STEP 2013 Solutions Watch

LShirley95
Badges: 0
Rep:
?
#21
Report 5 years ago
#21
(Original post by Brister)
How do you figure out the direction for friction? Sorry if you explained it already, but it is hard to read your attachments.
Sorry about the poor quality. The direction of the friction is always in the opposite direction to where the object would move if there were none. So as the system would just fall down without the friction, friction acts upwards on the top ball and downward (and to the right/left for the left/right ball respectively) for the bottom ones. Always perpendicular to the normal reaction, of course.

Posted from TSR Mobile
0
reply
mikelbird
Badges: 8
Rep:
?
#22
Report 5 years ago
#22
(Original post by jack.hadamard)
I had to practice some drawing on a software, so enjoy.

Question 4


Preliminary stuff

Spoiler:
Show

Attachment 228772

The line passing through (a,0) with gradient b has the equation y = b(x - a) and the unit circle has equation x^2 + y^2 = 1.

Hence,

x^2 + [b(x - a)]^2 = 1 \iff (1 + b^2)x^2 - 2ab^2x + (ab)^2 - 1 = 0

implies that

\displaystyle x = \frac{2ab^2 \pm \sqrt{(-2ab^2)^2 - 4(1 + b^2)[(ab)^2 - 1]}}{2(1 + b^2)} = \frac{ab^2 \pm \sqrt{1 + b^2(1 - a^2)}}{1 + b^2}.

It follows that the midpoint has x-coordinate \displaystyle \frac{ab^2}{1 + b^2}, and from that we see that its y-coordinate is

\displaystyle y = b(x - a) = ab\left[\frac{b^2}{1 + b^2} - 1\right] = -\frac{ab}{1 + b^2}.

Therefore, the midpoint is \displaystyle M\left(\frac{ab^2}{1 + b^2}, -\frac{ab}{1 + b^2} \right).


First part

Spoiler:
Show

Since b is fixed, the locus of M is given by the parametric equations

\displaystyle x(a) = a\frac{b^2}{1 + b^2} and \displaystyle y(a) = a \frac{-b}{1 + b^2}.

Clearly, -by = x.

The above equation shows that, since b is fixed, the locus is a straight line.

Now, the length is

\displaystyle \sqrt{[x(1) - x(-1)]^2 + [y(1) - y(-1)]^2} = \sqrt{[x(1) - x(-1)]^2 + \frac{1}{(-b)^2}[x(1) - x(-1)]^2}

which is just \displaystyle \frac{2b^2}{1 + b^2} \sqrt{\frac{1 + b^2}{b^2}} = \frac{2b}{\sqrt{1 + b^2}}.

The sketch looks like the following.

Attachment 228756


Second part (a)

Spoiler:
Show

Now a is fixed, so the parametric equations are x(b) and y(b).

Letting b = \tan(t), for -\frac{\pi}{2} < t < \frac{\pi}{2}, we see that

\displaystyle x = a\frac{\tan^2(t)}{1 + \tan^2(t)} = a\sin^2(t) = \frac{a}{2}[1 - \cos(2t)]}
and
\displaystyle y = a\frac{-\tan(t)}{1 + \tan^2(t)} = -a\sin(t)\cos(t) = -\frac{a}{2}\sin(2t)

It follows that (2x - a)^2 + (2y)^2 = a^2 \iff (x - \frac{a}{2})^2 + y^2 = \left(\frac{a}{2}\right)^2.

Bearing the interval in mind, we see that this traces out a complete circle. The sketch below.

Attachment 228757


Second part (b)

Spoiler:
Show

Now, both a and b vary, but 0 < b \leq 1.

Using ab = 1, we can let a = \cot(t), in the previous part, and notice that 0 < t \leq \frac{\pi}{4}.

Hence,
\displaystyle x = \frac{a}{2}[\sin^2(t)] = \frac{\cot(t)}{2}\sin(2t) = \sin(t)\cos(t) = \frac{1}{2}\sin(2t)
and
\displaystyle y = -a\sin(t)\cos(t) = -\cot(t)\sin(t)\cos(t) = -\cos^2(t).

Therefore, (2x)^2 + (2y + 1)^2 = 1 \iff x^2 + (y + \frac{1}{2})^2 = \left(\frac{1}{2}\right)^2.

Again, bearing the interval we see that it is a quarter of the circle. Sketch below.

Attachment 228778

When drawing it, notice that y = b(x - a) = bx - 1.
for the prelim stuff why not simply use the sum of the roots and then half it for the x coordinate of M?

In part (ii)(a) the point (a,0) in NOT included!!

In part (ii) (b) the point (0,-1) is not included
1
reply
jack.hadamard
Badges: 4
Rep:
?
#23
Report 5 years ago
#23
(Original post by mikelbird)
for the prelim stuff why not simply use the sum of the roots and then half it for the x coordinate of M?
Because STEP II is based on A-level Maths content.

In part (ii)(a) the point (a,0) in NOT included!!
You are right. I forgot to mention it.

In part (ii) (b) the point (0,-1) is not included
I didn't claim it was included.
0
reply
mikelbird
Badges: 8
Rep:
?
#24
Report 5 years ago
#24
Because STEP II is based on A-level Maths content.

Playing around with the roots of quadratic equations IS part of A level!
0
reply
mathsymathsy
Badges: 2
Rep:
?
#25
Report 5 years ago
#25
(Original post by DJMayes)
Q7 – STEP II

Part i):
Spoiler:
Show


 (3p+4q)^2-2(2p+3q)^2 = 1

 (9p^2+12pq+16q^2)-(8p^2+12pq+18q^2)=1

 p^2-2q^2 = 1

Hence the second solution reduces to the first and is therefore a solution if the first is a solution.

Part ii):
Spoiler:
Show


If x is odd and y is even, then write  x=2m+1 and  y = 2n :

 (2m+1)^2-2(2n)^2=1

 4m^2+4m+1-8n^2=1

 4m^2+4m = 8n^2

 n^2 = \frac{1}{2}m(m+1)



Part iii):

Spoiler:
Show


 b^3=c^4-a^2

 b^3 = (c^2-a)(c^2+a)

Now, b is a prime number meaning that  b^3 has four possible factors – 1, b,  b^2 and  b^3 . Let’s examine each of these in turn:

 c^2-a = b^3 ,  c^2+a = 1

This solves to get  2a = 1-b^3 which is not possible as a and b are both positive integers.

 c^2-a = 1 ,  c^2+a = b^3

 c^2 = \frac{1}{2}(b^3+1) ,  a = \frac{1}{2}(b^3-1)

 c^2-a = b^2 ,  c^2+a = b

This solves for  2a = b –b^2 which is not possible as a and b are both positive integers, yet b squared is greater than b implying that a = 0.

 c^2-a = b ,  c^2+a = b^2

 c^2 = \frac{1}{2}b(b+1) ,  ] a = \frac{1}{2}b(b-1)

Now, our final part – we need to find integer solutions of these equations for a, b and c where b is not prime. The easiest place to start is with b and c. From our final solution, we are after a square number which, when doubled, gives us the product of two consecutive numbers. Trying the obvious ones (1-10) and doubling we find that if c = 6, 2c^2 = 72 = 8x9, giving b=8 as a non prime solution, which is what we want. Finally, plugging in b to the equation with a yields a = 28, so a possible set of integer solutions is a=28, b=8, c=6.
I think part I is incomplete and the solution could be used in part III
0
reply
mathsymathsy
Badges: 2
Rep:
?
#26
Report 5 years ago
#26
Nobody is writing up a solution for step II Q1?
0
reply
mikelbird
Badges: 8
Rep:
?
#27
Report 5 years ago
#27
(Original post by mathsymathsy)
Nobody is writing up a solution for step II Q1?
OK...If you want one....
Attached files
1
reply
mikelbird
Badges: 8
Rep:
?
#28
Report 5 years ago
#28
(Original post by mikelbird)
Because STEP II is based on A-level Maths content.

Playing around with the roots of quadratic equations IS part of A level!
And it is in the Step specification!!
0
reply
metaltron
Badges: 11
Rep:
?
#29
Report 5 years ago
#29
(Original post by mikelbird)
And it is in the Step specification!!
For STEP III I would imagine, it is in FP1 on OCR.
0
reply
mathsymathsy
Badges: 2
Rep:
?
#30
Report 5 years ago
#30
(Original post by mikelbird)
OK...If you want one....
Step II is complete now
1
reply
mikelbird
Badges: 8
Rep:
?
#31
Report 5 years ago
#31
(Original post by metaltron)
For STEP III I would imagine, it is in FP1 on OCR.
no..for Step 2...it talks about a cubic and knowing that the constant is the product of the roots....but yes you are right about OCR!
0
reply
jack.hadamard
Badges: 4
Rep:
?
#32
Report 5 years ago
#32
(Original post by mikelbird)
no..for Step 2...it talks about a cubic and knowing that the constant is the product of the roots....but yes you are right about OCR!
What's this rant about? If you don't like the quadratic formula, then don't use it. The specification of STEP I and II is the same and, in particular, you are not required to know the Viète formulae. The example you gave is likely to instead emphasise identities and comparing coefficients. I tried to produce a solution that resembles official solutions, which I recall use the formula.
0
reply
mikelbird
Badges: 8
Rep:
?
#33
Report 5 years ago
#33
Oh dear...this was never meaning to be a rant. All I wanted to point out is a quick and easy ( and something well within the capabilities of a good A level student) method of getting a result. The fact that I pointed out the fact seems to have enraged you.. I apologise profusely but still believe that it was a better method... You don't have to. Let's leave it there.
0
reply
jack.hadamard
Badges: 4
Rep:
?
#34
Report 5 years ago
#34
(Original post by mikelbird)
The fact that I pointed out the fact seems to have enraged you.. I apologise profusely but still believe that it was a better method...
I apologise then. It is a better method -- there is no question about that. It didn't 'enrage' me, I just didn't see a reason why you would ask.
0
reply
DJMayes
Badges: 16
Rep:
?
#35
Report 5 years ago
#35
(Original post by mathsymathsy)
I think part I is incomplete and the solution could be used in part III
Damn. When I did the question I missed the "hence find". However, I had x as 3 and y=2 as my first solution, so would this likely lose marks?
0
reply
GG.TintiN
Badges: 0
Rep:
?
#36
Report 5 years ago
#36
(Original post by mathsymathsy)
I think part I is incomplete and the solution could be used in part III

Although it is incomplete, I think they won't give u the credit for writing down the solution for x^2-2y^2=1...in the previous year there was one question asking you to write down the cube from 1 to 10...the report said no mark is given.
0
reply
GG.TintiN
Badges: 0
Rep:
?
#37
Report 5 years ago
#37
(Original post by DJMayes)
Damn. When I did the question I missed the "hence find". However, I had x as 3 and y=2 as my first solution, so would this likely lose marks?
I missed the writing down part...and used nonsense logic to argue ii)...throw away easy marks sigh...
0
reply
Jkn
Badges: 11
Rep:
?
#38
Report 5 years ago
#38
(Original post by cpdavis)
Question 5:
To see when f(x)>\frac{343}{36}, we use our diagram as an aid and you see that x<-2, -\frac{1}{2}<x<\frac{1}{3}, \frac{2}{3}<x<\frac{3}{2}, x>3
You are forgetting to adjust your ranges such that they do not include the values for which f(x) is undefined
0
reply
MW24595
Badges: 0
Rep:
?
#39
Report 5 years ago
#39
(Original post by DJMayes)
Damn. When I did the question I missed the "hence find". However, I had x as 3 and y=2 as my first solution, so would this likely lose marks?
Ouch. Perhaps. No more than 1 or 2, though, provided the rest of the question is good. Because, if that's all good, then the last part is just trivial experimentation.
Meh, no issues. From what I see, I reckon you can still gun for a near-perfect score.
0
reply
Slowbro93
  • PS Reviewer
Badges: 20
Rep:
?
#40
Report Thread starter 5 years ago
#40
(Original post by Jkn)
You are forgetting to adjust your ranges such that they do not include the values for which f(x) is undefined
Why yes, I knew I forgot something will update solution later today in addition to the original OP. and will also stick in a sketch too

also, did someone do a complete solution to Q3? I've not had a chance to look at solutions as I'm finishing off my work for supervisor :ahee:
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • Arts University Bournemouth
    Art and Design Foundation Diploma Further education
    Sat, 25 May '19
  • SOAS University of London
    Postgraduate Open Day Postgraduate
    Wed, 29 May '19
  • University of Exeter
    Undergraduate Open Day - Penryn Campus Undergraduate
    Thu, 30 May '19

How did your AQA GCSE Physics Paper 1 go?

Loved the paper - Feeling positive (340)
30.38%
The paper was reasonable (451)
40.3%
Not feeling great about that exam... (193)
17.25%
It was TERRIBLE (135)
12.06%

Watched Threads

View All