Ask & Answer: India's Largest Education Community

ESEDS School of Design was established in 2015 with a vision to promote sustainable and globally relevant design education. The college is located in Kolkata and is one of the renowned colleges in West Bengal to offer quality education to students. Further, EcoAvid School of Ethical Design Studies is approved by the UGC. 

Yes, Biyani Girls B.Ed. College accepts 60 percent in Class 12 to pursue UG courses. In case of BEd, the applicants will have to qualify PTET as well. Most courses require the bare minimum score which is enclosed by your score in Class 12. Fill online form and follow merit list schedule.

(92-94) percentile for special rounds or lesser branches in nit shrinagar 

It also depends on your category 

Save zone is 98 percentile.Strong chance to get machanical engineering in earlier rounds.

Yes, there are merit based admission courses such as BA, MA in Biyani Girls B.Ed. College. But Post graduate course like MEd needs PTET, and BEd needs MPET. Make sure to check on the requirements per course before submitting.

ESEDS, also known as EcoAvid School of Ethical Design Studies, is a private design college and is based in Kolkata, West Bengal. The college offers sustainable fashion and interior design courses to students in collaboration with Techno India University. Further, ESEDS is recognised as India's first sustainable Fashion and Interior Design college. 

$\frac{{\left(x-3\right)}^2}{16}+\frac{{\left(y-4\right)}^2}{9}\le 1,x,y\in N,{\left(x-7\right)}^2+{\left(y-4\right)}^2\ge 36,x,y\in R$

Total number of common point = 1 + 5 + 7 + 5 + 5 + 3 + 1 = 27

Given : ${\left|\overrightarrow{a}+\overrightarrow{b}\right|}^2={\left|\overrightarrow{a}\right|}^2+2{\left|\overrightarrow{b}\right|}^2\&\overrightarrow{a}\cdot \overrightarrow{b}=3$

$\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|cos\theta =3$

$\left|\overrightarrow{a}\right|cos\theta =\frac{3}{\sqrt[]{6}}=\sqrt[]{\frac{9}{6}}=\sqrt[]{\frac{3}{2}}$

${\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^{2}si{n}^2\theta =75$

$\left|\overrightarrow{a}\right|sin\theta =\sqrt[]{\frac{75}{6}}=\frac{5}{\sqrt[]{2}}$

$\left|\overrightarrow{a}\right|cos\theta =\sqrt[]{\frac{3}{2}}$

${\left|\overrightarrow{a}\right|}^2=\frac{25}{2}+\frac{3}{2}=\frac{28}{2}=14$

Let C be the centre and M be the mid point of AB

$\Delta APC:sin\theta =\frac{13/2}{pc}=\frac{5}{13}\Rightarrow pC=\frac{169}{10}$

$\Delta AMC:cos\theta =\frac{6}{13/2}=\frac{12}{13}$

PC = $\frac{169}{10},MC=\frac{13}{2}sin\theta =\frac{13}{2}\cdot \frac{5}{13}$

PM = PC – MC = $\frac{169}{10}-\frac{5}{2}=\frac{144}{10}$

5PM = 72

y = 5x² + 2x – 25

P(2, -1)

T_(p) : T = 0

Þ y – 1 = 10x(2) + 2(x + 2) – 50

$\Rightarrow y=22x-45$ is also tangent to y = x³ – x² + x at point (a, b)

For y = x³ -x² + x

$\frac{dy}{dx}=3x^2-2x+1>0$               

y = 22x $-\frac{1939}{27}$  which is not tangent to the curve.

$3a^2-2a+1=22$ (slope of tangent)

$\Rightarrow 3a^2-2a-21=0\to a=\frac{2\pm \sqrt[]{4+252}}{6}=\frac{2\pm 16}{6}=3,\frac{-7}{3}$               

b = 27 – 9 + 3 = 21

tangent : y – 21 = 22(x – 3)

 ⇒ y = 22x – 45

a = 3, b = 21

2a + 9b = 6 + 189 = 195

Also, $a^3-a^2+a=b$  

For a = $\frac{-7}{3}$  

b = $\frac{-343}{27}-\frac{49}{9}-\frac{7}{3}$  

= $\frac{-343-147-63}{27}=\frac{-553}{27}$  

$f\left(x\right)=4\left|2x+3\right|+9\left[x+\frac{1}{2}\right]-12\left[x+20\right],-20<x<20$ doubtful points for differentiability : $x=\frac{-3}{2},$

$f\left(x\right)=4\left(2x+3\right)+9\left(-1\right)-12\left(-2\right)-240=8x-213forx=\frac{-3}{2}+h$

= $-8x-230$ for x = $\frac{-3}{2}-h$

Not diff. at x = $\frac{-3}{2}$

other doubtful points : $x+\frac{1}{2}=integer$

$-20+\frac{1}{2}<x+\frac{1}{2}<20+\frac{1}{2}$

$x+\frac{1}{2}=-19,-18,....,19,20$

$x=-19.5,-18.5,-17.5,.......,18.5,19.5\to$ total 40 numbers.

No. of number = 19.5 – $\left(-19.5\right)+1=40\left(-1.5\right)included$

$-20<x<90\Rightarrow x=-19,-18,.....,18,15\to 39points$

No. of number = 19 (19) + 1 = 39

Total : 40 + 39 = 79

• It appears the college does not disclose the fee structure online, possibly because it's a government?aided institution with regulated or subsidised pricing.

• Fee amounts may vary by category (General/OBC/SC/ST/EWS), and extra charges (library, development, exam, etc.) may apply.

• Government-aided colleges in Haryana sometimes do not publicly display detailed fee data to avoid frequent updates.

$r{\cdot }^n\text{?}{C}_r=n{\cdot }^{n-1}\text{?}{C}_{r-1}$

${\displaystyle \sum _{k=1}^{10}{\left(k{\cdot }^{10}\text{?}{C}_k\right)}^2}={\displaystyle \sum _{k=1}^{10}{\left(10{\cdot }^9\text{?}{C}_{K-1}\right)}^2}$

= $100{\cdot }^{18}\text{?}{C}_9$

22000L = $100{\cdot }^{18}\text{?}{C}_9$

L =

$18!=2^{16}\cdot 3^8\cdot 5^3\cdot 7^2\cdot 11^1\cdot 13\cdot 17$

$9+4+2+1$ $9!=2^7\cdot 3^4\cdot 5^1\cdot 7^1$

6 + 24 + 2 + 1 $\frac{18!}{{\left(9!\right)}^2}=2^2\cdot 5\cdot 11\cdot 13\cdot 17$

3 +3 + 1

= 221

1012 $\le$  Number in the question $\le$  23421

Also, the number has to use digits {2, 3, 4, 5, 6} without repetition and the number has to be divisible by $\frac{55}{11\times 5}$  

As the number has to be divisible by both 5 and 11,

5 → once place

Let us make 4-digit such numbers first:

{2, 3, 4, 6} (digits are not be repeated)

A number is divisible by 11 it difference of sum of its digits at even places and sum of digits at odd place is 0 or multiple of 11.

$A=\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]$

$A^2=\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]$

$=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=I+B,B=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$

$A^4=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]B^2\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]=0$

$A^{2n}={\left(I+B\right)}^n$

I + nB + 0 + 0 +……….

$A^{2n}=I+\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 6n\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 6n\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$X\text{'}{A}^{k}x=\left[33\right]$

$=\left[111\right]A^k\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]$

$\left[111\right]A^{2n}\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]\left(k=2n\right)$

= $\left[1+1+6n+1\right]=\left[6n+3\right]=\left[33\right]n=5$

k = 10

x² – x – 4 = 0

$P_n={\alpha }^n-{\beta }^n$

$?=\frac{P_{15}{P}_{16}-P_{14}{P}_{16}-P_{15}^2+P_{14}{P}_{15}}{P_{13}{P}_{14}}$

$=\frac{\left(P_{15}-P_{14}\right)\left(P_{16}-P_{15}\right)}{P_{13}{P}_{14}}$

$=\frac{4P_{13}\cdot 4P_{14}}{P_{13}\cdot P_{14}}=16$

P_n =  αⁿ - βⁿ

$={\alpha }^{n-1}\cdot \alpha -{\beta }^{n-1}\cdot \beta$

$={\alpha }^{n-1}\left({\alpha }^2-4\right)-{\beta }^{n-1}\left(\beta -4\right)$

$P_n={\alpha }^{n+1}-{\beta }^{n+1}-4\left({\alpha }^{n-1}-{\beta }^{n-1}\right)$

$P_n=P_{n+1}-4P_{n-1}\Rightarrow P_{n+1}-P_n=4P_{n-1}$