(Does anyone else agree that this is a really easy question considering it's a STEP paper? Perhaps Stats was considered new fangled in those days :wink:

)

Reply 49

bogstandardname

STEP 1 Q 1 does not appear to be solved so here goes.

It isn't immediately obvious where the inequalities come from.
we have 4x²+16xy+y²+24x = 0
Complete the square for y
(y+8x)²-(8x)²+4x²+24x=0
so (y+8x)²=(64-4)x²-24x
(y+8x)²=60x²-24x
LHS ≥ 0, so RHS must too be ≥ 0,
This implies 60x²-24x≥ 0, If we look at the graph (roots are 0 and 2/5), this is clearly where 0≥x and x≥2/5 as required.

For the next part complete square for x
4x²+(16y+24)x+y²=0
4(x²+(4y+6)x)+y²=0
4((x+(2y+3)²-(2y+3)²)+y²=0
4(x+(2y+3))²-4(2y+3)²+y²=0
4(x+(2y+3))²=4(2y+3)²-y²
4(x+(2y+3))²=4(4y²+12y+9)-y²
4(x+(2y+3))²=15y²+48y+36
LHS ≥ 0 so RHS must too be ≥ 0,
15y²+48y+36≥ 0
roots are -2 and -1.2
so restrictions on y are
-2≥y y≥-1.2

b is quite simple so can't be bothered, aplogies for mistakes, hard to remember what I did exactly

Reply 50

Lord of the Flies

Wow this is noticeably easier than the actual exam papers - especially I. 4. :eek:

That's all I have to say... :creep:

Reply 51

bogstandardname

Another solution for an uncompleted question, STEP I Question 8 is attached

STEP I 1987 SPECIMEN PAPER Q8.docx18.9KB

Reply 52

Sophie M-J

Original post by Lord of the Flies

Wow this is noticeably easier than the actual exam papers - especially I. 4. :eek:

That's all I have to say... :creep:

Just did this question in about 5 minutes ... was feeling so happy... until I read this haha!!

Reply 53

bogstandardname

solution for STEP II Q7

STEP II 1987 Specimen Q7.docx22.6KB

Reply 54

bogstandardname

Original post by Farhan.Hanif93

STEP III Q7

Solution

They call that a STEP question...

When you get to

\implies v'' + v' =0

taking auxillary equations gives roots 0 and -1, so why isn't the solution in the form

{y=Axe^{-x}+Bx}

Reply 55

Farhan.Hanif93

Original post by bogstandardname

When you get to

\implies v'' + v' =0

taking auxillary equations gives roots 0 and -1, so why isn't the solution in the form

{y=Axe^{-x}+Bx}

It is; my mistake. It seems I forgot to edit this solution last time so thanks for the reminder :smile:

(see here). Turns out that solving the problem whilst typing it in LaTeX lead to a typo (I wrote v, instead of v' and accordingly missed the second term).

Reply 56

username937760

STEP III - Question 5:

Part i):

Spoiler

Part ii):

Spoiler

And that's all of the STEP III questions for the specimen completed.

Reply 57

Genesis2703

Original post by ben-smith

STEP I Q3

\frac{d^2y}{dx^2}+2\frac{dy}{dx}-3y=2e^x[br]\Rightarrow \frac{d^2y}{dx^2}-\frac{dy}{dx}+3(\frac{dy}{dx}-y)=2e^x[br]\Rightarrow \frac{dz}{dx}+3z=2e^x[br]\Rightarrow \frac{dz}{dx}e^{3x}+3e^{3x}z=2e^{4x}[br]\Rightarrow \frac{d}{dx}[ze^{3x}]=2e^{4x} \Rightarrow z=\frac{e^x}{2}+C_1e^{-3x}

Using the initial conditions z(0)=1 so C_1=1/2. Now, to solve for y:

\frac{dy}{dx}-y=\frac{e^x}{2}+C_1e^{-3x}[br]\Rightarrow \frac{d}{dx}[ye^{-x}]=1/2+C_1e^{-4x}[br]\Rightarrow y=\frac{xe^x}{2}-C_1e^{-3x}+C_2e^x[br]y(0)=1 \Rightarrow C_2=9/8[br]

On the penultimate line (where you have y = ....) surely it should be a -1/4 * C1 lots of e^x, I know you changed it to a negative sign which is fine, but in my final answer I have 1/8 instead of 1/2 because you divide by the 4 when integrating?

Reply 58

abra-cad-abra

Original post by ben-smith

STEP I Q6
Consider the sector formed by the points A,D and E and let Q be the centre of the whole coin. We want to find the area of the segment subtended on A and the area of the triangle QDE. Let QD=QE=x and 2y=DE:

Sin(\pi/14)=y/a,Sin(\pi/7)=y/x[br]\Rightarrow x=a\dfrac{sin(\pi/14)}{sin(\pi/7)}

Adding the two areas we mentioned earlier we get:

[br]a^2/2(\pi/7-sin(\pi/7))+a^2/2 \dfrac{sin^2(\pi/14)sin(2\pi/7)}{sin^2(\pi/7)}[br]=a^2/2(\frac{\pi}{7}-sin(\pi/7)+\dfrac{sin^2(\pi/14)2sin(\pi/7)cos(\pi/7)}{sin^2(\pi/7)}[br]=a^2/2(\frac{\pi}{7}+\frac{2sin^2(\pi/7)cos(\pi/7)-sin^2(\pi/7)}{sin(\pi/7)})[br]=a^2/2(\frac{\pi}{7}+\frac{cos(\pi/7)-1}{sin(\pi/7)})[br][br]=a^2/2(\frac{\pi}{7}-Tan(\pi/14)

Multiplying by 7 gives the required result.

how did you know that that first angle was pi/7. i just guessed it as its a 7sided shape so makes sense and looking at the answer i was working towards it seemed right