By Kapil D Joshi, IIT Bombay
Read Online or Download Educative JEE PDF
Similar nonfiction_5 books
Management: Meeting and Exceeding Customer Expectations, 9th Edition
Administration: assembly AND EXCEEDING shopper expectancies, 9th variation is a finished survey of the foundations and practices of administration as they're being utilized worldwide. The content material and contours are dependent to augment carrying on with subject matters which are woven into the chapters' narratives: 1) the unending attempt by way of managers and corporations to fulfill or exceed shopper wishes, and a pair of) the necessity of firms and their humans to be guided by way of powerful management.
- Suffering in Luke's Gospel (Abhandlungen zur Theologie des Alten und Neuen Testaments 81)
- The Darkest Powers Trilogy 03 The Reckoning
- Discourse Description: Diverse Linguistic Analyses of a Fund-raising Text
- The Toy Wizard's Favorite Wood Toy Plans for Toy Cars and Trucks
Additional resources for Educative JEE
Example text
But taking logarithms simplifies things practically if not conceptually. In fact, the reason logarithms are so useful in computations is precisely that they convert products to sums. It is definitely less time-consuming to add than to multiply two numbers. The logarithms (and the exponentials) have many other uses besides conversion of progressions. The foremost is probably that certain natural growths satisfy the law that the rate of growth at any time is proportional to the quantity existing at that time.
Comment No. 10: By far the most celebrated result about polynomials in two variables is the binomial theorem which says that for every positive integer n, n (x + y)n = r=0 = n n−r r x y r n n n n−1 n n−2 2 x + x y+ x y + ... 0 1 2 n n n n + x2 y n−2 + xn−1 y + y n−2 n−1 n (25) where the coefficients nr (also often denoted by n Cr or sometimes by n Cr ) are called the binomial coefficients and are defined combinatorially as the n! (n − r)! with the understanding that 0! = 1. After cancelling the common factors, the n(n − 1)(n − 2) .
1 ak 38 Educative JEE = = ka1 ( 1r − 1) ( r1 )k − 1 a1 krk−1 (r − 1) rk − 1 Putting (13), (14) and (15) together, we have Gk = 1, 2, . . , n. It now follows that (15) √ Ak Hk for k = M = (G1 G2 . . Gn )1/n = (A1 H1 A2 H2 . . An Hn )1/2n . This is probably the answer the papersetters had in mind. P. (without which assumption, the relation Gk = Ak Hk would be true only for k = 1 and 2, but not for any higher values of k). Secondly, it involves all of A1 , A2 , . . , An and H1 , H2 , . . ) Comment No.