18th VTRMC, 1996, Solutions

1. Let I = ∫01∫√y-y2√1-y2xe(x4+2x2y2+y4) dxdy. We change to polar coordinates to obtain
   I = ∫0π/2∫sinθ1r cosθ er4 rdrdθ = ∫0π/2∫sinθ1r2er4cosθ drdθ.
   Now we reverse the order of integration; also we shall write t = θ. This yields

   I = ∫₀¹∫₀^sin-1rr²e^r4cos t dtdr = ∫₀¹[r²e^r4sin t]₀^sin-1r dr = ∫₀¹r³e^r4 dr = [e^r4/4]₀¹ = (e - 1)/4.

   
   

2. Write r₁ = m₁/n₁ and r₂ = m₂/n₂, where m₁, n₁, m₂, n₂ are positive integers and gcd(m₁, n₁) = 1 = gcd(m₂, n₂). Set Q = ((m₁ + m₂)/(n₁ + n₂), 1/(n₁ + n₂)). We note that Q is on the line joining (r₁, 0) with P(r₂), that is the line joining (m₁/n₁, 0) with (m₂/n₂, 1/n₂). This is because
   (m₁ + m₂)/(n₁ + n₂) = m₁/n₁ + (m₂/n₂ - m₁/n₁)(n₂/(n₁ + n₂)).
   Similarly Q is on the line joining P(r₁) with (r₂, 0). It follows that (m₁ + m₂)/(n₁ + n₂), 1/(n₁ + n₂) is the intersection of the line joining (r₁, 0) to P(r₂) and the line joining P(r₁) and (r₂, 0). Set
   P = P((r₁f (r₁) + r₂f (r₂))/(f (r₁) + f (r₂))).
   Since
   P = ((m₁ + m₂)/(n₁ + n₂),/f ((m₁ + m₂)/(n₁ + n₂))),
   we find that P is the point of intersection of the two given lines if and only if f ((m₁ + m₂)/(n₁ + n₂)) = n₁ + n₂. We conclude that the necessary and sufficient condition required is that gcd(m₁ + m₂, n₁ + n₂) = 1.

3. Taking logs, we get dy/dx = y ln y, hence dx/dy = 1/(y ln y). Integrating both sides, we obtain x = ln(ln y) + C where C is an arbitrary constant. Plugging in the initial condition y = e when x = 1, we find that C = 1. Thus ln(ln y) = x - 1 and we conclude that y = e^(ex-1).

4. Set g(x) = x²f (x). Then the given limit says lim_{x--> ∞}g''(x) = 1. Therefore lim_{x--> ∞}g'(x) = lim_{x--> ∞}g(x) = ∞. Thus by l'Hôpital's rule,
   lim_{x--> ∞}g(x)/x² = lim_{x--> ∞}g'(x)/(2x) = lim_{x--> ∞}g(x)/2 = 1/2.
   We deduce that lim_{x--> ∞}f (x) = 1/2 and lim_{x--> ∞}(xf'(x)/2 + f (x)) = 1/2, and the result follows.

5. Set

   f (x) = a₁ + b₁x + 3a₂x² + b₂x³ +5a₃x⁴ + b₃x⁵ +7a₄x⁶,

   g(x) = a₁x + b₁x²/2 + a₂x³ + b₂x⁴/4 + a₃x⁵ + b₃x⁶/6 + a₄x⁷.

   
   
   Then g(1) = g(- 1) because a₁ + a₂ + a₃ + a₄ = 0, hence there exists t∈(- 1, 1) such that g'(t) = 0. But g'(x) = f (x) and the result follows.

6. We choose the n line segments so that the sum of their lengths is as small as possible. We claim that no two line segments intersect. Indeed suppose A, B are red balls and C, D are green balls, and AC intersects BD at the point P. Since the length of one side of a triangle is less than the sum of the lengths of the two other sides, we have AD < AP + PD and BC < BP + PC, consequently
   AD + BC < AP + PC + BP + PD = AC + BD,
   and we have obtained a setup with the sum of the lengths of the line segments strictly smaller. This proves that the line segments can be chosen so that no two intersect.

7. We have f_{n, j+1}(x) - f_{n, j}(x) = √x/n, hence
   f_{n, j}(x) = f_{0, j}(x) + j√x/n = x + (j + 1)√x/n.
   Thus in particular f_{n, n}(x) = x + (n + 1)√x/n and we see that lim_{n--> ∞}f_{n, n}(x) = x + √x.