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    ∫₀^xdθ/(2 + tanθ) = (2x + ln(2 cos x + sin x) - ln 2)/5.
   Plugging in x = π/4, we conclude that
   ∫₀^π/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=1^∞n/(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 = e^u where u is a function of x to be determined. Then y' = u'e^u and plugging into the given differential equation, we find that u'e^u = e^uu + e^ue^x, hence u' - u = e^x. 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 = xe^x + Ce^x, where C is an arbitrary constant. We are given y = 1 when x = 0, and then u = 0. Therefore u = xe^x and we conclude that y = e^xex.

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 1×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 x^{1/(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/(n^{1/(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.