Math IOQM

Q1) (a,b,c) are real numbers such that a+b+c=0 a^2+b^2+c^2=1 Show that a^2b^2c^2 <= 1/54. S1) Let's construct a cubic equation with roots a,b,c. (a+b+c)^2 = 1 + 2(ab+bc+ca) => ab + bc + ca = -1/2 x^3 -x/2 - r = 0 where r = abc Since all 3 roots are real, the product of values of function at critical points(maxima,minima) should be <= 0. Find critical points, derivative = 3x^2 - 1/2 = 0 => x = -+1/sqrt(6) f(1/rt(6)) = 1/6.rt(6) - 1/2.rt(6) - r = 1/3.rt(6) - r f(-1/rt(6) = -1/3.rt(6) - r multiply: r^2 - 1/54 <= 0 H.P. Q2. Text from the image: P2) If 1/x + 1/y + 1/z = 1 for x,y,z > 0 pr. th. (x-1)(y-1)(z-1) >= 8 S2. let a = 1/x and similarly others. a + b + c = 1 x-1 = 1/a - 1 = (1-a)/a = (b+c)/a b+c >= 2.rt(bc) So expr. becomes: (b+c)(c+a)(a+b)/abc>= 8.abc/abc = 8 H.P.

  Homework(proofs): Q1. S1. n = 1 => (2k+1)^(2^1) = 4k^2 + 4k + 1 = 4k(k+1) + 1 = 8p + 1 = 1 mod 2^3(since k(k+1) is even) So works for n = 1 Assume (2k+1)^(2^n) = 1 mod 2^(2^(n+2)) _____[1] and using this prove for n+1: (2k+1)^(2^(n+1)) = 1 mod 2^(2^(n+3)) LHS = (2k+1)^(2^n.2^1) = [(2k+1)^(2^n)]^2 = [2^(n+2).m + 1]^2 using [1] = 1 + 2.m.2^(n+2) + m^2.[2^(n+2)]^2 = 1 + m.2^(n+3) + m^2.2^(2n+4) If you take mod 2^(n+3), it's clearly 1. H.P. Q2. Call a natural number n convenient, if n^2 + 1 is divisible by 1000001 . Prove that among the numbers   1,2,…,1000000, there are evenly many "convenient" numbers. S2. So essentially we need to find some sort of complement for each 'n' satisfying this. If we can show that for each 'n' satisfying this, there is another number satisfying this, we are done. Simplest 'complement' here is -n but that's negative. So let's think circular and try N-n where N = 1000,001 (N-n)^2 +1 = N^2 + n^2 - 2N.n + 1 = n^2...

 Homework multiple choice: (done) Q1. True/False For a polynomial f(x) with integer coefficients and integers a,b,l  we have a = b mod l => f(a) = f(b) mod l? S1. True Q2. The 2-digit integers from 19 to 92 are written consecutively to form the integer  N = 192021⋯9192. Suppose that 3^k is the highest power of 3 that is a factor of N. What is k? S2. Method 1: Since 10^m = 1 mod 9 for any m => 192021⋯9192 = 19 + 20 .. 92 mod 9 = 74/2[19 + 92] = 37*111 mod 3 = 0  mod 9 = 1 * 3 = 3 So k = 1. Method 2: Sum of digits =  (2+3+4+5+6+7+8)*10 + (1+2... 9)*7 + sum_digits(19 + 90 + 91 + 92) = 35*10 + 45*7 + 40 Doing mod 3 gives 2 + 0 + 1 = 0 mod 3 Doing mod 9 gives 8 + 0 + 4 = 3 mod 9 So 3 divides it but 9 doesn't. k = 1 = answer. Q3. Remainder of 3^89*7^86 mod 17? S3. Method 1: = 21^86*27 21 mod 17 = 4 21^2 mod 17 = 16 = -1 (-1)^43 = -1 => -1*27 = -27 mod 17 = -10 mod 17 = 7 mod 17 Method 2: 3^4 = 81 = -4 m 17 3^8 = 16 m 17 = -1 3^89 = 3^88.3 = (-1)^11.3 mod 17 =...

double derivative for concavity check. single derivative for slope(rate of change check). Rolle's theorem: f(a) = f(b) => for some c in (a,b) f'(c) = 0 Mean value theorem, for some c, f'(c) = (f(b) - f(a)/(b-a)

Q1. The position of an object is given by s(t) = sin⁡(3t) − 2t + 4. Determine where in the interval [0,3] the object is moving to the right and moving to the left. Given arccos(2/3) = 0.841 radians. and 2.pie = 6.283 radians. S1. s'(t) = 3cos(3t) - 2 First find critical points: 3cos(3t) - 2 = 0 => cos(3t) = 2/3 => Since cos(3t) is positive, there are 2 solutions (1st and 4th quadrant). 3t = acos(2/3) + 2.pie.k = 0.841 + 2.pie.k and 3t = [2.pie - acos(2/3)] + 2.pie.k = 5.442 + 2.pie.k How many valid solutions are there for 3t in [0,9] since t in [0,3]. Apart from the 2 above, there is only 1 more when 3t = .841 + 6.283 = 7.124 So these 3 are our critical points. Let's evaluate at each of them. Case 1: 3t between [0,.841) s'(0) = 1 => +ive => moving right Case 2: 3t between (.841, 5.442), try 3t = 3(t = 1) => 3cos(3t) - 2 = 3cos3 - 2 cos3 is close to cos(3.14), i.e. -1 so value is negative, moving left. Case 3: 3t between (5.442,7.124), try 3t = 6.283 = 2.pi =...

remainders/residues/residue classes Q1. If a = b mod m and c = d mod m, then: Prove that: (1) a +- c = b +- d mod m (2) ac = bd mod m (3) ax + cy = bx + dy mod m Q2. if a = b mod m, pr. th. a^n = b^n mod m. Q3. If a = b mod m then a = b mod d if d | m. Q4.  Remainder of 13^73 + 14^3 mod 11. Q5. Pr. th. ax = ay mod m iff x = y mod (m/gcd(a,m)) S5. m = k.q1, a = k.q2 where k = gcd(a,m) m/k = q1 x = y mod q1 => ax = ay mod q1 => ax = ay mod m since q1| m. ----------------------------------------------------------------