STEP 2017 Solutions

STEP I (paper here)
1: Solution by Zacken
2: Solution by Zacken
3: Solution by Rishab_01
4: Solution by Zacken
5: Solution by Rishab_01
6: Solution by Zacken
7: Solution by harlem_basket
8: Solution by Zacken
9: Solution by FractalSteinway
10: Solution by FractalSteinway
11: Solution by JohnCraig
12: Solution by Farhan.Hanif93
13: Solution by 13 1 20 8 42

STEP II (paper here)
1: Solution by Zacken
2: Solution by Farhan.Hanif13
3: Solution by Slice of Pi
4: Solution by Slice of Pi and SolC
5: Solution by Rishab_01
6: Solution by IrrationalRoot
7: Solution by Zacken
8: Solution by Farhan.Hanif13
9: Solution by atsruser
10: Solution by Forecast
11: Solution by DFranklin
12: Solution by DFranklin
13: Solution by 13 1 20 8 42

STEP III (paper here):
1: Solution by Zacken
2: Solution by SliceofPi
3: Solution by Jamvicious
4: Solution by DFranklin
5: Solution by ThatPerson
6: Solution by DFranklin
7: Solution by DFranklin
8: Solution by IrrationalRoot
9: Solution by Stelios98
10: Solution by DFranklin
11: Solution by DFranklin
12: Solution by Zacken
13: Solution by Zacken

(edited 6 years ago)

Reply 1

Zacken

OP

22

STEP I, Question 1:

Let

u=x\sin x+\cos x

so that

\frac{\mathrm{d}u}{\mathrm{d}x}= x\cos x

. Hence the integral is

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\int \frac{x\cos x}{x\sin x + \cos x} \, \mathrm{d}x = \int \frac{\mathrm{d}u}{u} = \log u +\mathcal{C} = \log (x\sin x + \cos x) + \mathcal{C} \end{equation*}

Let

u = x\cos x - \sin x

so that

\frac{\mathrm{d}u}{\mathrm{d}x} = -x\sin x

. Hence the integral is

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\int \frac{x\sin x}{x\cos x - \sin x} \, \mathrm{d}x = -\int\frac{\mathrm{d}u}{u} = -\log u + \mathcal{C}= -\log (x\cos x - \sin x) + \mathcal{C}\end{equation*}

Let

u = x\sec^2 x - \tan x

so that

\frac{\mathrm{d}u}{\mathrm{d}x} = 2x\tan x \sec^2 x

. Hence the integral is

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\int \frac{x\sec^2 x \tan x}{x\sec^2 x - \tan x} \, \mathrm{d}x = \frac{1}{2}\int \frac{\mathrm{d}u}{u} = \frac{1}{2}\log u + \mathcal{C} = \frac{\log |x\sec^2 x - \tan x|}{2} + \mathcal{C}\end{equation*}

Let

u = x\sec^2 x - \tan x

so that

\frac{\mathrm{d}u}{\mathrm{d}x} = 2x\tan x \sec^2 x

. Hence the integral is

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\int \frac{x\sin x \cos x}{(x-\sin x \cos x)^2} \, \mathrm{d}x = \int \frac{x\sec^2 x\tan x}{(x\sec^2 x - \tan x)^2} \, \mathrm{d}x = \frac{1}{2}\int \frac{\mathrm{d}u}{u^2} = -\frac{1}{2u} + \mathcal{C} = \frac{1}{2(\tan x - x\sec^2 x)} + \mathcal{C}\end{equation*}

This was embarrassingly easy for a STEP question, so it's hard to predict the spread of marks but something like: 3/4/5/8 or 3/4/6/7 is what I'd guess.

(edited 6 years ago)

Reply 2

Student403

19

1.

STEP IV (paper here):
1:
2:
3:
4:
5:
6:
7:
8:
9:
10:
11:
12:
13:

Reply 3

username1162972

18

Original post by Zacken

Question 1:

Let

u=x\sin x+\cos x

so that

\frac{\mathrm{d}u}{\mathrm{d}x}= x\cos x

. Hence the integral is

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\int \frac{x\cos x}{x\sin x + \cos x} \, \mathrm{d}x = \int \frac{\mathrm{d}u}{u} = \log u +\mathcal{C} = \log (x\sin x + \cos x) + \mathcal{C} \end{equation*}

Let

u = x\cos x - \sin x

so that

\frac{\mathrm{d}u}{\mathrm{d}x} = -x\sin x

. Hence the integral is

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\int \frac{x\sin x}{x\cos x - \sin x} \, \mathrm{d}x = -\int\frac{\mathrm{d}u}{u} = -\log u + \mathcal{C}= -\log (x\cos x - \sin x) + \mathcal{C}\end{equation*}

Let

u = x\sec^2 x - \tan x

so that

\frac{\mathrm{d}u}{\mathrm{d}x} = 2x\tan x \sec^2 x

. Hence the integral is

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\int \frac{x\sec^2 x \tan x}{x\sec^2 x - \tan x} \, \mathrm{d}x = \frac{1}{2}\int \frac{\mathrm{d}u}{u} = \frac{1}{2}\log u + \mathcal{C} = \frac{\log |x\sec^2 x - \tan x|}{2} + \mathcal{C}\end{equation*}

Let

u = x\sec^2 x - \tan x

so that

\frac{\mathrm{d}u}{\mathrm{d}x} = 2x\tan x \sec^2 x

. Hence the integral is

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\int \frac{x\sin x \cos x}{(x-\sin x \cos x)^2} \, \mathrm{d}x = \int \frac{x\sec^2 x\tan x}{(x\sec^2 x - \tan x)^2} \, \mathrm{d}x = \frac{1}{2}\int \frac{\mathrm{d}u}{u^2} = -\frac{1}{2u} + \mathcal{C} = \frac{1}{2(\tan x - x\sec^2 x)} + \mathcal{C}\end{equation*}

This was embarrassingly easy for a STEP question, so it's hard to predict the spread of marks but something like: 3/4/5/8 or 3/4/6/7 is what I'd guess.

zain set me link for good copy for the paper

Question 3

Do you have the paper?

Reply 6

Rishabh_01

6

Original post by A Slice of Pi

Do you have the paper?

No but i remember this question, it was pretty standard. If you happened to study parabola then its a piece of cake else I would rate it moderate.

Reply 7

Zacken

OP

22

Question 2:

Note that, since integration preserves rank, we have

\int_1^x \frac{\mathrm{d}t}{t^2} \leqslant \int_1^x \frac{\mathrm{d}t}{t} \leqslant \int_1^x \,\mathrm{d}t

for all

x\geqslant 1

. This gives

Unparseable latex formula:

\displaystyle[br]\begin{equation*}1 - \frac{1}{x} \leqslant \log x \leqslant x-1\end{equation*}

for all

x \geq 1

. If

0 < x \leqslant 1

then replace the above with

x\mapsto \frac{1}{x}\geqslant 1

to get

Unparseable latex formula:

\displaystyle[br]\begin{equation*}1-x\leqslant \log (1/x) \leqslant \frac{1}{x}-1 \implies 1-\frac{1}{x} \leqslant \log x \leqslant x-1\end{equation*}

by multiplying both sides by

-1

. This completes (i) and (ii). Again. Integration preserves rank so for all

y > 1

we have

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\int_1^y 1-\frac{1}{x} \, \mathrm{d}x \leqslant \int_1^y \log x \, \mathrm{d}x \leqslant \int_1^y x-1 \, \mathrm{d}x \end{equation*}

Hence

Unparseable latex formula:

\displaystyle[br]\begin{equation*}y- 1- \log y \leqslant y \log y - y + 1 \leqslant \frac{1}{2}(y-1)^2\end{equation*}

so that

y \log y \leqslant \frac{(y-1)(y+1)}{2}

from above and

Unparseable latex formula:

\logy \geqslant 2(y-1)/(y+1)

from below by some trivial rearranging. Hence we get

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\frac{2}{y+1} \leqslant \frac{\log y}{y-1}\leqslant \frac{y+1}{2y}\end{equation*}

for all

y > 1

. For

0<y<1

simply invert

y \mapsto \frac{1}{y}

and you get the same thing after some trivial rearranging, completing (iii).

I'd predict a split of about 5/7/8.

Reply 8

Zacken

OP

22

Original post by solC

Question 11: (Credit to @JohnCraig)

The attachment doesn't seem to be working for me - could you try uploading it so something like imgur and then inject the image here using the "insert image via URL"?

Original post by A Slice of Pi

Do you have the paper?

Yep, but just a bunch of badly taken pictures - I'll rotate/crop and pdf them at some point today.

Original post by Rishabh_01

...

If anybody is willing to LaTeX this up, please do!

Reply 9

username1162972

18

Original post by Zacken

The attachment doesn't seem to be working for me - could you try uploading it so something like imgur and then inject the image here using the "insert image via URL"?

Yep, but just a bunch of badly taken pictures - I'll rotate/crop and pdf them at some point today.

If anybody is willing to LaTeX this up, please do!

dished why u blanking me

Reply 10

solC

14

Original post by Zacken

The attachment doesn't seem to be working for me - could you try uploading it so something like imgur and then inject the image here using the "insert image via URL"?

Oh right, sorry. I'll try to fix it now.

What are your opinions on the paper thus far, from the questions you've seen?

Question 5

What do you thing you be the marks boundries for grade 1 and grade S

Reply 13

solC

14

Question 11: (credit to @JohnCraig)

2017 Paper I Q11.jpg

If this doesn't work I give up

(edited 6 years ago)

Reply 14

Zacken

OP

22

STEP I, Question 8:

Base case is trivial. Assume

a_n^2 + 2a_n b_n - b_n^2 = 1

then

Unparseable latex formula:

\displaystyle [br]\begin{align*}a^2_{n+1} + 2a_{n+1}b_{n+1}- b^2_{n+1} &= a_n^2 + 4a_n b_n + 4b_n^2 + 2(2a_n^2 + 9a_nb_n+10b_n^2) - 4a_n^2 - 20a_nb_n - 25b_n^2 \\ &=a_n^2 + 2a_nb_n -b_n^2 = 1.\end{align*}

(i) Assume that

a_n \geqslant 0

and

b_n \geqslant 2\cdot 5^{n-1}

. Then

a_{n+1} = a_n + 4\cdot 5^{n-1} \geqslant a_n \geqslant 0

and

b_{n+1} = 2a_n + 5b_n \geqslant 5b_{n} \geqslant 2 \cdot 5^n.

Now let

c_n = \frac{a_n}{b_n}

so that

c_n^2 + 2c_n - 1 = 1/b_n^2

, so that as

n \to \infty

we have

b_n

unbounded above so

1/b_n^2 \to 0

and assuming that

c_n \to c

gives

c^2+ 2c - 1 = 0

, i.e

c=-1+\sqrt{2}

since

c_n \geqslant 0

.

(ii) Assume that

c_n > \sqrt{2}-1

, then since

c_{n+1} = \frac{a_{n+1}}{b_{n+1}} = \frac{a_n + 2b_n}{2a_n + 5b_{n}} = \frac{c_n + 2}{2c_n + 5} = \frac{1}{2}\left(1 - \frac{1}{2c_n + 5}\right)

we have

\frac{1}{2c_n + 5} < \frac{1}{2\sqrt{2} + 3} = 3 - 2\sqrt{2}

so

c_{n+1} > \frac{1}{2}(1-(3-2\sqrt{2})) = \sqrt{2} - 1

.

Hence since

c_n + 1 > \sqrt{2}

we have

\frac{2}{c_n + 1} < \frac{2}{\sqrt{2}} = \sqrt{2}

giving

\frac{2}{c_n + 1} < \sqrt{2} < c_n + 1

as required.

Now note that from the recurrence for

c_n

we can compute

c_3 = \frac{29}{70}

so

\frac{140}{99} < \sqrt{2} < \frac{99}{70}

.

(edited 6 years ago)

Reply 15

A Slice of Pi

13

If anyone has a copy of the paper can you send it to me? I could write up a few solutions.

OP

Paper uploaded.

OP

STEP I, Question 6:

(i) Assume that

f(x)

takes only (wlog) non-negative values in the interval

[0,1]

. Let

x_0

be such that

f(x_0) > 0

. Then by continuity we have an interval

[0,1] \cap (x_0 - \delta, x_0 + \delta)

where

|f(x)- f(x_0)| < f(x_0)/2 \Rightarrow f(x) > 0

. But then

\int_0^1 f > \int_{x_0 - \delta}^{x_0 + \delta} f > 0

. Contradiction. (really contraposition)

(ii) We have

\int_0^1 (x-\alpha)^2 g = \int_0^1 x^2 g - 2\alpha \int_0^1 xg + \alpha^2 \int_0^1 g = 0

. Hence

(x-\alpha)^2 g

must take both positive and negative values on

[0,1]

(or

g

identically

0\,

and we are done anyway). But since squares are non-negative this means that

g

must take positive and negative values on

[0,1]

hence must hit

0\,

at some point.

We get

2a+b = 1, 3a + 2b = 6\alpha

and

4a + 3b = 12\alpha^2

so two possible solutions are

(1 \pm \sqrt{3}, \mp 2\sqrt{3}, \frac{1}{2}\mp \frac{1}{2\sqrt{3}})

. Both of which hit 0 at some point in (0,1) which you can check using obvious inequalities. e.g

x = \frac{1+ \sqrt{3}}{2\sqrt{3}} = \frac{1}{2} + \frac{1}{2\sqrt{3}} < 1

.

(iii) Set

g = h'

then

\int_0^1 h' = h(1) - h(0) = 1

. Also

\int_0^1 xh' = xh]_0^1 - \int_0^1 h = 1 - \beta

. And finally

\int_0^1 x^2 h' = x^2h]_0^1 - 2\int xh = 1 - \beta(2-\beta) = 1 - 2\beta + \beta^2 = (1-\beta)^2

.

So apply (ii) with

g = h'

and

\alpha = 1-\beta

get

g = h' = 0

somewhere in

[0,1]

.

(edited 6 years ago)

Reply 18

Rishabh_01

6

Could you add my solution of question 5 to pinned thread

Reply 19

Zacken

OP

22

Original post by Rishabh_01

Could you add my solution of question 5 to pinned thread

I already have.

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