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STEP 2013 Solutions

Unofficial STEP Solutions 2013

Here is a thread of all of the STEP Solutions (I, II and III) that have been created by TSR users. I will update (or the relevant mod) when new solutions are posted :smile:

STEP I (Sat 25th June 2013)

Question 1: cpdavis
Question 2: mikelbird
Question 3: FJacob
Question 4: Pyoro, cpdavis
Question 5: mikelbird
Question 6: mikelbird
Question 7: Mastermind2
Question 8: metaltron
Question 9: DJMayes
Question 10: DJMayes
Question 11: DJMayes
Question 12: HCIO
Question 13: DFranklin

STEP II (Sat 19th June 2013)

Question 1: mikelbird
Question 2: Smaug123 Alternative to part ii - Fuzzy12345
Question 3: mikelbird Part ii without considering turning points: Nebula
Question 4: jack.hadamard
Question 5: cpdavis
Question 6: metaltron
Question 7: DJMayes
Question 8:MW24595
Question 9: LShirley95
Question 10: DJMayes
Question 11: DJMayes
Question 12: Smaug123
Question 13: Nebula

STEP III (Sat 26th June 2013)

Question 1: Pyoro
Question 2: borealis72
Question 3: jack.hadamard
Question 4: IrrationalNumber, Blutooth
Question 5: jack.hadamard
Question 6: jack.hadamard
Question 7: DJMayes
Question 8: jack.hadamard
Question 9: Brammer
Question 10: DJMayes
Question 11: bananarama2
Question 12: Blutooth, ukdragon37 - only part iic, MAD Phil
Question 13: DFranklin

STEP I Paper: http://www.thestudentroom.co.uk/showpost.php?p=43320667&postcount=1205
STEP II Paper: http://www.thestudentroom.co.uk/showpost.php?p=43252399&postcount=671
STEP III Paper: http://www.thestudentroom.co.uk/showpost.php?p=43334052&postcount=1384

(edited 9 years ago)

Scroll to see replies

Reply 1

LShirley95

You wouldn't happen to have the full paper would you? :tongue:

Reply 2

Slowbro93

Original post by LShirley95

You wouldn't happen to have the full paper would you? :tongue:

Yep, thanks Jack! http://www.thestudentroom.co.uk/showpost.php?p=43252399&postcount=672

Reply 3

Smaug123

Question 2:

Spoiler

(edited 10 years ago)

Reply 4

Fuzzy12345

Alternatively to Q2 ii

Spoiler

(edited 10 years ago)

Reply 5

username937760

Q7 – STEP II

Part i):

Spoiler

Part ii):

Spoiler

Part iii):

Spoiler

(edited 10 years ago)

Reply 6

username937760

Q10 - STEP II:

Part i):

Spoiler

Part ii):

Spoiler

Part iii):

Spoiler

Reply 7

username937760

Q11 – STEP II
Part i):

Spoiler

Part ii):

Spoiler

(edited 10 years ago)

Reply 8

LShirley95

Got Q9 written with diagrams, unfortunately don't have a scanner so this is the best I can do:

Posted from TSR Mobile

Reply 9

mimx

In case anyone wants a higher res version of the (II) paper to work from:

http://www.thestudentroom.co.uk/showpost.php?p=43258896&postcount=829

Reply 10

Nebula

13 (i)

Spoiler

13 (ii)

Spoiler

13 (iii)

Spoiler

Alternate to 3ii without considering turning points:

Spoiler

(edited 10 years ago)

Reply 11

Smaug123

Q12:

i)

Spoiler

ii)

Spoiler

Expression for Var(Y):

Spoiler

Do non-zero

\lambda

exist?

Spoiler

(edited 10 years ago)

Reply 12

metaltron

Question 6:

Part i)

Spoiler

ii)

Spoiler

iii)

Spoiler

(edited 10 years ago)

Reply 13

Zorgz

Original post by Smaug123

Q12:

i)

Spoiler

ii)

Spoiler

Expression for Var(Y):

Spoiler

Do non-zero

\lambda

exist?

Spoiler

I got very similar but ended up having to show that a^2+b^2/(a+b)^2= 1, which of course isn't possible for non zero values. Not sure but think you might have missed out a lambda squared term when cancelling.

Reply 14

Smaug123

Original post by Zorgz

Not sure but think you might have missed out a lambda squared term when cancelling.

Thanks for that - I lost a minus sign and a lambdasquared. Fixed :smile:

that's what you get for trying to do it all in LaTeX without using paper first!

Reply 15

Slowbro93

Getting ready to type up a solution to question 5 :yep:

Also going to update OP

Reply 16

Slowbro93

Question 5:

Spoiler

(edited 10 years ago)

Reply 17

MW24595

Question 8

i- Let

A(x, t)

be the area of any rectangle with sides parallel to the co-ordinate axes, with one side on

y= f(t)

and another on the y-axis, and with one corner on the curve f(x). It is clear that, the rectangle with the maximum area will be one of this type.

Looking at a rudimentary diagram, it is evident that:

A(x,t)= x(f(x)-f(t))[br][br]\Rightarrow \displaystyle \frac{dA(x,t)}{dx} = (f(x)-f(t)) + x f'(x)[br]

Now, as we're concerned with maximising area, let

x_0

be the value of x for which

A(x,t)

is greatest. So,

\displaystyle \frac{dA(x,t)}{dx} = 0[br][br]\Rightarrow (f(x_0)-f(t)) + x f'(x_0) = 0[br][br]\Rightarrow x_0 f'(x_0)+ f(x_0) = f(t)[br][br]

By definition,

A_0 (t)

is the maximum value of

A(x,t)

so,

A_0(t) = x_0 (f(x_0)-f(t))

ii- So, we have,

tg(t) = \displaystyle \int_0^t f(x)\ dx [br][br]\Rightarrow \frac{d(t g(t)}{dt} = \frac{d}{dt} (\displaystyle \int_0^t f(x)\ dx )[br][br]

Now, notice that on integrating, we have a difference between a function of t and a constant. On differentiating this with respect to t, we just get

f(t)

on the RHS. To see this, let

h(x)

be the antiderivative of

f(x)

.

Then,

tg(t) = h(t) - h(0)[br][br]\Rightarrow \frac{d(t g(t))}{dt} = \frac{d(h(t) - h(0))}{dt} = \frac{d h(t)}{dt} = f(t)[br][br]\Rightarrow t g'(t) + g(t) = f(t)[br][br]\Rightarrow tg'(t) = f(t)- g(t)[br]

Now, consider that:

\displaystyle \int^t_0 (f(x)-f(t))\ dx = \displaystyle \int^{x_0}_0 (f(x)-f(t))\ dx + \displaystyle \int^t_{x_0} (f(x)-f(t))\ dx

Further, as

f(x)

is a decreasing function, we have, for all

0 \le x< x_0

f(x) > f(x_0)[br][br]\Rightarrow \displaystyle \int^{x_0}_0 f(x)\ dx > \displaystyle \int^{x_0}_0 f(x_0)\ dx = f(x_0) \displaystyle \int^{x_0}_0\ dx = x_0 f(x_0)[br][br]\Rightarrow \displaystyle \int^{x_0}_0 f(x)\ dx - x_0(f(t)) = x_0 (f(x_0)- f(t))[br][br]\Rightarrow \displaystyle \int^{x_0}_0 f(x)-f(t)\ dx = x_0 (f(x_0)- f(t))[br][br]

Further, for

x_0 \le x < t

, we have,

f(x) > f(t)[br][br]\Rightarrow \displaystyle \int_{x_0}^t f(x)\ dx >\displaystyle \int_{x_0}^t f(t)\ dx[br][br]\Rightarrow \displaystyle \int_{x_0}^t f(x)- f(t)\ dx >0

Therefore, we clearly have,

\displaystyle \int^{x_0}_0 (f(x)-f(t))\ dx + \displaystyle \int^t_{x_0} (f(x)-f(t))\ dx > x_0 (f(x_0) - f(t))[br][br][br]\Rightarrow \displaystyle \int^t_0 (f(x)-f(t))\ dx > A_0(t)[br][br]

Now, observe that,

Unparseable latex formula:

\displaystyle \int^t_0 (f(x)-f(t))\ dx[br][br]\beginaligned = \displaystyle \int^t_0 f(x)\ dx - t(f(t))[br][br]\beginaligned = t g(t) - t f(t)[br][br]\beginaligned = -t(f(t) - g(t))[br][br]\beginaligned = -t(tg'(t)) = -t^2 g'(t)[br][br]\Rightarrow -t^2 g'(t) >A_0 (t)[br][br]

iii-

f(x) = \frac{1}{1+x} \Rightarrow f'(x) = \frac{-1}{(1+x)^2}

Now, if order to find

x_0

, we simply plug these into the condition for

x_0

, namely:

x_0 f'(x_0)+ f(x_0) = f(t)[br][br]\Rightarrow \displaystyle \frac{-x_0}{(1+x_0)^2} + \frac{1}{1+x_0} = \frac{1}{1+t}[br][br]\Rightarrow x_0 = -1+ \sqrt {1+t}[br][br]

Now,

g(t) = \frac{1}{t} \displaystyle \int_0^t \frac{1}{1+x}\ dx = \frac{1}{t} (ln(1+t))[br][br]\Rightarrow tg'(t) = f(t) - g(t) = \frac{1}{1+t} - \frac{1}{t} (ln(1+t))[br][br]\Rightarrow -t^2 g'(t) = ln(1+t) - \frac{t}{1+t}[br][br]

Similarly, we have,

Unparseable latex formula:

A_0(t) = x_0 (f(x_0) - f(t))[br][br]\beginaligned = \displaystyle (-1+ \sqrt {1+t})(\frac{1}{ \sqrt {1+t}} - \frac{1}{1+t})[br][br]\beginaligned \Rightarrow A_0(t) = 1- \frac{2}{ \sqrt {1+t}} + \frac{1}{1+t}[br][br]