STEP I, II, III 2000 solutions

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Here's I/8:

The graph of y = xe^x will cut at x = 0. When x > 0, both x and e^x will be positive so it will lie above the x-axis hence a positive integral. When x < 0, x will be negative and e^x positive so it lie beneath the x-axis hence a negative integral. So by subtracting the negative integral, we will get the absolute value which will be given by the modulus of xe^x.

[br]\displaystyle \int_0^1 xe^x \,dx = \left[xe^x\right]^1_0 - \int_0^1 e^x \,dx.

Unparseable latex formula:

[br]= \left[xe^x\right]^1_0 - \left[e^x\right]^1_0 \\[br]= e - (e - 1) \\[br]= 1. \\ \\[br]

Unparseable latex formula:

[br]\displaystyle[br]\int_{-1}^0 xe^x \,dx = \left[xe^x\right]^0_{-1} - \left[e^x\right]^0_{-1} \\[br]= 1/e - (1 - 1/e) \\[br]= 2/e - 1. \\[br][br]1 - (2/e - 1) = 2 - 2/e

i) Solving f(x) = x^3-2x^2 - x + 2 = 0. f(1) = 0, hence x-1 is a factor. x^3 - 2x^2 - x + 2 = (x-1)(x^2 - x - 2) = (x-1)(x-2)(x+1). Roots at x = -1, 1 and 2.

A quick sketch shows the graph is above the axis between 0 and 1, below it between 1 and 2 and above it between 2 and 4.

So

\displaystyle \int_0^4 |f(x)| \,dx = \int_0^1 f(x) \,dx - \int_1^2 f(x) \,dx + \int_2^4 f(x) \,dx

\displaystyle \int_a^b f(x) = \left[1/4x^4 - 2/3x^3 - 1/2x^2 + 2x\right]^b_a = G(x)

Unparseable latex formula:

\displaystyle \int_0^1 f(x) \,dx = \left[G(x)\right]^1_0 = 13/12. \\[br]\int_1^2 f(x) \,dx = \left[G(x)\right]^2_1 = (4-16/3 - 2 + 4) - (1/4 - 2/3 + 1/2 + 2) = (2/3) - (13/12) = -5/12. \\[br]\int_2^4 f(x) \,dx = \left[G(x)\right]^4_2 = (64/3) - (2/3) = 62/3. \\[br][br][br]62/3 + 13/12 - (-5/12) = 133/6.

ii) Solving sin x + cos x = 0. sin x = - cos x. tan x = -1, x =

-\dfrac{\pi}{4}, \dfrac{3\pi}{4}

.

A quick sketch shows the graph of y = sin x + cos x is below the x axis between -pi and -1/4pi, and above it between -1/4pi and 3/4pi, and below it between 3/4pi and pi.

Letting f(x) = sin x + cos x:

\displaystyle \int_{-\pi}^{\pi} |f(x)| \, dx = -\int_{-\pi}^{-1/4\pi} f(x) \,dx + \int_{-1/4\pi}^{3/4\pi} f(x) \,dx - \int_{3/4\pi}^{\pi} f(x) \, dx

\displaystyle \int_a^b f(x) \,dx = \left[\sin x - \cos x\right]^b_a.

\left[\sin x - \cos x \right]^{-1/4\pi}_{-\pi} = \left(-\dfrac{\sqrt{2}}{2} - \dfrac{\sqrt{2}}{2}\right) - 1 = -\sqrt{2} - 1

Unparseable latex formula:

\left[\sin x - \cos x]^{3/4\pi}_{-1/4pi} = \left(\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2}\right) - (-\sqrt{2}) = 2\sqrt{2}

Unparseable latex formula:

\left[\sin x - \cos x]^{\pi}_{3/4\pi} = 1 - \sqrt{2}

2\sqrt{2} - (-\sqrt{2} - 1) - (1 - \sqrt{2}) = 4\sqrt{2}

Reply 21

Glutamic Acid

I/1:

i)log5 = log10 - log2 = 1 - 0.3010.. = 0.699...
log6 = log3 + log2 = 0.3010 + 0.477 = 0.778

Suppose it true:

\log(5 \times 10^{47}) < \log 3^{100} < \log(6 \times 10^{47})

\Leftarrow 47 + \log 5 < 100 \log 3 < 47 + \log 6

\Leftarrow 47.699... < 47.7 < 47.778...

, which is correct.
First digit of 3^100 = 5.

ii) The notation k* will mean the first digit of k, where k is a value for which the inequality holds true. The exponent of 10 we use will not matter since it won't alter the first digit.

k \times 10^{100} < 2^{1000} < (k+1) \times 10^{100}

\log k + 100 < 1000 \log 2 \Rightarrow \log k < 201.029996...

\log (k+1) + 100 > 1000 \log 2 \Rightarrow \log (k+1) > 201.029996...

Considering the decimal part of 201.02996, we know that log k* < 0.029996 and log (k+1)* > 0.029996. As you can see k* = 1. Hence the first digit of 2^1,000 is 1.

k \times 10^{1000} < 2^{10000} < (k+1) \times 10^{1000}

\log k + 1000 < 10000 \log 2 \Rightarrow \log k < 2010.29996...

\log (k+1) + 1000 > 10000 \log 2 \Rightarrow \log (k+1) > 2010.29996

Considering the decimal part again.

\log k^* < 0.29996

\log (k+1)^* > 0.29996

. As you can see, k* = 1. Hence the first digit of 2^10,000 is 1.

Same process, same answer?

k \times 10^{10000} < 2^{100000} < (k+1) \times 10^{10000}

\log k + 10000 < 100000\log 2 \Rightarrow \log k < 20102.9996...

\log (k+1) + 10000 < 100000 \log 2 \Rightarrow log (k+1) > 20102.999...

Decimal part again...

\log k^* < 0.9996...

\log (k+1)^* > 0.9996...

By calculating log 9 as 2 log 3 = 0.954..., you can see that k* = 9. Hence the first digit of 2^100,000 is 9.

Reply 22

DFranklin

Glutamic Acid

I/1:

i)log5 = log10 - log2 = 1 - 0.3010.. = 0.699...
log6 = log3 + log2 = 0.3010 + 0.477 = 0.778

Suppose it true:

\log(5 \times 10^{47}) < \log 3^{100} < \log(6 \times 10^{47})

47 + \log 5 < 100 \log 3 < 47 + \log 6

47.699... < 47.7 < 47.778...

, which is correct.As it stands, you've assumed the conclusion. If you're going to argue like this you must use

\Leftarrow

connectives to show the direction of implication.

ii) The notation k* will mean the first digit of k. The exponent of 10 we use will not matter since it won't alter the first digit.

k \times 10^{100} < 2^{1000} < (k+1) \times 10^{100}

\log k + 100 < 1000 \log 2 \Rightarrow \log k < 201.029996...

\log (k+1) + 100 > 1000 \log 2 \Rightarrow \log (k+1) > 201.029996...

Considering the decimal part of 201.02996, we know that log k* < 0.029996 and log (k+1)* > 0.029996. As you can see k* = 1. Hence the first digit of 2^1,000 is 1.

I reallly can't follow what you're trying to argue here. It doesn't help that you never say what k is supposed to be.

I'm also very unconvinced by the way you've used 'the exponent of 10 we use will not matter since it won't alter the first digit'. That may be true, but it doesn't mean you should therefore get the exponent wrong. In other words, arguing "2^10 = 1024000 so the first digit is 1" is a flawed argument that would lose you marks. I'm not entirely sure whether you've done this because you don't say what k is, but I think you have.

Not withstanding all of that, you've done all the calculations right, so you'd still get most of the marks. I think you'd drop 4 or 5 marks here just for the way it's written up, though.

Reply 23

Glutamic Acid

DFranklin

As it stands, you've assumed the conclusion. If you're going to argue like this you must use

\Leftarrow

connectives to show the direction of implication.

Yes, that was a bit of laziness there.

k is a value, for which the inequality holds true. If you wanted to find the first digit of 2^10, then whether you said 10 x 10^2 < 2^10 < 11 x 10^2. then the first digit is a 1, or whether 1 x 10^3 < 2^10 < 2 x 10^3 so the first digit is a one shouldn't matter. I couldn't really see an alternative, as I didn't see a way of getting the exponent correct.

Reply 24

DFranklin

Glutamic Acid

k is a value, for which the inequality holds true.

In that case, you need to show that k exists and is > 0 (since you take the log of it). But it was sloppy reading on my part as well, I admit.

So now I know what k is supposed to be, let's look again:

k \times 10^{100} < 2^{1000} < (k+1) \times 10^{100}

\log k + 100 < 1000 \log 2 \Rightarrow \log k < 201.029996...

\log (k+1) + 100 > 1000 \log 2 \Rightarrow \log (k+1) > 201.029996...

Considering the decimal part of 201.02996, we know that log k* < 0.029996 and log (k+1)* > 0.029996. As you can see k* = 1. Hence the first digit of 2^1,000 is 1.

Two problems here: firstly 201.029996... isn't really a number, (or if it is, I don't know what number it's supposed to be) and yet you're treating it like it is. That's not just me being picky - all they've given you is an approximation to log 2, and for full marks you would need to make sure you only treat it as an approximation and not an exact value.

But the killer if k is big (as it is here), it's almost certain that k and k+1 start with the same digit, so k* = (k+1)*. And yet you argue that log k* < 0.029996 and log (k+1)* > 0.029996. I don't see any reason to think this will be true.

I couldn't really see an alternative, as I didn't see a way of getting the exponent correct.

log 2^1000 = 1000 log 2. So log 2^1000 > 301, so 2^1000 > 10^301. And log 2^1000 < 301.1 < 301 + log 2, so 2^1000 < 2 * 10^301.

So 10^301 < 2^1000< 2*10^301 and so the first digit is 1.

Reply 25

Glutamic Acid

DFranklin

So now I know what k is supposed to be, let's look again:

Two problems here: firstly 201.029996... isn't really a number, (or if it is, I don't know what number it's supposed to be) and yet you're treating it like it is. That's not just me being picky - all they've given you is an approximation to log 2, and for full marks you would need to make sure you only treat it as an approximation and not an exact value.

I see what you mean, but I didn't think it'd make much difference. Would writing ~201.02996 take care of this?

But the killer if k is big (as it is here), it's almost certain that k and k+1 start with the same digit, so k* = (k+1)*. And yet you argue that log k* < 0.029996 and log (k+1)* > 0.029996. I don't see any reason to think this will be true.

Oh yes, I've made a big error there. I suppose I was thinking of log(k*+1) > 0.029996 in my head.

log 2^1000 = 1000 log 2. So log 2^1000 > 301, so 2^1000 > 10^301. And log 2^1000 < 301.1 < 301 + log 2, so 2^1000 < 2 * 10^301.

So 10^301 < 2^1000< 2*10^301 and so the first digit is 1.

Ah yes, that's very nice. I'd seen the inequality used in part i), so used a similar method hoping it'd work.

Reply 26

DFranklin

Glutamic Acid

I see what you mean, but I didn't think it'd make much difference. Would writing ~201.02996 take care of this?

Not really - because your argument seems to is relying quite heavily on the value actually being exact. I say "seems to" because I'm still not clear exactly what you mean when you say 201.029996...

If you're doing it "properly", you'll typically have an "extra layer" of inequality. That is, I want to argue 1000 log 2 < log(2*10^301). But I don't know exactly what log 2 is, so I argue 1000 log 2 < 301.1 (as log 2 < 0.3011), and log (2*10^301) = 301 + log 2 > 301.2 (as log 2 > 0.2).

So we have 1000 log 2 < 301.1 < 301.2 < log(2*10^301).

So I've proved what we wanted without ever having to know exactly what either side actually is.

Ah yes, that's very nice. I'd seen the inequality used in part i), so used a similar method hoping it'd work.

But (as long as you argue in the right direction), part i) is exactly the same:

log 3^100 = 100 log 3 < 47.726
log 6*10^47 = log 6 + 47 > 47.75

So log 3^100 < 47.726 < 47.75 < log 6 * 10^47.

And log 3^100 > 47.7, while log 5*10^47 = 48 - log 2 < 47.7

So 5*10^47 < 10^47.7 < 3^100 < 10^47.75 < 6*10^47.

(This isn't really any different from how you did it except I'm arguing in the opposite direction and using that "extra layer" I mentioned).

Reply 27

Glutamic Acid

DFranklin

The ... bit was to indicate that I only knew the value of 1000 log 2 to that degree of accuracy. I could see why it'd have an effect if I was finding 2^(10^10), since the decimal part of 10^10 log 2 would be unknown so I couldn't proceed, but since we know that 1000 log 2 = 301.029996 to 6dp, we can be certain that log 1 < 0.029996 < log 2.

Reply 28

Glutamic Acid

I/2:
I think my answers are right, but the method is rather tedious so I've most likely missed a trick somewhere along the line.

Considering the expansion of (x^4-1/x^2)^5 =

(x^4)^5 + \dbinom{5}{1}(x^4)^4(-1/x^2) + \dbinom{5}{2}(x^4)^3(-1/x^2)^2 + \dbinom{5}{3}(x^4)^2(-1/x^2)^3 + \dbinom{5}{1}(x^4)(-1/x^2)^4 + (-1/x^2)^5

= x^{20} - 5x^{14} + 10x^8 - 10x^2 + 5x^{-4} - x^{-10}

Considering the expansion of (x-1/x)^6 =

x^6 + \dbinom{6}{1}(x^5)(-1/x) + \dbinom{6}{2}(x^4)(-1/x)^2 + \dbinom{6}{3}(x^3)(-1/x)^3 + \dbinom{6}{4}(x^2)(-1/x)^2 + \dbinom{6}{5}(x)(-1/x)^5 + (-1/x)

= x^6 - 6x^4 + 15x^2 - 20 + 15x^{-2} - 6x^{-4} + x^{-6}

Consider the multiplication of these, the exponents of x that will add to give -12 are -x^-10 and 15x^-2. Therefore the coefficient of x^-12 is -1 x 15 = -15.

The exponents of x that will add to give 2, are 10x^8 and x^-6, -10x^2 and -20, and 5x^-4 and x^6. So the coefficient of x^2 is (10 x 1) + (-10 x -20) + (5 x 1) = 10 + 200 + 5 = 215.

Second part:
I couldn't see a systematic way of dealing with the trinomial and making sure I got all the terms correct other than just expanding it out. I can write in the coefficient given by the initial expansion as they'll be the same as in the first expansion.

(x^4 + (x^2+1))^5 = x^{20} + 5(x^{16})(x^2+1) + 10(x^{12})(x^2+1)^2 + 10(x^8)(x^2+3)^3 + 5(x^4)(x^2+1)^4 + (x^2+1)^5

= x^{20} + 5x^{16}(x^2+1) + 10x^{12}(x^4+2x^2+1) + 10x^8(x^6 +3x^4+3x^2+1) + 5x^4(x^8+4x^6+6x^4+4x^2+1) + (x^{10}+5x^8+10x^6+10x^4+5x^2+1)

= x^{20} + 5x^{18} + 5x^{16} + 10x^{16} + 20x^{14} + 10x^{12} + 10x^{14} + 30x^{12} + 30x^{10} + 10x^8 + 5x^{12} + 20x^{10} + 30x^8 + 20x^6 + 5x^4 + x^{10} + 5x^8 + 10x^6 + 10x^4 + 5x^2 + 1

= x^{20} + 5x^{18} + 15x^{16} + 30x^{14} + 45x^{12} + 51x^{10} + 45x^8 + 30x^6 + 15x^4 + 5x^2 + 1

.

Now, the expansion of (x^2-1)^11 will have exponents of 22,20,18 .... 4,2,0. These which add to 4 are 4 and 0, 2 and 2 and 0 and 4. So this is given by

(1 \times \dbinom{11}{9} \times (-1)^9) + (5 \times \dbinom{11}{10} \times (-1)^{10}) + (15 \times (-1)^{11}) = -55 + 55 -15 = -15.

The exponents adding to 38 are 20 and 18, 18 and 20, 16 and 22. So this is given by

(1 \times \dbinom{11}{2} \times (-1)^2) + (5 \times \dbinom{11}{1} \times (-1)) + (15 \times 1) = 55 - 55 + 15 = 15.

Reply 29

DFranklin

Glutamic Acid

The ... bit was to indicate that I only knew the value of 1000 log 2 to that degree of accuracy.

I could see why it'd have an effect if I was finding 2^(10^10), since the decimal part of 10^10 log 2 would be unknown so I couldn't proceed, but since we know that 1000 log 2 = 301.029996 to 6dp, we can be certain that log 1 < 0.029996 < log 2.

But what does

log 1 < 0.029996 < log 2

tell you about 1000 log 2 - 301?

Forgive me for putting words in your mouth, but it seems your argument here is essentially "log 1 < 0.029996 < log 2 and 1000 log 2 - 301 = 0.029996..., so log 1 < 1000 log 2 - 301 < log 2".

That's not a valid argument. It might be that log 2 = 0.0299960001 and 1000 log 2 - 301 = 0.0299960002, in which case your conclusion would be false.

I don't want to keep going over this. The basic point is that you only have approximations to log 2. So you don't know that log 2 = 0.301029996, you only know that 0.3010299955 < log 2 < 3.010299965. Your argument has to take account of that to get full marks. I've seen examiners' reports on similar questions, and that's essentially what they say.

Reply 30

DFranklin

Glutamic Acid

I think my answers are right, but the method is rather tedious so I've most likely missed a trick somewhere along the line.

Second part:
I couldn't see a systematic way of dealing with the trinomial and making sure I got all the terms correct other than just expanding it out. I can write in the coefficient given by the initial expansion as they'll be the same as in the first expansion.

Spoiler

Reply 31

Glutamic Acid

DFranklin

The thing is, I still don't know what you actually mean by 301.029996...

If you mean 1000 log 2, you should write "1000 log 2". You can't replace it by 301.029996... (or 301.029996, for that matter).

So once you want to get down to comparing actual numbers, you have to use inequalities. You can't say log 2 = 301.029996...., but you can say log 2 < 301.03.

But what does

log 1 < 0.029996 < log 2

tell you about 1000 log 2 - 301?

Forgive me for putting words in your mouth, but it seems your argument here is essentially "log 1 < 0.029996 < log 2 and 1000 log 2 - 301 = 0.029996..., so log 1 < 1000 log 2 - 301 < log 2".

That's not a valid argument. It might be that log 2 = 0.0299960001 and 1000 log 2 - 301 = 0.0299960002, in which case your conclusion would be false.

I don't want to keep going over this. The basic point is that you only have approximations to log 2. So you don't know that log 2 = 0.301029996, you only know that 0.3010299955 < log 2 < 3.010299965. Your argument has to take account of that to get full marks. I've seen examiners' reports on similar questions, and that's essentially what they say.

Ahhh, thanks for that. I get it now. So it wouldn't have any affect on the answer, but it could've done for a different number I should've taken care of it that way.

Reply 32

Glutamic Acid

I/3:

I'm useless at graph sketching, but I've tried.

\displaystyle \int_0^5 [x] \,dx

= (4x1) + (3x1) + (2x1) + (1x1) = 10, as can be seen from the graph.

2nd Part:

By looking at the graph, you can say:

\displaystyle \int_{0}^{\ln n} [e^x] \,dx = \int_{\ln (n-1)}^{\ln n} [e^x] \,dx + \int_{\ln(n-2)}^{\ln (n-1)} [e^x] \,dx + ... + \int_0^{\ln 2} [e^x] \,dx

.

Considering a single strip of [e^x] from ln n to ln (n-1), the height is given by e^(ln (n-1)) = n-1, and the width is given by ln n - ln (n-1). Therefore:

\displaystyle \int_0^{\ln n} [e^x] \,dx = (\ln n - \ln (n-1)) \times (n-1) + (\ln (n-1) - \ln (n-2)) \times (n-2) + ... + (\ln 2 - \ln 1) \times 1)

\displaystyle \Rightarrow \int_0^{\ln n} [e^x] \,dx = (n-1)\ln \left(\frac{n}{n-1}\right) + (n-2)\ln \left(\frac{n-1}{n-2}\right) + ... + \ln \dfrac{2}{1}

\displaystyle \Rightarrow \int_0^{\ln n} [e^x] \,dx = \ln \left(\frac{n}{n-1}\right)^{n-1} + \ln \left(\frac{n-1}{n-2}\right)^{n-2} + ... + \ln \dfrac{2}{1}

\displaystyle \Rightarrow \int_0^{\ln n} [e^x] \,dx = \ln \left(\frac{n^{n-1} \times (n-1)^{n-2} \times (n-2)^{n-3} \times ... \times 2}{(n-1)^{n-1} \times (n-2)^{n-2} \times (n-3)^{n-3} \times ... \times 1}\right)

Using the power rules, terms of the type (n-a)^(n-a-1)) / (n-a)^(n-a) will cancel to (n-a)^(-1) = 1/(n-a), because n - a - 1 - (n - a) = -1.

\displaystyle \Rightarrow \int_0^{\ln n} [e^x] \,dx = \ln \left(\frac{n^{n-1}}{(n-1)!}\right)

.

Multiplying both top and bottom by n.

\displaystyle \int_0^{\ln n} [e^x] \,dx = \ln \left(\frac{n^n}{n!}\right) = \ln n^n - \ln (n!) = n \ln n - \ln (n!)

Shaky attempt at I/4.

i) Substituting in 0, you get 0. Substituting in 1, you get 1/2^4 = 1/16. But we must check for stationary points.

\dfrac{d}{dx} \dfrac{x^6}{(x^2+1)^4} = \dfrac{6x^5(x^2+1)^4 - 8x^7(x^2+1)^3}{(x^2+1)^8}

= \dfrac{6x^5(x^2+1) - 8x^7}{(x^2+1)^5}

\Rightarrow 6x^5(x^2+1) - 8x^7 = 0 \Rightarrow 6x^5 - 2x^7 = 0

\Rightarrow 2x^5(3-x^2) = 0 \Rightarrow x = 0 or \pm \sqrt{3}

.

So no other stationary points, hence 1/16 is the largest value.

ii) Differentiating the second term gives

\dfrac{(x^2+1)^3(5Ax^4+3Bx^2+C) - 6x(x^2+1)^2(Ax^5+Bx^3+Cx)}{(x^2+1)^6}

= \dfrac{(x^2+1)(5Ax^4+3Bx^2+C) - 6x(Ax^5+Bx^3+Cx)}{(x^2+1)^4}

Combining this with the last term gives the identity:

(5Ax^4+3Bx^2+C)(x^2+1) - 6x(Ax^5+Bx^3+Cx) + Dx^6 \equiv 1

Big-ass expansion:

5Ax^4+3Bx^4+Cx^2+5Ax^4+3Bx^2+C-6Ax^6 - 6Bx^4 - 6Cx^2+Dx^6 \equiv 1

Factorizing:

(D-A)x^6 + (5A-3B)x^4+(3B-5C)x^2+C \equiv 1

Letting x =0, C = 1.
So

(D-A)x^6 + (5A-3B)x^4 + (3B-5)x^2 \equiv 0

. Making the coefficients of x = zero, A = 1, B = 5/3, C = 1 and D = 1.

iii). Integrate both sides of ii) to get:

\displaystyle \int_0^1 \frac{1}{(x^2+1)^4} \,dx = \int_0^1 \frac{d}{dx} \left(\frac{x^5+5/3x^3+x}{(x^2+1)^3} \right)\,dx + \int_0^1 \frac{x^6}{(x^2+1)^4} \,dx

\displaystyle \int_0^1 \frac{1}{(x^2+1)^4} \,dx = \left[ \frac{x^5 + 5/3x^3 + x}{(x^2+1)^3}\right]^1_0 + \int_0^1 \frac{x^6}{(x^2+1)^4} \,dx

\displaystyle \int_0^1 \frac{1}{(x^2+1)^4} \,dx = 11/24 + \int_0^1 \frac{x^6}{(x^2+1)^4} \,dx

.

Considering y = x^6/(x^2+1)^4, the maximum area under the curve would occur when y = 1/16 for x = 0 to 1, (since we know 1/16 is the maximum value) giving an area of 1 x 1/16 = 1/16.
As y would be positive for all values of x (as the numerator and denominator are always positive), the minimum integral would occur when y = 0 for 0 <= x < 1. giving an area of 0.

Hence you can say

\displaystyle \frac{11}{24} \le \int_0^1 \frac{1}{(x^2+1)^4}\,dx \le \frac{11}{24} + \frac{1}{16}

Reply 34

Square

I think I've just solved STEP I/5.

Which is strange as my geometry sucks.

Reply 35

Glutamic Acid

II/2.

Write it as

p(x) = (x-a)^2q(x)

Differentiating using the product rule:

p'(x) = 2(x-a)q(x) + q'(x)(x-a)^2

p'(x) = (x-a)\left( 2q(x) + q'(x)(x-a)\right)

x-a is a factor of p'(x), hence p'(a) = 0.

p(x) = (x-a)^4q(x)

p'(x) = 4(x-a)^3q(x) + q'(x)(x-a)^4

p'(x) = (x-a)^3\left( 4q(x) + q'(x)(x-a)\right)

x-a is a factor of p'(x), hence p'(a) = 0.

If (x-a) is a factor, then substituting x = a into the equation will give zero, so:

a^6 + 4a^5 - 5a^4 - 40a^3 - 40a^2 + 32a + k = 0

And using the result proved earlier, substituting x = a into the derivative will give zero, so:

6a^5 + 20a^4 - 20a^3 - 120a^2 - 80a + 32 = 0

3a^5 + 10a^4 - 10a^3 - 60a^2 - 40a + 16 = 0

Trying a few integer values of a, a = -2 satisfies the equation.
Substituting a = -2 into the first equation.

64 - 128 - 80 + 320 -160 - 64 + k = 0

k = 48

Reply 36

DFranklin

Glutamic Acid

II/2.

Write it as

p(x) = (x-a)^2q(x)

Differentiating using the product rule:

p'(x) = 2(x-a)q(x) + q'(x)(x-a)^2

p'(x) = (x-a)\left( 2q(x) + q'(x)(x-a)\right)

x-a is a factor of p'(x), hence p'(a) = 0.

p(x) = (x-a)^4q(x)

p'(x) = 4(x-a)^3q(x) + q'(x)(x-a)^4

p'(x) = (x-a)^3\left( 4q(x) + q'(x)(x-a)\right)

x-a is a factor of p'(x), hence p'(a) = 0.Not what they were looking for. The result you state is just a special case of the first one: if p(x)=(x-a)^4q(x), then p(x)=(x-a)^2r(x), where r(x)=(x-a)^2q(x).

The result they were looking for is that p'''(a) = 0.

I have no idea what marks they'd give you for this: what you've done is very different from what they wanted, but all you've done "wrong" is make a bad decision about what "corresponding result" means. Personally, I wouldn't dock you more than a mark or two (but if I was the question setter I'd be feeling very embarrassed I didn't make it clearer).

Reply 37

Glutamic Acid

DFranklin

Not what they were looking for. The result you state is just a special case of the first one: if p(x)=(x-a)^4q(x), then p(x)=(x-a)^2r(x), where r(x)=(x-a)^2q(x).

The result they were looking for is that p'''(a) = 0.

I have no idea what marks they'd give you for this: what you've done is very different from what they wanted, but all you've done "wrong" is make a bad decision about what "corresponding result" means. Personally, I wouldn't dock you more than a mark or two (but if I was the question setter I'd be feeling very embarrassed I didn't make it clearer).

Oh, missed that completely, it does seem very ambiguous though. I'll have a look to see if I can find the examiners' report to see if it was a common error.

Reply 38

Zhen Lin

Tried 2001 paper III earlier today, and that was nasty... Hrm.

Anyway, I also tried a mechanics question for the first time - Q9 on 2000 paper III is fairly straightforward, once one learns the required laws...

III-2000-Q09.pdf29.8KB

Reply 39

johnheyes91

Zhen just curious, how do you type maths and transfer it to a pdf? Cheers!

Also, is paper I, II and III all for further maths students? Or is there a paper just for the basic A Level maths students?

Finally, is AEA or STEP easier?