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Candidates can apply online for the course of their choice at the School of Law, Pandit Deendayal Energy University. Below are the documents required for registration:

• Class 10, Class 12 and graduation marksheet
• School Leaving Certificate
• Transfer Certificate
• Entrance exam result
• Recent photograph
• Signature

The institute has been providing good placements to its recruiters since the very beginning. For more information on placement packages and recruiters for the Justice K S Hegde Institute of Management placements during 2021-2023 for MBA, refer to the table below:

There are eight semesters in the curriculum of BTech course of IIIT Ranchi. The entire course is four years long, offered in full time basis only. Each specialisation of BTech course at IIIT Ranchi has eight semesters.

Yes, Poona College of Pharmacy does offer a BPharm course for the duration of four years. Further, the course is offered in full-time mode. The college is a constituent college of Bharati Vidyapeeth-Deemed-to-be-University. The minimum eligibility requirement for admission to the BPharm course is that candidates must complete their Class 12 from the PCM/PCB stream with an aggregate of 50% (45% for SC/ST candidates).

 $\frac{dy}{dx}-y=2-e^{-x}$

$I.F.=e^{-{\displaystyle \int dx}}=e^{-x}$

$\therefore$ soln

$y{e}^{-x}={\displaystyle \int \left(2e^{-x}-e^{-2x}\right)dx}$

$y=-2+\frac{e^{-x}}{2}+C{e}^x$

as for x $\infty$ y finite c = 0

$\therefore y=\frac{e^{-x}}{2}-2$

$\begin{array}{l}\Rightarrow x+2y=-3\\ \Rightarrow a=-3b=\frac{-3}{2}\end{array}$

$\therefore a=4b=-3+6=3$

k_{eq= $\frac{L_1+L_2}{\frac{L_1}{k_1}+\frac{L_2}{k_2}}=\frac{4+2.5}{\frac{4}{k}+\frac{2.5}{2k}}$}

_{keq= $\frac{6.5}{\frac{8+2.5}{2k}}=\frac{6.5}{10.5}\times 2k=\frac{13\times 2}{21}k$}

_{= $\left(1+\frac{5}{51}\right)k$}

_{= $\left(1+\frac{5}{\alpha }\right)$ =21}

(D)

$\sim \left(p\iff \sim q\right)\wedge q=\left(p\iff \sim q\right)\wedge qis:$

$\therefore \left(\sim \left(P\iff \sim Q\right)\right)\wedge$ q is equivalent to $\left(p\Rightarrow q\right)\wedge$ p

School of Law, Pandit Deendayal Energy University offers BA LLB and LLB courses. Admission to the School of Law, Pandit Deendayal Energy University, is entrance-based. Candidates who have completed Class 12 or graduation are eligible to apply for the School of Law, Pandit Deendayal Energy University courses.

$L_1:\mathcal{l}x-y+3\left(1-z\right)=1,x+2y-z=2$

plane containing the line P : 3x – 8y + 7z = 4

If $\overrightarrow{n}$ be vector parallel to L.

then $\overrightarrow{n}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill \mathcal{l}\hfill & \hfill -1\hfill & \hfill 3\left(1-\mathcal{l}\right)\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=\left(6\mathcal{l}-5\right)\widehat{i}+\left(3-2\mathcal{l}\right)\widehat{j}+\left(2\mathcal{l}+1\right)\widehat{k}$ as P containing the line

$\therefore 3\left(6\mathcal{l}-5\right)-8\left(3-2\mathcal{l}\right)+7\left(2\mathcal{l}+1\right)=0$

$\Rightarrow \mathcal{l}=\frac{2}{3}$

If be the acute angle between line L & Y axis then cos = $\frac{5/3}{\sqrt[]{1+\frac{25}{9}+\frac{49}{9}}}=\frac{5}{\sqrt[]{83}}$

$\therefore 415co{s}^2\theta =125$

 $tan\left(2ta{n}^{-1}\frac{1}{5}+se{c}^{-1}\frac{\sqrt[]{5}}{2}+2ta{n}^{-1}\frac{1}{8}\right)tan\left(2ta{n}^{-1}\frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5}\times \frac{1}{8}}+se{c}^{-1}\frac{\sqrt[]{5}}{2}\right)$

$=tan\left(ta{n}^{-1}\frac{3}{4}+ta{n}^{-1}\frac{1}{2}\right)=tan\left(ta{n}^{-1}\frac{\frac{3}{4}+\frac{1}{2}}{1-\frac{3}{8}}\right)$

$=tan\left(ta{n}^{-1}\frac{\frac{5}{4}}{\frac{5}{8}}\right)=2$

 $\overline{x}=\frac{\sum xi}{40}=30\Rightarrow \sum xi=1200$ ………. (i)

${\alpha }^2=\frac{1}{40}\sum x{i}_i^2-{\left(30\right)}^2=25$

$\Rightarrow \sum x_i^2=37000$

after omitting two wrong observations

$\sum y_i^2=37000-144-100=36756$

$\therefore 38a^2=36756-36158=238$

The BTech curriculum of IIIT Ranchi has been designed to allow students to easily understand the difficult concepts by creating a strong understanding of relevant subjects. For example, in the BTech in Computer Science and Engineering, the curriculum includes subjects like mathematics in the early semesters before moving to complex subjects like Python programming.

Ajeenkya DY Patil University Seamedu offers a variety of courses to candidates ranging from both UG and PG levels. These include- BSc, BBA, BA, BTech, BCA, MCA, MBA, etc. The university also offers Diploma courses to candidates in multiple specialisations, such as- Film Making, Cloud Computing, etc. Candidates can apply online on the official website of the institute. 

Yes, it is a best choice to pursue BTech degree from Netaji Subhas University of Technology. The university offers BTech specialisation in Computer Engineering, Mathematics & Computing, Electrical Engineering, Mechanical Engineering, Bio-Technology, Information Technology, and others. NSUT is a well-regarded institution with a strong reputation for academics, placements, and a vibrant campus life. Many reputable companies recruit from NSUT, offering good placement opportunities for B.Tech graduates. 

 $S=\left\{\theta \in \left[0,2\pi \right]:8^{2si{n}^{2}x}+8^{2co{s}^{2}x}=16\right\}$

Now apply AM $\ge GM$ for $\frac{8^{2si{n}^{2}x}+8^{2co{s}^{2}x}}{2}\le {\left(8^{2si{n}^{2}x+2co{s}^{2}x}\right)}^{\frac{1}{2}}\Rightarrow 8^{2si{n}^{2}x}=8^{2co{s}^{2}x}$

$si{n}^2\theta =co{s}^2\theta$ $\therefore \theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$

$=4+\left[cosec\left(\frac{\pi }{2}+\pi \right)+cosec\left(\frac{\pi }{2}+3\pi \right)+cosec\left(\frac{\pi }{2}+5\pi \right)+cosec\left(\frac{\pi }{2}+7\pi \right)\right]$

$=4-2\left(4\right)=-4$

$2si{n}^2\theta =cos2\theta =1-2si{n}^2\theta$

$\Rightarrow 4si{n}^2\theta =1\Rightarrow sin\theta =\pm \frac{1}{2}$

$\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}$

$Sum\frac{7\pi }{6}+\frac{11\pi }{6}=3\pi \Rightarrow k=3$

$\frac{B_0^2}{2{\mu }_0}C$

$B_0^2=\frac{I2{\mu }_0}{C}$

$B_0^2=\frac{0.22\times 2\times 4\times 10^{-7}}{9\times 108}$

B₀ = 42.9 × 10^-9

Yes. Candidates filling the Online UPTAC Application Form are required to upload scanned images of his/her the latest color photograph and signature. Application forms having blurred photographs are liable to be rejected. Therefore candidates must be careful while uploading the scaneed images of their photo and signature in the UPTAC 2025 registration form.

 ${\displaystyle \sum _{k=1}^{10}\frac{k}{k^4+k^2+1}}$

$=\frac{1}{2}{\displaystyle \sum _{k=1}^{10}\left[\frac{1}{k^2-k+1}-\frac{1}{k^2+k+1}\right]}$

$=\frac{1}{2}\left[1-\frac{1}{111}\right]=\frac{110}{222}=\frac{55}{111}=\frac{m}{n}$

$\therefore m+n=166$