Oxford MAT 2013/2014

Original post by Noble.

I take it as meaning the second in your original post (i.e. to differentiate it using the chain rule on the 2x as opposed to just on some t and subbing in t=2x at the end). But you're right, the question is ambiguous.

The "exam savvy" part of me agrees - (particularly since my I think many people would find the first solution by "accident", so it would only be stronger candidates who made the mistake of using the second second solution), but the consensus on here seemed to be that the first interpretation was correct. (I haven't read that much of the thread though).

Reply 1304

Noble.

17

Original post by Khallil

I just did the first 5 multiple choice questions and got BCCBA.
Could you guys compare your answers against mine?
(Also I didn't do this paper, I'm just curious)

Edit, I just did G and got D as my answer.

I'm pretty sure for the first one the answer is (a) - unless I can't even solve quadratic equations properly anymore.

For distinct real roots the discriminant must be greater than 0, so

a^2 - 4(a-1) > 0 \implies a^2 - 4a + 4 > 0 \implies (a-2)^2 > 0 \implies a \neq 2

Reply 1305

user2020user

16

Original post by Noble.

I'm pretty sure for the first one the answer is (a) - unless I can't even solve quadratic equations properly anymore.

For distinct real roots the discriminant must be greater than 0, so

a^2 - 4(a-1) > 0 \implies a^2 - 4a + 4 > 0 \implies (a-2)^2 > 0 \implies a \neq 2

Oh yea! I completed neglected the fact that

a

can be less than 2 :rolleyes:

Reply 1306

Original post by kapur

assume you have f(x)=x
then f'(x) =1 , right?
the f(2x)=2x and f'(2x)=2 . this is the first and correct interpretation.
using the other interpretation , you get derivative of f(2x)= f'(2X) *2= 2*2=4 which is obv wrong

Neither of these really correspond to what I wrote (and I can't really make sense of your reasoning, to be honest),

Let's make a concrete example of 1C:

Let h(x) = 2, g(x) = 2(x-1), f(x) = x^2. Then f'(x) = 2x = g(x+1), g'(x) = h(x-1).

So I think we can all agree that f''(x) = 2, and f(2x) = 4x^2. So the question is,
does f''(2x) = 2 (because f''(x) = 2 everywhere, and so when we evaluate it at 2x we still get 2).

Or does f''(2x) = 8, because if q(x) = f(2x), then q(x) = 4x^2 and so q''(x) = 8.

Looking around, I believe the standard notation for the 2nd interpretation would be f(2x)'', and so f''(2x) implies the 1st case. Which does make a certain amount of sense, but it's pretty darn subtle.

Note that as I understand these standards, with your example the correct answer is f'(2x) = 1.

Reply 1307

kapur

Original post by DFranklin

Neither of these really correspond to what I wrote (and I can't really make sense of your reasoning, to be honest),

Let's make a concrete example of 1C:

Let h(x) = 2, g(x) = 2(x-1), f(x) = x^2. Then f'(x) = 2x = g(x+1), g'(x) = h(x-1).

So I think we can all agree that f''(x) = 2, and f(2x) = 4x^2. So the question is,
does f''(2x) = 2 (because f''(x) = 2 everywhere, and so when we evaluate it at 2x we still get 2).

Or does f''(2x) = 8, because if q(x) = f(2x), then q(x) = 4x^2 and so q''(x) = 8.

Looking around, I believe the standard notation for the 2nd interpretation would be f(2x)'', and so f''(2x) implies the 1st case. Which does make a certain amount of sense, but it's pretty darn subtle.

Note that as I understand these standards, with your example the correct answer is f'(2x) = 1.

really? you think the derivative of 2x is 1?

Reply 1308

Noble.

17

Original post by kapur

really? you think the derivative of 2x is 1?

lol you do realise that

a) the person you're talking to is a Cambridge maths graduate

b)

f'(2x)

does not mean

\dfrac{d}{dx}(2x)

Reply 1309

kapur

Original post by Noble.

lol you do realise that

a) the person you're talking to is a Cambridge maths graduate

b)

f'(2x)

does not mean

\dfrac{d}{dx}(2x)

unless f(x)=x ... as i said

Reply 1310

Original post by kapur

really? you think the derivative of 2x is 1?

No, I think the derivative of f(t) = t, evaluated when t = 2x, is 1.

Reply 1311

Original post by Tarquin Digby

Nah, pretty sure I'm right. And (c) doesn't have multiple interpretations, the answer is d. (remember chain rule)

You might want to note that the way you write the chain rule using dashes f'(x) etc. looks like:

If

F(x) = f(g(x))

, then

F'(x) = f'(g(x)) g'(x)

(see for example http://math.kennesaw.edu/~sellerme/sfehtml/classes/math1190/chainrule.pdf)

note that this implies f'(g(x)) is NOT the derivative of f(g(x)), since if it was, f'(g(x)) would equal F'(x) by definition.

That said, it seems confusion about a subtle issue in notation is a rotten way of losing all the marks on a question (even if it's just a multiple choice one).

Reply 1312

Original post by henpen

Another possibility was to notice that

f(\frac{1}{2}+t)-f(\frac{1}{2}-t))=8t^3

,thus

f(t)=4(t-\frac{1}{2})^3

is a solution (although I forgot to undo the

t \rightarrow t+\frac{1}{2}

transformation: I'm happy that part wasn't multiple choice!).

The stars and bars method killed question 5 very quickly.

I was under the impression that

f'(u)=\frac{df(u)}{dx}

. I dislike Leibniz notation.

You mean Lagrange, right? I made the same mistake, I got confused by the notation as well.
And yes, there are infinitely many solutions for f(t) for every g(t) that follows the condition we deduced in 2.ii.b.

Original post by DFranklin

I doubt it, since the fact they gave both "possible" answers implies they wanted you to choose the "correct" one.

I'm just arrogant enough to think that if I'm not sure what the correct answer is, it's not a very fair question. [Although I could just be being stupid or forgetful about a standard definition].

I'm interested to see what RichE has to say about it.

Someone actually checked Wolfram Alpha. :tongue:

Differentiating f(t) twice, then putting t=2x seems to be the standard interpretation. This is probably the only thing that I got completely wrong on the paper. I'm feeling terribly annoyed.

Reply 1313

Original post by souktik

Differentiating f(t) twice, then putting t=2x seems to be the standard interpretation. This is probably the only thing that I got completely wrong on the paper. I'm feeling terribly annoyed.

Well, if you weren't happy with that question I'd say you're in pretty good company!

And of course, if that's the only major mistake you made, you don't have much to worry about! :smile:

Reply 1314

Original post by DFranklin

Well, if you weren't happy with that question I'd say you're in pretty good company!

And of course, if that's the only major mistake you made, you don't have much to worry about! :smile:

What makes it more annoying is that I had initially interpreted the notation the right way. Thinking about it too much screwed me. Ah well, nothing I can do about it now. Hey, have you looked at 1. I, by any chance? That's the only other MCQ for which my answer didn't match with someone. I got (b) 28. I looked at it in binary. If the first two digits of a number are 10 in binary, its function is +1. If 11, then -1. So equal +1's and -1's from 2 till 63. 1 is +1, 64-95 are +1, 96-100 are -1. +28 total. Anything I messed up?

Posted from TSR Mobile

Reply 1315

Original post by souktik

What makes it more annoying is that I had initially interpreted the notation the right way. Thinking about it too much screwed me. Ah well, nothing I can do about it now. Hey, have you looked at 1. I, by any chance? That's the only other MCQ for which my answer didn't match with someone. I got (b) 28. I looked at it in binary. If the first two digits of a number are 10 in binary, its function is +1. If 11, then -1. So equal +1's and -1's from 2 till 63. 1 is +1, 64-95 are +1, 96-100 are -1. +28 total. Anything I messed up?Yes I agree with 28. (Also verified by writing a small computer program, for that 3rd unbiased opinion...)

Reply 1316

Original post by DFranklin

Yes I agree with 28. (Also verified by writing a small computer program, for that 3rd unbiased opinion...)

Thank you, that's a relief! :smile:

Posted from TSR Mobile

Reply 1317

GMU10

4

Here's how I did C, and I got that the answer was (c).

f'(x) = g(x+1) ((given))
We can call f'(x)=F(x) to avoid confusion, so F(x)=g(x+1)
Therefore F(2x)=g(2x+1)
Differentiate both sides, using the chain rule, to get 2F'(2x)=2g'(2x+1)
F'(2x)=g'(2x+1)
F'(2x)=h((2x+1)-1)=h(2x)

Reply 1318