STEP Maths I, II, III 1988 solutionsWatch

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2 months ago
#121
another way of doing the last part is to consider the range [0,1] and [1,infinite], and we have x^2/(1+x)<x^2 for 0<x<1, and x^2/(1+x)<x^2/x=x for x>1 , and these integral can be calculated, so give the answer.

(Original post by kabbers)
III/1:

Sketch I'm not going to bother posting most of the differentiating legwork, there arent too many tricks, its just an algebra bash.

Differentiating (quotient rule is your friend), we get So we have turning points at x = 0, and x = Differentiating again, we get Substituting in values of x for the turning points, we find that x = 0 is a minimum, and the other two maximums.

Noticing the denominator of y, we find that there is an asymptote at x = -1.

And because of the exponential properties of e^-x, y tends to zero as y tends to as since the numerator remains positive and given the exponenential properties of e^x much greater than the denominator, while the denominator becomes negative.

So the graph will look http://www.thestudentroom.co.uk/show...8&postcount=14

Prove First note that Hence we may split the integral into Consider    Now consider I posit the inequality for x > 0   So our inequality holds.

So:   Thus And therefore, Now notice that the graph of is greater than 0 for x > 0, and hence so is the infinite integral.

So, please point out any mistakes Last edited by cxs; 2 weeks ago
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2 months ago
#122
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2 months ago
#123
the induction is godd, but there is another better way of doing that
for the part 2, Sn+1=（1+k）Sn -d, so Sn+1-d/k=(1+k)(Sn-d/k), take it as an geometric sequence we have
ST+m=(1+k)m(ST-d/k)+d/k , the result will soon get.
(Original post by kabbers)
relatively straightforward q, pointing out any mistakes will be appreciated as always Ok first note that at the end of day n, the firm will owe everything it owed the previous day, the interest on this plus everything it has borrowed that day.

This may be expressed as By evaluating the first few terms of this sequence, it becomes easy to 'guess' and then prove a total value for Sn:    ok my guess is that (by geometric series)

this fits the questions, but we have to prove it (induction!)

Basis case: , therefore the basis case works.

Inductive step:  This is in the same form as the above, so it works.

for part 2, firstly we can assert that (this is the initial amount owed)

the firm must pay interest on what it owes then pay back what it has earned. Again, let's try a few values:   hypothesis:  (again by geometric)

we need to prove this, again by induction.

basis case: when m = 0, basis works

inductive step:    Hence the sum works. However the question did not ask for us to express S_T, so we must substitute in (see our above def. for S_T)   for part 3, note that if the sequence is decreasing, the firm will pay off its debt. so the conditions are:  Letting x = k+1  taking everything on to one side   k+1 > 0, so both x^m and (x-1) are greater than zero. Therefore, for the inequality to be satisfied,    as required.
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1 month ago
#124
instead of expressing tan ，we can express sin directly by vx、vy（original resolving speed of the first ball），then write t as vx/vy，we get an expression of sin in t .next step is quite easy，whether to differentiate or use the Cauchy inequality
(Original post by Dirac Spinor)
STEP III Q12
Let be the velocities/speeds of ball 1 and 2 with u being before the collision and v after.
Momentum is conserved in a closed system so: The coefficient of restitution is < 1 so kinetic energy is not conserved amongst the two balls, i.e: Combining the two: Where is the angle between the two velocities as desired.
For the next part, let be the angle between .Separating momentum into two components and and considering restitution in the direction of we obtain 3 equations, sufficient to eliminate the three speeds and obtain an equation in terms of our two angles, thus confirming the intuitive idea that the initial velocity of the first ball is irrelevant: By adding the first and last and substituting, we obtain: The angle of deflection is which we shall call k. Consider Notice that the maximum of Tank will occur for the same value of theta as the maximum of k so, differentiating both sides w.r.t. theta we get: Which, if we equate it to zero, gives us: So: as required.

Comment: I found this Q particularly difficult and it has taken me many attempts to get it right. Comments are welcome.
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2 weeks ago
#125
I think the derivative of what braineverit get is right? f'(b)=pi-2/a*√a2+b2 rather than having the 2/a*a√2 term for we should differentiating w.r.t. b
but for the final result, pi2/4-1>1 so we should choose a???
(Original post by mikelbird)
I am afraid I disagree with this result...see attached file...
Last edited by cxs; 2 weeks ago
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2 weeks ago
#126
I think you are absolutely right...cannot recall what on earth I was thinking about!!
(Original post by cxs)
I think the derivative of what braineverit get is right? f'(b)=pi-2/a*√a2+b2 rather than having the 2/a*a√2 term for we should differentiating w.r.t. b
but for the final result, pi2/4-1>1 so we should choose a???
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2 weeks ago
#127
(Original post by mikelbird)
I think you are absolutely right...cannot recall what on earth I was thinking about!!
🤨😀
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2 weeks ago
#128
only trouble is that the given answer makes b greater then a!!!!
(Original post by cxs)
🤨😀
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2 weeks ago
#129
(Original post by mikelbird)
only trouble is that the given answer makes b greater then a!!!!
Is that an official answer of just someone did it?
(P.S. I think this question is a good idea.Begin daydreaming ope this year STEP questions will have this kind of questions haha.stop daydreaming))
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2 weeks ago
#130
Ihave corrected my version, pointed out the contradion of the answer and offered an alternative solution....see if you like it..!
(Original post by cxs)
Is that an official answer of just someone did it?
(P.S. I think this question is a good idea.Begin daydreaming ope this year STEP questions will have this kind of questions haha.stop daydreaming))
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2 weeks ago
#131
This ALT. is brilliant!
(Original post by mikelbird)
Ihave corrected my version, pointed out the contradion of the answer and offered an alternative solution....see if you like it..!
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