OCR MEI FP2 Thread - AM 27th June 2016

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1/2 ln (2x/5 + root(4x^2/25 + 1) is my answer.

The answer in the guidance section is ln (2x/5 + root(4x^2/25 + 1)

Isn't this my answer without the 1/2 infront of it?

Oh yeah sorry I didn't notice that, I'm not sure, maybe make a thread asking the question if no one else responds as someone will surely know the answer :frown:

Reply 41

Connorbwfc

Original post by Alex621

Oh yeah sorry I didn't notice that, I'm not sure, maybe make a thread asking the question if no one else responds as someone will surely know the answer :frown:

Please could you attempt the question and see what answer you get...

I've asked my friend and they also get the same answer as me.

Reply 42

Alex621

I'm probably just being really stupid but how do you integrate the 1/(pi + theta)^2 part? Thanks

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Reply 43

Alex621

Original post by Connorbwfc

Please could you attempt the question and see what answer you get...

I've asked my friend and they also get the same answer as me.

Usually I integrate into the form arsinh(x/a), then convert to ln, and that gave me the same answer as you, with the 1/2 at the front, but I just tried integrating straight into ln form, using the formula in the booklet in the integration table, rather than the formula on the next page and I got the answer in the mark scheme, I'm really confused too now...

Reply 44

Connorbwfc

Original post by Alex621

I'm probably just being really stupid but how do you integrate the 1/(pi + theta)^2 part? Thanks

It's just basic integration rules. So it's like integrating (Pi + theta)^-2, so you increase the power to -1 and then divide by the new power (-1).

Reply 45

Alex621

Original post by Connorbwfc

It's just basic integration rules. So it's like integrating (Pi + theta)^-2, so you increase the power to -1 and then divide by the new power (-1).

Oh right ok thanks :lol:

I haven't done any basic C4 integration for ages

Reply 46

username2640523

Could anyone help me with june 13 4IV, I tried intergrating by parts but I just end up going in a loops :s-smilie:

really confused

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Reply 47

Alex621

Original post by HFancy1997

Could anyone help me with june 13 4IV, I tried intergrating by parts but I just end up going in a loops :s-smilie:

really confused

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If you write out the whole integration by parts formula, on the LHS you have the original integral, and on RHS you have the negative of it, add it onto the LHS, then you have 2x(the original integral) so divide both sides by two, and then continue

Reply 48

username2640523

Original post by Alex621

I dont quite understand what you mean

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Reply 49

Alex621

Original post by HFancy1997

I dont quite understand what you mean

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Reply 50

groovy_q

In the eigenvector question in jan 13, i got all the 3 eigenvalues correct and one eigenvector correct but the other two i got the ones in the mark scheme but with the signs switched? Does the signs matter in the eigenvector even if theyre compatible with sim equations you derived?

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Reply 51

-Gifted-

Original post by groovy_q

Doesn't matter if they work.

Reply 52

username2640523

Original post by Alex621

Oh wow didnt know this was a thing! Thanks man

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Reply 53

Aph

Original post by groovy_q

Generally signs don't matter but you should give the. In the simplest form. Thsi means in a 3x3 matrix you should always have at least 2 positives in the eigenvalues vector but in a 2x2 matrix it doesn't matter.

Reply 54

username2640523

Original post by Alex621

I'm probably just being really stupid but how do you integrate the 1/(pi + theta)^2 part? Thanks

I used substitution for this, dont understand what the mark scheme did but substitution worked :smile:

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Reply 55

groovy_q

Original post by Aph

In my answer i had two positive values in the eigenvector and the answer in the mark scheme only had 1? Theres also nothing from that mark scheme that suggests ignore the sign so im pretty worried haha

Reply 56

Aph

Original post by groovy_q

I wouldn't worry then. Both are generally correct but my guess would be that examiner didn't think about simplification. You'd get the marks :h:

Reply 57

username2640523

So worried about this exam, need an A for this...any predictions?

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Reply 58

Wunderbarr

Original post by Connorbwfc

1/2 ln (2x/5 + root(4x^2/25 + 1) is my answer.

The answer in the guidance section is ln (2x/5 + root(4x^2/25 + 1)

Isn't this my answer without the 1/2 infront of it?

I'm guessing you did something like factor out

\frac{1}{\sqrt{25}}

from the original integral, and then had

a^2 = 1

and

x = \frac{2x}{5}

?

So

\frac{1}{5} \displaystyle\int \frac{1}{\sqrt{1+\frac{4x^2}{25}}} dx

Reply 59

TheBBT

Quick question: When plotting points on an Argand diagram do you have to draw lines to the point from the origin? This is usually when plotting the roots of a complex number.