This discussion is now closed.

Check out other Related discussions

Trinity Admissions Test Solutions

Scroll to see replies

My recollection of looking at this was that this question was in a completely different league from the other items on the list and even a half-way rigourous answer was extremely tricky to obtain.

I'm really not sure what they are after on this one.

I enjoyed doing the question, quite analysis-y, in a certain sense, but formalising it was a real pain and I just couldn't quite finish it off.

Reply 41

Krollo

Test 2, Question 3

Hint

Consider the relationship between the coefficients of a polynomial and its roots.

Solution

Suppose four points on the curve are collinear along line y=mx+c. Then their x-coordinates are the roots of the equation

2x^4 +7x^3 + 3x -5 -mx - c =0

or

2x^4 +7x^3 + (3-m)x - (5+c) = 0

The sum of the roots of this polynomial is hence -7/2, giving the roots an average, k, of -7/8.

Posted from TSR Mobile

Reply 42

DFranklin

Original post by -Gifted-

Thanks a ton, it makes sense to me. However, I used conservation of mechanical energy, but that yielded a different result where the velocity doesn't cancel out. Let me scan it.

To confirm other comments: in this scenario momentum is conserved but energy isn't.

Reply 43

Zacken

Specimen Test 2, Question 1:

Hints:

Spoiler

Solution:

Spoiler

(edited 7 years ago)

Reply 44

joostan

Another solution to 9:

Spoiler

Reply 45

Zacken

Test 2, Question 4:

Hints:

Spoiler

Solution:

Spoiler

(edited 8 years ago)

Reply 46

DFranklin

Original post by Zacken

Specimen Test 2, Question 1:

Hints:

Spoiler

Solution:

Spoiler

Alternative:

Spoiler

Reply 47

Zacken

Original post by DFranklin

Alternative:

Spoiler

That's interesting, thanks for that. :smile:

It's a good technique as well, for inequalities that have one quantity being much bigger/smaller than the other, you should make use of all the extra room to your advantage. Although I always have trouble knowing what to do initially that will get me what I want down the road.

Reply 48

edothero

Specimen Test 1, Question 10:

Hints:

Spoiler

Solution Part 1:

Spoiler

Solution Part 2:

Spoiler

(edited 8 years ago)

Reply 49

DFranklin

Original post by edothero

Specimen Test 1, Question 10:

Spoiler

Reply 50

Zacken

Test 2, Question 2:

Hints:

Spoiler

Solution:

Part 1:

Spoiler

Part 2:

Spoiler

Reply 51

username1763791

Specimen Test 2, Question 6

Hint:

Spoiler

Solution -

First Part:

Spoiler

Second Part:

Spoiler

(edited 8 years ago)

Reply 52

Krollo

Specimen 2, Q7

Hint

(x-1) is a factor of x^n -1 always.
When is (x+1) a factor of x^n+1?

Solution

Unless otherwise stated, assume all new constants are positive integers.

Suppose p = a^n - 1.
Then p = (a-1)(a^n-1 + ... + 1)

So p is composite unless a-1 = 1 (as the second bracket is greater or equal to the first), so a=2.

So p = 2^n - 1.

Suppose n is composite, i.e. n=cd, with neither c or d less than 2.

Then p =l 2^(cd) -1 = (2^c)^d -1 = (2^c -1)(2^(cd-c) + ... +1)

As c>1, p is thus composite. So n cannot be composite - in other words, n is prime.

- - -

Let q = a^n + 1. We seek the conditions in which q is prime.

If a is 1, then q is clearly always prime.

If a is an odd integer greater than 1, then a^n is odd, so q is even (and not 2) hence it is never prime in this case.

If a is an even integer, then if n is odd and greater than 1, q=(a+1)(a^n-1 - ... +1). q is thus never prime as a+1 > 1.

If a is an even integer, then if n is even but has an odd factor r where r>1, and n=rs, then q = a^(rs) +1 = (a^s)^r + 1 = (a^s+1)(a^rs-s... +1) and so q is never prime.

This leaves the only possibility as when a is even, and n is even and has no odd factors > 1, i.e. n is a non-negative integer power of two. Examples of this would be
2^4 +1 =17, 6^1+1 = 7, 6^2 +1 =37.

Posted from TSR Mobile

Reply 53

edothero

Specimen Test 1, Question 1:

Solution 1:

Spoiler

Solution 2:

Spoiler

(edited 8 years ago)

Reply 54

Renzhi10122

Specimen 2, q8

Hint:

Spoiler

Solution:

Spoiler

(edited 8 years ago)

Reply 55

Krollo

Test 2, Q10

Hints

Might be worth considering energy.

Solution

For the particle to rest, friction must exceed the restoring force. So F>kx, i.e. x< F/k.

For the second part, consider energy. Call the excursion at the start of the halfcycle x0, and the excursion at the end x1. We want to find x0 - x1.

Initially, the energy in the spring is 0.5k(x0)^2. It then loses F*(x0+x1) of energy to friction. It finally has 0.5k(x1)^2 of energy.

So k(x0)^2 = 2F(x0+x1) + k(x1)^2
k(x0-x1)(x0+x1) = 2F(x0+x1)
k(x0-x1)=2F
x0-x1 = 2F/k
, as required.

To discuss the motion, you could say that the particle oscillates about the point at which the spring is anchored, with decreasing amplitude as time goes on until it comes to a stop (as soon as the amplitude falls below F/k).

(edited 8 years ago)

Reply 56

Zacken

Original post by Krollo

Test 2, Q10
x

My mechanics is a little rusty, but would the 'discuss' basically just be damped SHM?

Edit: Stop doing maths and enjoy your vacation! :tongue:

(edited 8 years ago)

Reply 57

Krollo

Original post by Zacken

My mechanics is a little rusty, but would the 'discuss' basically just be damped SHM?

Edit: Stop doing maths and enjoy your vacation! :tongue:

Yes, I think, but I can't imagine it would be expected.

And I am enjoying my holiday - I'm doing maths, of course. :-)

Reply 58

joostan

Specimen Paper 2, Question 9:
Caveat - my probability is a bit weak and I'm not convinced about the last part.
Also succinct notation was a pain.

Spoiler

(edited 8 years ago)

Reply 59

atsruser

Original post by Krollo

Test 2, Q10

Hints

Might be worth considering energy.

Solution

For the particle to rest, friction must exceed the restoring force. So F>kx, i.e. x< F/k.

For the second part, consider energy. Call the excursion at the start of the halfcycle x0, and the excursion at the end x1. We want to find x0 - x1.

Initially, the energy in the spring is 0.5k(x0)^2. It then loses F*(x0+x1) of energy to friction. It finally has 0.5k(x1)^2 of energy.

So k(x0)^2 = 2F(x0+x1) + k(x1)^2
k(x0-x1)(x0+x1) = 2F(x0+x1)
k(x0-x1)=2F
x0-x1 = 2F/k
, as required.

To discuss the motion, you could say that the particle oscillates about the point at which the spring is anchored, with decreasing amplitude as time goes on until it comes to a stop (as soon as the amplitude falls below F/k).

Some further thoughts (I don't have a complete solution):

Spoiler

(edited 8 years ago)