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#21
STEP III, Question 2

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Set p = 1/2 to get the identity in the question

Differentiating with respect to x and multiplying by 2x and adding the original expression

Putting x = 1/5 we get the required identity
1
10 years ago
#22
(Original post by tommm)
n.b. I don't know how to LaTex the dots denoting differentiation w.r.t to time, so I'm just using x for displacement, v for speed and a for acceleration.
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Do you mean like this: and ?

The codes are \dot{x} and \ddot{x} respectively.
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#23
STEP III, Question 6

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A trivial consequence of the product to sum formulae

Let S be the sum

So

Therefore or
3
10 years ago
#24
(Original post by Horizontal 8)
STEP II Q3

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As

As n increases without bound.

(Therefore the given integer m is one.)

I grant you it's not 100% clear, but I would interpret this part as "given an integer m, give an example of a sequence of irrational numbers that converges to m". (In other words, you have to produce a sequence that works for every m).
1
10 years ago
#25
(Original post by DFranklin)
I grant you it's not 100% clear, but I would interpret this part as "given an integer m, give an example of a sequence of irrational numbers that converges to m". (In other words, you have to produce a sequence that works for every m).
Define a sequence of which all the terms are clearly irrational since U_n is irrational for all n and a rational number + irrational number is always irrational. Yet it converges to m as

Is this what you mean?
0
10 years ago
#26
STEP II 2003 Q1

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i)

when a = 0,

(1)
(2)
(3)

(1) - (2) gives
and also

Therefore, a solution exists iff

ii)

(1)
(2)
(3)

(2) - 2.(1) gives

Substituting this into (1) gives

Substituting this values into (3) shows that the solution is consistent no matter what equals.

iii)
When

therefore

This is least when the squared part equals 0, i.e.

Therefore

iv)

If , then

and

Therefore we can take something like to give , and this has a solution which satisfies both conditions.

1
10 years ago
#27
(Original post by GHOSH-5)
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Do you mean like this: and ?

The codes are \dot{x} and \ddot{x} respectively.
Thank you, I shall remember that
0
10 years ago
#28
(Original post by tommm)
STEP II 2003 Q1

Spoiler:
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i)

when a = 0,

(1)
(2)
(3)

(1) - (2) gives
and also

Therefore, a solution exists iff

ii)

(1)
(2)
(3)

(2) - 2.(1) gives

Substituting this into (1) gives

Substituting this values into (3) shows that the solution is consistent no matter what equals.

iii)
When

therefore

This is least when the squared part equals 0, i.e.

Therefore

iv)

If , then

and

Therefore we can take something like to give , and this has a solution which satisfies both conditions.

For implications I'd use \implies and \iff since your \rightarrow is coming out as a 'tends to' (or \to) on my screen atleast.
0
#29
You could us \Rightarrow (note the capital)
0
10 years ago
#30
(Original post by Horizontal 8)
Define a sequence of which all the terms are clearly irrational since U_n is irrational for all n and a rational number + irrational number is always irrational. Yet it converges to m as

Is this what you mean?
Yes. I doubt it would lose you much here, because it's so trivial to fix it to work for the general case. But the standard interpretation of "show XXX for a given integer n" means "it has to work for any n, but you can consider n as fixed".
0
10 years ago
#31
(Original post by Horizontal 8)
For implications I'd use \implies and \iff since your \rightarrow is coming out as a 'tends to' (or \to) on my screen atleast.
Cheers, I'll clean that up.
0
10 years ago
#32
(Original post by DFranklin)
Yes. I doubt it would lose you much here, because it's so trivial to fix it to work for the general case. But the standard interpretation of "show XXX for a given integer n" means "it has to work for any n, but you can consider n as fixed".
ok thanks
0
10 years ago
#33
STEP II 2003 Q2

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Let

We find, using the usual method, that

Therefore

Now remembering that , and multiplying both sides by 3, we achieve the required result.

For the second part, by comparison with the first, we want to find an equation of a similar form that leads to

Using a bit of guesswork, we find that this equation is

which is satisfied by

Rearranging , we get:

and remembering that leads to the required result.

0
10 years ago
#34
STEP I question 2

First part:
Spoiler:
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Considering the discriminant:
In order for a to be real
Which is impossible for real values of b
Similarly: In order for a to be real
Which is impossible for real values of a

Therefore there are no values of a and b for which this will give the correct answer.

Second part:
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Putting it all under a common denominator:

Multiplying out:

Hence the answer will be correct if and only if one of the three equations hold.
3
10 years ago
#35
Question 11, STEP I, 2003
First Part
Let the particle starting at A be labelled P and the particle starting at B be labelled Q. Their masses respectively are and . Let A be a distance l away from where P and Q meet. Let the plane be inclined at angle to the horizontal. Let T be the time when they reach after setting off from A and B.

We can find their accelerations:

Similarly Q accelerates down the slope.

Using SUVAT parallel to the plane:

For P:

For Q:

Second Part
We can find their speeds just before the collision using SUVAT:

For P:

For Q:

We can now find the kinetic energy before the collision:

We need to find the speeds of P and Q after the collision. Let their speeds after the collision be and respectively. Let us assume their directions of motions change after the collision.
By Newton's law of restitution, we find that:

And by the conservation of linear momentum, we find that:

Subtracting these two equations, we find that:

And adding those two equations, we find that:

We can now find the KE after the collision:

And we can now find the loss in KE due to the collision:

0
#36
STEP II, Question 7

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If then

since we are integrating a strictly positive, function, the conclusion follows.

8
10 years ago
#37
STEP II 2003 Q4

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The required area is given by

, which can be seen from drawing a diagram[/latex]

Upon using the substitution , and some manipulation with the cosine double-angle formula, we get

(N.B. that represents limits of integration)

The difficulty in evaluating this lies in the evaluation of . First, using the sine double angle formula, we get

By drawing a suitable right-angled triangle, we find that and from this we can find the required result.

To find the area in the three listed cases, we note that i) and ii) are both specific versions of iii), therefore we need only find the answer to iii) and i) and ii) follow easily.

From a diagram, we see that we must replace d with the perpendicular distance from the line to the furthest point on the circle from the line. This gives us:

, where D is the shortest distance from the origin to the line. The distance from a point (x, y) to the centre is given by

But the point (x, y) lies on the line y = mx + c, therefore

To find the minimum value of D, we must differentiate this expression and set it equal to 0. This gives us

substituting this back into our expression for D^2, we get

and therefore, to find the new area, we must modify (*) such that

The answer to i) follows by letting m = 0; the answer to ii) follows by letting a = b = 0.

0
10 years ago
#38
Question 6, STEP I, 2003
(i)
Case 1: a = 2

Case 2: a ≠ 2

(ii)
With the substitution we find this second integral is exactly the same as the first, and so we have the same two cases with the same two answers.
0
10 years ago
#39
(Original post by DFranklin)
I grant you it's not 100% clear, but I would interpret this part as "given an integer m, give an example of a sequence of irrational numbers that converges to m". (In other words, you have to produce a sequence that works for every m).
Sorry to pester you, but when it says "converges to a given integer m", does that mean that its sum converges to m or does that mean the general term tends to m?
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#40
(Original post by tommm)
Sorry to pester you, but when it says "converges to a given integer m", does that mean that its sum converges to m or does that mean the general term tends to m?
If a sequence converges to a value, then the general term converges to that value
If a series or sum converges to a value then the sum of the series converges to that value
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