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/2sinθ1r cosθ er4 rdrdθ = ∫0π/2sinθ1r2er4cosθ drdθ.

Now we reverse the order of integration; also we shall write t = θ. This yields

 I = ∫01∫0sin-1rr2er4cos t dtdr = ∫01[r2er4sin t]0sin-1r dr = ∫01r3er4 dr = [er4/4]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 limx--> ∞g''(x) = 1. Therefore limx--> ∞g'(x) = limx--> ∞g(x) = ∞. Thus by l'Hôpital's rule,

limx--> ∞g(x)/x2 = limx--> ∞g'(x)/(2x) = limx--> ∞g(x)/2 = 1/2.

We deduce that limx--> ∞f (x) = 1/2 and limx--> ∞(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 fn, j+1(x) - fn, j(x) = √x/n, hence

fn, j(x) = f0, j(x) + jx/n = x + (j + 1)√x/n.

Thus in particular fn, n(x) = x + (n + 1)√x/n and we see that limn--> ∞fn, n(x) = x + √x.

Peter Linnell 2007-05-20