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Trinity Admissions Test Solutions

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Specimen Paper 4: Question 6.
Hint:

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.
Solution:

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(edited 8 years ago)
Downing Test Question 6 Solution:
Answer:

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Full Solution:

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(edited 8 years ago)
There are more practice questions in the following:

https://www1.maths.leeds.ac.uk/~read/TQ2.pdf
https://www1.maths.leeds.ac.uk/~read/TQ3.pdf

(Mostly the first couple seem to be different, but haven't done a proper read of them. Have fun! :biggrin:)

If you're done with all of those, then you might want to try:

https://www1.maths.leeds.ac.uk/~read/TQ1.pdf

Spoiler

Trinity Paper 4 - Question 7

Hint

How can you find a volume without using geometry?


Solution

I shall assume that the plane cuts at a perpendicular distance a from the centre, otherwise the question doesn't make much sense.

The volume of the smaller part can be found using the volume integral:

[br]π∫arr2−x2dx[br][br]=2πr33−πar2+πa33[br][br]\pi \int_{a}^{r} r^2 - x^2 dx[br][br]= \dfrac{2\pi r^3}{3} - \pi ar^2 + \dfrac{\pi a^3}{3}[br]

and since the volume of the whole sphere is 4/3pir^3, the volume of the larger part is

[br]2πr33+πar2−πa33[br][br]\dfrac{2\pi r^3}{3} + \pi ar^2 - \dfrac{\pi a^3}{3}[br]

As a quick check, plugging in a=0 gives two equal hemispheres, whereas a=r gives a single sphere, as we would expect.

Specimen Test 3, Question 4:

Hint:

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Solution:

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Specimen Test 3, Question 9:

Hint:

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Solution:

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Original post by Jordan\
SPECIMEN TEST 1, QUESTION 5

Hints

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Solution

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lol first probability question I have ever got right.


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Original post by joostan
Specimen Paper 4: Question 6.
Hint:

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.
Solution:

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That's not quite correct. With A = 11, n = 4 yields a larger product than n = 5.
2.75^4 is roughly 57 which is larger than 2.2^5 which is roughly 52.

The maximum product is the maximum of {(A/n)^n : n is A/e rounded up/down} rather than {n^n : n is A/e rounded up/down} - the latter (which you've said) leads you to incorrectly always pick n as the round up of A/e

On a side note, you've done a good job narrowing the problem down, but you haven't given any method to determine which n of the two choices to pick to maximise (A/n)^n: the round up or the round down of A/e.

Surely once you've established that each a{i} must be equal, narrowing n down to two choices (round up/round down of A/e) is a bit pointless if there's no good way of determining which to pick.

Why not just say take n in X = {1,2,...,A} (it's easier to justify that n won't be greater than A than to narrow it down as you've done) to be such that (A/n)^n > (A/m)^m for all m in X\{n}? This solution gets stuck in the same way yours - we can't actually pick between these n without a calculator - but it takes a lot less work to get there! :biggrin:
(edited 8 years ago)
Original post by studentro
That's not quite correct. With A = 11, n = 4 yields a larger product than n = 5.
2.75^4 is roughly 57 which is larger than 2.2^5 which is roughly 52.

The maximum product is the maximum of {(A/n)^n : n is A/e rounded up/down} rather than {n^n : n is A/e rounded up/down} - the latter (which you've said) leads you to incorrectly always pick n as the round up of A/e

On a side note, you've done a good job narrowing the problem down, but you haven't given any method to determine which n of the two choices to pick to maximise (A/n)^n: the round up or the round down of A/e.

Surely once you've established that each a{i} must be equal, narrowing n down to two choices (round up/round down of A/e) is a bit pointless if there's no good way of determining which to pick.

Why not just say take n in X = {1,2,...,A} (it's easier to justify that n won't be greater than A than to narrow it down as you've done) to be such that (A/n)^n > (A/m)^m for all m in X\{n}? This solution gets stuck in the same way yours - we can't actually pick between these n without a calculator - but it takes a lot less work to get there! :biggrin:


This is why I need a calculator o.O - been a bit dim somewhere, though I've thrown my working away so I've no idea why I said that 2.25>2.7542.2^5>2.75^4, though I strongly suspect it's cos I'm rubbish at arithmetic.

You're right about the (An)n\left(\dfrac{A}{n} \right)^n that's just a typo from being too lazy with copy and pasting latex will edit the post.
I'd say that narrowing it down to two cases is better than having AA cases to check, particularly if AA is large, and the amount of work it takes isn't huge, certainly less than it would be to even check each value for say A=100A=100.
Finding which of the two maximises for any given number is, I imagine, an unnecessary amount of work, if it's even possible. I couldn't be bothered at the time, and I'm not that bothered now.
Downing paper - Question 4
Hint 1:

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Hint 2:

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Solution:

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(edited 8 years ago)
I'm a bit iffy about this, but I think it works...

Paper 4, Question 10

Hint

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Solution

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Original post by Krollo
I'm a bit iffy about this, but I think it works...

Paper 4, Question 10

Hint

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Solution

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That's the answer I got! Krollo agrees with me! :ahee:
Though you haven't answered this question:

Assuming both startat the top with zero speed, and that friction plays a negligible role in the second case, which will get to the bottom faster, the man who slides or the log that rolls?

(Log, obviously, since lower translational speed when reaches bottom ∧\wedge same translational acceleration   ⟹  \implies accelerated for lower period of time )

P.S. Why you use Tex rather than LaTex?
Original post by 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.


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Precisely how I done it. Very nice question.


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Original post by Johann von Gauss
That's the answer I got! Krollo agrees with me! :ahee:
Though you haven't answered this question:

Assuming both startat the top with zero speed, and that friction plays a negligible role in the second case, which will get to the bottom faster, the man who slides or the log that rolls?

(Log, obviously, since lower translational speed when reaches bottom ∧\wedge same translational acceleration   ⟹  \implies accelerated for lower period of time )

P.S. Why you use Tex rather than LaTex?


Sorry, that was just me being dopey and interpreting the question as which one reached the bottom with the greater speed (the question seems a tad ambiguous, though I'm fairly sure you're right on this). I agree with your logic.

In relation to your other point, bad habit.
Original post by Krollo
Test 2, Question 3

Hint

Spoiler


Solution

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

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How do you work out the sum of the roots of the polynomial?
Original post by AmarPatel98
How do you work out the sum of the roots of the polynomial?


Vietas formulas, it is a common result.
Or (x-a)(x-b)(x-c)(x-d) where abcd are the roots. Expand that and eqaute coefficients.


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Original post by physicsmaths
Vietas formulas, it is a common result.
Or (x-a)(x-b)(x-c)(x-d) where abcd are the roots. Expand that and eqaute coefficients.


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How did he come up with the polynomial?
Original post by Jordan\
How did he come up with the polynomial?


If points are collinear they lie on a line.
So that means there are four solutions to that polynomial=mx+c
Solving for 0 we get a new polynomial but see that sum of the roots is unaffected by the change in in polynomial as coefficient of x^3 is unchanged so -2(sum)=7 sum=-7/2
We require k=sum/4=-7/8


Posted from TSR Mobile
Original post by physicsmaths
If points are collinear they lie on a line.
So that means there are four solutions to that polynomial=mx+c
Solving for 0 we get a new polynomial but see that sum of the roots is unaffected by the change in in polynomial as coefficient of x^3 is unchanged so -2(sum)=7 sum=-7/2
We require k=sum/4=-7/8


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No what I'm asking is where did all the numbers come from? The 2, the 7 the 5 etc.
Has he just made them up?
Original post by Johann von Gauss

P.S. Why you use Tex rather than LaTex?


Easier to type on this site + barely any typesetting difference. (on this site)

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