16th VTRMC, 1994, Solutions
I = ∫01∫0x∫01-x2e(1-z)2 dzdydx.
We change the order of integration, so we write
I = ∫∫∫Ve(1-z)2 dV, where V is the region of integration.
It can be
described as the cylinder with axis parallel to the z-axis and
cross-section A, bounded below by
z = 0 and bounded above by z = 1 - x2. This region can also be
described as the cylinder with axis parallel to the y-axis and
cross-section B, bounded on the left by y = 0 and on the right by
y = x. Therefore
||= ∫01∫0√1-z∫0xe(1-z)2 dydxdz
||= ∫01∫0√1-zxe(1-z)2 dxdz = ∫01(1 - z)e(1-z)2/2 dz
||= [- e(1-z)2/4]01 = (e - 1)/4.
- We need to prove that
pq≤∫0pf (t) dt + ∫0qg(t) dt. Either
q≤f (p) or
q≥f (p) and without loss of
generality we may assume that
q≥f (p) (if
q≤f (p), then we
interchange x and y; alternatively just follow a similar argument
to what is given below). Then we have the following diagram.
We now interpret the quantities in terms of areas: pq is the area
∫0pf (t) dt is the area of C, and
∫0qg(t) dt is the area of
A∪B. The result follows.
- Differentiating both sides with respect to x, we obtain
2ff' = f2 - f4 + (f')2. Thus
f4 = (f - f')2, hence
f - f' = ±f2 and we deduce that
dx/df = 1/(f±f2).
We have two cases to consider; first we consider the + sign, that
dx/df = 1/f - 1/(f + 1) and we obtain
x = ln| f| - ln| f + 1| + C, where C is an arbitrary constant. Now we have the initial
f (0) = ±10.
If f (0) = 10, we find that
C = ln(11/10) and
x = ln(11/10) - ln|(f + 1)|/f|. Solving this for x,
we see that
f (x) = 10/(11e-x - 10). On the other hand if
f (0) = - 10, then
C = ln(9/10), consequently
x = ln(9/10) - ln|(f + 1)/f|. Solving this for x, we conclude that
f (x) = 10/(9e-x - 10).
Now we consider the - sign, that is
dx/df = 1/f - 1/(f - 1) and we
x = ln| f| - ln| f - 1| + D, where D is an arbitrary
constant. If the initial condition f (0) = 10, we find
D = ln(9/10) and consequently
x = ln| f /(f - 1)| + ln(9/10). Solving this for x, we deduce that
f (x) = 10/(10 - 9e-x). On the other hand if the initial condition is
f (0) = - 10, then
D = ln(11/10) and hence
x = ln| f /(f - 1)| + ln(11/10). Solving for x, we conclude that
f (x) = 10/(10 - 11e-x).
Summing up, we have
f (x) = ±10/(10 - 9e-x) or ±10/(10 - 11e-x).
f (x) = ax4 + bx3 + x2 + bx + a = 0.
We will show that the maximum value of a + b is -1/2; certainly
-1/2 can be obtained, e.g. with a = 1 and b = - 3/2, because in
this case 1 is a root of f. Furthermore
f (1) = 2(a + b) + 1, hence
we may assume that 1 is not a root of f. Note that if
a root of f, then so is
1/α. Thus we may assume that two of
the roots of f are
1 < α is a real
We claim that the other two roots of f cannot be positive. Indeed
β is another positive root of f, which we may suppose
is not 1 (though could be
1/α). Since the product
of the roots of f is a/a = 1, we see that the fourth root of f is
1/β. It now follows that the sum of the products of roots of
f two at a time is greater than
α(1/α) + β(1/β) = 2; however this sum is also 1/a and we have contradicted the
hypothesis a > 1/2.
Thus f has exactly two positive roots, namely
and by considering the graph of f, we see that f (1) < 0. Thus
a + b + 1 + b + a < 0 and the result follows.
- We follow the hint given, so suppose
β is another eigenvector
corresponding to 1 with components
Choose t such that
all i (if
βi = 0, then we interpret
αi/|βi| = + ∞; note that
βt≠ 0). Multiplying
αt/βt, we may assume that
βt = αt, and then
βi≤αi for all i. Since
∑i=1natiαi = αt = βt = ∑i=1natiβi,
we see that
βi = αi for all i and the result follows.
an = f (n)2. Then squaring
f (n + 1) = 2√f (n)2+n, we
an+1 = 4an + 4n. We solve this recurrence relation in a
similar way to solving the corresponding first order differential
y' = 4y + 4t. The solution to
an+1 = 4an is an = C4n for some constant n. Then we look for a solution to
an+1 = 4an + 4n in the form
an = An + B, where A, B are constants to
be determined. Plugging this into the recurrence relation, we obtain
A(n + 1) + B = 4An + 4B + 4n, and then equating the coefficients of
n and the constant term, we find that A = - 4/3, B = - 4/9.
an = C4n - 4n/3 - 4/9, and then plugging in a1 = 1,
we see that C = 25/36 and we conclude that
an = 25·4n/36 - 4n/3 - 4/9. We now need to calculate
∑n=1Nan. This is
25(4N -1)/27 - 2N2/3 - 10N/9.
- We have
xn+3 = 19xn+2/(94n+1) = 192/(942xn)
and we deduce that
xn+6 = xn for all nonnegative integers n.
It follows that
∑n=0∞x6n/2n = ∑n=0∞10/2n = 20.