Math IOQM

 Q1. S1. f(0+0) = f(0) + f(0) = 2f(0) => f(0) = 0. f(t-t) = f(0) = 0 = f(t) + f(-t) => f(-t) = -f(t) => odd function. f(1+1) = 2f(1) = f(2) Similarly, f(3) = 3.f(1) In fact, simply using (a) we can prove f(x) = cx for all rational numbers. How? Assume f(1) = a For any positive integer 'n': f(n. 1/n) = f(1) = f(1/n) + f(1/n) ... n times. using (a) f(1) = n.f(1/n) => f(1/n) = a/n For another non zero positive integer 'm': f(m/n) = f(m.1/n) = m.f(1/n) = m.a/n = a.(m/n) = f(1).(m/n) f(-m/n) = -f(m/n) as we showed earlier. So any rational number m/n can be expressed as a multiple of f(1). Hence for any rational number 'x': f(x) = a.x H.P. But how do we prove it for irrational numbers? Partially we can. 2.f(sqrt(3)/2) = f(sqrt(3)) using (a) Similarly, f(2.sqrt(3)) = 2.f(sqrt(3)) So f(q.t) = q.f(t) where q is rational and t is real. But we need to prove f(x) = cx. How? We need to use the relation (b). f(x) = (x^2).f(1/x) Earlier we assumed f(1) = a We need ...

Q1. S1. f(2) = 3 f(2^2) = f(4) = 3.3 = 9 2 < 3 < 4 => f(2) < f(3) < f(4) => 3 < f(3) < 9 => f(3) can be 4,5,6,7,8 Let f(3) = a. Now, f(3^2) = f(9) = f(3).f(3) 8 < 9 < 16 => f(8) < f(9) < f(16) => 27 < f(9) = a^2 < 81 => 27 < a^2 < 81 => a is 6,7,8 (Since 4,5 doesn't work) Now 16 < 27 < 32 f(16) < f(27) < f(32) 81 < a^3 < 243 => a = 6 (7,8 don't work) So f(3) = 6 Now, Powers of 2: 2,4,8,16,32,64,128,256 Powers of 3: 3,9,27,81,243 f(243) < f(256) 6^5 < 3^8 3^5.2^5 < 3^8 2^5 < 3^3 32 < 27 contradiction. H.P. Q2. S2. Let S be the set of all values of f(n) for n >= 1. From well ordering principle, there will be a minimum value in this set(made up of non zero integers). Let that value be d. Let that value occur at m => f(m) = d Now, there are 2 cases, m = 1 or m >= 1. If m >= 1, we can write the given inequality as: f(m) > f(f(m-1)) If f(m-1) = t then f(m) > f(t) =>...

Q1.  S1.

 

Q1. Let a, b, c ∈ ℝ. Prove the inequalities: 3(ab + bc + ca) ≤ (a + b + c)² ≤ 3(a² + b² + c²) S1. First see the RHS one. Cauchy shwartz. (1^2 + 1^2 + 1^2)(a^2 + b^2 + c^2) >= (1.a + 1.b + 1.c)^2 Since (a1.b1 + a2.b2 + a3.b3)^2 <= (a1^2 + a2^2 + a3^2).(b1^2 + b2^2 + b3^2) And equality happens when: a1/b1 = a2/b2 = a3/b3 => a/1 = b/1 = c/1 => a = b = c Now the LHS one, we need to do effectively this: ab + bc + ca <= a^2 + b^2 + c^2 WLOG a <= b <= c Which is simple using rearrangement inequality. And in RE also a = b = c gives the equality. Q2. S2. Now expanding it will give us the required inequality. Q3. S3. Q4. S4. Q5. S5. Method 2: