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Oxford MAT 2013/2014

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Original post by kapur
I'm very bad at these graph sketching questions. Any suggestions for where I can practice such questions from? :P


I'm not that good at those either to be honest. My exam board syllabus doesn't contain graph sketching at all so I pretty much taught myself.:P I solved the c1 c2 c3 module textbook lessons for graphs to get the basics. I used to do this thing where my friend gave me a function,I used to sketch the graph and then check using a graph plotter. That's good for practice, you could do that.

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Reply 1821
Original post by revelry26
I'm not that good at those either to be honest. My exam board syllabus doesn't contain graph sketching at all so I pretty much taught myself.:P I solved the c1 c2 c3 module textbook lessons for graphs to get the basics. I used to do this thing where my friend gave me a function,I used to sketch the graph and then check using a graph plotter. That's good for practice, you could do that.

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i thinnk i got the 15th one .i'll just make sure its right and type it here
Reply 1822
Original post by revelry26
The calculus ones are pretty easy so far. The others are interesting. I'm still not convinced with my solution for the fifteenth one. :-/

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Since the fifteenth one was being discussed so much, I guess I'll give it a shot. I don't feel all that comfortable integrating, to be honest.
Original post by kapur
i thinnk i got the 15th one .i'll just make sure its right and type it here


Awesome!

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Reply 1824
Original post by TheFuture001
Hey, I'm trying the Mathematics Interview Questions, was wondering if I could get a hint at differentiating x^x. I'm fairly sure you can't differentiate as normal, I'm tempted to differentiate in the same manner you differentiate exponential functions; however there are two variables. I can see that as x increases the function increases very ​quickly, similar to an exponential but far faster. Tip?


Also, just a question relating to the MAT. When marking the multiple choice, are any marks at all given for working?


the way I did it when I first saw the question was to rewrite x^x as e^ln(x^x) and then rewrite it as e^xln(x) which isn't too hard to differentiate using the product and chain rules.
Reply 1825
q .15 1/(1-lnx) =-x/-x(1-lnx)
take 1-lnx=t
-1/x=dt
also since 1-lnx=t.. 1-t=lnx thus e^(1-)t=x
then youre left with the integral of -e^(1-t) which is e^(1-t) +constant
replace t with 1-lnx
you have e^lnx=x
ANS x + constant :biggrin:
(edited 10 years ago)
Original post by kapur
1/(1-lnx) =-x/-x(1-lnx)
take 1-lnx=t
-1/x=dt
also since 1-lnx=t.. 1-t=lnx thus e^(1-)t=x
then youre left with the integral of -e^(1-t) which is e^(1-t) +constant
replace t with 1-lnx
you have e^lnx=e
ANS e + constant :biggrin:


I did the same thing but I made a stupid mistake with the value of x . Silly me:colondollar:
Shouldn't it be -e + c though?

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Reply 1827
Original post by revelry26
I did the same thing but I made a stupid mistake with the value of x . Silly me:colondollar:
Shouldn't it be -e + c though?

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it becomes positive after integrating it
when isi your interview btw?
Original post by kapur
it becomes positive after integrating it
when isi your interview btw?


Oh yes!
I haven't received one and I'm expecting a rejection ti be honest:colondollar: I'm just doing these for fun. When is yours?

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Reply 1829
Original post by kapur
q .15 1/(1-lnx) =-x/-x(1-lnx)
take 1-lnx=t
-1/x=dt
also since 1-lnx=t.. 1-t=lnx thus e^(1-)t=x
then youre left with the integral of -e^(1-t) which is e^(1-t) +constant
replace t with 1-lnx
you have e^lnx=e
ANS e + constant :biggrin:


Original post by revelry26
I did the same thing but I made a stupid mistake with the value of x . Silly me:colondollar:
Shouldn't it be -e + c though?

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Weird. That's the first substitution I tried but I got something else. :confused:
Let me check my working once, must have messed something up.
Reply 1830
Original post by revelry26
Oh yes!
I haven't received one and I'm expecting a rejection ti be honest:colondollar: I'm just doing these for fun. When is yours?

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i hope you dont mind cross checking a couple of
answers? :P
how did you do 9? and id you get 3/5 for 8?
Reply 1831
Original post by kapur
q .15 1/(1-lnx) =-x/-x(1-lnx)
take 1-lnx=t
-1/x=dt
also since 1-lnx=t.. 1-t=lnx thus e^(1-)t=x
then youre left with the integral of -e^(1-t) which is e^(1-t) +constant
replace t with 1-lnx
you have e^lnx=e
ANS e + constant :biggrin:


Your reasoning is a bit off, it looks like you missed the original u that you had substituted. It actually should be,

xx(1lnx)dx=1xtdt\int \frac{-x}{-x(1-\ln x)}dx = \int \frac{1}{-xt} dt

Substituting for x=e^(1-t) thus gives,

t1et1du\int t^{-1}e^{t-1} du

Which is a non-elementary integral, as it simplifies to form with the exponential integral. Also on a related note, the second last-line is also incorrect as e^{ln x} = x.
Reply 1832
Original post by kapur
q .15 1/(1-lnx) =-x/-x(1-lnx)
take 1-lnx=t
-1/x=dt
also since 1-lnx=t.. 1-t=lnx thus e^(1-)t=x
then youre left with the integral of -e^(1-t) which is e^(1-t) +constant
replace t with 1-lnx
you have e^lnx=e
ANS e + constant :biggrin:

(e + constant)'=e I do not understand your method.....
Reply 1833
Original post by ctrls
Your reasoning is a bit off, it looks like you missed the original u that you had substituted. It actually should be,

xx(1lnx)dx=1xtdt\int \frac{-x}{-x(1-\ln x)}dx = \int \frac{1}{-xt} dt

Substituting for x=e^(1-t) thus gives,

t1et1du\int t^{-1}e^{t-1} du

Which is a non-elementary integral, as it simplifies to form with the exponential integral. Also on a related note, the second last-line is also incorrect as e^{ln x} = x.

So when we are interviewed those type of questions, we do not have to give the final answer, right?
(edited 10 years ago)
Reply 1834
Original post by ctrls
Your reasoning is a bit off, it looks like you missed the original u that you had substituted. It actually should be,

xx(1lnx)dx=1xtdt\int \frac{-x}{-x(1-\ln x)}dx = \int \frac{1}{-xt} dt

Substituting for x=e^(1-t) thus gives,

t1et1du\int t^{-1}e^{t-1} du

Which is a non-elementary integral, as it simplifies to form with the exponential integral. Also on a related note, the second last-line is also incorrect as e^{ln x} = x.

derivative of 1-lnx = -1/x
so answer is X + c

haha that was silly mistake ! spry . e^lnx=x my bad
(edited 10 years ago)
Would CS applicants be asked 'graph sketching' or 'calculus' problems?
Reply 1836
Original post by yxcai
(e + constant)'=e I do not understand your method.....

made a mistake in the last line my bad ! ans is x + constant
Reply 1837
Original post by kapur
derivative of 1-lnx = -1/x
so answer is X + c

haha that was silly mistake ! spry . e^lnx=x my bad

What? How do you get x+c? I still don't follow.
Reply 1838
Original post by souktik
What? How do you get x+c? I still don't follow.

which step exactly do you think is incorrect?
Reply 1839
Original post by kapur
which step exactly do you think is incorrect?

You're integrating -e^(1-t).t^(-1)dt, right? What do you do after that?
(edited 10 years ago)

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