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To be eligible for admission at Zell Education, one must ensure that they meet the minimum educational qualifications set by the institute. The institute mainly accepts students based on academic merit secured in the last qualifying exam. Further, the criteria vary form course to course. The table below mentions the eligibility criteria for some of the popular programmes offered at the college.

$\Delta =4\Rightarrow \frac{1}{2}\left|\begin{array}{ccc}\hfill 1\hfill & \hfill \alpha \hfill & \hfill 1\hfill \\ \hfill \alpha \hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill \alpha \hfill & \hfill 1\hfill \end{array}\right|=4$

$\Rightarrow \alpha =\pm 8$

$\left(\alpha ,-\alpha \right),\left(-\alpha ,\alpha \right)and\left({\alpha }^2,\beta \right)$ are collinear.

$\therefore \left|\begin{array}{ccc}\hfill \alpha \hfill & \hfill -\alpha \hfill & \hfill 1\hfill \\ \hfill -\alpha \hfill & \hfill \alpha \hfill & \hfill 1\hfill \\ \hfill {\alpha }^2\hfill & \hfill \beta \hfill & \hfill 1\hfill \end{array}\right|=0\Rightarrow \beta =-64$

 $\frac{x^2}{a^2}+\frac{y^2}{4}=1$

$\therefore \Delta =\frac{1}{2}\times a\left(1+cos\theta \right).4sin\theta$

$\therefore {\Delta }_{max}^{\Delta }=6\sqrt[]{3}\Rightarrow a=4$

$\therefore e=\frac{\sqrt[]{3}}{2}$

To get admission in Zell Education, one must go through several steps, including application submission, online assessment/ counselling if any and admission confirmation. Students wishing to join the institute may do so by applying online and proceeding accordingly. Someone from the counselling team will reach out to guide you through the remaining process. Once an admission offer is made, pay the admission fees to secure the spot at the institute.

Equation of tangent at P (x, y) is Y = $\frac{dy}{dx}\left(X-x\right)$

$A/q,2x-y\frac{dx}{dy}=0\Rightarrow 2\frac{dy}{y}=\frac{dx}{x}$

$\begin{array}{l}\Rightarrow 2\mathcal{l}ny=\mathcal{l}nx+\mathcal{l}nc\\ y^2=xc\end{array}$

It passes through (3, 3), c = 3

$\therefore y^2=3x\therefore$ Length of latus rectum = 3

No, Zell Education does not offer direct admissions for any of its courses. Students seeking to enrol into the institute need to ensure that they satisfy both the basic eligibility requirements in addition to clearing the selection rounds (if any). Prospective students can reach out to the admission desk of the college to learn more about the Zell Education admission chances.

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

$\frac{\pi }{8}+\frac{1}{4}\mathcal{l}n2$

Admissions at Zell Education are ongoing. Students who are aiming for admission can open the official website of the institute and apply for the course they want. An enquiry form is also available for students on the official website. 

$l={\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{dx}{\left(1+e^x\right)\left(si{n}^{6}x+co{s}^{6}x\right)}}$ ……. (i)

$=2{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{dt}{4+t^2}=2{\left(ta{n}^{-1}\left(\frac{t}{2}\right)\right)}_0^{\infty }}=\pi$

$f\left(x\right)=\{\begin{array}{l}\frac{sin\left(x+2\right)}{x+2},x\in \left(-2,-1\right)\\ 0,x\in \left(-1,0\right]\\ 2x,x\in \left(0,1\right)\\ 1,otherwise\end{array}$

$LHD=\underset{h\to 0}{Lt}\frac{f\left(0-h\right)-f\left(0\right)}{-h}=0$

$RHD=\underset{h\to 0}{Lt}\frac{f\left(0+h\right)-f\left(0\right)}{h}=2$

Hence f (x) is not differentiable at x = 1, 0, 1

$\therefore m=2,n=3$

 $x_{1}y>0and{x}^3{y}^2=2^{15}$

$AM\ge GM$

$\frac{3x+2y}{5}\ge {\left(x^3{y}^2\right)}^{\frac{1}{5}}$

$\Rightarrow 3x+2y\ge 40$

x + y + az = 2………… (i)

3x + y + z = 4 ………… (ii)

x + 2z = 1 ……………. (iii)

x = 1, y = 1, z = 0

(for unique solution $a\ne -3$  

$\left(\alpha ,-1\right),\left(1,\alpha \right)and\left(1,-1\right)$ are collinear.

$\frac{\alpha -1}{1-\alpha }=\frac{-1-\alpha }{1-1}\Rightarrow 1-{\alpha }^2=0\Rightarrow \alpha =\pm 1$                

Sum of absolute value = 2

$e^{2x}-4=0or6e^{2x}-5e^x+1=0$

$\Rightarrow x=\mathcal{l}n2x=\mathcal{l}n\left(\frac{1}{3}\right),\mathcal{l}n\left(\frac{1}{2}\right)$

= $-\mathcal{l}n3,-\mathcal{l}n2$

$\left(x^2+1\right)*1=x*2$

$\Rightarrow {\left(x^2+1\right)}^2+1=x^2+8\Rightarrow x^2=2$

$2si{n}^{-1}\left(\frac{x^4+x^2-2}{x^4+x^2+2}\right)=2si{n}^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{3}$

There are about 156 Ethical Hacking colleges in India. Out of these,  32 colleges are owned by the Private sector and the Public sector owns 7. Some of the popular ethical hacking colleges in India are S.S. Jain Subodh Pg College, Gujarat University, IANT- Institute of Advance Network Technology, IZee Business School, Rayat Bhara University, Pillai College of Arts, Commerce and Arts, and many more. About 67 colleges ethical hacking course fee ranges < INR 1 lakh. 

Panache Academy Vadodara offers direct admission to candidates based on their merit. It is important for the aspirants to meet the eligibility requirements for the preferred course. Eligible candidates have to apply online or directly visit the campus for the application form. Once selected, aspirants can confirm their seat by paying the programme fee.

Candidates who want to get admission into the BCom course through an entrance test are required to appear and secure valid scores in LUACMAT. Moreover, if applicants want to get admission through the management quota, they can enrol in the programme via merit, released by the college after evaluating applicants' Class 12/ last qualifying scores. 

Admission to the BCom course at SRMCEM Lucknow is both merit- and entrance-based. For merit-based admission, applicants are required to secure good scores in their Class 12. The merit-based admissions are usually given to management quota students at SRMCEM. Moreover, for entrance-based admission, applicants are required to appear and secure good scores in LUACMAT stands for Lucknow University Associated Colleges Management Admission Test. 

Option A. Geography+ Environmental science+Statistics

Option B. Geography + political Science+ Sociology 

Option 3. Geography+ Economic+ Statistics