STEP 2016 Solutions
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For the f(x), g(x), h(x) would an acceptable answer to the final part be the observation that it can be arranged to have (h(x)+x)+(h(1/1-x)+(1/1-x))=1 and so each of these pairs of terms are 'something' +the function of that 'something' and so it is clear that h(x)=1/2-x would be the solution as the -x part would cancel the 'something' and then the two +1/2 would add to give the one. Note: the 'something' is a function of x
(edited 7 years ago)
Original post by fefssdf
wow everyone here is so clever. have fun at oxbridge you guys 
thanks fam ill try ot enjoy it
Original post by Ewanclementson
For the f(x), g(x), h(x) would an acceptable answer to the final part be the observation that it can be arranged to have (h(x)+x)+(h(1/1-x)+(1/1-x))=1 and so each of these pairs of terms are 'something' +the function of that 'something' and so it is clear that h(x)=1/2-x would be the solution as the -x part would cancel the 'something' and then the two +1/2 would add to give the one. Note: the 'something' is a function of x
Nice! Yes I think it would be.
Original post by Zacken
STEP III: Question 4
Sum to N
Sum to infinity
Hyperbolic sum
OMG OMG OMG OMG I Totally forgot that I had done most of this question too!! I think I'm gonna get 13 marks out of it because I used x=e^2y and didn't do the lastest bit. You know what i might actually get a 1 out of that messy exam.
Original post by Hext
You would not be deducted more than 4 marks.
Try quote people
Posted from TSR Mobile
Quite liked the simple harmonic question (q9) so I thought I'd write up and post a solution:
Original post by L-Tyrosine
Quite liked the simple harmonic question (q9) so I thought I'd write up and post a solution:
How many marks do you think I'd get for getting all the way to the F=ma part and then stoping because I was too scared of the algebra hahaha ?
Original post by matthewdjones
How many marks do you think I'd get for getting all the way to the F=ma part and then stoping because I was too scared of the algebra hahaha ?
Dunno, I don't do step (I'm a bio natsci lmfao) , but judging from my instincts I think that'll be at least 8-12 marks off because that seemed to be the majority of the work (had to spot x^2=0 approximation and binomial approximation, trigonometric manipulation etc)
(edited 7 years ago)
Seems no one has posted a solution to STEP III Q2, so here it is.
(i) y2 = 4ax in parametric form is: x = at2, y = 2at, where t is our parameter.
Hence t = p at P, t = q at Q, and t = r at R.
Gradient of curve at the point with parameter t is:
dy/dx = dy/dt * dt/dx = 2a / 2at = t -1.
So the gradient of the normal to the curve at the point with parameter t is -t.
Hence, the normal to the curve at Q has gradient -q, so has equation
y - 2aq = -q (x - aq2).
Since this line passes through P, we substitute y = 2ap and x = ap2. Dividing through by a, we have
2 (p - q) = -q (p2 - q2) = -q (p + q) (p - q), so 2 = -q (p + q) and hence q2 + qp + 2 = 0.
Similarly, r2 + rp + 2 = 0.
(ii) Line QR has gradient (2aq - 2ar) / (aq2 - ar2) = 2(q - r) / [ (q - r) (q + p) ] = 2 (p + q) -1.
r2 + rp + 2 = 0 = q2 + qp + 2
=> r2 - q2 = (r - q) (r + q) = p (q - r)
=> r + q = -p
=> gradient of line QR is -2 p -1
=> equation of QR is y = -2 p -1 (x - ar2) + 2ar.
Let y = 0. Then x = a (r2 + rp) = -2a.
So QR passes through ( -2a, 0). This point is clearly dependent solely on a and not on the positon of P.
(iii) Line OP has equation y = 2 p -1 x.
So, at T, where OP and QR intersect,
2 p -1 x = -2 p -1 (x - ar2) + 2ar, which implies that
x = 1/2 a (r2 + 2rp) = 1/2 a (-2) = -a.
So T lies on the line x = -a. This is clearly dependent solely on A and not on the positon of P.
Substituting x = -a into the equation of OP, we find that T has y co-ordinate -2ap -1. So the distance of the x-axis from T is the modulus of this.
Since q and r are distinct real numbers satisfying u2 + up + 2 = 0, this quadratic must have positive discriminant, i.e. p2 > 8, so the modulus of p is greater than 2 (2)1/2.
Hence the modulus of -2ap-1 is less than a / 21/2.
(i) y2 = 4ax in parametric form is: x = at2, y = 2at, where t is our parameter.
Hence t = p at P, t = q at Q, and t = r at R.
Gradient of curve at the point with parameter t is:
dy/dx = dy/dt * dt/dx = 2a / 2at = t -1.
So the gradient of the normal to the curve at the point with parameter t is -t.
Hence, the normal to the curve at Q has gradient -q, so has equation
y - 2aq = -q (x - aq2).
Since this line passes through P, we substitute y = 2ap and x = ap2. Dividing through by a, we have
2 (p - q) = -q (p2 - q2) = -q (p + q) (p - q), so 2 = -q (p + q) and hence q2 + qp + 2 = 0.
Similarly, r2 + rp + 2 = 0.
(ii) Line QR has gradient (2aq - 2ar) / (aq2 - ar2) = 2(q - r) / [ (q - r) (q + p) ] = 2 (p + q) -1.
r2 + rp + 2 = 0 = q2 + qp + 2
=> r2 - q2 = (r - q) (r + q) = p (q - r)
=> r + q = -p
=> gradient of line QR is -2 p -1
=> equation of QR is y = -2 p -1 (x - ar2) + 2ar.
Let y = 0. Then x = a (r2 + rp) = -2a.
So QR passes through ( -2a, 0). This point is clearly dependent solely on a and not on the positon of P.
(iii) Line OP has equation y = 2 p -1 x.
So, at T, where OP and QR intersect,
2 p -1 x = -2 p -1 (x - ar2) + 2ar, which implies that
x = 1/2 a (r2 + 2rp) = 1/2 a (-2) = -a.
So T lies on the line x = -a. This is clearly dependent solely on A and not on the positon of P.
Substituting x = -a into the equation of OP, we find that T has y co-ordinate -2ap -1. So the distance of the x-axis from T is the modulus of this.
Since q and r are distinct real numbers satisfying u2 + up + 2 = 0, this quadratic must have positive discriminant, i.e. p2 > 8, so the modulus of p is greater than 2 (2)1/2.
Hence the modulus of -2ap-1 is less than a / 21/2.
(edited 7 years ago)
Original post by riquix
Seems no one has posted a solution to STEP III Q2, so here it is.
(i) y2 = 4ax in parametric form is: x = at2, y = 2at, where t is our parameter.
Hence t = p at P, t = q at Q, and t = r at R.
Gradient of curve at the point with parameter t is:
dy/dx = dy/dt * dt/dx = 2a / 2at = t -1.
So the gradient of the normal to the curve at the point with parameter t is -t.
Hence, the normal to the curve at Q has gradient -q, so has equation
y - 2aq = -q (x - aq2).
Since this line passes through P, we substitute y = 2ap and x = ap2. Dividing through by a, we have
2 (p - q) = -q (p2 - q2) = -q (p + q) (p - q), so 2 = -q (p + q) and hence q2 + qp + 2 = 0.
Similarly, r2 + rp + 2 = 0.
(ii) Line QR has gradient (2aq - 2ar) / (aq2 - ar2) = 2(q - r) / [ (q - r) (q + p) ] = 2 (p + q) -1.
r2 + rp + 2 = 0 = q2 + qp + 2
(i) y2 = 4ax in parametric form is: x = at2, y = 2at, where t is our parameter.
Hence t = p at P, t = q at Q, and t = r at R.
Gradient of curve at the point with parameter t is:
dy/dx = dy/dt * dt/dx = 2a / 2at = t -1.
So the gradient of the normal to the curve at the point with parameter t is -t.
Hence, the normal to the curve at Q has gradient -q, so has equation
y - 2aq = -q (x - aq2).
Since this line passes through P, we substitute y = 2ap and x = ap2. Dividing through by a, we have
2 (p - q) = -q (p2 - q2) = -q (p + q) (p - q), so 2 = -q (p + q) and hence q2 + qp + 2 = 0.
Similarly, r2 + rp + 2 = 0.
(ii) Line QR has gradient (2aq - 2ar) / (aq2 - ar2) = 2(q - r) / [ (q - r) (q + p) ] = 2 (p + q) -1.
r2 + rp + 2 = 0 = q2 + qp + 2
Finish it then lol.
Posted from TSR Mobile
Original post by physicsmaths
lol shhh
Original post by riquix
Seems no one has posted a solution to STEP III Q2, so here it is.
(i) y2 = 4ax in parametric form is: x = at2, y = 2at, where t is our parameter.
Hence t = p at P, t = q at Q, and t = r at R.
Gradient of curve at the point with parameter t is:
dy/dx = dy/dt * dt/dx = 2a / 2at = t -1.
So the gradient of the normal to the curve at the point with parameter t is -t.
Hence, the normal to the curve at Q has gradient -q, so has equation
y - 2aq = -q (x - aq2).
Since this line passes through P, we substitute y = 2ap and x = ap2. Dividing through by a, we have
2 (p - q) = -q (p2 - q2) = -q (p + q) (p - q), so 2 = -q (p + q) and hence q2 + qp + 2 = 0.
Similarly, r2 + rp + 2 = 0.
(ii) Line QR has gradient (2aq - 2ar) / (aq2 - ar2) = 2(q - r) / [ (q - r) (q + p) ] = 2 (p + q) -1.
r2 + rp + 2 = 0 = q2 + qp + 2
=> r2 - q2 = (r - q) (r + q) = p (q - r)
=> r + q = -p
=> gradient of line QR is -2 p -1
=> equation of QR is y = -2 p -1 (x - ar2) + 2ar.
Let y = 0. Then x = a (r2 + rp) = -2a.
So QR passes through ( -2a, 0). This point is clearly dependent solely on a and not on the positon of P.
(iii) Line OP has equation y = 2 p -1 x.
So, at T, where OP and QR intersect,
2 p -1 x = -2 p -1 (x - ar2) + 2ar, which implies that
x = 1/2 a (r2 + 2rp) = 1/2 a (-2) = -a.
So T lies on the line x = -a. This is clearly dependent solely on A and not on the positon of P.
Substituting x = -a into the equation of OP, we find that T has y co-ordinate -2ap -1. So the distance of the x-axis from T is the modulus of this.
Since q and r are distinct real numbers satisfying u2 + up + 2 = 0, this quadratic must have positive discriminant, i.e. p2 > 8, so the modulus of p is greater than 2 (2)1/2.
Hence the modulus of -2ap-1 is less than a / 21/2.
(i) y2 = 4ax in parametric form is: x = at2, y = 2at, where t is our parameter.
Hence t = p at P, t = q at Q, and t = r at R.
Gradient of curve at the point with parameter t is:
dy/dx = dy/dt * dt/dx = 2a / 2at = t -1.
So the gradient of the normal to the curve at the point with parameter t is -t.
Hence, the normal to the curve at Q has gradient -q, so has equation
y - 2aq = -q (x - aq2).
Since this line passes through P, we substitute y = 2ap and x = ap2. Dividing through by a, we have
2 (p - q) = -q (p2 - q2) = -q (p + q) (p - q), so 2 = -q (p + q) and hence q2 + qp + 2 = 0.
Similarly, r2 + rp + 2 = 0.
(ii) Line QR has gradient (2aq - 2ar) / (aq2 - ar2) = 2(q - r) / [ (q - r) (q + p) ] = 2 (p + q) -1.
r2 + rp + 2 = 0 = q2 + qp + 2
=> r2 - q2 = (r - q) (r + q) = p (q - r)
=> r + q = -p
=> gradient of line QR is -2 p -1
=> equation of QR is y = -2 p -1 (x - ar2) + 2ar.
Let y = 0. Then x = a (r2 + rp) = -2a.
So QR passes through ( -2a, 0). This point is clearly dependent solely on a and not on the positon of P.
(iii) Line OP has equation y = 2 p -1 x.
So, at T, where OP and QR intersect,
2 p -1 x = -2 p -1 (x - ar2) + 2ar, which implies that
x = 1/2 a (r2 + 2rp) = 1/2 a (-2) = -a.
So T lies on the line x = -a. This is clearly dependent solely on A and not on the positon of P.
Substituting x = -a into the equation of OP, we find that T has y co-ordinate -2ap -1. So the distance of the x-axis from T is the modulus of this.
Since q and r are distinct real numbers satisfying u2 + up + 2 = 0, this quadratic must have positive discriminant, i.e. p2 > 8, so the modulus of p is greater than 2 (2)1/2.
Hence the modulus of -2ap-1 is less than a / 21/2.
Y coordinate is wrong mate for T. Should -2a/p iirc.
Posted from TSR Mobile
Original post by physicsmaths
hence why I wrote -2ap-1 mate
STEP III Q1 attached.
Very nice starter question to the paper in my opinion! Will do some of the others tomorrow if I get some time
(Edit: there's a pair of limits, a pair of brackets and a power of n missing from the denominator of the first term after "integration by parts" missing but it's still very obvious what should be there - sorry, I did it quite quickly)
Very nice starter question to the paper in my opinion! Will do some of the others tomorrow if I get some time
(Edit: there's a pair of limits, a pair of brackets and a power of n missing from the denominator of the first term after "integration by parts" missing but it's still very obvious what should be there - sorry, I did it quite quickly)
(edited 7 years ago)
STEP III Q5
Solution
(edited 7 years ago)
Original post by Alex_Aits
STEP III Q1 attached.
Very nice starter question to the paper in my opinion! Will do some of the others tomorrow if I get some time
(Edit: there's a pair of limits and a pair of brackets missing but it's still very obvious what should be there - sorry, I did it quite quickly)
Very nice starter question to the paper in my opinion! Will do some of the others tomorrow if I get some time
(Edit: there's a pair of limits and a pair of brackets missing but it's still very obvious what should be there - sorry, I did it quite quickly)
I did the paper earlier so for anyone who maintained their belief that the suggested expression would make the entire question easier here is a (very) slightly different solution (:P)
Attachment not found
Attachment not found
Edit: these pictures aren't coming out very clearly at all, but here is q2 and q3
Attachment not found
Seems like my wifi isn't playing ball.. looks like I will have to latex the rest. Although 7,9,10 all need diagrams to do the solution justice
(edited 7 years ago)
STEP III Q6
Latex here doesn't seem to work so i'm uploading a pdf (+latex source +geogebra but I had to add .txt to the filename otherwise TSR doesn't let me upload)
I couldn't do it all correctly. In part (iii), I forgot to consider the case b≥√(1+a²), and in part (iv) I did one of the derivations wrong. How much marks do U think I will lose?
Latex here doesn't seem to work so i'm uploading a pdf (+latex source +geogebra but I had to add .txt to the filename otherwise TSR doesn't let me upload)
I couldn't do it all correctly. In part (iii), I forgot to consider the case b≥√(1+a²), and in part (iv) I did one of the derivations wrong. How much marks do U think I will lose?
Original post by Mathemagicien
STEP III Question 3
Haha, I was also proud of my notation for the first part. I didn't use the partial derivative notation though, I just used D of something to denote the degree.
Posted from TSR Mobile
Original post by Farhan.Hanif93
STEP III Q5
Solution
For the last part shouldnt you also say that p1,2m+2=p1,2m+1?
Original post by Ewanclementson
For the f(x), g(x), h(x) would an acceptable answer to the final part be the observation that it can be arranged to have (h(x)+x)+(h(1/1-x)+(1/1-x))=1 and so each of these pairs of terms are 'something' +the function of that 'something' and so it is clear that h(x)=1/2-x would be the solution as the -x part would cancel the 'something' and then the two +1/2 would add to give the one. Note: the 'something' is a function of x
No, that is not showing that it is the only solution.
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