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OCR C4 (not mei) 18th June 2013 revision

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I'd use partial fractions, but in C4 you are not expected to be able to do partial fractions where the order of the denominator is the same as the numerator...

You are expected to be able to do partial fractions when the numerator is larger than or equal to the denominator. Check for example page 310 of the Core 4 book

Reply 161

ninuzu

Original post by jassi1

You are expected to be able to do partial fractions when the numerator is larger than or equal to the denominator. Check for example page 310 of the Core 4 book

Not equal to - that's FP2? Or if they do they will tell you how to structure it, since you have to have an 'A' term at the front with no denominator when the degrees are equal.
Actually, that's not partial fractions entirely - if you look at the specification, they will only ever ask you partial fractions where the degree of the numerator < degree of the denominator (anything else is FP2). If they do have the degree greater than or equal to the denominator, you will need to do algebraic division first.

(edited 10 years ago)

Reply 162

Vip3rgt9

Original post by Hello_ImJess

How would you integrate this?

ImageUploadedByStudent Room1371470840.451832.jpg

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I'd use integration by parts of X(X+2)^-1

Reply 163

saberahmed786

Hey guys, would we need to know the following integrations or are they given in the formula booklet? I can't find my copy of the FB :frown:

sec^2(x) \ dx \ = tan(x)+c[br]cosec^2(x) \ dx \ = -cot(x) + c[br]cosec(x)cot(x) \ dx \ = -cosec(x)+c[br]

any help would be appreciated!

Reply 164

Namige

Original post by Vip3rgt9

I'd use integration by parts of X(X+2)^-1

No, integration by parts would take too long as you have to then integrate ln(x+2).

Use substitution u=x+2 or turn the numerator into (x + 2) -2 and divide by (x+2).

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Reply 165

Namige

Original post by saberahmed786

Hey guys, would we need to know the following integrations or are they given in the formula booklet? I can't find my copy of the FB :frown:

sec^2(x) \ dx \ = tan(x)+c[br]cosec^2(x) \ dx \ = -cot(x) + c[br]cosec(x)cot(x) \ dx \ = -cosec(x)+c[br]

any help would be appreciated!

They are in the formula booklet. Look on ocr website and download a copy if you need to.

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Reply 166

Vip3rgt9

Original post by Namige

No, integration by parts would take too long as you have to then integrate ln(x+2).

Use substitution u=x+2 or turn the numerator into (x + 2) -2 and divide by (x+2).

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Thanks. For OCR they always tell us which questions to do substitutions in though don't they?

Reply 167

Hello_ImJess

Thanks everyone :smile:

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Reply 168

ninuzu

Original post by Vip3rgt9

Thanks. For OCR they always tell us which questions to do substitutions in though don't they?

If they are deliberately looking for substitution, they will tell you in the question use the substitution u=blah.
If the question is ambiguous i.e. they don't tell you what to do - you can use various different methods including substitution.

Reply 169

daniel_vincent

JAN 2009

(1+2x)^0.5
(1+x)^3

Q3) iii) State the set of values of x for which the expansion is valid. [1]

I know the answer is: -0.5 < x < 0.5

but why is this?

Reply 170

stirkee

Original post by daniel_vincent

JAN 2009

(1+2x)^0.5
(1+x)^3

Q3) iii) State the set of values of x for which the expansion is valid. [1]

I know the answer is: -0.5 < x < 0.5

but why is this?

|2x| < 1

so -0.5 < x < 0.5

Reply 171

ninuzu

Original post by daniel_vincent

JAN 2009

(1+2x)^0.5
(1+x)^3

Q3) iii) State the set of values of x for which the expansion is valid. [1]

I know the answer is: -0.5 < x < 0.5

but why is this?

Because the formula (1+x)^n = 1 + nx + (n(n-1))/2! *x^2 + ... is only valid when |x|<1 so -1<x<1.
In the previous part of the question, you have expanded (1+2x)^0.5 using this formula. Replacing x in the values for which the formula is valid, you get |2x|<1, so |x|<0.5 ---> -0.5<x<0.5

Reply 172

AlesanaWill

Original post by Vip3rgt9

Thanks. For OCR they always tell us which questions to do substitutions in though don't they?

Write x / (x+2) as (x + 2 - 2) / (x+2), then you have (x+2)/(x+2) - 2/(x+2).

Which is just 1 - 2(x+2).

Reply 173

a10

just finished the last paper Jan 13, I have to say that second part of the differential equation ****ed me over. Make sure u read the questions carefully guys they often give wayyy too much information to hide the very subtle hints, can very easily catch you out and unfortunately with DE if one part is wrong your pretty much screwed for the rest of it.

Reply 174

SexyAndIKnowIt.

How to determine if vector lines are
a) parallel, b) intersect, c) skew ??

Reply 175

a10

Original post by SexyAndIKnowIt.

How to determine if vector lines are
a) parallel, b) intersect, c) skew ??

a) same direction vector (usually the other one is a multiple).

b) sub in and solve should be equal on both sides.

c) sub in and solve and shouldn't be equal on both sides.