14th VTRMC, 1992, Solutions

1. 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 6≠ 0 at x = 0, so x = 0 is indeed a point inflection).

2. 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.

3. (i)
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, then | sin y|≤| y| with equality if and only if y = 0. It follows that 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.

(ii)
From (i), we have xn = 1 for all n. Thus we need to find n=01/3n. 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.

4. Clearly 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, because 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.

5. First we find the eigenvalues and eigenvectors of A. The eigenvalues satisfy x(x - 3) + 2 = 0, so the eigenvalues are 1,2. To find the eigenvectors corresponding to 1, we need to solve the matrix equation

(
 -1 -2 1 2
) (
 u v
) = (
 0 0
)

One solution is u = 2, v= - 1 and we see that

(
 2 -1
)

is an eigenvector corresponding to 1.

To find the eigenvectors corresponding to 2, we need to solve

(
 -2 -2 1 1
) (
 u v
) = (
 0 0
)

One solution is u = 1, v = - 1 and we see that

(
 1 -1
)

is an eigenvector corresponding to 2.

We now know that if

T = (
 2 1 -1 -1
)

then

T-1AT = (
 1 0 0 2
)

We deduce that

T-1A100T = (
 1 0 0 2100
)

Therefore

A100 = (
 2 - 2100 2 - 2101 -1 + 2100 -1 + 2101
)

6. 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.

7. 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.

Since 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.

8. 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.

In general, 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 N≥2,

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 for all N≥2. We also have G(1) = 2.

Peter Linnell 2009-06-12