IIT JEE/MAINS MATH (CHANDRA KANT SIR)

A wonderful question from Stanford: Q)a,b,c >0 such that a+b+c=12 & bc+ ca+ ab=21. Find (abc)(max). Discussion: We are tempted to use the AM/GM inequality on a, b, c. Thus cube root (abc) <= (a+b+c)/3 = 4, giving abc <= 64. However, in the AM/GM inequality, equality takes place iff a=b=c So, in our case, each of a,b,c = 4 which is untenable with bc+ ca+ab= 21. Our investigation, so far, only shows that abc < 64. Our next line of thought is to investigate the polynomial whose zeros are a,b,c. Thus, let P(x) = (x-a)(x-b)(x-c) = x^3- 12x^2+21x- p, where p= abc. Analyse P(x) for local optima. You will get x=1( local maxima) & x= 7 ( local minima). P(1)= 10-p ( local maximum value) P(7)= -98-p ( local minimum value). Now the unexpected reasoning. ( Draw a graph of y= P(x) ). If 10- p were negative, the P(x) has only one real zero, which is a contradiction, since P(x) has three real zeros, a,b,c. Hence 10-p >= 0, giving p<=10 Thus (abc)(max)=10 Is,this attained? p(max)= 10 iff two of a,b,c are equal. This gives 1,1,10 as the values of a,b,c (in any order).

Classical Number Theory is crazy : Q) 2^29 is a number of nine different digits. Which digit is missing? (Romania). Let 2^29 = abcdefghi...(*) 2^29 = ((2^6)^4)(2^5) = 5, mod 9. 0+1+2+...+9 = 45 = 0, mod 9. To bring equality, mod9, the missing digit is 4. Reach for your calculator. 2^29 = 536, 870, 912

" An arithmetic progression, whose nth term is an+b, where a & b are relatively prime, contains infinitely many primes." ( Dirichlet). In Crux Mathematicorum one finds a brilliant application of Dirichlet's theorem. Prove that infinitely many positive integers cannot be written in the form 3ab+a+b, for any positive integers a & b. Put n=3ab+a+b, so that 3n+1=(3a+1)(3b+1). This step comes after a great deal of thought, though it looks simple now. Thus, if n were of form 3ab+a+b, then 3n+1 is composite. The contra positive is, " If 3n+1 were prime, then n cannot be expressed as 3ab+a+b." Now, appeal to Dirichlet. There are, indeed, infinitely many primes of form 3n+1. We are done.

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p,q,r are unequal primes such that 1) p^2+ q^2+ r^2= 2331 and 2) (1+p)(1+q)(1+r)= 10,560. Find p,q,r. Disc:1) implies each of p, q, r is < 48 Now, 11 | 10,560. The only prime < 48, which is 1 less than a multiple of 11, is 43. Take r=43. Go back to 1). Thus p^2+ q^2= 482. Let p < q , q^2 < 482 < 2q^2, giving 15< q< 21. Hence qε { 17,19 }. Check that q=19 works, giving p = 11. Finally, check that p=11, q=19, r= 43 satisfy 2).

Cryptography : A cryptographer encodes the digits 0,1,2,3,4 by setting up a one-to-one correspondence with the letters V,W,X,Y,Z, in some order. The integer to be encoded is first expressed in base 5.If VYZ, VYX, VVW are 3 consecutive integers in increasing order, find XYZ, base 10. ( AIME). Start with VYZ +1= VYX. Since the 2nd place, Y, remains unchanged, there was no " carry over" when Z+1 was done. Hence X= Z+1. Now, VYX+1=VVW. Here, the 2nd place,Y, has changed to V. Hence there is a carry over when X+1 was done. So X=4 & W=0. Z=X-1=3. To fix V & Y, resort to a trial. If Y=2 & V=1, then VYZ=123, VYX= 124, VVW=110 & VYX , VVW are not consecutive in base 5. If Y=1, V=2, then VYZ=213, VYX= 214, VVW=220, which are, indeed, consecutive in base 5. We have laid bare the cryptographer: V,W,X,Y,Z are 2, 0,4, 1, 3, respectively. Ans: XYZ= 413, base 5 = 108, base 10.

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Berkeley: f : [0,1] to R is a twice differentiable function such that f (x)-2f'(x)+f"(x) > e^x ,for all x ε [0,1]. If f (0)=f (1)=0,show that f < 0 in (0,1). Disc: Fiddling with the inequality you realise that, shifting e^x to the lhs, you get an expression which snugly fits the derivative of (e^(-x))f (x). Seizing the opportunity, take g(x)=(e^(-x))f (x), xε [0,1]. The given inequality now reads g"(x)>1, for all xε [0,1]...(*) We wish to show that f, hence g, is < 0, for all xε (0,1). One of the powerful tools is "Reductio-ad-absurdum." ( The indirect method). If g were 0 everywhere in (0.1), then so would f be, which goes against the inequality given. If g were positive at some point, then, since g(0)=g (1)=0, g attains a local maxima at some point, say c, in (0,1). However, this implies g"(c) < 0. This ontradicts g"(x) > 1, for all x ε [0,1].

What is a "parity " argument? Here's an example: a, b, c are odd integers .Can the equation ax^2+bx+c=0 have a rational root ? This is called an "Open" question. If the reply is " Yes", then an example of such an equation should be given. If, on the other hand, the reply is "No", then nothing short of a proof is required. We assert that the answer is "No". Our proof is " indirect". Let x = p/q, a rational in its "lowest" terms, be a root. So p,q cannot both be even. We have ap^2+ bpq+ cq^2= 0...(*) Case1: p,q are odd. Case2: p,q have opposite parties. In either case, the lhs of (*) is odd, showing that (*) does not hold. Hence, the equation cannot have a rational root. The indirect method, called " Reductio-ad-absurdum ", was used by Euclid with remarkable success.

Q) f is the function from positive reals to positive reals such that f(xf(y)) = (x^p)(y^q), for all x, y > 0 & for some p & q > 0. If p+q = 20, find p.(Olympiad). Disc: f(xf(y))=(x^p)(y^q))...(*). Let f(1)= c, for some c>0. y=1 in (*) gives f(cx)= x^p. Replace x by x/c to get f(x)= (x/c)^p, for all x>0 & for some c, p >0.(**). Can we determine c? In (*), take y = x to get f(xf(x))= x^(p+q)= x^20...(***). The lhs of this equation is (xf(x)/c)^p = (x(x/c)^p/c)^p=(x/c)^( p+p^2). In particular, x=1 gives c^(p+p^2)= 1, forcing c=1. Hence f(x)= x^p, for all x>0 and for some p>0. Finally, (***) gives x^20= f(x^(p+1))= x^(p(p+1)), for all x>0, so p(p+1)=20, giving p= 4. Solving a mathematical question is akin to exploring a new terrain. Some trials are inevitable.

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