16th VTRMC, 1994, Solutions

1. Let I = ∫010x01-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

 I = ∫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.

2. We need to prove that pq≤∫0pf (tdt + ∫0qg(tdt. Either qf (p) or qf (p) and without loss of generality we may assume that qf (p) (if qf (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 of AC, 0pf (tdt is the area of C, and 0qg(tdt is the area of AB. The result follows.

3. 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 is 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 condition f (0) = ±10. If f (0) = 10, we find that C = ln(11/10) and consequently 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 obtain x = ln| f| - ln| f - 1| + D, where D is an arbitrary constant. If the initial condition f (0) = 10, we find that 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).

4. Set 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 α is a root of f, then so is 1/α. Thus we may assume that two of the roots of f are 1/α,α, where 1 < α is a real number.

We claim that the other two roots of f cannot be positive. Indeed suppose β is another positive root of f, which we may suppose is not 1 (though could be α or 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 1/α,α, and by considering the graph of f, we see that f (1) < 0. Thus a + b + 1 + b + a < 0 and the result follows.

5. We follow the hint given, so suppose β is another eigenvector corresponding to 1 with components βi ( 1≤in). Choose t such that αt/|βt|≤αi/|βi| for all i (if βi = 0, then we interpret αi/|βi| = + ∞; note that βt≠ 0). Multiplying β by αtt, we may assume that βt = αt, and then we have βi≤αi for all i. Since

i=1natiαi = αt = βt = ∑i=1natiβi,

we see that βi = αi for all i and the result follows.

6. Set an = f (n)2. Then squaring f (n + 1) = 2√f (n)2+n, we obtain an+1 = 4an + 4n. We solve this recurrence relation in a similar way to solving the corresponding first order differential equation 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. Therefore 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.

7. 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=0x6n/2n = ∑n=010/2n = 20.

Peter Linnell 2007-06-16