Dd Form 2400: Fill & Download for Free

Fill the MCI application form and submit required documents in check list together with application.Required documentsPassportBirth certificate10th certificate and mark list12th certificate and mark listAdmission letter from universityInvitation letter copyVisa copy if obtained alreadyDegree certificate if after the courseCatogory certificate if obc/se2 self attested copy of all above documentsEtc…A lso need to take a Bank DD in the name of MCI of Rs 2400 RS.You can find the application form and document check list in MCI website..All the bestRegards Dr. Patrick Royson

At first sight I thought that this was just yet another variant of one of the most repeated “which is greater?” questions on Quora (e.g. Which one is greater 15^2400 or 2400^15 ? )But no, this question is much more interesting. So how do you prove that if [math]0 \leq a \leq b \leq c \leq d[/math], then [math][/math][math] a^bb^cc^dd^a \geq b^ac^bd^ca^d[/math]?We’ll prove this by contradiction, by assuming that there is a solution [math]0 \leq a \leq b \leq c \leq d[/math] such that[math]a^b b^c c^d d^a < b^a c^b d^c a^d[/math]Applying log to both sides of the inequality give us:[math]b\ln(a) [/math][math][/math][math]+ c\ln(b) [/math][math][/math][math]+ d\ln(c) [/math][math][/math][math]+ a\ln(d) < a\ln(b) [/math][math][/math][math]+ b\ln(c) [/math][math][/math][math]+ c\ln(d) [/math][math][/math][math]+ d\ln(a) \implies[/math][math](b-d)\ln(a) [/math][math][/math][math]+ (c-a)\ln(b) [/math][math][/math][math]+ (d-b)\ln(c) [/math][math][/math][math]+ (a-c)\ln(d) < 0[/math]Now, let’s find where this formula is minimum. We’ll do that using the fact that at the minimum all the partial differentials become zero:[math]0 = \frac{\partial }{\partial a}(b-d)\ln(a) [/math][math][/math][math]+ (c-a)\ln(b) [/math][math][/math][math]+ (d-b)\ln(c) [/math][math][/math][math]+ (a-c)\ln(d) = [/math][math][/math][math][/math][math]\frac{b-d}{a} - \ln(b) [/math][math][/math][math]+ \ln(d) \implies[/math][math]\ln(b) - \ln(d) = \frac{b-d}{a} \implies a = \frac{b-d}{\ln(b) - \ln(d)}[/math][math]0 = \frac{\partial }{\partial b}(b-d)\ln(a) [/math][math][/math][math]+ (c-a)\ln(b) [/math][math][/math][math]+ (d-b)\ln(c) [/math][math][/math][math]+ (a-c)\ln(d) = [/math][math][/math][math][/math][math]\frac{c-a}{b} [/math][math][/math][math]+ \ln(a) - \ln(c) \implies[/math][math]\ln(c) - \ln(a) = \frac{c-a}{b} \implies b = \frac{c-a}{\ln(c) - \ln(a)}[/math][math]0 = \frac{\partial }{\partial c}(b-d)\ln(a) [/math][math][/math][math]+ (c-a)\ln(b) [/math][math][/math][math]+ (d-b)\ln(c) [/math][math][/math][math]+ (a-c)\ln(d) = [/math][math][/math][math][/math][math]\frac{d-b}{c} [/math][math][/math][math]+ \ln(b) - \ln(d)\implies[/math][math]\ln(d) - \ln(b) = \frac{d-b}{c} \implies c = \frac{d-b}{\ln(d) - \ln(b)}[/math][math]0 = \frac{\partial }{\partial d}(b-d)\ln(a) [/math][math][/math][math]+ (c-a)\ln(b) [/math][math][/math][math]+ (d-b)\ln(c) [/math][math][/math][math]+ (a-c)\ln(d) = [/math][math][/math][math][/math][math]\frac{a-c}{d} - \ln(a) [/math][math][/math][math]+ \ln(c) \implies[/math][math]\ln(a) - \ln(c) = \frac{a-c}{d} \implies d = \frac{a-c}{\ln(a) - \ln(c)}[/math]The first thing to notice is that: [math]a=c[/math] and [math]b=d[/math]However, we’re told that the variables are interleaved, so:[math]a \leq b \leq c \leq d \implies a=b=c=d[/math]From this we must conclude that at all its extremum points,[math](b-d)\ln(a) [/math][math][/math][math]+ (c-a)\ln(b) [/math][math][/math][math]+ (d-b)\ln(c) [/math][math][/math][math]+ (a-c)\ln(d) = 0[/math]With a little more analysis you can show that these extremum points are all minima. And therefore we cannot have a negative value, in contradiction with our assumption.[math]\blacksquare[/math][math][/math][math]\ [/math][math][/math][math][/math]EDIT 1:In the proof above I glossed over the analysis showing that the extremums are always a minima. It’s a minor point, but a necessary one. In this particular case I couldn’t find a way of showing this that is not cumbersome, but for completion I will describe here the lease ugly proof I could find.We showed that the minima occur when [math]a=b=c=d[/math]. So let’s see what happens when we perturb these variables by a small amount:[math]b = a [/math][math][/math][math]+ \beta [/math][math][/math][math]\ [/math][math][/math][math][/math] ; [math][/math][math] [/math][math][/math][math]\ c = a [/math][math][/math][math]+ \gamma [/math][math][/math][math]\ [/math][math][/math][math][/math] ; [math][/math][math] [/math][math][/math][math]\ d = a [/math][math][/math][math]+ \delta[/math]And we’ll use a second order approximation of the logarithms:[math]\ln(a+x) \approx \ln(a) [/math][math][/math][math]+ \frac{x}{a} - \frac{x^2}{2a^2} [/math][math][/math][math][/math]Putting this together:[math](b-d)\ln(a) [/math][math][/math][math]+ (c-a)\ln(b) [/math][math][/math][math]+ (d-b)\ln(c) [/math][math][/math][math]+ (a-c)\ln(d) =[/math][math](\beta-\delta)\ln(a) [/math][math][/math][math]+ \gamma\ln(a+\beta) [/math][math][/math][math]+ (\delta-\beta)\ln(a+\gamma) - \gamma\ln(a+\delta) \approx[/math][math](\beta-\delta)\ln(a) [/math][math][/math][math]+ \gamma (\ln(a) [/math][math][/math][math]+ \frac{\beta}{a} - \frac{\beta^2}{2a^2}) [/math][math][/math][math]+ (\delta-\beta)(\ln(a) [/math][math][/math][math]+[/math][math][/math][math] \frac{\gamma}{a} - \frac{\gamma^2}{2a^2}) - \gamma(\ln(a) [/math][math][/math][math]+ \frac{\delta}{a} - \frac{\delta^2}{2a^2}) =[/math][math]-\gamma \frac{\beta^2}{2a^2} - (\delta-\beta)\frac{\gamma^2}{2a^2} [/math][math][/math][math]+ \gamma\frac{\delta^2}{2a^2} =[/math][math]\frac{\gamma}{2a^2}(-\beta^2 - \delta\gamma [/math][math][/math][math]+ \beta\gamma [/math][math][/math][math]+ \delta^2) = [/math][math][/math][math][/math][math]\frac{\gamma}{2a^2}\left(\delta(\delta - \gamma) [/math][math][/math][math]+ \beta(\gamma - \beta) \right) \geq 0[/math]This final step must always be positive (or zero) as a direct consequence of the variable ordering, which tells us that: [math]0\leq \beta \leq \gamma \leq \delta[/math][math][/math][math]\ [/math][math][/math][math][/math]EDIT 2:The simpler version of this theorem is also true:If [math]0 \leq a \leq b \leq c [/math][math][/math][math][/math], then: [math]a^b b^c c^a \geq b^a c^b a^c[/math]The proof of this is very much analogous to the proof above, though proving that the extremums of[math](b-c)\ln(a) [/math][math][/math][math]+ (c-a)\ln(b) [/math][math][/math][math]+ (a-b)\ln(c)[/math]always occur when [math]a=b=c[/math] , is a bit trickier.And you can’t simplify the theorem any further, because trivially:[math]a^b b^a = b^a a^b\ [/math][math][/math][math]\ [/math][math][/math][math][/math] ; [math][/math][math]\ [/math][math][/math][math]\ a^a = a^a[/math]Going the other direction, it would not surprise me if you can generalize this theorem to arbitrarily large numbers of parameters.

No one gives out such numbers. Not the US, not the Russians or Indian military.We can guestimate it though.When the production was started in 2005, the news was that about 2000 would be made in 10 years. That should have been complete by now.Now, the nation has added 2 more production lines bringing the capacity to 3 lines while increasing the indigenization levels. We can assume 3 times the capacity. This means the last 5 years we should have made another 3000 of new generation variants.So a total of 4500 - 5500 would be a good estimate. Most of these would be land based variants. Only 20% of that might be naval variants.In 2016 there was a news of ramping production to 100 missiles a month. By 2018 the third production line was added. So current production should be 150 - 200 missiles a month. That's 1800 - 2400 missiles a year. In 5 years that will translate to 9000 - 12000 missiles. That's the second most aggressive manufacturing in Indian defense second only to Pinaka MLRS at 5000 rockets a year since 2005. India must have produced over 80,000 rockets by now, but most of the initial rockets are either used or expired. Rocket propellant as well as warhead are primarily chemicals that means they come with an expiration date i.e a use by dd/mm/yyyy. Usually 8 - 10 years for open units like 80 mm rockets and MLRS rockets while those in casister like brahmos have 15 + years.There was also news last year of “service life extension” for brahmos which means a test and replace of expired chemicals and batteries. Warhead I believe are also replaced.