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Deen Dayal Upadhyaya College, Delhi University provides the flagship programme of Bachelor of Science (BSc), which is a popular choice for students in Engineering. Deen Dayal Upadhyaya College BSc admissions are held on the basis of entrance test, and accepted examis CUET. Interested candidates can visit the official website and begin the application process for Deen Dayal Upadhyaya College BSc.

Let the equation of normal is Y – y = - $\frac{1}{m}\left(X-x\right)$  

where m is slope of tangent to the given curve then

  $Y-y=-\frac{dx}{dy}\left(X-x\right)$         

It passes through (a, b) so b – y = $\frac{-dx}{dy}\left(a-x\right)$

=> (a – x) dx = (y – b) dy

On integration     $ax-\frac{x^2}{2}=\frac{y^2}{2}-by+c.........\left(i\right)$  

(ii) passes through (3, -3) &  then

3a – 3b – c = 9       .(ii)

& 4a - $2\sqrt[]{2}b$ - c = 12           .(iii)

also given  $a-2\sqrt[]{2}b=3............(iv)$

Solve (ii), (iii) & (iv) b = 0, a = 3

Hence a² + b² + ab = 9

The number of companies visited the campus in the past three years during FSM Delhi placement is presented below:

The number of PPOs / PPIs made during FSM Delhi placements over the past three years is presented below:

Quality factor =   $\frac{\omega L}{R}=\frac{2\times 3.14\times 10\times 10^6\times 2\times 10^{-4}}{6.28}=2000$

The Columbia University MBA acceptance rate stands at 12% for 2024, per data curated from official sources. The university received around 7,487 applications, out of which, only 972 applications were accepted. 

The Columbia University MBA average GMAT score stood at 732, while the Columbia University MBA average GPA, stood at 3.6. 

Source: Columbia University Business School

The placement trend witnessed during FORE School of Management in the past three years is presented below:

Given $l_{m,n}={\displaystyle {\int }_0^1{x}^{m-1}{\left(1-x\right)}^{n-1}dx............\left(i\right)}$  

put 1 - x = $t\{\begin{array}{l}x=0,t=1\\ x=1,t=0\end{array}$

dx = -dt

From (i) $l_{m,n}={\displaystyle {\int }_1^0{\left(1-t\right)}^{m-1}.t^{n-1}\left(-dt\right)}$

$l_{m,n}={\displaystyle {\int }_0^1{t}^{n-1}{\left(1-t\right)}^{m-1}dt={\displaystyle {\int }_0^1{x}^{n-1}{\left(1-x\right)}^{m-1}dx..........\left(ii\right)}}$    

$Putx=\frac{1}{1+y}in\left(i\right)thendx=-\frac{1}{{\left(1+y\right)}^2}dy$                          

(i)  $l_{m,n}={\displaystyle {\int }_{\infty }^0\frac{1}{{\left(1+y\right)}^{m-1}}}{\left(1-\frac{1}{1+y}\right)}^{n-1}\left(\frac{-dy}{{\left(1+y\right)}^2}\right)={\displaystyle {\int }_0^{\infty }\frac{y^{n-1}}{{\left(1+y\right)}^{m+n}}dy}..........\left(iii\right)$

Similarly by (ii) $l_{m,n}={\displaystyle {\int }_0^{\infty }\frac{y^{m-1}}{{\left(y+1\right)}^{m+n}}dy..........\left(iv\right)}$

Adding (iii) & (iv) $2l_{m,n}={\displaystyle {\int }_0^{\infty }\frac{y^{n-1}+y^{m+1}}{{\left(y+1\right)}^{m+n}}}$

Putting  $\frac{1}{z}\{\begin{array}{l}y=1,z=1\\ y=\infty ,z=0\end{array}$ $\Rightarrow dy=\frac{-1}{z^2}dz$  

Hence $l_{m,n}={\displaystyle {\int }_0^1\frac{x^{m-1}+x^{n-1}}{{\left(1+x\right)}^{m+n}}}$ dx = α lm, n

=> α = 1

f = W = 0.5 × 10 = 5 N

N = F

For block not to slide,

  $f\le \mu N$          

$\begin{array}{l}\Rightarrow 5\le 0.2F\\ \Rightarrow F\ge 25N\end{array}$

Ratio of masses on two pistons of the hydraulic lift equals to that of their cross- section area.

$\frac{A_1}{A_2}=\frac{100}{m}$            

Now,     $\frac{4^2{A}_1}{A_2/4^2}=\frac{M}{m}?M=256\frac{A_1}{A_2}.m=25600kg.$

Given   ${\displaystyle \sum _{i=1}^{18}\left(x_i-\alpha \right)=36i.e}{\displaystyle \sum _{i=1}^{18}{x}_i-18\alpha =36..........\left(i\right)}$

&       ${\displaystyle \sum _{i=1}^{18}{\left(x_i-\beta \right)}^2=90i.e{\displaystyle \sum _{i=1}^{18}{x_i}^2}}-2\beta {\displaystyle \sum _{}^{}{x}_i+18{\beta }^2=90.............\left(ii\right)}$         

(i) & (ii)  ${\displaystyle \sum _{i=1}^{18}{x_i}^2}=90-18\beta +36\beta \left(\alpha +2\right).............\left(iii\right)$

Now variance ${\sigma }^2=\frac{\sum {x_i}^2}{n}-{\left(\frac{\sum x_i}{n}\right)}^2$ = 1 given

=> (α - β) (α - β + 4) = 0

Since   $\alpha \ne \beta so\left|\alpha -\beta \right|=4$

No, there is no entrance exam required for admission to various courses at Sewa Community College, Andheri. Admission to all the offered courses at the college is completely merit-based. Students only need to meet the basic eligibility requirements defined by the college for the respective programmes.

Lady Shri Ram College offers the flagship BA course in various specialisations. LSR BA course admissions are held on the basis of entrance tests. The total tuition fees is INR 60.81 K - 76.47 K. Accepted examination is CUET. Students can visit the official website of Lady Shri Ram College to learn more and apply online.

Percentage Modulation =   $\frac{A_m}{A_c}*100=\frac{20}{80}*100=25\mathrm{\%}$

International students can choose from more than 100 degree programs on the bachelor's as well as master's levels at Columbia University. According to the historical data of Columbia University alumni, some of the Columbia University Popular Degrees are: Economics, Computer Science, Political Science, Psychology and Finance. Here's a detailed breakdown of the same:

Given $f\left(x\right)=2x^5+5x^4+10x^3+10x^2+10x+10$ ${}^{}{}^{}{}^{}{}^{}$

$f\left(-1\right)=3>0\&f\left(-2\right)=-34<0$

So at least one root will lie in (2, 1)

now $f\text{'}\left(x\right)=10x^4+20x^3+30x^2+20x+10$ ${}^{}{}^{}{}^{}$

$=10\left[x^4+2x^3+3x^2+2x+1\right]$

$=10x^2{\left(x+\frac{1}{x}+1\right)}^2>0\forall x\in R$

So, f (x) be purely increasing function so exactly one root of f (x) that will lie in (-2, 1). Hence |a| = 2

IHM Gwalior BSc ranks competitively in hospitality education. In The Week 2023 rankings, it was the 2nd best in Madhya Pradesh. In Times Higher Education 2023 'Impact' rankings, it shares the #1001 position with IHM Goa and IHM Ahmedabad, while institutes like IMI Bhubaneswar rank higher at #50.

The institute offers BVoc and MVoc courses. Apart from the UG and PG courses, the institute also offers Certificate and Diploma courses. Candidates must complete Class 12 as the basic eligibility criteria. Once selected, candidates must pay the programme fee to confirm their seats in the institute. Admission to all the courses is merit-based.

Potential difference across 2k $\Omega$  is 5V, thus current through it,

  $i=\frac{5}{2\times 10^3}=25\times 10^{-4}A.$          

$Given\left|z+5\right|\le 4..........\left(i\right)$

->Represent a circle ${\left(x+5\right)}^2+y^2\le 16$

$=z\left(1+i\right)+\overline{z}\left(1-i\right)\ge -10..........\left(ii\right)$

->Represent a line X – y $\ge -5$

So max |z + 1|² = AQ²

$={\left(-4-2\sqrt[]{2}\right)}^2+8$

$\begin{array}{l}=32+16\sqrt[]{2}\\ =\alpha +\beta \sqrt[]{2}\end{array}$  

Hence α + β) = 48