STEP 2000 Paper 1 Q4,Q6,Q8

By Anonymous on Monday, October 23, 2000 - 01:31 pm :

Hi there,

I'm having trouble with Q.4(i) of STEP 2000 Paper 1 - how can I show that the gradient is always positive over the given range? Neil Morrison mentioned something about SPs in his solution to the question, something which I have never come across before (what are they)?
I'm also having trouble with the factorisation in question 6 - it's easy enough to multiply out the brackets to check the answer, but could someone show me the steps in the factorisation please?

Thanks!

By The Editor :

Relevant bits of questions:

4.(i) Show that, for 0< =0< =1, the largest value of x⁶ /(x² +1)⁴ is 1/16.

6. Show that x² -y² +x+3y-2=(x-y+2)(x+y-1) ...

By Kerwin Hui (Kwkh2) on Monday, October 23, 2000 - 01:44 pm :

SP is a non-standard abbreviation for stationary point.

For question 6, first note that x² -y² =(x-y)(x+y), so we propose that

x² -y² +x+3y-2=(x-y+A)(x+y+B)

and obtain 3 equations for A and B, with any luck, are consistent.

Kerwin

By William Astle (Wja24) on Monday, October 23, 2000 - 01:58 pm :

For question 4(i):

Differentiate the expression. At x = 0 the gradient is 6 (I hope - I did it quickly). Then confirm that the differentiated expression has no roots in the interval. So there are no turning points in the interval. Since the function is continuous on the interval you can conclude it must always have positive gradient.

I am not sure what you are getting at with question 6. Factorisation is the inverse operation to multiplying out. So reverse your argument and you have done the factorisation :-)

By Anonymous on Monday, October 23, 2000 - 01:59 pm :

To show that there are no SPs, do I need to differentiate the equation (which looks utterly horrible) and compare it to 0, or is there some kind of logical reasoning behind it? The differentiation would probably not look as bad if I had done the chain and product rules at school - I only have a passing familiarity with them.

Thanks

By Kerwin Hui (Kwkh2) on Monday, October 23, 2000 - 02:09 pm :

To show there are no SPs, just consider the numerator when you differentiate the fraction. If it is non-zero for all x in the open interval, then there are no SPs that needs worrying.

The numerator is 6x⁵ (x² +1)⁴ -8x⁷ (x² +1)³ =2x⁵ (x² +1)³ (3-x² ), so no SP in the open interval (0,1).

Kerwin

By Olof Sisask (P3033) on Monday, October 23, 2000 - 04:21 pm :

On the same paper: Q.8 - how do you integrate x.e^x ?

By Michael Doré (Md285) on Monday, October 23, 2000 - 04:24 pm :

This is done by integration by parts. Are you familiar with this? I'll assume yes - write back if not!

$\int [u dv / dx dx] = uv - \int [v du / dx dx]$

where $u$ , $v$ are functions of $x$ .

So let $u = x$ and $v = e^{x}$ . You then get:

$\int [x e^{x} dx] = x e^{x} - \int [e^{x} 1 dx]$

because $dv / dx = e^{x}$ , $du / dx = 1$ .

So:

$\int [x e^{x} dx] = x e^{x} - e^{x} + C$

where $C$ is the arbitrary constant of integration.

By Olof Sisask (P3033) on Monday, October 23, 2000 - 04:32 pm :

I'm afraid I'm not familiar with integration by parts - nor have we done the chain or product rule in my school's (strange) syllabus yet - laws which seem to crop up everywhere (hope no-one asks me any questions at my Uni. interviews :)). I'd be very grateful if you could explain what integrating by parts is!

/Olof

By James Lingard (Jchl2) on Monday, October 23, 2000 - 04:36 pm :

Personally, I prefer the following (although essentially the same) method for remembering how to integrate

x . e^{x}

$(d / dx) (x . e^{x}) = x . e^{x} + e^{x}$ (by the product rule), so

$(d / dx) (x . e^{x} - e^{x}) = x . e^{x}$ .

Therefore $\int (x . e^{x}) dx = x . e^{x} - e^{x} (+ c)$ .

James.

By Olof Sisask (P3033) on Monday, October 23, 2000 - 06:04 pm :

Thanks for your help Kerwin, William, Michael and James, I'm gonna tackle STEP II now!

/Olof