33rd VTRMC, 2011, Solutions

1. Write I = ∫14(x - 2)/((x2 +4)√xdx 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 an = n2 - 1; let us prove this by induction on n. The result is certainly true if n = 0 or 1. So suppose ar = r2 - 1 for rn. Then

 an+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 an = n2 - 1 for all nN and we deduce that a100 = 9999.

3. Let S = ∑k=1n(k2 - 2)/((k + 2)!) where nN. 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 limn-> ∞1/n! = 0, it follows that the required sum is 0.

4. We repeatedly use the Chinese remainder theorem without further comment. Let bZ. We claim that |{[br] | rZ}| = mn/(mn, b). Set k = (mn, b). Then {[kr] | rZ} = {[kr] | r = 1, 2,..., mn/k} so |{[kr] | rZ}| = mn/k. It follows that |{[br] | rZ}|≤mn/k. On the other hand there exists cZ such that and bck mod mn. We conclude that |{[br] | rZ}|≥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 sZ; clearly (s, m) = 1. Now if tZ, then (s + tm)na 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) = x1+1/x -x - ln x. We first show that limx-> ∞f (x) = 0. Note that limx-> ∞x1/x = 1 because limx-> ∞ln(x1/x) = limx-> ∞(ln x)/x = 0. Thus limx-> ∞f (x)/x = 0. Now we apply l'hôpital's rule. We obtain

 limx-> ∞f (x) = limx-> ∞(f (x)/x)/(1/x) = limx-> ∞((f (x)/x)')/(- 1/x2) = limx-> ∞ -x2(x1/x -1 - (ln x)/x)' = limx-> ∞ -x2(x1/x/x2 - (x1/xln x)/x2 -1/x2 + (ln x)/x2) = limx-> ∞(x1/x -1)(ln x - 1).

Thus we need to prove limx-> ∞(x1/x -1)ln x = 0. Again we apply l'hôpital's rule.

 limx-> ∞(x1/x -1)ln x = limx-> ∞(x1/x -1)/(1/ln x) = limx-> ∞(x1/x(1 - ln x)/x2)/(- 1/(x(ln x)2)) = limx-> ∞((ln x)3)/x = 0.

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

 f (2x) -f (x) = (2x)1+1/(2x) -2x - ln 2x -x1+1/x + x + ln x = (2x)1+1/(2x) -x1+1/x -x - ln 2

and we deduce that limx-> ∞(2x)1+1/(2x) -x1+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=1100xi = -20 and i < jxixj = 198. Therefore i=1100xi2 = (- 20)2 - 2*198 = 4 and we deduce that (∑i=1100xi)2 = 100∑i=1100xi2. Now apply the Cauchy-Schwartz triangle inequality, that is (∑ixiyi)2≤∑ixi2iyi2 with equality if and only if one of (xi),(yi) is a scalar multiple of the other, in particular taking yi = 1 for all i, we obtain (∑i=1100xi)2≤100∑i=1100xi2 with equality if and only if (xi) is a scalar multiple of (1,..., 1), i.e. all the xi are equal. We deduce that all the xi are equal. But the product of the roots is 1, consequently all the xi are 1 or all the xi are -1, which contradicts the fact that i=1100xi = -20.

Peter Linnell 2011-10-31