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Students can get admission to the top MBA colleges in Chennai only through entrance exams. The table given below offers some details about the commonly accepted exams at MBA colleges in Chennai.

Disclaimer: This information is sourced from the official website and may vary.

The median package offered during Vignan's Foundation for Science, Technology and Research placements 2024 for BTech and M.Tech programs is presented below:

132. Given, ${(x-a)}^2+{(y-b)}^2=c^2.$

Differentiating w r t ‘x’ we get

$\frac{d}{dx}{(x-a)}^2+\frac{d}{dx}{(y-b)}^2=\frac{d}{dx}{c}^2$

$\Rightarrow 2(x-a)+2(y-b)\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=-\frac{2(x-a)}{2(y-b)}=-\frac{(x-a)}{(y-b)}$

Again, $\frac{d^{2}y}{d{x}^2}=-\left\{\frac{(y-b)\frac{d}{dx}(x-a)-(x-a)\frac{d}{dx}(y-b)}{{(y-b)}^2}\right\}$

$=-\left\{\frac{(y-b)-(x-a)\frac{dy}{dx}}{{(y-b)}^2}\right\}$

$=-\left\{\frac{(y-b)+(x-a)\frac{(x-a)}{(y-b)}}{{(y-b)}^2}\right\}$

$=-\left\{\frac{{(y-b)}^2+{(x-a)}^2}{{(y-b)}^3}\right\}$

$=-\frac{c^2}{{(y-b)}^3}\left\{?{(x-a)}^2+{(y-b)}^2=c^2\right\}$

Then, L.H.S = $\frac{{\left\{1+{\left(\frac{dy}{dx}\right)}^2\right\}}^{3/2}}{\frac{d^{2}y}{d{x}^2}}=\frac{{\left\{1+\frac{{(x-a)}^2}{{(y-b)}^2}\right\}}^{3/2}}{\frac{-c^2}{{(y-b)}^2}}$

$=\frac{{\left\{{(y-b)}^2+{(x-a)}^2\right\}}^{3/2}}{{(y-b)}^3}\times \frac{{(y-b)}^3}{-c^2}$

$=\frac{c^{2\times 3/2}}{-c^2}=\frac{c^3}{-c^2}=-c$ Where c is a constant and is independent of a and b.

The placement trend witnessed during Vignan's Foundation for Science, Technology and Research placements over the past three years is presented below:

13. The general term of the expansion ${\left(3+ax\right)}^9$ is

T_r+1 = ⁹C_r $3^{\left(9-r\right)}$ ${\left(ax\right)}^r$

= ⁹C_r $3^{\left(9-r\right)}{a}^r{x}^r$

At r = 2,

T₂₊₁ = ⁹C₂ $3^{\left(9-2\right)}{a}^2{x}^2$

= $\frac{9!}{2!\left(9-2\right)!}$ 3⁷a²x²

= $\mathrm{9\prime }$$\mathrm{8\prime }$$7!$/$\mathrm{2\prime }$$\mathrm{1\prime }$$7!\mathrm{}$ 3⁷a²x²

= 36 ×3⁷a²x²

At r = 3,

T₃₊₁ = ⁹C₃ $3^{\left(9-3\right)}{a}^3{x}^3$

= $\frac{9!}{3!\left(9-3\right)!}$ 3⁶a³x³

= 9'8'7'6!/3'2'1'6! 3⁶a³x³

= 84 ×3⁶a³x³

Given that,

Co-efficient of $x^2$ = co-efficient of $x^3$

=> 36 × $3^7{a}^2$ = 84 × $3^6{a}^3$

=> $\frac{a^3}{a^2}$ = 36' 3^{7  /84'}3⁶

=> $a$ = $\mathrm{3\prime 3/7}$

= $\frac{9}{7}$

The key highlights of Vignan's Foundation for Science, Technology and Research placements 2024 for BTech and M.Tech programs are presented below:

Sure! Take a look at the table below for some details about the top MBA colleges in Chennai:

Disclaimer: This information is sourced from the official website and may vary.

BAU Sabour has a state-of-the-art campus with multiple facilities available at the university. The university has a big library with a large collection of books and journals. The below-mentioned are the Library Resource Developments:

131. Kindly go through the solution

Here are the course fees of the top MBA colleges in Chennai:

130. Kindly go through the solution

Scroll down to know the steps to fill out the application form for the CDS exam:

• Go to the official website for application
• Log in to your registered account or register for the OTR (the steps to register for UPSC OTR have been given below in the article)
• Click on the CDS exam link
• Fill out the details asked
• Upload the photograph, signature, and Photo Identity card
• Make the application fee for the CDS exam 
• Choose the CDS exam centre
• Finally, submit the application form 
• A confirmation message will be sent to the candidate's registered mobile number and email id 

129. Given, $y=12(1-cost). x=10(t-sint).$

Differentiating w r t ‘t’ we get,

$\frac{dy}{dt}=12\frac{d}{dt}(1-cost)=12(0-(-sint))=12sint.$

$\frac{dx}{dt}=10\frac{d}{dt}(t-sint)=10(1-cost).$

$\therefore \frac{dy}{dx}=\frac{dy/dt}{dx/dt}$

$=\frac{12sint}{10(1-cost)}=\frac{12sint}{10[1-cost]}$

$=\frac{12\times 2sint/tcost/2}{10\times 2si{n}^{2}t/2}$

$\left\{\begin{array}{cc}\hfill ?sin2\theta =2sin\theta cos\theta \hfill & \hfill cos2\theta =1-2si{n}^2\theta \hfill \end{array}\right\}$ $=\frac{6}{5}\frac{cost/2}{sint/2}=\frac{6}{5}cott/2$

Listed below are the top MBA colleges in Chennai along with their NIRF rankings of 2023, 2024, and 2025:

Disclaimer: This information is sourced from the official website and may vary.

11. The general term of the expansion ${\left(1+x\right)}^m$ is given by,

T_r+1 = ^mC_r $1^{m-r} x^r$

= ^mC_rx^r

At r = 2,

T₂₊₁ = ^mC₂x²

Given that, co-efficient of x² = 6

=>^mC₂ = 6

=> $\frac{m!}{2!\left(m-2\right)!}$ = 6

=>m² – m = 12

=>m² – m – 12 = 0

=>m² + 3m – 4m – 12 = 0

=>m (m + 3) – 4 (m+ 3) = 0

=> (m – 4) (m + 3) = 0

=>m = 4 and m = –3

Since, we need a positive value of m we have,  m = 4