Year 12s: what will be the first way you research your future options? >>

STEP 2006 Solutions Thread

Scroll to see replies

For paper II question 3 part ii, can some one explain why (N+rt(N^2-1)^k + (1/(N+rt(N^2-1)))^k would be the integer nearest to (N+rt(N^2-1)^k? I know that it is definitely integer, but why is it the nearest one? Thanks

Well, what's its distance to (N+rt(N^2-1)^k?

So, can any other integer be nearer?

Reply 101

michaelxukong

Original post by DFranklin

Well, what's its distance to (N+rt(N^2-1)^k?

So, can any other integer be nearer?

Ahhhh, sorry for the silly question :P

Reply 102

michaelxukong

Original post by sonofdot

STEP II 2006 Question 4

First Part

(i)

(ii)

(iii)

Hi sonofdot, for the First Part of your solution, when you did the substitution, shouldn't there be a sign change because dx=-dt?

Reply 103

DFranklin

Original post by michaelxukong

Hi sonofdot, for the First Part of your solution, when you did the substitution, shouldn't there be a sign change because dx=-dt?

No, the limits swap direction at the same time, do there are actually two sign changes that cancel.

Reply 104

michaelxukong

Ahhh! Tbh, I never considered this scenario. I see now, thanks :biggrin:

Reply 105

Omghacklol

Original post by DFranklin

For 2006 II Q6 last part, is it necessary once we've found the solutions to our equality to substitute back into the original equations to check validity? In other words, does the logic of solving for p,q,r hold both ways? Why or why not? Thanks a lot in advance

Reply 106

Wu Yizhou2007

Mistake in working in last few lines of second last part

Reply 107

Student1256

Original post by SimonM

STEP II, Question 7

Spoiler

In the case k=0 won’t pq be undefined as the denominator becomes 0?

Reply 108

Student1256

Original post by Student1256

In the case k=0 won’t pq be undefined as the denominator becomes 0?

Never mind this lol I can’t delete it on my phone

Reply 109

cxs

Original post by Daniel Freedman

STEP III 2006, Question 1

Spoiler

Great! Initially I use lny=ln2+lnx+ln(x^2-5)-ln(x^2-4) to differentiate and find the tangent, but your method is more effieient.

Reply 110

Wynelf

I got a different answer for u using your equation obtained from comparing coefficients, could you be so kind as to check it again?

Reply 111

varinosky

Original post by Daniel Freedman

STEP II 2006, Question 8

Spoiler

Am I wrong or OD should be

Unparseable latex formula:

\boxed{\vec{OD} = \frac{ \bold{c}}{\1-lambda}}

Reply 112

varinosky

Original post by varinosky

Am I wrong or OD should be

Unparseable latex formula:

\boxed{\vec{OD} = \frac{ \bold{c}}{\1-lambda}}

c / 1 - lambda*
(I suck with computers :smile:

)

Reply 113

kjena

Has anyone realised that Step 3 2006 is just repeated questions from old format papers?!
Q1 = 1989 Step 3 Q4
Q2 = 1987 Step 2 Q6
Q9 = 1987 Step 2 Q12
Q12 = 1988 Step 1 Q14
Q3 = 1989 Step 2 Q2
Q6 = 1989 Step 2 Q5
Q11 = 1989 Step 2 Q11
Q10 = 1990 Step 2 Q11
Q13 = 1989 Step 2 Q15
Q7 = 1991 Step 2 Q8
Q8 = 1992 Step 2 Q4
Q14 = 1997 Step 3 Q14

(edited 3 years ago)

Reply 114

DFranklin

Original post by kjena

Has anyone realised that Step 3 2006 is just repeated questions from old format papers?!
Q1 = 1989 Step 3 Q4
Q2 = 1987 Step 2 Q6
Q9 = 1987 Step 2 Q12
Q12 = 1988 Step 1 Q14
Q3 = 1989 Step 2 Q2
Q6 = 1989 Step 2 Q5
Q11 = 1989 Step 2 Q11
Q10 = 1990 Step 2 Q11
Q13 = 1989 Step 2 Q15
Q9 = 1991 Step 2 Q8
Q8 = 1992 Step 2 Q4
Q14 = 1997 Step 3 Q14

That's really interesting - I haven't quite checked every question, but the pure seems identical (up to minor wording changes), there are small variations in the mechanics and stats.

I wonder what was going on with the question setters - if they were lazy or just experimenting to see what happened. The examiners' report doesn't mention anything about the questions having been asked previously.

Reply 115

RuairidhW

For question 3i, to show it’s not a necessary condition, just use the condition you’ve made in 3ii. 3ii asks to determine necessary AND suffiicient conditionS for the roots to be real and POSITIVE, There fore you have to make the quadratic formula greater than 0 and set up an inequality, and then also use the inequality used to prove that the condition wasn’t necessary in 3i. Therefore you have 2 conditions which are sufficient and necessary (<=>)for the function to have real and positive roots. For question 3iii, one of the y-values of the turning point is always negative- the one on the right- while the other (the one on the left) depends on the value of q. Therefore you can create inequalities and an equality using the value of -q. When these are rearranged you get different inequalities and equalities for 4p^3 27q^2 which show the number, nature and value (positive or negative) of the roots. These deduction are identified based on the negative value of the right turning point as well as the negative gradient when x=0.

Reply 116

aramis8

Original post by michaelxukong

For paper II question 3 part ii, can some one explain why 􀀋 􀀌(N+rt(N^2-1)^k + (1/(N+rt(N^2-1)))^k would be the integer nearest to (N+rt(N^2-1)^k? I know that it is definitely integer, but why is it the nearest one? Thanks

I have the same question, can you please explain why?

Thanks

Reply 117

DFranklin

Original post by aramis8

Can you please explain the implication in your very last line?

Thanks

SimonM hasn't posted for years, so I'll answer on his behalf:

Writing

A_k = (N+\sqrt{N^2-1})^k, B_k = 1/A_k

, we've shown that

A_k+B_kl

is an integer. So the difference between

A_k

and the nearest integer is

B_k

. So we need to show

B_k < (2N-\frac{1}{2})^{-k}

, which is going to be true if

(N+\sqrt{N^2-1}) > 2N-\frac{1}{2}

.

If this isn't clear, please try to explain *exactly* which part you're unsure about.

Reply 118

aramis8

Original post by DFranklin

SimonM hasn't posted for years, so I'll answer on his behalf:

Writing

A_k = (N+\sqrt{N^2-1})^k, B_k = 1/A_k

, we've shown that

A_k+B_kl

is an integer. So the difference between

A_k

and the nearest integer is

B_k

. So we need to show

B_k < (2N-\frac{1}{2})^{-k}

, which is going to be true if

(N+\sqrt{N^2-1}) > 2N-\frac{1}{2}

.

If this isn't clear, please try to explain *exactly* which part you're unsure about.

Thank you ever so much, as a follow up. How do we know that the above integer is infact the nearest one?

(sorry, it might be obvious)

It is it because (2N-0.5)^(-k) is less than half?

(edited 6 months ago)

Reply 119

aramis8

Original post by DFranklin

SimonM hasn't posted for years, so I'll answer on his behalf:

Writing

A_k = (N+\sqrt{N^2-1})^k, B_k = 1/A_k

, we've shown that

A_k+B_kl

is an integer. So the difference between

A_k

and the nearest integer is

B_k

. So we need to show

B_k < (2N-\frac{1}{2})^{-k}

, which is going to be true if

(N+\sqrt{N^2-1}) > 2N-\frac{1}{2}

.

If this isn't clear, please try to explain *exactly* which part you're unsure about.

Sorry to keep bothering you and I promise this will be the last question, I don't understand the recurrence relation SimonM wrote. How does k fit into the recurrence relation?

Thanks