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Mega A Level Maths Thread MK II

Reply 920

Revisionbug

1

Im stuck on this c4 question normlyy i just find t and then sub in but for this one im stuck can anyone help plss

i cant do it :frown:

Reply 921

joostan

13

Original post by annaridgway95

1 - 2sin^2x but I don't understand where that fits in?

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\dfrac{\cos(x) - 2\sin^2(x)\cos(x)}{\sin(x)}

Now you need to work in a

1 - 2\sin^2(x)

Can you see how?

Reply 922

annaridgway95

Original post by joostan

\dfrac{\cos(x) - 2\sin^2(x)\cos(x)}{\sin(x)}

Now you need to work in a

1 - 2\sin^2(x)

Can you see how?

Yeah I get it now, I'm being stupid! Hahaha.
Thank you :smile:

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Reply 923

joostan

13

Original post by annaridgway95

Yeah I get it now, I'm being stupid! Hahaha.
Thank you :smile:

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No problem :smile:

Reply 924

joostan

13

Original post by Revisionbug

Im stuck on this c4 question normlyy i just find t and then sub in but for this one im stuck can anyone help plss

i cant do it :frown:

Add x and y then substitute in :smile:

Negging me is somewhat unnecessary, no?

(edited 10 years ago)

Reply 925

Lamalam

3

Hi! I found something called the "common mistake "section in the c2 maths CD , there's one part that i dont understand

Whats meant by "not actually a mistake"?

I didnt know how to tackle this question ( to use the integration and equation L )

Can someone explain? Thanks loads!!!

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Reply 926

Reyoru

1

Original post by Revisionbug

Im stuck on this c4 question normlyy i just find t and then sub in but for this one im stuck can anyone help plss

i cant do it :frown:

x=(t+\frac{2}{t})

y=(t-\frac{2}{t})

x^2=(t+\frac{2}{t})^2=t^2+4+ \frac{4}{t^2}

y^2=(t-\frac{2}{t})^2=t^2-4+\frac{4}{t^2}

x^2 - y^2 = 8

(edited 10 years ago)

Reply 927

ScarlettFierce

2

GAAH nevermind! Blonde moment there! sorry for wasting your time, people!

(edited 10 years ago)

s1 question.PNG19.7KB

Reply 928

Fuzzy12345

10

Original post by ScarlettFierce

GAAH nevermind! Blonde moment there! sorry for wasting your time, people!

i)

\mathbb{P} (X = 5) = \displaystyle\binom{11}{5} \left(\dfrac{3}{4} \right)^{5} \left(1-\dfrac{3}{4} \right)^{6}

ii)

\mathbb{P} (X = 0) = \displaystyle\binom{11}{0} \left(p \right)^{0} \left(1-p \right)^{11} = 0.05

Var(X) for a binomial distribution = np(1-p)

Reply 929

Moon1ight

15

2 weird but urgent questions but how do you know when to solve a quadratic equation using two brackets - also in a log table - is x at the top row or is it the other way round ?

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Reply 930

BobLoblawLawBlog

16

Original post by krishkmistry

2 weird but urgent questions but how do you know when to solve a quadratic equation using two brackets - also in a log table - is x at the top row or is it the other way round ?

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For the first one, if you are able to factorise it, then factorise it. But not all quadratics will be able to be solved that way. I'm not sure what you mean by the next part.

Reply 931

Moon1ight

15

Original post by brittanna

For the first one, if you are able to factorise it, then factorise it. But not all quadratics will be able to be solved that way. I'm not sure what you mean by the next part.

Basically I'm asking how many terms do you need to solve quadratically - via brackets of qp - I'm also asking that when you get a table on a logs graph q - is the top row the x value or the y value when plotting it on a graph ?

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Reply 932

BobLoblawLawBlog

16

Original post by krishkmistry

Basically I'm asking how many terms do you need to solve quadratically - via brackets of qp - I'm also asking that when you get a table on a logs graph q - is the top row the x value or the y value when plotting it on a graph ?

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If you have something like

x^2+5x+4=0

, you can write this as

(x+4)(x+1)=0

\Rightarrow

x=-4, x=-1

. However, if you have something like

x^2+2x+2=0

, you would need to solve this by completing the square or by substituting into the quadratic formula. Only quadratics with at least two terms can be factorised.

Does it not tell you in the table which row is which?

(edited 10 years ago)

Reply 933

username937760

16

Original post by MathsNerd1

..

Following on from my suggestion,, have you tried any STEP DE's/Induction? If so, how have you found it?

Reply 934

MathsNerd1

17

Original post by DJMayes

Following on from my suggestion,, have you tried any STEP DE's/Induction? If so, how have you found it?

I'm going to be honest with you and say I haven't yet as I've been practising on my Maths and Chemistry for the mocks I have but I'm devoting 3 hours tomorrow to STEP either for a mock or some questions and I'll make sure that I'll do one of those topics and let you know how I found them.

Reply 935

shyamshah

12

How do you work out normal distribution with quartiles?

Reply 936

NJam

2

Quick question:

Given

x=\frac{1}{5}(2sint-cost+e^{-2t})

show that x varies between

-\frac{1}{5}\sqrt 5

and

\frac{1}{5}\sqrt 5

for large positive values of t

I understand that the term involving e will tend to 0, and I understand what they have done to obtain the answer

Spoiler

Why do you do this? Apologise for the vague question, I just can't wrap my head around it.

(edited 10 years ago)

Reply 937

furryface12

21

I've suddenly started getting email updates whenever someone posts on this thread, any idea how to stop it?

Reply 938

username937760

16

Original post by NJam

Quick question:

Given

x=\frac{1}{5}(2sint-cost+e^{-2t})

show that x varies between

-\frac{1}{5}\sqrt 5

and

\frac{1}{5}\sqrt 5

for large positive values of t

I understand that the term involving e will tend to 0, and I understand what they have done to obtain the answer

Spoiler

Why do you do this? Apologise for the vague question, I just can't wrap my head around it.

This in an idea called the harmonic functions. Essentially, we can express the sum of any sine and cosine as a single trig function:

asinx+bcosx = Rsin(x+c)

In order to do this, expand the RHS:

asinx+bcosx = Rsinxcosc+Rcosxsinc

Knowing your values of a and b, you can then solve for values of R and c in order to express it as a single function:

Rcosc = a

Rsinc = b

You can then use a bit of re-arrangement to find a and b:

R^2(cos^2c+sin^2c) = a^2 + b^2

So

R = \sqrt{a^2+b^2}

And you can work out C by dividing the two equations to get something involving tan only. This is how they've got the root 5; by expressing the function as a single trigonometric function.

Reply 939

NJam

2

Original post by DJMayes

This in an idea called the harmonic functions. Essentially, we can express the sum of any sine and cosine as a single trig function:

asinx+bcosx = Rsin(x+c)

In order to do this, expand the RHS:

asinx+bcosx = Rsinxcosc+Rcosxsinc

Knowing your values of a and b, you can then solve for values of R and c in order to express it as a single function:

Rcosc = a

Rsinc = b

You can then use a bit of re-arrangement to find a and b:

R^2(cos^2c+sin^2c) = a^2 + b^2

So

R = \sqrt{a^2+b^2}

And you can work out C by dividing the two equations to get something involving tan only. This is how they've got the root 5; by expressing the function as a single trigonometric function.

Fantastic, exactly what I needed! +rep :smile:

Mega A Level Maths Thread MK II

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