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Manipal University Jaipur has an impressive placement record. Refer to the data tabulated below to know more:

The placement rate for Manipal University Jaipur stood at 94%. However, Manipal University Jaipur provides good placement to its students. The data tabulated below provides Manipal University Jaipur placement highlights for 2024:

Here, $K_{eq}=2k$ ${}_{}$

$T=2\pi \sqrt[]{\frac{m}{2k}}$

$\therefore f=\frac{1}{2\pi }\sqrt[]{\frac{2k}{m}}$

BAOU offers admission to UG and PG programmes based on their merit in the last qualifying exam. Hence, aspirants do not require passing an entrance test. The selection is done purely based on merit. Aspirants only need to meet the eligibility requirement and apply for the preferred course to get admission in the university. Once the selection is done, candidates have to pay the course fee to confirm their seat in the university.

$\left[\begin{array}{cc}\hfill 0\hfill & \hfill -tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 0\hfill \end{array}\right],I_2+A=\left[\begin{array}{cc}\hfill 1\hfill & \hfill tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill -tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 1\hfill \end{array}\right],I_2-A=\left[\begin{array}{cc}\hfill 1\hfill & \hfill tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill -tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 1\hfill \end{array}\right]$           

  $\left(I_2+A\right){\left(I_2-A\right)}^{-1}=\left[\begin{array}{cc}\hfill a\hfill & \hfill -b\hfill \\ \hfill b\hfill & \hfill a\hfill \end{array}\right]$          

  $a^2+b^2=\left|\left(I_2+A\right){\left(I_2-A\right)}^{-1}\right|=se{c}^2\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times co{s}^2\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.=1$          

$\therefore 13\left(a^2+b^2\right)=13\times 1=13$

Students can evaluate the ROI, or the Return on Investment, on an AI & Robotics course from Tamil Nadu colleges by comparing the fees and placement salary at the popular colleges:

Disclaimer: This info. is taken from the official website and may differ.

Critical point of function are x = $\frac{-1}{2},$ 1 and -2 but x = -2 is making zero .

$\therefore nondifferentiableatx=\frac{-1}{2},1$      

$\Delta E=hv\to \left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)$

(1) n₁ = 3, n₂ = 2

    $\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\frac{9-4}{36}=\frac{5}{36}$         

(2) n₁ = 4, n₂ = 3

$\left(\frac{1}{3^2}-\frac{1}{4^2}\right)=\frac{16-9}{144}=\frac{7}{144}$        

(3) n₁ = 2, n₂ = 1

$\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{4-1}{4}=\frac{3}{4}$  

(4) n₁ = 5, n₂ = 4

$\left(\frac{1}{4^2}-\frac{1}{5^2}\right)=\frac{25-16}{400}=\frac{9}{400}$

$A={\displaystyle \underset{\pi /4}{\overset{5\pi /4}{\int }}\left(sinx-cosx\right)dx={\left[-cosx-sinx\right]}_{\pi /4}^{5\pi /4}}=\left(\frac{1}{\sqrt[]{2}}+\frac{1}{\sqrt[]{2}}\right)-\left(-\frac{1}{\sqrt[]{2}}-\frac{1}{\sqrt[]{2}}\right)$

A =    $2\sqrt[]{2}$

A⁴ = 64

The selection process for BTech / BE admissions at Adichunchanagiri University is entirely online. Candidates are shortlisted based on their performance in accepted entrance exams such as the ACU Entrance Test, COMEDK UGET, or KCET. Students can apply with the scorecard of any one of these exams. Only those who meet the required eligibility criteria will be considered for admission.

 $\underset{n\to \infty }{lim}{\displaystyle \sum _{r=0}^{n-1}\frac{n}{{\left(n+r\right)}^2}}$

$\underset{n\to \infty }{lim}{\displaystyle \sum _{r=0}^{n-1}\frac{1}{n}.\frac{1}{{\left(1+\frac{r}{n}\right)}^2}={\displaystyle \underset{0}{\overset{1}{\int }}\frac{1}{{\left(1+x\right)}^2}={\left[-\frac{1}{1+x}\right]}_0^1}=-\frac{1}{2}-\left(-1\right)=\frac{1}{2}}$

$F_{net}=\frac{2kee}{\left(d^2+x^2\right)}sin\theta =\left(\frac{2k{e}^2}{\left(d^2{+}^2\right)}\right)\theta$

[for small displacement i.e, x is small so|, sin $\approx$ $\approx$ $$ $$ tan ]

$\frac{{}^{}}{}$ $=\frac{2k{e}^2}{\left(d^2+x^2\right)}.\frac{x}{d}=2.\frac{1}{4\pi {\epsilon }_0}{q}^2\frac{x}{d^3}$   [q = e]

$=\frac{q^2}{2\pi {\epsilon }_0{d}^3}x=m{\omega }^{2}x$

$\therefore \omega =\sqrt[]{\frac{q^2}{2\pi {\epsilon }_0{d}^{3}m}}$

Manipal University Jaipur packages recorded in the 2024 placement season are present in the table below:

$F=-\alpha x^2\Rightarrow \frac{mvdv}{dx}=-\alpha x^2$

$\Rightarrow {\displaystyle \underset{v_0}{\overset{0}{\int }}vdv=-\frac{\alpha }{m}{\displaystyle \underset{0}{\overset{x}{\int }}{x}^{2}dx\Rightarrow \frac{0^2}{2}-\frac{v_0^2}{2}=\frac{-\alpha }{3m}{x}^3}}$

$\Rightarrow \frac{-v_0^2}{2}=-\frac{\alpha }{3m}{x}^3\Rightarrow x={\left(\frac{3v_0^{2}m}{2\alpha }\right)}^{1/3}$

$A_{3\times 3}$

$det\left(A\right)=42A=\left[\begin{array}{ccc}\hfill 2a_1\hfill & \hfill 2a_2\hfill & \hfill 2a_3\hfill \\ \hfill 2b_1\hfill & \hfill 2b_2\hfill & \hfill 2b_3\hfill \\ \hfill 2c_1\hfill & \hfill 2c_2\hfill & \hfill 2c_3\hfill \end{array}\right]$

For R₂ →2R₂ + 5R₃

$det\left(B\right)=\left|\begin{array}{ccc}\hfill 2a_1\hfill & \hfill 2a_2\hfill & \hfill 2a_3\hfill \\ \hfill 4b_1+10c_1\hfill & \hfill 4b_2+10c_2\hfill & \hfill 4b_3+10c_3\hfill \\ \hfill 2c_1\hfill & \hfill 2c_2\hfill & \hfill 2c_3\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill 2a_1\hfill & \hfill 2a_2\hfill & \hfill 2a_3\hfill \\ \hfill 4b_1\hfill & \hfill 4b_2\hfill & \hfill 4b_3\hfill \\ \hfill 2c_1\hfill & \hfill 2c_2\hfill & \hfill 2c_3\hfill \end{array}\right|=16det\left(A\right)=64$

Adichunchanagiri University offers multiple specialisations in the BTech / BE programme. The course is offered in full-time mode and for a period of four years. Popular options include Civil Engineering, Computer Science Engineering, Electronics Engineering, Mechanical Engineering, and Artificial Intelligence & Machine Learning. These specialisations allow students to focus on a specific niche.

A tristor is formed by doping a semi conductor water from one side by N-type dopant (high concentration while comparatively lower concentration on other side thus forming NPN tristor & vice-versa.

So, statement (I) is false & statement (II) is true.

Number divisible by 3;

(a) sum of the digit must be divisible by 3

(i)  $\boxed{1}\boxed{2}\boxed{3}$  3! = 6

(ii) 1, 3, 5 -> 3! = 6

(iii) 2, 3, 4 -> 3! = 6

(iv) 3, 4, 5 = 3! = 6

Total = 24

(b) Divisible by

$\boxed{}\boxed{}\boxed{5}$  4 × 3 = 12

(c) Now common divisible by both

$\boxed{1}\boxed{3}\boxed{5}$ 2! = 2

For 3, 4  $\boxed{}\boxed{}\boxed{5}$  2! = 2

$\therefore$ Total ways = 24 + 12 – 4 = 32