Trinity College Admissions Quiz (Mathematics) 2

1. Which is greater?

Solution 1. Consider taking natural logarithms. This gives and . Since the former is a (translated) exponential function while the latter is a polynomial, as gets large, so .

Solution 2. if , which is clearly true for using induction.

2. Recursive Integrals

Solution. Integration by parts.

which gives , i.e., as required.

Now let which gives and which gives . Also, This gives us and hence .

3. Four Collinear Points

Solution. Suppose the four points on the curve are collinear along line . Then their -coordinates are the roots of the equation or . By Viète's formula, the sum of the roots of this polynomial is , giving the roots an average of .

4. Graphs of "Exponential" Functions

Solution. We sketch and . They both have a root at so that's one solution. To help us sketch the graphs, we look at only to find that's a solution as well. Since our graph clearly depicts only two solutions, all our solutions are given by or .

Noe consider the graph. First thing to note it that the domain of our function is .

  1. An easy root (by inspection) is since we get . (Note that the graph touches the -axis here, which will be justified in 4.)

  2. Next, we look at the -intercepts: yields .

  3. We can also see that when , as well, so we know that the tail of the graph will look exponential (which is clear since dominates for large ).

  4. The stationary points are when , upon which taking logarithms yields . Thus, or , with and respectively.

Putting all of this together yields the sketch.

5. Coins in a Box

Solution. Using Bayes' rule.

(a) Let be the event the final coin selected is biased and let be the event the coin comes up heads three times.

(b) Let be the event the biased coin is among the three selected and let be the event the coin comes up heads three times.

6. Chebyshev Polynomial

Solution 1. so , giving where . The second part is equivalent to where and , so , giving eventually or by solving the cubic equation.

Solution 2. The first part is identical. Now use , giving . Solving for gives the same roots.

7. Primes

Solution. Suppose . Then , so is composite unless , giving .

Suppose is composite, i.e. for some with neither or less than 2. Then . As , is composite, so cannot be composite.

Let . If is 1, then is clearly always prime. If is an odd integer greater than 1, then is odd, so is even (and not 2); hence, it is never prime in this case. If is an even integer, then if n is odd and greater than 1, , with never being a prime as . If is an even integer and is even but has an odd factor where , we can write in the form for some . Consequently, and so is never prime. This leaves the only possibility as when is even, and is even and has no odd factors , i.e., is a non-negative integer power of two.

8. A Regular Pentagon

Solution. Let , , and . We have , , , where is the center of the pentagon. Then , giving by the double-angle formula. Also, . Noting that , we have .

Now, , so . We have , so .

We know that , so , implying . Hence, the area of the pentagon is .

9. Thirteen Cards

Solution. Let be the number of cards of type where . Then, assuming ,



c) , so Thus,

10. A Mass on a Table

Solution. For the particle to rest, friction must exceed the restoring force, so , i.e., .

For the second part, consider energy. Call the excursion at the start of the halfcycle , and the excursion at the end . We want to find .

Initially, the energy in the spring is . It then loses of energy to friction. It finally has of energy. Applying the conservation of energy,