18th VTRMC, 1996, Solutions
I = ∫01∫√y-y2√1-y2xe(x4+2x2y2+y4) dxdy. We change to polar coordinates to
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
||= ∫01∫0sin-1rr2er4cos t dtdr = ∫01[r2er4sin t]0sin-1r dr
||= ∫01r3er4 dr = [er4/4]01 = (e - 1)/4.
r1 = m1/n1 and
r2 = m2/n2, where
m1, n1, m2, n2
are positive integers and
gcd(m1, n1) = 1 = gcd(m2, n2).
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
(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))).
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.
- 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
y = e(ex-1).
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
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.
||= a1 + b1x + 3a2x2 + b2x3 +5a3x4 + b3x5 +7a4x6,
||= a1x + b1x2/2 + a2x3 + b2x4/4 + a3x5 + b3x6/6 + a4x7.
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
- 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,
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.
- We have
fn, j+1(x) - fn, j(x) = √x/n, hence
fn, j(x) = f0, j(x) + j√x/n = x + (j + 1)√x/n.
Thus in particular
fn, n(x) = x + (n + 1)√x/n and we see
limn--> ∞fn, n(x) = x + √x.