STEP 2018 Solutions Watch

DFranklin
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#41
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(Original post by Aditya33)
Here is my attempt at the third question:
I think what you've done is fine, but for the record, for (ii), from the point where you've deduced \tan 2\alpha = \beta, I think it's fine to simply claim from the diagram that for the "right hand" curve \alpha, \beta < \pi/2 (and so 2\alpha = \beta), while for the "left hand" curve \frac{\pi}{2} < \alpha < \pi so \pi < 2\alpha < 2\pi and therefore \tan 2\alpha = \tan \beta \implies 2\alpha - \pi = \beta.

Edit: Just to say that to my "difficulty spider sense", this looked like being the worst of the pure questions, but was probably actually one of the easiest questions on the paper. Will be interesting to see how many attempted it!
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Aditya33
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(Original post by DFranklin)
I think what you've done is fine, but for the record, for (ii), from the point where you've deduced \tan 2\alpha = \beta, I think it's fine to simply claim from the diagram that for the "right hand" curve \alpha, \beta < \pi/2 (and so 2\alpha = \beta), while for the "left hand" curve \frac{\pi}{2} < \alpha < \pi so \pi < 2\alpha < 2\pi and therefore \tan 2\alpha = \tan \beta \implies 2\alpha - \pi = \beta.

Edit: Just to say that to my "difficulty spider sense", this looked like being the worst of the pure questions, but was probably actually one of the easiest questions on the paper. Will be interesting to see how many attempted it!
Indeed. That would be simpler . Also, it is easy to see that twice of alpha lies between pi and 3pi/2 since tan2a is positive so my argument to show a bound for alpha was completely unnecessary.
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Notnek
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(Original post by DFranklin)
Edit: Just to say that to my "difficulty spider sense", this looked like being the worst of the pure questions, but was probably actually one of the easiest questions on the paper. Will be interesting to see how many attempted it!
I don't know if you've looked at all the questions but you have a pick for hardest question yet?
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gidon_996
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How many marks do you think question 2 part iii will be?
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DFranklin
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(Original post by Orwellian49)
Could anyone enlighten me with their solutions for Q13? When I answered it on Tuesday I felt that it went alright, but now I think I may have missed out some parts in my working.

On a side note, I only managed to complete 3 & a quarter questions, which seemed pretty little compared to others. Not sure if I could meet the grades needed for a 2. ((
Here's an attempt: I have to say I found this a massive (and boring) algebra grind. I couldn't see anything interesting (or any real way of reducing the grind). Possibly I missed something. Anyhow...

For ease of writing, define p = 1/n as the probability of getting any particular question correct. Note that p <= 1/2.

(i) Need to attempt at least 3 questions to score >=5. Define
P3 = p(Pass | answered 3 questions) = p^3.
P4 = p(Pass | 4 questions) = 4p^3(1-p) + p^4 (must either get 3 questions right or 4)
P5 = p(Pass | 5 questions) = 5p^4(1-p) + p^5.

Note P4 > 4p^3(1-p) > p^3 = P3 since 4(1-p) > 1 since p <=1/2.
Note also p^5 < p^4 and 4p^3 > 5p^4 (since 4 > 5p) so P4 > P5. So P4 is the highest probability and she should choose k=4.

(ii) p(Pass AND answered 3 questions) = P3 / 6. p(Pass AND answered 4) = P4/6. p(Pass AND answered 5) = P5/6.
so p(pass) = (P3+P4+P5)/6.
So p(answered 4 | pass) = p(Pass AND answered 4) / p(pass) = P4/(P3+P4+P5) = \dfrac{4p^3(1-p)+p^4}{p^3+4p^3(1-p)+p^4+5p^4(1-p)+p^5}

 = \dfrac{4 -4p +p}{1+4-4p + p+5p(1-p)+p^2} = \dfrac{4-3p}{5 - 3p + 5p - 5p^2 + p^2}

 = \dfrac{4-3p}{5+2p-4p^2} = \dfrac{4-3/n}{5+2/n-4/n^2} = \dfrac{4n^2-3n}{5n^2+2n-4}

(iii) Candidate has probability (1-p) of answering a question. So p(answers 3) = \binom{5}{2} p^2(1-p)^3 = 10 p^2(1-p)^3. So p(Pass AND answers 3) = 10p^2(1-p)^3 P3 = 10p^2(1-p)^3 p^3 = 10p^5(1-p)^3.
Similarly, p(Pass AND answers 4) = 5 p(1-p)^4 P4 = 20p^4(1-p)^5 + 5p^5(1-p)^4
And p(Pass AND answers 5) = (1-p)^5 P5 = 5p^4(1-p)^6 + p^5(1-p)^5

So p(Pass) is the sum of these, that is:

p^4(1-p)^3(10p + 20(1-p)^2+5p(1-p)+5(1-p)^3+p(1-p)^2)

=\frac{1}{n^4}\frac{(n-1)^3}{n^3}\left(\frac{10}{n} + 20\frac{(n-1)^2}{n^2} + 5\frac{n(n-1)}{n^2} + 5 \frac{(n-1)^3}{n^3} + \frac{(n-1)^2}{n^3}\right)

=\frac{(n-1)^3}{n^{10}}(10n^2 +20n(n-1)^2+5n^2(n-1)+5(n-1)^3+(n-1)^2)

=\frac{(n-1)^3}{n^{10}}(10n^2 + 20n^3-40n^2 + 20n +5n^3-5n^2+5n^3-15n^2+15n-5+n^2-2n+1)

=\frac{(n-1)^3}{n^{10}}(30n^3-49n^2+33n-4)

(Wouldn't be surprised at all if there are algebra errors in this - direct to LaTeX and to be honest I really found it hard to care given the messiness of it all).
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DFranklin
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(Original post by Notnek)
I don't know if you've looked at all the questions but you have a pick for hardest question yet?
Q9 looks vile, and I had a quick go and decided it was vile.

One thing I'd say, however, is that I found quite a common theme with this paper was "unpleasant notation". That is, questions where you keep having to write something fairly complicated (e.g. \left(\dfrac{\pi}{2n}\right) in Q6).

Since I generally just go "straight to LaTeX" when trying to answer questions, I think I see a very exaggerated effect from this. (Writing \left(\dfrac{\pi}{2n}\right) is slightly annoying, but writing \left(\dfrac{\pi}{2n}\right) is rather a lot more so, let alone looking at something like:

A_n+B_n = \dfrac{\pi}{n}\left(1+\cos\left( \dfrac{\pi}{2n}\right)\right) / \sin\left(\dfrac{\pi}{2n}\right) and trying to keep any kind of mental model of what you're looking at.

In Q9, it gets very messy, but I think if you initially defined P = g sin alpha and Q = g sin beta, it would become a lot less painful. Of course, often the best choice of notation isn't obvious until you've already got neck deep in a world of pain.
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Notnek
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(Original post by DFranklin)
Q9 looks vile, and I had a quick go and decided it was vile.

One thing I'd say, however, is that I found quite a common theme with this paper was "unpleasant notation". That is, questions where you keep having to write something fairly complicated (e.g. \left(\dfrac{\pi}{2n}\right) in Q6).

Since I generally just go "straight to LaTeX" when trying to answer questions, I think I see a very exaggerated effect from this. (Writing \left(\dfrac{\pi}{2n}\right) is slightly annoying, but writing \left(\dfrac{\pi}{2n}\right) is rather a lot more so, let alone looking at something like:

A_n+B_n = \dfrac{\pi}{n}\left(1+\cos\left( \dfrac{\pi}{2n}\right)\right) / \sin\left(\dfrac{\pi}{2n}\right) and trying to keep any kind of mental model of what you're looking at.

In Q9, it gets very messy, but I think if you initially defined P = g sin alpha and Q = g sin beta, it would become a lot less painful. Of course, often the best choice of notation isn't obvious until you've already got neck deep in a world of pain.
That one was next on my list to try
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Superdjman1
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(Original post by JeremyC)
Question 7 is an absolute monster of algebra.

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7(i)
Make the substitution and wade through a page of algebra. The equation transforms into:

(p^3+pq^2 -2p^2 q)z^3 + (q^3 + p^2 q - 2 p q^2) = 0

(Note that p and q are distinct - if they were equal then this would just be 0 = 0)

7(ii)
This requires solving the simultaneous equations
c = pq
d = pq(p+q)

Solving them leads to a quadratic formula (either cp^2 -dp + c^2 = 0 or the same equation with p replaced by q, depending on which variable you eliminate).

Solving with the quadratic equation, you find that:
p, q = \frac{d \pm \sqrt{d^2-4c^3}}{2c}
The condition d^2 &gt; 4c^3 follows from the requirement that p and q are real and distinct.

7(iii)
We now apply our results from parts (ii) and (i). We see that we need to pick c = -2, d = -2 (noting that the condition in (ii) is satisfied).

Using our solution to the quadratic equation in (ii) we find that p, q = -1, 2 (or the other way round).

Using our result in (i), we find that substituting
x = \frac{-z+2}{z+1}
into the equation leads to -9z^3+18 = 0, with solution z = \sqrt[3]{2}. Therefore, our answer is
 x = \frac{2-\sqrt[3]{2}}{\sqrt[3]{2}+1}

7(iv)
We cannot use the result from (i), as p and q are now equal. Using the remainder theorem we see that (x-p) is a factor of the equation x^3-3p^2x+2p^3=0, and we can then factorise the equation as (x-p)^2(x+2p) = 0, from which the roots immediately follow.

In the case d^2 = 4c^3 we get repeated roots from the quadratic equation we found in (ii). Therefore, p = q = \frac{d}{2c}, and so we can use the first part of (iv) to see that the roots are now x = \frac{d}{2c}, -\frac{d}{c}



In 7(i) can you not just take (p-q)^2 out as a factor and then deduce that a = p and b =q ?
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Orwellian49
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(Original post by DFranklin)
Here's an attempt: I have to say I found this a massive (and boring) algebra grind. I couldn't see anything interesting (or any real way of reducing the grind). Possibly I missed something. Anyhow...

For ease of writing, define p = 1/n as the probability of getting any particular question correct. Note that p <= 1/2.

(i) Need to attempt at least 3 questions to score >=5. Define
P3 = p(Pass | answered 3 questions) = p^3.
P4 = p(Pass | 4 questions) = 4p^3(1-p) + p^4 (must either get 3 questions right or 4)
P5 = p(Pass | 5 questions) = 5p^4(1-p) + p^5.

Note P4 > 4p^3(1-p) > p^3 = P3 since 4(1-p) > 1 since p <=1/2.
Note also p^5 < p^4 and 4p^3 > 5p^4 (since 4 > 5p) so P4 > P5. So P4 is the highest probability and she should choose k=4.

(ii) p(Pass AND answered 3 questions) = P3 / 6. p(Pass AND answered 4) = P4/6. p(Pass AND answered 5) = P5/6.
so p(pass) = (P3+P4+P5)/6.
So p(answered 4 | pass) = p(Pass AND answered 4) / p(pass) = P4/(P3+P4+P5) = \dfrac{4p^3(1-p)+p^4}{p^3+4p^3(1-p)+p^4+5p^4(1-p)+p^5}

 = \dfrac{4 -4p +p}{1+4-4p + p+4+5p(1-p)+p^2} = \dfrac{4-3p}{9 - 3p + 5p - 5p^2 + p^2}

 = \dfrac{4-3p}{9+2p-4p^2} = \dfrac{4-3/n}{9+2/n-4/n^2} = \dfrac{4n^2-3n}{9n^2+2n-4}

(iii) Candidate has probability (1-p) of answering a question. So p(answers 3) = \binom{5}{2} p^2(1-p)^3 = 10 p^2(1-p)^3. So p(Pass AND answers 3) = 10p^2(1-p)^3 P3 = 10p^2(1-p)^3 p^3 = 10p^5(1-p)^3.
Similarly, p(Pass AND answers 4) = 5 p(1-p)^4 P4 = 20p^4(1-p)^5 + 5p^5(1-p)^4
And p(Pass AND answers 5) = (1-p)^5 P5 = 5p^4(1-p)^6 + p^5(1-p)^5

So p(Pass) is the sum of these, that is:

p^4(1-p)^3(10p + 20(1-p)^2+5p(1-p)+5(1-p)^3+p(1-p)^2)

=\frac{1}{n^4}\frac{(n-1)^3}{n^3}\left(\frac{10}{n} + 20\frac{(n-1)^2}{n^2} + 5\frac{n(n-1)}{n^2} + 5 \frac{(n-1)^3}{n^3} + \frac{(n-1)^2}{n^3}\right)

=\frac{(n-1)^3}{n^{10}}(10n^2 +20n(n-1)^2+5n^2(n-1)+5(n-1)^3+(n-1)^2)

=\frac{(n-1)^3}{n^{10}}(10n^2 + 20n^3-40n^2 + 20n +5n^3-5n^2+5n^3-15n^2+15n-5+n^2-2n+1)

=\frac{(n-1)^3}{n^{10}}(30n^3-49n^2+33n-4)

(Wouldn't be surprised at all if there are algebra errors in this - direct to LaTeX and to be honest I really found it hard to care given the messiness of it all).
Thanks for the help mate!
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Electric-man7
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(Original post by Aditya33)
Here is my attempt at the third question:
How many marks do you think I would get, If I proved the first part completely, graphed the curve for second part and began considering the cases but not getting much further into the problem?
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JeremyC
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(Original post by Superdjman1)
In 7(i) can you not just take (p-q)^2 out as a factor and then deduce that a = p and b =q ?
Yes, that would be much nicer.
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123Master321
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(Original post by DFranklin)
Here's an attempt: I have to say I found this a massive (and boring) algebra grind. I couldn't see anything interesting (or any real way of reducing the grind). Possibly I missed something. Anyhow...

For ease of writing, define p = 1/n as the probability of getting any particular question correct. Note that p <= 1/2.

(i) Need to attempt at least 3 questions to score >=5. Define
P3 = p(Pass | answered 3 questions) = p^3.
P4 = p(Pass | 4 questions) = 4p^3(1-p) + p^4 (must either get 3 questions right or 4)
P5 = p(Pass | 5 questions) = 5p^4(1-p) + p^5.

Note P4 > 4p^3(1-p) > p^3 = P3 since 4(1-p) > 1 since p <=1/2.
Note also p^5 < p^4 and 4p^3 > 5p^4 (since 4 > 5p) so P4 > P5. So P4 is the highest probability and she should choose k=4.

(ii) p(Pass AND answered 3 questions) = P3 / 6. p(Pass AND answered 4) = P4/6. p(Pass AND answered 5) = P5/6.
so p(pass) = (P3+P4+P5)/6.
So p(answered 4 | pass) = p(Pass AND answered 4) / p(pass) = P4/(P3+P4+P5) = \dfrac{4p^3(1-p)+p^4}{p^3+4p^3(1-p)+p^4+5p^4(1-p)+p^5}

 = \dfrac{4 -4p +p}{1+4-4p + p+4+5p(1-p)+p^2} = \dfrac{4-3p}{9 - 3p + 5p - 5p^2 + p^2}
The second '4' on the denominator shouldnt be there I think (in the last line quoted)
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DFranklin
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(Original post by 123Master321)
The second '4' on the denominator shouldnt be there I think
Yeah, looks a bit whiffy, doesn't it? I'll edit the post.
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123Master321
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(Original post by DFranklin)
Here's an attempt: I have to say I found this a massive (and boring) algebra grind. I couldn't see anything interesting (or any real way of reducing the grind). Possibly I missed something. Anyhow...
(iii) Candidate has probability (1-p) of answering a question. So p(answers 3) = \binom{5}{2} p^2(1-p)^3 = 10 p^2(1-p)^3. So p(Pass AND answers 3) = 10p^2(1-p)^3 P3 = 10p^2(1-p)^3 p^3 = 10p^5(1-p)^3.
Similarly, p(Pass AND answers 4) = 5 p(1-p)^4 P4 = 20p^4(1-p)^5 + 5p^5(1-p)^4
And p(Pass AND answers 5) = (1-p)^5 P5 = 5p^4(1-p)^6 + p^5(1-p)^5

So p(Pass) is the sum of these, that is:

p^4(1-p)^3(10p + 20(1-p)^2+5p(1-p)+5(1-p)^3+p(1-p)^2)
The probability of answering a question is \frac{n}{n+1} not 1-p=\frac{n-1}{n}

So easy to make a mistake here, I wrote a computer program yesterday which ran 10 million cycles and compared it to my algebraic answers as a check
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DFranklin
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(Original post by 123Master321)
The probability of answering a question is \frac{n}{n+1} not 1-p=\frac{n-1}{n}

So easy to make a mistake here, I wrote a computer program yesterday which ran 10 million cycles and compared it to my algebraic answers as a check
Oh, flip. That's unrecoverable, of course (although on one level I'm like "the question's so crap anyhow who really cares?"). Just post your solution and I'll delete mine.

(In my defense, this was largely caused by "I can't be bothered to LaTeX \dfrac{n}{n+1} all over the place" - not a mistake I'd make doing the question for real...)
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123Master321
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(Original post by DFranklin)
Oh, flip. That's unrecoverable, of course (although on one level I'm like "the question's so crap anyhow who really cares?". Just post your solution and I'll delete mine.

(In my defense, this was largely caused by "I can't be bothered to LaTeX \dfrac{n}{n+1} all over the place" - not a mistake I'd make doing the question for real...)
I would write my solution but I really cant be bothered to wade through all that LaTeX, plus I have FP3 tomorrow, so ill probably have to do it tomorrow(if someone hasnt done it first).
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Aditya33
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I think about 8-10 marks depending on how far you have progressed into the cases. I feel the second part would carry greater weigtage { Some of which would be for analysing the bounds for the angles graphically } . Don't take my word for it though. It could be more.
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DFranklin
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(Original post by Aditya33)
I think about 8-10 marks depending on how far you have progressed into the cases. I feel the second part would carry greater weigtage { Some of which would be for analysing the bounds for the angles graphically } . Don't take my word for it though. It could be more.
I really don't think the second part is worth that much. Hard to be sure, of course.
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etothepiiplusone
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Zacken I've done Q12, it's post #12.
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Notnek
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Q9:

First part

Spoiler:
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Call the first section of motion AB and the second BC. The vertical distance travelled by the particle in AB is x\sin \alpha and in BC it is d\sin \beta. The initial KE at point A is 0 and if the particle reaches the point C then its KE at C must be \geq 0. So using conservation of energy, the GPE at A must be \geq the GPE at C:

mgx\sin \alpha \geq mgd\sin \beta \Rightarrow x\sin \alpha \geq d\sin \beta.

Second part

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The particle travels down AB with acceleration g\sin \alpha and distance x then up BC with acceleration -g\sin \beta and distance d. Both of these accelerations will be used a lot in the working plus the final result contains one of these terms so we'll simplify the algebra by letting p=g\sin \alpha and q=g\sin \beta. Also since the velocity will always be positive we can use these equations of motion:

\displaystyle v=\sqrt{u^2+2as}

\displaystyle t=\frac{\sqrt{u^2+2as}-u}{a}

It's also worth noting that \displaystyle k=\frac{\sin \alpha}{\sin \beta} = \frac{p}{q}

For the AB motion we'll need the time and the final velocity so we can use it for the BC motion.

\displaystyle v_{AB} = \sqrt{2px}

\displaystyle t_{AB} = \frac{\sqrt{2px}}{p}

Then for BC:

\displaystyle t_{BC} = \frac{\sqrt{2px-2qd} - \sqrt{2px}}{-q}

which is valid since x\sin \alpha \geq d\sin \beta \Rightarrow px-qd\geq 0 so the expression inside the first square root above is non-negative.

Looking at the final result, we will now consider the LHS which is

\displaystyle \begin{aligned} \left(\frac{g\sin\alpha}{2} \right)^{\frac{1}{2}} T &= \sqrt{\frac{p}{2}}\left(t_{AB} + t_{BC}\right) \\ &=\frac{\sqrt{p}}{\sqrt{2}}\left  (\frac{\sqrt{2}\sqrt{p}\sqrt{x}}  {p} + \frac{\sqrt{2}\sqrt{p}\sqrt{x}-\sqrt{2}\sqrt{px-qd}}{q}\right) \\ &= \sqrt{x}+\frac{p}{q}\sqrt{x} - \frac{\sqrt{p}}{q}\sqrt{px-qd} \\ &= \left(1+\frac{p}{q}\right)\sqrt{  x} - \sqrt{\frac{p^2x}{q^2}-\frac{p}{q}d} \\ &= (1+k)\sqrt{x} - \sqrt{k^2x-kd}

Third part

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\displaystyle \begin{aligned} \frac{dT}{dx} = 0 &\Rightarrow \frac{1+k}{2\sqrt{x}} - \frac{k^2}{2\sqrt{k^2x-kd}}=0 \\ &\Rightarrow 2(1+k)\sqrt{k^2x-kd}=2k^2\sqrt{x} \\ &\Rightarrow (1+k)^2(k^2x-kd)=k^4x \\ &\Rightarrow k(1+k)^2x-k^3x = d(1+k)^2 \\ &\Rightarrow x=\frac{d(1+k)^2}{k(1+k)^2-k^3} = \frac{d(k+1)^2}{k(2k+1)}
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