i}$

= ${\left(\frac{10\times 10^{-3}}{100\times 10^{-6}}\right)}^2\times \frac{2}{1}$

= 2 × 104 = x × 104

= 2

Taking admission to top ranked BBA colleges in Mumbai is somewhat expensive as most colleges have tuition fees of more than INR 5 lakh. Listed below is the fee range along with their various colleges:

Disclaimer: This information is sourced from the official website.

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 $\frac{dy}{dx}=\frac{y}{x}+\frac{\sqrt[]{y^2+16x^2}}{x}$

Put y = $V\cdot x$

differentiable worst = x

$\frac{dy}{dx}=V+x\cdot \frac{dV}{dx}$

V + $x\frac{dV}{dx}=V+\sqrt[]{V^2+16}$

$x\frac{dV}{dx}=\sqrt[]{V^2+16}$

apply variable separable method

${\displaystyle \int \frac{dV}{\sqrt[]{V^2+16}}={\displaystyle \int \frac{dx}{x}+\mathcal{l}nC}}$

$\Rightarrow \mathcal{l}n\left|V+\sqrt[]{V^2+16}\right|=\mathcal{l}nCx$

$\frac{y}{x}+\frac{\sqrt[]{y^2+16x^2}}{x}=Cx$

Given y(1) = 3 C = 8

Now at x = 2

$\frac{y}{2}+\frac{\sqrt[]{y^2+16.4}}{2}=8.2$

y2 + 64 = (32 – y)2

y = 15

Yes, the application forms for Takshashila University M.Tech courses can be filled out online. To do so, interested students can visit the university's official website and follow the steps listed below:

1. Navigate to the admission portal.

2. Register to make a new student account.

3. Log in to the account and fill out the application form.

4. Pay the application fee.

5. Click on the submit button.

Set of first 10 prime numbers

= {2,3,5,7,11,13,17,19,23,29,31}

So sample space = 104.

Favourable cases

So required probability

$=\frac{10+4{\times }^{10}\text{?}{C}_2}{10^4}=\frac{10+\frac{4\times 10.9}{2}}{10^4}=\frac{19}{1000}$

For SRM IST BTech Admission 2025 the application process usually begins around November each year. The university conducts SRMJEEE in three phases, with Phase 1 registration typically closing in April, followed by Phase 2 and Phase 3 starting in June and July, respectively. All application steps are completed entirely online. Counselling for seat allotment begins soon after the SRMJEEE results are announced. Keep checking official SRMJEEE portal for the latest updates.

The admission process for Takshashila University M.Tech courses is ongoing. The university is accepting the applications for 2025 admissions. Those who wish to take enroll into any of the M.Tech courses can apply for the preferred course by filling out the application form. Moreover, the deadline to apply has not been announced yet. Thus, students are advised to register timely.