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Year 13 Maths Help Thread

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9x + 12 ??

I still feel like this is wrong and if i had a similar question i wouldnt know how to get it into the correct form. I understood what you said but it 'didnt hit me' (i wouldnt have thought og answering it like this) :biggrin:

I'm not sure what your 9x+12 is referring to, the ax + b? I haven't done this myself so I don't know what the answer is.

(x+1)^3 6 (x+2)^5 + (x+2)^6 3(x+1)^2

But thats not in the form (x+1)^2 (x+2)^5 (ax + b) D:

given c = (x+2), d = (x+1), the differentiated output is

6d^3c^5 + 3c^6d^2

and the required form is

d^2c^5(ax+b)

... how have the two terms been turned into one?

Reply 921

kiiten

Original post by SeanFM

I'm not sure what your 9x+12 is referring to, the ax + b? I haven't done this myself so I don't know what the answer is.

given c = (x+2), d = (x+1), the differentiated output is

6d^3c^5 + 3c^6d^2

and the required form is

d^2c^5(ax+b)

... how have the two terms been turned into one?

Oh yeah, sorry i mean is ax + b = 9x + 12.

Yeah thats what i did - kinda.... Screenshot 2016-10-05 19.21.53.png

Reply 922

Kevin De Bruyne

Original post by kiiten

Oh yeah, sorry i mean is ax + b = 9x + 12.

Yeah thats what i did - kinda.... Screenshot 2016-10-05 19.21.53.png

Looks right to me :smile:

does it look right to you?

Reply 923

kiiten

Original post by SeanFM

Looks right to me :smile:

does it look right to you?

Yes? Yayyy finally im so glad i finally answered this question now. Thanks for your help :h:

- and sorry if i was annoying.

Reply 924

Kevin De Bruyne

Original post by kiiten

Yes? Yayyy finally im so glad i finally answered this question now. Thanks for your help :h:

- and sorry if i was annoying.

When you are stuck it is important to look at the definitions and similar examples, then think about what topics the question is testing you on, and if a form is given, how you can get to it (rather than looking at it and wondering why your answer isn't in the correct form when you think you have done everything right)

Reply 925

jamestg

Right so the question is to express:

f(x)=(2x+3)/(x-1) as f(x)=a + b/(x-1)

Was I right to do this...

(2x+3)/(x-1)=a + b/(x-1)

(2x+3)=a(x-1) + b

sub x=1

5=b

sub b=5 and then end up a=2

so f(x) = 2 +5/(x-1)

Was this the correct method? I just used the partial fractions method, but this question is for C3, although it did work.

Reply 926

B_9710

Original post by jamestg

It's correct. But it may be easier to realise that

\displaystyle \frac{2x+3}{x-1} \equiv \frac{2(x-1)+5}{x-1}

and split the fraction up.
You can also just do it using algebraic division as you normally would

Reply 927

jamestg

Original post by B_9710

It's correct. But it may be easier to realise that

\displaystyle \frac{2x+3}{x-1} \equiv \frac{2(x-1)+5}{x-1}

and split the fraction up.
You can also just do it using algebraic division as you normally would

Damn, algebraic division would have been much easier...

At least I now know this before the exam :colondollar:

Thanks for your help!

Reply 928

Palette

Hi, are there any relatively easy STEP I papers from recent years? I was told that 2007 was relatively easy but I could only manage 4 questions out of that paper.

Reply 929

Zacken

Original post by Palette

Hi, are there any relatively easy STEP I papers from recent years? I was told that 2007 was relatively easy but I could only manage 4 questions out of that paper.

2010

Reply 930

jamestg

Can someone help me with implicit differentiation?

Find the gradient of

3y=\frac{8x^2}{4x^2-3y^3}

at the point (3,2)

I got the answer to be

\frac{4}{9}

but it's wrong.

Reply 931

jamestg

Working out maybe helpful

Reply 932

MathMoFarah

Original post by jamestg

Can someone help me with implicit differentiation?

Find the gradient of

3y=\frac{8x^2}{4x^2-3y^3}

at the point (3,2)

I got the answer to be

\frac{4}{9}

but it's wrong.

Multiply both sides by the denom

12x^2y-9y^4=8x^2
differentiate with respect to x
24xy+12x^2(dy/dx)-36y^3(dy/dx)=16x
(dy/dx)(12x^2-36y^3)=16x-24xy
put in the numbers
(dy/dx)(108-288)=48-144
dy/dx=-96/-180=16/30=8/15

(edited 7 years ago)

Reply 933

jamestg

Whoops realised when I did the quotient rule I didn't divide by

v^{2}

Reply 934

jamestg

Original post by MathMoFarah

Damn I even considered doing it that way initially :frown:

Thanks for the help!

Reply 935

medhelp

Q1.7
How many zeros are there at the end of the number 100! ?
I don't get what that exclamation mark means, and 2 isn't the answer. I remember using it as symbol in AS maths but not anything else.

Q1.11
A salesmen drives from L to B. The first half of the distance he drives at a constant speed of 80 mph. The second half he drives at a constant speed of 120 mph. What is his average speed for the total journey?

I don't understand why the answer is 96 mph and not 100?

(edited 7 years ago)

Reply 936

DylanJ42

Original post by medhelp

100! just means 100 x 99 x 98 x 97 x ... x 1.

A smaller example to show you in full; 5! is 5 x 4 x 3 x 2 x 1 = 120

the second one is deceptive, it tricks you into just finding the average of the speeds but you have to remember that the first half of the journey took longer to complete than the second half. Work out the time taken for each half separately and go from there

Reply 937

kiiten

How would you find the period in radians of y=cosec4x (does the period tell you the intervals where the graph repeats?)

My book says y=cosecx repeats after 2pi intervals. How do you find this for transformed graphs?

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Reply 938

mik1a

cosec(x) is 1/sin(x).

sin(x) repeats every 2pi, therefore so does 1/sin(x).

regarding transformed graphs: consider x slowly increasing from 0 to 2*pi. Now consider what happens to 2x in the same time. It reaches twice the distance, from 0 to 4pi. So in the time it takes sin(x) to complete one cycle, sin(2x) will have completed two cycles. The period of the function decreases by a factor of two when the argument of the function is doubled. This is true of all functions, including sin(x) and cosec(x).