29th VTRMC, 2007, Solutions

1. Let I = ∫dθ/(2 + tanθ) We make the substitution y = tanθ. Then dy = sec2θdθ = (1 + y2)dθ and we find that

I = ∫dy/((1 + y2)(2 + y)).

Since 5/((1 + y2)(2 + y)) = 1/(2 + y) - y/(1 + y2) + 2/(1 + y2), we find that

5I = ∫dy/(y + 2) - ∫ydy/(1 + y2) + ∫2dy/(1 + y2) = ln(2 + y) - (ln(1 + y2))/2 + 2 tan-1y.

Therefore 5I = ln(2 + tanθ)/secθ +2θ = ln(2 cosθ + sinθ) + 2θ and we deduce that

I = (2θ + ln(2 cosθ + sinθ))/5,

hence    ∫0xdθ/(2 + tanθ) = (2x + ln(2 cos x + sin x) - ln 2)/5.

Plugging in x = π/4, we conclude that

0π/4dθ/(2 + tanθ) = (π +2 ln(3/√2) - 2 ln 2)/10 = (π + ln(9/8))/10.

2. Let A = 1 + ∑n=1(n + 1)/(2n + 1)! and B = ∑n=1n/(2n + 1)!, so A and B are the values of the sums in (a) and (b) respectively. Now

 A + B = ∑n=0∞1/(2n)! = (e + e-1)/2, A - B = ∑n=0∞1/(2n + 1)! = (e - e-1)/2.

Therefore A = e/2 and B = 1/(2e).

3. We make the substitution y = eu where u is a function of x to be determined. Then y' = u'eu and plugging into the given differential equation, we find that u'eu = euu + euex, hence u' - u = ex. This is a first order linear differential equation which can be solved in several ways, for example one method would be to multiply by the integrating factor e-x. We obtain the general solution u = xex + Cex, where C is an arbitrary constant. We are given y = 1 when x = 0, and then u = 0. Therefore u = xex and we conclude that y = exex.

4. Ceva's theorem applied to the triangle ABC shows that (AR/RB)(BP/PC)(CQ/QA) = 1. Since RP bisects BRC, we see that BP/PC = BR/RC. Therefore AR/RC = AQ/QC, consequently ARQ = ∠QRC and the result follows.

5. Let

 A = (2 + √5)100((1 + √2)100 + (1 + √2)-100) B = (√5 -2)100((1 + √2)100 + (1 + √2)-100) C = (√5 +2)100 + (√5 -2)100 D = (√2 +1)100 + (√2 -1)100

First note that C and D are integers; one way to see this is to use the binomial theorem. Also √2 -1 = (√2 +1)-1. Thus A + B = CD is an integer. Now √5 -2 < 1/4, √2 +1 < 2.5 and √2 -1 < 1. Therefore 0 < B < (5/8)100 + (1/4)100 < 10-4. We conclude that there is a positive number ε < 10-4 such that A + ε is an integer, and the result follows.

6. Suppose det(A2 + B2) = 0. Then A2 + B2 is not invertible and hence there exists a nonzero n×1 matrix (column vector) u with real entries such that (A2 + B2)u = 0. Then u'A2u + u'B2u = 0, where u' denotes the transpose of u, a n matrix. Therefore (Au)'(Au) + (Bu)'(Bu) = 0 and we deduce that u'A = u'B = 0, consequently u'(AX + BY) = 0. This shows that det(AX + BY) = 0, a contradiction and the result follows.

7. We claim that x1/(ln(ln x))2 > (ln x)2 for large x. Indeed by taking logs, we need (ln x)/(ln(ln x))2 > 2 ln(ln(x)), that is ln x > 2(ln(ln x))3. So by making the substitution y = ln x, we want y > 2(ln y)3, which is true for y large. It now follows that for large n,

n-(1+1/(ln(ln n))2) = (1/n)1/(n1/(ln(ln n))2) < 1/(n(ln n)2).

However ∑1/(n(ln n)2) is well known to be convergent, by using the integral test, and it now follows from the basic comparison test that the given series is also convergent.

Peter Linnell 2007-11-07