33rd VTRMC, 2011, Solutions

1. Write I = ∫₁⁴(x - 2)/((x² +4)√x) dx and make the substitution y = √x. Then dx = 2y dy and I = ∫₁²(2y² -4)/(y⁴ +4) dy. Now y⁴ +4 = (y² -2y + 2)(y² + 2y + 2) and using partial fractions, we find
   (2y² -4)/(y⁴ +4) = (y - 1)/(y² -2y + 2) - (y + 1)/(y² + 2y + 2).
   It follows that 2I = [ln(y² -2y + 2) - ln(y² +2y + 2)]₁² = ln 2 - ln 1 - (ln 10 - ln 5) = 0, so the answer is 0.

2. The first few terms (starting with a_-1) are -1, 0, 3, 8, so it seems reasonable that a_n = n² - 1; let us prove this by induction on n. The result is certainly true if n = 0 or 1. So suppose a_r = r² - 1 for r≤n. Then

   a_n+1 = (n² -1)² - (n + 1)²((n - 1)² - 1) - 1 = n⁴ -2n² +1 - (n⁴ -2n² +1) + n² + 2n + 1 - 1 = n² +2n = (n + 1)² - 1.

   
   
   and the induction is complete. Thus a_n = n² - 1 for all n∈N and we deduce that a₁₀₀ = 9999.

3. Let S = ∑_k=1ⁿ(k² - 2)/((k + 2)!) where n∈N. We have
   (k² - 2)/((k + 2)!) = 1/k! - 1/(k + 2)! - 3(1/(k + 1)! - 1/(k + 2)!).
   By telescoping series, we find that S = 1 + 1/2 - 3/2 - 1/(n + 1)! - 1/(n + 2)! + 3/(n + 2)! = -1/(n + 1)! + 2/(n + 2)!. Since lim_{n-> ∞}1/n! = 0, it follows that the required sum is 0.

4. We repeatedly use the Chinese remainder theorem without further comment. Let b∈Z. We claim that |{[br] | r∈Z}| = mn/(mn, b). Set k = (mn, b). Then {[kr] | r∈Z} = {[kr] | r = 1, 2,..., mn/k} so |{[kr] | r∈Z}| = mn/k. It follows that |{[br] | r∈Z}|≤mn/k. On the other hand there exists c∈Z such that and bc≡k mod mn. We conclude that |{[br] | r∈Z}|≥m and the claim is established.

   Therefore mn/(a, mn) = m. Thus (a, mn) = n, hence n | a and we may write a = sn where s∈Z; clearly (s, m) = 1. Now if t∈Z, then (s + tm)n≡a mod mn, i.e. [n(s + tm)] = [a]. Since (s, m) = 1, we may choose t so that (s + tm, mn) = 1. The result follows by setting q = s + tm.

5. Set f (x) = x^1+1/x -x - ln x. We first show that lim_{x-> ∞}f (x) = 0. Note that lim_{x-> ∞}x^1/x = 1 because lim_{x-> ∞}ln(x^1/x) = lim_{x-> ∞}(ln x)/x = 0. Thus lim_{x-> ∞}f (x)/x = 0. Now we apply l'hôpital's rule. We obtain

   lim_{x-> ∞}f (x) = lim_{x-> ∞}(f (x)/x)/(1/x) = lim_{x-> ∞}((f (x)/x)')/(- 1/x²) = lim_{x-> ∞} -x²(x^1/x -1 - (ln x)/x)' = lim_{x-> ∞} -x²(x^1/x/x² - (x^1/xln x)/x² -1/x² + (ln x)/x²) = lim_{x-> ∞}(x^1/x -1)(ln x - 1).

   
   
   Thus we need to prove lim_{x-> ∞}(x^1/x -1)ln x = 0. Again we apply l'hôpital's rule.

   lim_{x-> ∞}(x^1/x -1)ln x = lim_{x-> ∞}(x^1/x -1)/(1/ln x) = lim_{x-> ∞}(x^1/x(1 - ln x)/x²)/(- 1/(x(ln x)²)) = lim_{x-> ∞}((ln x)³)/x = 0.

   
   
   Thus lim_{x-> ∞}f (x) = 0 and we deduce that lim_{x-> ∞}f (2x) -f (x) = 0. But

   f (2x) -f (x) = (2x)^1+1/(2x) -2x - ln 2x -x^1+1/x + x + ln x = (2x)^1+1/(2x) -x^1+1/x -x - ln 2

   
   
   and we deduce that lim_{x-> ∞}(2x)^1+1/(2x) -x^1+1/x -x = ln 2.

6. Let < indicate the usual order on Q; thus < is an asymmetric relation on Q. Define T = S×Q and let ≺ denote the lexicographic asymmetric relation on T, that is (a, p)≺(b, q) if and only if a < b or a = b and p < q. It is easily checked that ≺ is asymmetric. Also we may identify S with S× 0 via s |--> (s, 0), and then the restriction of ≺ to S is <. Now suppose (a, p)≺(b, q). If a < b, then (a, p)≺(a, p + 1)≺(b, q). On the other hand if a = b, then p < q and (a, p)≺(a,(p + q)/2)≺(a, q) and the result is proven.

7. Assume that the roots x₁,..., x₁₀₀ are all real. By the Viete relations we have ∑_i=1¹⁰⁰x_i = -20 and ∑_{i < j}x_ix_j = 198. Therefore ∑_i=1¹⁰⁰x_i² = (- 20)² - 2*198 = 4 and we deduce that (∑_i=1¹⁰⁰xⁱ)² = 100∑_i=1¹⁰⁰x_i². Now apply the Cauchy-Schwartz triangle inequality, that is (∑_ix_iy_i)²≤∑_ix_i²∑_iy_i² with equality if and only if one of (x_i),(y_i) is a scalar multiple of the other, in particular taking y_i = 1 for all i, we obtain (∑_i=1¹⁰⁰x_i)²≤100∑_i=1¹⁰⁰x_i² with equality if and only if (x_i) is a scalar multiple of (1,..., 1), i.e. all the x_i are equal. We deduce that all the x_i are equal. But the product of the roots is 1, consequently all the x_i are 1 or all the x_i are -1, which contradicts the fact that ∑_i=1¹⁰⁰x_i = -20.