Edexcel A2 C4 Mathematics June 2016 - Official Thread

I think so

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Actually there would be no a.

Posted from TSR Mobile

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Could someone please explain part 8b fully, including how you integrate 3^2x, what method was used as I don't get it at all, where did the ln3 come from etc., also how did they take the triangle area into consideration to minus it away from the total area under the curve, I'm co confused?

Consider how you'd differentiate 3^x, you'd get 3^x.ln3
The integral from part a uses this idea to give 3^x.1/ln3

You could then solve 3^2x by using a substitution of u=2x and finding that with the part a answer.

You then have the triangle (I've labelled this T), when the shape from 3^2x from 0 to 9 is rotated, it also rotates the triangle into a cone (with it's top at Q) with radius r and height h. You'll then need to take this area away, to get the volume of R rotated 360 degrees.

I differentiated in part a to help find the gradient and equation of the tangent I didn't integrate anything in part a, which part a integral do you mean? Also how do you use the solution of differentiating 3^x to get the integral of 3^x, I get how to differentiate this function but I don't understand how it helps me find the integral of it? I'm just really confused could you please show me step by step?

Sorry I misread the question, thought part a was integrating.

Consider if you were asked to differentiate 3^x.1/ln3
As the 1/ln3 is just a number, you do nothing to that. As for 3^x, the differential of that is 3^x.ln3
This means that overall, the differential of 3^x.1/ln3 is 3^x, as the ln3's would cancel out with eachother.
That means that, if you are asked to integrade 3^x, you would be able to do the reverse and get 3^x.1/ln3 + c.

If we have cotx = 0 are we supposed to know that tanx = 1/0 and we treat 1/0 as infinity? Just saw this in a solomon and it really threw me off

Guys why do we set 3/2root(2) = 543/256 for part c of this question?



MS:

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Because that's what you get when you plug 1/10 into the approximation/expansion up to x^3

What does expanding up to X^3 have to do with it? We only went up to x^2?

likely it wouldn't come up but now you know it anyways?

If we have cotx = 0 are we supposed to know that tanx = 1/0 and we treat 1/0 as infinity? Just saw this in a solomon and it really threw me off

Yeah I meant x^2 sorry
The point is that it's what you plug into the expansion you already have, this serves as your approximation
(well obvs you have to rearrange to get the root(2) approximation. I assume you can see that plugging 1/10 in to the actual expression gives you (3/2)root(2)

I'm sorry, I'm really confused as to what's going on... XD So in part A I did the expansion up to x^2. Got that.

In part b, I subbed x = 1/10 into (4 + 5x)^0.5 to get the exact answer of (4 + 5x)^0.5.

In part c, I subbed in (1/10) into my expansion to get 543/256.

So because the exact answer in part b was 3/2root(2),

3/2root(2) = 543/256. So to get an approximation for root(2), root(2) = 543/246 * 2/3.

Is that reasoning correct?

ok someone explain this
(all in degrees)
cot(90)= 1/tan(90) = Cos(90)/sin(90)

but 1 over tan 90 gives error

cos 90 over sine 90 gives 0

i get that tan 90 is at an asymptote but surely this dosent work?

If you look at the limiting value of 1/(tan x) as x tends to 90 (in ugly degrees), you get 1/(massive huge infinite number), which is basically 0.