OCR MEI C3 Maths June 2015

Is there a quick way or rule to know when to use integration by substitution or by parts? Thanks :smile:

Integration by substitution is usually done to fractions and it is also obvious if the question says something like 'using the substitution u=x+1 or otherwise.....' . Also, when it is a 'function of a function'. Integration by parts is used to integrate products.

Reply 41

Velocity_

18

Boom! after hours and hours spending till late at night trying to do well in C3 so i can accomplish a B (shh dont judge me :P ) i finally have mastered Section A to a grade A level... Just a C in section B and i should get that B but omg i do not like section B. Is it weird that i like C4 section A? sometimes....

Reply 42

papayarama

2

Thanks! Also, what is the period of a function / how do you find it?

Reply 43

Duskstar

13

Original post by papayarama

Thanks! Also, what is the period of a function / how do you find it?

The period is the amount of x before the function repeats. Eg, for f(x) = sinx the period is 360 degrees (or 2 pi radians). The same for cosx, while f(x) = tanx has a period of 180 degrees.

If the question asks for the period of another function they'll probably have a sketch, and then just take the difference between the start of a 'cycle' and end (like wavelength in Physics if that helps)

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Reply 44

emarshy97

Does anyone have a link for last years paper for more revision? Ant find it anywhere.

Reply 45

Hody421

15

These type of questions get me every time - can someone explain the answer this to me?

Reply 46

Duskstar

13

Original post by Hody421

These type of questions get me every time - can someone explain the answer this to me?

I'm assuming you understand the concept of the range? If not, it's the range (literally) of values that f(x) can take (always for a one - one function).

For this question, you want the range of f(x) = 0.5 atanx in terms of radians.

If g(x) = atanx then you can imagine it's simply the tanx graph on it's side. The domain of the one - one tan function is -pi to +pi and so you can just transform this for f(x) - the minimum and maximum are going to be 1/2 of g(x), so the range is -pi/2 <= f(x) <= pi/2

If you have any more questions just ask!

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Reply 47

Duskstar

13

Original post by emarshy97

Does anyone have a link for last years paper for more revision? Ant find it anywhere.

Your teacher should. I don't think the 2014 papers have been released publicly yet, although they probably will be so soon.

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Reply 48

Duskstar

13

Original post by papayarama

Is there a quick way or rule to know when to use integration by substitution or by parts? Thanks :smile:

Integration by substitution will always work when the integral looks like a derivative (think the chain rule - if something does look like this, it's also easy to do in your head).

Integration by parts is for when you've got a product and it doesn't look like a derivative/you can't see the substation easily.

Generally the question will tell you if you have to use substation - if it doesn't tell you then I guess think about the above.

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Reply 49

Hody421

15

Original post by Duskstar

I'm assuming you understand the concept of the range? If not, it's the range (literally) of values that f(x) can take (always for a one - one function).

For this question, you want the range of f(x) = 0.5 atanx in terms of radians.

If g(x) = atanx then you can imagine it's simply the tanx graph on it's side. The domain of the one - one tan function is -pi to +pi and so you can just transform this for f(x) - the minimum and maximum are going to be 1/2 of g(x), so the range is -pi/2 <= f(x) <= pi/2

If you have any more questions just ask!

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But from the mark scheme:
−π/2 < arctan x < π/2⇒ −π/4 < f(x) < π/4⇒ range is −π/4 to π/4

(edited 8 years ago)

Reply 50

JackB003

2

Really looking forward to getting this out of the way! My class finished the work for it about 3/4 months ago, so for 3/4 months all we've been doing is papers. Likewise for C4. I just want these to end... they're boring me now :/

Reply 51

Duskstar

13

Original post by Hody421

But from the mark scheme:
−π/2 < arctan x < π/2⇒ −π/4 < f(x) < π/4⇒ range is −π/4 to π/4

Oh yeah that's my fault sorry - that's correct; I haven't done maths in a while lol.

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Reply 52

iAmanze

18

Oh boy, with this M1 that idk if I'm looking at an A or B... I need to 90%+ this exam.

Reply 53

QueenAryela

8

Original post by lizard54142

x

Original post by Leechayy

x

http://www.mei.org.uk/files/papers/c307ja_02c76.pdf

Q1(ii) how does that work? x

Reply 54

lizard54142

6

Original post by QueenAryela

http://www.mei.org.uk/files/papers/c307ja_02c76.pdf

Q1(ii) how does that work? x

Q is the point where

|x| = |x-2| + 1

. It says verify so you can just use y=1.5

Reply 55

poorform

10

Original post by QueenAryela

http://www.mei.org.uk/files/papers/c307ja_02c76.pdf

Q1(ii) how does that work? x

Notice that the right hand branch of y=|x| is simply the line y=x.

Notice that the left hand branch of y=|x-2|+1 is the line y=2-x+1. (since x<2 in this region)

So you can solve easily for x and y to verify.

(edited 8 years ago)

Reply 56

QueenAryela

8

Original post by lizard54142

Q is the point where

|x| = |x-2| + 1

. It says verify so you can just use y=1.5

did that, didnt know how to open the modulus

Original post by poorform

Notice that the right hand branch of y=|x| is simply the line y=x.

Notice that the left hand branch of y=|x-2|+1 is the line y=2-x+1. (since x<2 in this region)

So you can solve easily for x and y to verify.

That makes sense. I was wondering why it became essentially x = -(x-2)+1

So it's not possible to work out without the graph?

Reply 57

lizard54142

6

Original post by QueenAryela

did that, didnt know how to open the modulus

That makes sense. I was wondering why it became essentially x = -(x-2)+1

So it's not possible to work out without the graph?

Notice at Q x > 0, so you can just use y=x to find x. You have to then verify this with the other equation

Reply 58

poorform

10

Original post by QueenAryela

did that, didnt know how to open the modulus

That makes sense. I was wondering why it became essentially x = -(x-2)+1

So it's not possible to work out without the graph?

Lizard's solution is quicker. You don't need to anything really. You know that the point of intersection must lie on the line y=x so consequently y=x=1.5 so just sub x=1.5 into both sides, show that the equation holds and you are done.

I suppose you could work it out without the graph if you wanted it's given but doesn't really help that much