# STEP 2018 SolutionsWatch

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11 months ago
#21
Heres 6: Zacken
Spoiler:
Show

i)

ii)

Note

iii)

iv)

1
11 months ago
#22
(Original post by Zacken)
Note that we have so that we can see that when .

Let , then for we have

For any with , we have

i) So

ii) Check on Desmos.
Would I lose marks if I defined my F(x) piecewise so it had ln(lnx)) for x>1 and ln(-lnx) for 0<x<1. Also how many marks would i lose if in my graph i had the curve shaping downards as x--> infinity. It still tends to infinity but the gradient is decreasing.
0
11 months ago
#23
(Original post by gnsbrg)
STEP I Q5

b)
implies

Taking into account , we find that there are 4 set of integers

This last bit is the one I'm unsure about (btw sorry for the bad latex first time ever)
I'm not following your logic; if each quarter of values is supposed to correspond to a,b,c,d then surely they need to multiply to 16?
But -1 x 1 X 2 X 8 = -16, for example.
0
#24
(Original post by 123Master321)
Would I lose marks if I defined my F(x) piecewise so it had ln(lnx)) for x>1 and ln(-lnx) for 0<x<1. Also how many marks would i lose if in my graph i had the curve shaping downards as x--> infinity. It still tends to infinity but the gradient is decreasing.
Piecewise F(x) is fine (I was just too lazy)

you’d lose a mark or two at most for the second slip up
0
11 months ago
#25
Question 7 is an absolute monster of algebra.

Spoiler:
Show

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

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

Superdjman1 kindly pointed out that this nasty expression has a factor of . Since a and b are distinct we can divide through by this factor to leave us with the much more pleasant expression

7(ii)
This requires solving the simultaneous equations

Solving them leads to a quadratic formula (either or the same equation with p replaced by q, depending on which variable you eliminate).

Solving with the quadratic equation, you find that:

The condition 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 (noting that the condition in (ii) is satisfied).

Using our solution to the quadratic equation in (ii) we find that (or the other way round).

Using our result in (i), we find that substituting

into the equation leads to , with solution . Therefore, our answer is

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

In the case we get repeated roots from the quadratic equation we found in (ii). Therefore, , and so we can use the first part of (iv) to see that the roots are now
1
11 months ago
#26
(Original post by Zacken)
Note that we have so that we can see that when .

Let , then for we have

For any with , we have

i) So

ii) Check on Desmos.

How many marks for the graph? I messed it up entirely
0
11 months ago
#27
(Original post by Notnek)
STEP I Q8 :

(i)

Spoiler:
Show

To show is constant we can show that the derivative of the expression is 0.

So we have shown that where is constant.

Then plugging in the initial conditions:

thus proving that

(ii)

Spoiler:
Show

Using the product rule:

Using the quotient rule:

(iii)

Spoiler:
Show

(iv)

Spoiler:
Show

(v)

Spoiler:
Show

Using the same substitution as above:

Then using the second result in (ii):

Again using the same substitution:

Rearranging the first result in (ii) gives

(Original post by Rohan77642)
How much do you (and other people) think I would get for the question if I did the parts highlighted in the solution?
How many marks would I lose if I forgot the 1/2 in the very final part of question 8?
0
11 months ago
#28
(Original post by Electric-man7)
How many marks for the graph? I messed it up entirely
I would have thought about 4 or 5 but then again, this is a complete guess
0
11 months ago
#29
Question 1.

Spoiler:
Show

Finding P & Q.
Set the equations equal to each other:

Expand out and factorise to be left with the quadratic equation .

Solving with the quadratic equation, we find that (with the minus value corresponding to the x-coordinate of P, and the plus value corresponding to the x-coordinate of Q).

The sketch has already been provided by igot99problems here.

Finding the tangent line at P.
We differentiate the curve to find .

We substitute in the x-coordinate at P (), and deal with some algebra, to find that the gradient of the curve at P is .

We also substitute this x-coordinate into the equation for y (), and deal with the algebra, to find that the y-coordinate at P is .

We substitute our results into the equation of a straight line through with gradient , namely
, and do some algebra to find that

as required.

Calculating the area of S
Here we need to integrate the area under the curve from 0 to the x-coordinate of P, minus the area under the line from 0 to the x-coordinate of P. Mathematically, we need to calculate:

Expanding the bracket and integrating, we get

Factorising out to make the algebra more bearable, and substituting in the limits we eventually find that the area of S is as required.

Showing the inequality
We use the fact that the area of a triangle is half its base times its height.

The height of triangle OPR equals the y-intercept of the tangent line, namely . The width of the triangle equals the x-coordinate of P, namely . We therefore find that the area of the triangle, T, equals .

For the final inequality, we note that

Now, since we have

Substituting this into the previous result, we find that

as required.

2
11 months ago
#30
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. ((
0
11 months ago
#31
Problem 6

(i)
ith strip is between and . So midpoint is . So

Set (n is a constant).

(ii) is similar but I'll type it later
1
11 months ago
#32
(Original post by DFranklin)
I'm not following your logic; if each quarter of values is supposed to correspond to a,b,c,d then surely they need to multiply to 16?
But -1 x 1 X 2 X 8 = -16, for example.
Absolutely, such a silly mistake (I think in the exam I did it right but I typed this so late at night)
Thanks for checking
0
11 months ago
#33
(Original post by gnsbrg)
Absolutely, such a silly mistake (I think in the exam I did it right but I typed this so late at night)
Thanks for checking
0
11 months ago
#34
(Original post by DFranklin)
Yeah it works; I forgot it but I hope they don’t take out many marks (I think there aren’t more possibilities??)
0
11 months ago
#35
(Original post by gnsbrg)
Yeah it works; I forgot it but I hope they don’t take out many marks (I think there aren’t more possibilities??)
I think it would only cost you a mark (and that's the only one you missed).
0
#36
(Original post by Electric-man7)
How many marks for the graph? I messed it up entirely
I'd say about 6 marks (in Tripos terms, not doing it would just about cost you the alpha).
0
11 months ago
#37
(Original post by JeremyC)
Question 1.

Spoiler:
Show

Finding P & Q.
Set the equations equal to each other:

Expand out and factorise to be left with the quadratic equation .

Solving with the quadratic equation, we find that (with the minus value corresponding to the x-coordinate of P, and the plus value corresponding to the x-coordinate of Q).

The sketch has already been provided by igot99problems here.

Finding the tangent line at P.
We differentiate the curve to find .

We substitute in the x-coordinate at P (), and deal with some algebra, to find that the gradient of the curve at P is .

We also substitute this x-coordinate into the equation for y (), and deal with the algebra, to find that the y-coordinate at P is .

We substitute our results into the equation of a straight line through with gradient , namely
, and do some algebra to find that

as required.

Calculating the area of S
Here we need to integrate the area under the curve from 0 to the x-coordinate of P, minus the area under the line from 0 to the x-coordinate of P. Mathematically, we need to calculate:

Expanding the bracket and integrating, we get

Factorising out to make the algebra more bearable, and substituting in the limits we eventually find that the area of S is as required.

Showing the inequality
We use the fact that the area of a triangle is half its base times its height.

The height of triangle OPR equals the y-intercept of the tangent line, namely . The width of the triangle equals the x-coordinate of P, namely . We therefore find that the area of the triangle, T, equals .

For the final inequality, we note that

Now, since we have

Substituting this into the previous result, we find that

as required.

Can anyone estimate how many marks I’ll get if I did everything up until the point you actually show S in the form they require? I even showed the integral and subtracted the triangle you get from the line OP. I just couldnt get it in the required form so I left the question from there.
0
#38
(Original post by Student1256)
Can anyone estimate how many marks I’ll get if I did everything up until the point you actually show S in the form they require? I even showed the integral and subtracted the triangle you get from the line OP. I just couldnt get it in the required form so I left the question from there.
12-ish?
0
11 months ago
#39
Here is my attempt at the third question:
2
11 months ago
#40
(Original post by Zacken)
12-ish?
Wow thank you, pleasantly surprised with that.
0
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