Some problems, tell me if they are easy! (1 Viewer)

[forstudents]

I found some of this quite difficult..tell me wat u think.

HATE CIRCLE GEOMETRY:burn:

Oh i got most of this just not the integration bit

ok another one because of popular demand..lol

 

[Riviet]

For the integral one, think of y=f(x) as the gradient function of your new curve.

For the first part in the circle geometry question, ^PQS = ^PRS (angles in the same segment theorem) and you are given ^SPR = ^QPK, therefore the triangles are similar.

 

[pLuvia]

Where did you get these questions?

Anyway, for the first one
Let P(x)=(x-k)²Q(x), since k is a double root. Then differentiate it and you should see that when you sub in x=k, the P'(x)=0. Hence k is a root of multplicity 2. For that second part, sub in y=1, and you should see P(y)=0, differentiate P(y), then sub in y=1 and it should also equal to 0

 

[housemouse]

Are these questions from the Prior exam cause I think Ive seen these in them especially that multiplicity question as well as the circle geometry?

 

[Riviet]

For the circle geometry part ii), ^PSQ = ^PRQ (angles in the same segment) and ^HPK is common to ^SPK and ^HPQ, therefore the triangles are similar.

 

[forstudents]

Interesting..thanx for replying guys appreciate the help..hope u gain something out of this aswell.

 

[forstudents]

i think the first two are from the trials..the ones i din get i posted up. The other one is from a revision sheet.

 

[Riviet]

Are these questions from the Prior exam cause I think Ive seen these in them especially that multiplicity question as well as the circle geometry?

I have never seen any of these questions other than the multiplicity one, which can be commonly asked in exams anyway.

For part iii) in the circle geometry question, we use matching sides of similar triangles in proportion from part i) and ii):
PQ/PR = PK/PS = QK/RS (i)

PQ/PK = QR/KS = PR/PS (ii)

RHS = PR.QS
=(PS.PQ/PK)(KS + QK)
=(PS.PQ/PK)(PK.QR/PQ + RS.PK/PS)
=PS.QR + PQ.RS as required.

 

[zeek]

Solutions to Question B

Solutions

 

[little master]

Re: Solutions to Question B

i was having trouble with this question .. was wondering if anyone could help me with this...

integral of Ln [(sqrt(x^2 +1) + x)^2]

thanks

 

[pLuvia]

Re: Solutions to Question B

Tried using integrating by parts?
Let u=ln[{sqrt(x²+1)+1}²] dv=dx

 

[little master]

Re: Solutions to Question B

oh i am sry the question was:
integral of Ln [(sqrt(x^2 +1) + <B>x</B>)^2]


i tried using integration by parts but .. i dont kno whether it gives a correct answer. i got an answer in the end but i think i am way off.

btw i got:

2x^2sqrt(x^2 +1) + sqrt(x^2 +1) + 2x(x^2 +1)

 

[vafa]

This is the solution for previous one the one with 1 not with x
View attachment 13611

View attachment 13612

 

[vafa]

This is the solution for the new one the one with x not with 1
View attachment 13613

 

[vafa]

are these solutions of two integrations acceptable and understandable?

 

[little master]

Yeah .. they are perfectly fine... i realise my mistake

thanks

 

[zeek]

lol i love the way how vafa is so formal