14th VTRMC, 1992, Solutions
- First make the substitution y = x3. Then
dF/dx = (dF/dy)(dy/dx) = ey23x2 = 3x2ex6 by the
chain rule. Therefore
d2F/dx2 = 6xex6 +18x7ex6. To find the point of
inflection, we set
d2F/dx2 = 0; thus we need to solve
6x + 18x7 = 0. The only solution is x = 0, so this is the point of
inflection (perhaps we should note that d3F/dx3 is
at x = 0, so x = 0 is indeed a point inflection).
- The shortest path will first be reflected off the x-axis, then be
reflected off the y-axis. So we reflect (x2, y2) in the
y-axis, and then in the x-axis, which yields the point
(- x2, - y2). Thus the length of the shortest path is the distance
from (x1, y1) to
(- x2, - y2), which is
((x1 + x2)2 + (y1 + y2)2)1/2.
- We have
f (f (x)) = 1 + sin(f (x) - 1) = 1 + sin(sin(x - 1)), so
f2(x) = x if and only if
x - 1 = sinsin(x - 1). If y is a real number,
| sin y|≤| y| with equality if and only if y = 0. It
y = sinsin(y) if and only if y = 0 and we deduce
that there is a unique point x0 such that
f2(x0) = x0, namely
x0 = 1.
- From (i), we have xn = 1 for all n. Thus
we need to find
∑n=0∞1/3n. This is a geometric
series with first term 1 and ratio between successive terms 1/3.
Therefore this sum is
1/(1 - 1/3) = 3/2.
tn≥1 for all n. Set
T = (1 + √5)/2 (the
positive root of x2 - x - 1).
Note that if
1≤x < T, then
x2 < 1 + x < T2. This shows that
tn < T for all n, and also that tn is an increasing sequence,
tn+12 - tn2 = 1 + + tn - tn2.
Therefore this sequence converges to a number between 1 and T.
Since the number must satisfy x2 = x + 1, we deduce that
limn--> ∞tn = T = (1 + √5)/2.
- First we find the eigenvalues and eigenvectors of A. The
x(x - 3) + 2 = 0, so the eigenvalues are 1,2.
To find the eigenvectors corresponding to 1, we need to solve the
One solution is u = 2, v= - 1 and we see that
is an eigenvector corresponding to 1.
To find the eigenvectors corresponding to 2, we need to solve
One solution is u = 1, v = - 1 and we see that
is an eigenvector corresponding to 2.
We now know that if
We deduce that
| ||A100 =
| ||2 - 2100
||2 - 2101
-1 + 2100
-1 + 2101
- Since p(r) = 0, we may write
p(x) = q(x)(x - r), where q(x) is
of the form
x2 + dx + e. Then
p(x)/(x - r) - 2p(x + 1)/(x + 1 - r) + p(x + 2)/(x + 2 - r) = q(x) - 2q(x + 1) + q(x + 2) = 2 as required.
- Assume that log means natural log. Note that log x is a positive
increasing function for x > 1. Therefore
∫1nx log x dx≤∑t=2nt log t < ∫2n+1x log x dx.
∫x log x = (x2log x)/2 - x2/4, we see that
(n2log n)/2 - n2/4≤∑t=2nt log t < ((n + 1)2log n)/2 - (n + 1)2/4 - 2 log 2 + 1.
Now divide by
n2log n and take
limn--> ∞. We obtain
1/2≤limn--> ∞(∑t=2nt log t)/(n2log n)≤1/2.
We conclude that the required limit is 1/2.
- For G(3) we have 8 possible rows of goblins, and by writing these
out we see that G(3) = 17. Similarly for G(4) we have 16
possible rows of goblins, and by writing these out we see that G(4) = 44.
let X be a row of N goblins. Then the rows with N + 1 columns
are of the form X2 or X3 (where X2 indicates adding a goblin
with height 2' to the end of X). If X ends in 3 or 32
2N-1 +2N-2 possible rows), then
X3 has 1 more LGG than X; on the other hand if X ends in 22,
then X3 has the same number of LGG's as X. If X ends in 2
(2N-1 possible rows),
then X2 has 1 more LGG than X, while if X ends in 3, then X2
has the same number of LGG's as X.
We conclude that for
G(N + 1) = 2G(N) + 2N-1 +2N-1 +2N-2 = 2G(N) + 5·2N-2.
We now solve this recurrence relation in a similar way to solving a
linear differential equation.
The general solution will be
G(N) = C·2N + aN2N-2,
where a is to be determined. Then
G(N + 1) = 2G(N) + 5·2N-2 yields a = 5/2, so
G(N) = C·2N +5N2N-3. The initial condition G(2) = 6 shows that
C = 1/4 and we conclude that
G(N) = 2N-2 +5N2N-3
N≥2. We also have G(1) = 2.