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rmo aakash lecture 11 pending 21 oct 2026 - functions - Q1,Q2 done

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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) =>...
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rmo aakash lecture 9 13 oct 2026 - yet to complete myself, Q1 done, pending

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Q1.  S1.
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Advanced inequalities pending

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why tangent is perpendicular to radius?

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rmo aakash lecture 10, 15th oct 2026 pending - done Q1,Q2,Q3,Q4

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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:
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p-adic valuation and legendre's formula

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 V_p(n) = k => highest power of prime p in number 'n' is k. This is known as p-adic valuation. For e.g. V2(12) = 2 => p-adic valuation of 2 wrt 12 is 2. Also it's additive, ie V_p(m.n) = V_p(m) + V_p(n) V2(36) = V2(18.2) = V2(18) + V2(2) = 1 + 1 = 2 Legendre's formula:
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Bertrand's postulate

 for any integer k > 1, there will always be a prime p s.t. k < p < 2k.
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