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Netaji Subhas University of Technology offers UG programme across various streams such as Engineerig, Computer Application, Design, Architecture, and others. To be eligible for undergraduate (UG) courses at Netaji Subhas University of Technology (NSUT), Delhi, candidates generally need to have passed their 10+2 (or equivalent) examination with a minimum of 60% marks in aggregate. The institute accepts national entrance exams such as JEE Main, JAC Delhi,  CUET, MAT, NATA, CAT,  CMAT, UCEED, and others. 

Yes, RGUKT: Rajiv Gandhi University of Knowledge Technologies Andhra Pradesh offers PhD course to candidates. The course is offered in full-time mode and for three years of duration. Admissions to this course are entrance-based. The exams accpeted for this course by RGUKT are GATE, UGC NET and CSIR NET. Additionally, candidates must ensure fulfilling the eligibility criteria before applying for admission in PhD cousrse at RGUKT.

 $f\left(x\right)=\{\begin{array}{cc}\hfill x^3-x^2+10x-7,\hfill & \hfill x\le 1\hfill \\ \hfill -2x+lo{g}_2\left(b^2-4\right),\hfill & \hfill x>1\hfill \end{array}$

If f (x) has maximum value at x = 1 then

$f\left(1\right)\le f\left(1\right)\Rightarrow -2+lo{g}_2\left(b^2-4\right)\le 1-1+10-7$

$\Rightarrow lo{g}_2\left(b^2-4\right)\le 5\Rightarrow 0<b^2-4\le 32$

$\Rightarrow b^2-4>0\Rightarrow b\in \left(-\infty ,-2\right)\cup \left(2,\infty \right)$ ……. (i)

$And{b}^2-4\le 32\Rightarrow b\in \left[-6,6\right]$ ……. (ii)

From (i) and (ii) we get $b\in [-6,-2)\cup (2,6]$

To make accurate IIIT Ranchi comparisons, you need to take into account the IIIT Ranchi fees, student review, infrastructure, curriculum, scholarships, rankings, and future career prospects. Comparing these factors with other similar institutes will give you accurate results of IIIT Ranchi.

Abscissae of PQ are roots of x² – 4x – 6 = 0

Ordinates of PQ are roots of y² + 2y – 7 = 0 and PQ is diameter

$\therefore$ Equation of circle is $x^2+y^2-4x+2y-13=0$   ……………. (i)

But, given  $x^2+y^2+2ax+2by+c=0$ ……………. (ii)

By comparison a = -2, b = 1, c = -13 Þ a + b – c = -2 + 1 + 13 = 12

The CENTAC NEET Counselling merit list is released in two stages- Draft and Final. The draft merit list is released to invite objections from students, based on which the final merit list is released. Candidates whose names feature on the final merit list can report to the institutes for admissions.

 $f\left(x\right)=\{\begin{array}{cc}\hfill x+a,\hfill & \hfill x\le 0\hfill \\ \hfill \left|x-4\right|,\hfill & \hfill x>0\hfill \end{array}andg\left(x\right)=\{\begin{array}{cc}\hfill x+1,\hfill & \hfill x<0\hfill \\ \hfill {\left(x-4\right)}^2+b,\hfill & \hfill x\ge 0\hfill \end{array}$

$?$ f (x) and g (x) are continuous on R $\therefore$ a = 4 and b = 1 – 16 = 15

then (gof) (2) + (fog) (2) = g (2) + f (-1) = -11 + 3 = -8

To apply for BITS Pilani Digital, the candidates must have their Class 10 and 12 marksheet and certificate, Bachelors degree (when applicable), a passport size photo, Aadhar Card, and caste certificate (when applicable) in readiness. They should be submitted together with application form and pay slip to be successfully submitted.

Yes students can download their AIIMS Paramedical 2026 scorecard from portal by using their roll number and date of birth. It is advisable to take printouts of several copies of the scorecard to be used at time of counselling and admission.

Application to BITS Pilani Digital entirely takes place online and begins on the official site. An applicant has to do registration, download the form, make payment, and provide all necessary details. The bank challan and documents together with filled form should either be sent by post or personally. The application which is successfully submitted will only be considered during the admission.

 $f\left(x\right)=\{\begin{array}{l}\frac{lo{g}_e\left(1-x+x^2\right)+lo{g}_e\left(1+x+x^2\right)}{secx-cosx},x\in \left(\frac{-\pi }{2},\frac{\pi }{2}\right)-\left\{0\right\}\\ k,x=0\end{array}$ for continuity at x = 0

$\underset{x\to 0}{lim}f\left(x\right)=k\therefore k=\underset{x\to 0}{lim}\frac{lo{g}_e\left(1+x^2+x^4\right)}{secx-cosx}\left(\frac{0}{0}form\right)=\underset{x\to 0}{lim}\frac{cosxlo{g}_e\left(1+x^2+x^4\right)}{si{n}^{2}x}=1$

The ties in AIIMS Paramedical results are done based on the following pointers.
Ties are resolved in the AIIMS Paramedical exam by prioritizing candidates with higher marks in Biology/Mathematics, followed by Chemistry, then Physics. If the tie persists, older candidates are given preference.

Candidates with a valid Bachelor's degree passed with at least 50% aggregate (45% for SC/ST candidates) can apply for admission and take the entrance test. In addition, the candidates appearing for the final year exams can also apply. It is advised that the candidates keep the important documents (original and copies) ready with them. These are required for verification purposes by the institute.         

a =   $\frac{qE}{m}=\frac{40*10^{?6}*10^5}{100*10^{?3}*10^{?3}}$

                      = 40 * 10³ m/sec²

              v² = u² + 2a s

$?0={\left(200\right)}^2?2*40*10^{3}s$

$s=\frac{200*200}{2*40*10^3}=0.5m$                

Equation of tangent of slope m to y = x² is y = mx $\frac{1}{4}{m}^2$ - …………. (i)

Equation of tangent of slope m to y = - (x - 2)² is y = m (x – 2) + $\frac{1}{4}{m}^2$  …………… (ii)

If both equation represent the same line therefore on comparing (i) and (ii) we get m = 0, 4

therefore equation of tangent is y = 4x – 4

Admission to the School of Law, Pandit Deendayal Energy University courses involves a few steps. Candidates must first check if they meet the eligibility criteria and then apply online through the official website. The application form should be filled in and submitted with the required documents. The selection criteria are entrance-based. 

Given G.P’s 2, 2², 2³, …60 term and 4, 4², 4³, … of 60

Now G.M. = ${\left(2\right)}^{\frac{225}{8}}\Rightarrow {\left(2,2^2,2^3,....\right)}^{\frac{1}{60+n}}={\left(2\right)}^{\frac{225}{8}}\Rightarrow n=\frac{57}{8},20son=20$

$\therefore {\displaystyle \sum _{k=1}^{n}k(n-k)}20\times \frac{20\times 21}{2}-\frac{20\times 21\times 41}{6}=1330$

Yes, TSSM College of Engineering Pune offers scholarships to students as financial assistance. The below-mentioned are the TSSM College of Engineering Pune scholarship details that are mentioned below:

The total cost of pursuing BTech at IIIT Ranchi can vary between candidates. Hoewever, the cost of pursuing BTech at IIIT Ranchi is going to be around INR 13.24 lakh. Candidates are advised to confirm the fees they need topay from the official website.