C4 partial fractions questions?

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For the first q, note that all but the second term, when differentiated term by term will be 0.

The second, sub y = 1 into the equation given and solve for x

For Q1, to clarify something, but the expansion (2 + 1/2x - 3/4x^2 + 17/8x^3) is essentially equal to y right? So you would just differentiate it as per usual?

For 2, so that means I have to do 1 = 2/[(x-1)(x-3)] ? Wouldn't I get a quadratic equation? Which means there would be several answers?

In essence, yues - sub in x=0 and all the terms of the expansion (except one) will be equal 0 too.

It would help if I could see the region R - I'm assuming that there are some limitations imposed upon it?

For the second question, I've written out all of the information that I am given. Unfortunately there isn't a graph to accompany it! So no limitations either.

There are two points where y=1. However there is an asymptote between one of the points where y =1, so we see that $\ x =2+\sqrt{3}$ for there to be a defined integral.

Also on your part c) you need to check your - signs

When you do 1/4 = 2/(x-1)(x-3), don't you get x^2 - 4x - 5 = 0 when simplified? So when you factorise, you get x = 5 or x = -1? I'm assuming we take x = -1 as it is the lower bound, but why can't you use the other value?

I'm a bit confused as I didn't get x = 2 square root 3 for either solution :/

Where did the $\frac{1}{4}$ come from?
Allow me to refer you to wolfram.

Oops, I just realised that I copied down the question wrong. The line is suppose to say, "The finite region bounded by the curve with equation y=... and the lines x=4 and y=1/4 is denoted by R". Sorry!

In this case, x=5 or -1.

In which case there are two asymptotes present between 4 and -1.
So the solution x = 5 must be the other limit

How do you know that there is an asymptote present in the graph? And how did you deduce that there are two asymptotes between 4 and -1?

You cannot divide by zero.
Inspect the function, and see where x causes you to divide by 0.

x

No problem
First q, you guessed corrctly, you just need to factorise the quartic, sub in x=10
Second q, consider what happens to $\dfrac{2x +3}{x^2 + x+1}$
when x gets very large

Er back to 27 (I just tried the integration), so I integrated 1/(x-3) - 1/(x-1) with limits of 5 and 4. So I ended up with [ln|x-3| - ln|x-1|]5 and 4 so...

ln|2| - ln|4| - (ln|1| - ln|3|) = ln1.5? Whereas in the question it says show that the area of R is ln3/2 - 1/4? Where does the 1/4 come from?

For 11, when you say factorise the quartic, do you mean x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2)? Surely that's the same as subbing in x=10 into the quartic?

For 18. I think I understand it now, as when x is very large, then 2x-3 / x^2+x+1 tends to infinity. As a result, the graph is equal to x^2 - 3x + 2?

One final thing, in binomial expansions, they sometimes say, "If x is small compared to a, expand ...", what does the "if x is small" bit mean?

27) You need to subtract $1 \times \frac{1}{4}$ can you see why?

11) In essence yes, I hadn't checked to see if it factorised further.

18) Not quite, the added bit tends to 0.

For the expansion to be justified:
$|x| <1$ I think that's basically what they're saying.

For the first one, I basically found R + 1/4 as the difference is 1/4 (in y) and 1 in x. So I subtract 1/4 to get R? ...This question is full of tricks! However, how do you know that the 'extra bit' is 1/4? What if the graph looks like this? :

It doesn't look like that, it looks like this.
Personally, I think it's mean to set a question like this, but it should be obvious that it's underneath, as the function is positive here, so the area must be too.
$\ (5-4) \times \frac{1}{4}$

(I was just drawing a random curve by the way!) That makes much more sense now, I suppose if I can do these harder questions, then I should be able to tackle the harder exam questions!

I'm having a bit of trouble with part c. So with this, it is just a typical volume of revolution question, so you do pi multiplied by the integral of y^2? When it says rotated by 2pi radians about the x-axis, is this basically saying rotation by 360o? So I shouldn't need to factor that in my working?

Here's my working, I'm not sure where I have gone wrong...as the answer is supposed to be pi(25/48 - ln3/2)