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2u Mathematics Marathon v1.0 (1 Viewer)

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pLuvia

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Nothing to do might as well answer it:
2ln(x)-ln(14x+1)=0
lnx2-ln(14x+1)=0
ln[x2/(14x+1)]=0
x2/(14x+1)=1
x2=14x+1
x2-14x-1=0
x=[14+sqrt{200}]/2
=[14+10sqrt{2}]/2
=7+5sqrt{2}
 

word.

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one thing to be careful of: ln(x) is undefined for x <= 0
 

Sober

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word. said:
one thing to be careful of: ln(x) is undefined for x <= 0
Actually, to be pedantic:

ln(-x) = ln(x) + π(2k+1)i

Where k is some integer and i2= -1
 

Riviet

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You can count on it that the examiners won't put anything like that in a 2u paper. Anything to do with complex numbers and i is explicitly in the 4u course.
 
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i got it
y = e<SUP>xffice:eek:ffice" /><O:p></O:p></SUP>
using first principles, <O:p></O:p>
y’ = lim e<SUP>x+h </SUP>- e<SUP>x<O:p></O:p></SUP>
<SUP>(h->0) </SUP>h<O:p></O:p>
= lim e<SUP>x</SUP>(e<SUP>h </SUP>– 1)<SUP><O:p></O:p></SUP>
<SUP>(h->0) </SUP>h<O:p></O:p>
= e<SUP>x</SUP> * lim e<SUP>h</SUP> - 1<O:p></O:p>
<SUP>(h->0) </SUP>h<O:p></O:p>
(not finished)
 
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but y’ = e<SUP>x<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:eek:ffice:eek:ffice" /><o:p></o:p></SUP>
therefore <o:p></o:p>
e<SUP>x</SUP> = e<SUP>x</SUP> * lim e<SUP>h</SUP> - 1<o:p></o:p>
<SUP>(h->0) </SUP> h<o:p></o:p>
therefore <o:p></o:p>
1 = lim e<SUP>h</SUP> - 1<o:p></o:p>
<SUP>(h->0) </SUP> h<o:p></o:p>
 
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there u go, soulsearcher, thank u for a good question, and a sensible 1, can i remind u that this is a 2unit forum, not 4 unit, so complex numbers won't be examined, and plz don't post 4 unit questions, i can't answer them...<?xml:namespace prefix = v ns = "urn:schemas-microsoft-com:vml" /><v:shapetype id=_x0000_t75 stroked="f" filled="f" path="m@4@5l@4@11@9@11@9@5xe" o:preferrelative="t" o:spt="75" coordsize="21600,21600"> :(,</v:shapetype> oh well<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:eek:ffice:eek:ffice" /><o:p></o:p>
<o:p></o:p>
here goes,<o:p></o:p>
y=x4 + 2x3-3x2+4x-1. Solve finding roots, turning points and inflection points. Also, graph the eqn (graph is optional)<o:p></o:p>
<o:p></o:p>
have fun!!:)
 

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A big bump for this marathon.

New Question:


(i) Show that (6x+4)/(x-1) = 6 + 10/(x-1)
aaaaaaaaaaaaaaa.e
(ii) Hence evaluate ∫(6x+4)/(x-1) dx.
aaaaaaaaaaaaaaa2
 
P

pLuvia

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(i)
(6x+4)/(x-1)=6+10/(x-1)
LHS=(6x+4)/(x-1)
=6x/(x-1)+4/(x-1)
=6(x-1)/(x-1)+(4+6)/(x-1)
=6+10/(x-1)
=RHS
QED
(ii)
int.{e to 2}(6x+4)/(x-1)dx
=int.{e to 2}[6+10/(x-1)]dx
=[6x+10ln(x-1)]{e to 2}
=[6e+10ln(e-1)]-[12+10ln1]
=6e+10ln(e-1)-12

Next Question
int.sqrt{1+sin2x}dx
 

Templar

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Sqrt[1+Sin[2x]]=Sqrt[Sin2[x]+2Sin[x]Cos[x]+Cos2[x]]
=Sqrt[(Sin[x]+Cos[x])2]
=Sin[x]+Cos[x]

Integrate[Sqrt[1+Sin[2x]]]=Integrate[Sin[x]+Cos[x]]
=Sin[x]-Cos[x]+C
 

Riviet

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Also don't forget to post up a new question, Templar. ;)
 
Last edited:

Templar

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You know I don't post questions Riviet, I only solve them.
 

angmor

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Find the gradient of the secant AB to the curve y= 3x^2 - 5, if, at A, x=1, and at B, x=3.
 

SoulSearcher

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angmor said:
Find the gradient of the secant AB to the curve y= 3x^2 - 5, if, at A, x=1, and at B, x=3.
At x = 1,
y = 3 - 5
= -2, therefore point A is (1, -2)
At x = 3,
y = 27 - 5
= 22, therefore point B is (3, 22)
From here, using gradient formula,
m = (y2-y1) / (x2-x1), where point A is (x1, y1) and point B is (x2, y2)

= 22-(-2) / 3-1
= 24/2
= 12
Note: I don't really know what a secant is :p
 

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