2u Mathematics Marathon v1.0 (2 Viewers)

[pLuvia]

Nothing to do might as well answer it:

Spoiler

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.]

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

 

[Sober]

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

 

[SoulSearcher]

Actually, to be pedantic:

ln(-x) = ln(x) + π(2k+1)i

Where k is some integer and i²= -1

How often would you see that in a 2 unit question though?

 

[Mc_Meaney]

never!!!...i hope

 

[Riviet]

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.

 

[Mc_Meaney]

Next question...?

 

[SoulSearcher]

Here's a question to start this thread up again.

Differentiate y = e^x from first principles and use the result y' = e^x to show that lim (h->0) (e^h - 1)/ h = 1

 

[the pospinator]

i got it
y = e<SUP>xfficeffice" /><O></O></SUP>
using first principles, <O></O>
y’ = lim e<SUP>x+h </SUP>- e<SUP>x<O></O></SUP>
<SUP>(h->0) </SUP>h<O></O>
= lim e<SUP>x</SUP>(e<SUP>h </SUP>– 1)<SUP><O></O></SUP>
<SUP>(h->0) </SUP>h<O></O>
= e<SUP>x</SUP> * lim e<SUP>h</SUP> - 1<O></O>
<SUP>(h->0) </SUP>h<O></O>
(not finished)

 

[the pospinator]

but y’ = e<SUP>x<?xml:namespace prefix = o ns = "urn:schemas-microsoft-comfficeffice" /><o></o></SUP>
therefore <o></o>
e<SUP>x</SUP> = e<SUP>x</SUP> * lim e<SUP>h</SUP> - 1<o></o>
<SUP>(h->0) </SUP> h<o></o>
therefore <o></o>
1 = lim e<SUP>h</SUP> - 1<o></o>
<SUP>(h->0) </SUP> h<o></o>

 

[the pospinator]

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" oreferrelative="t" o:spt="75" coordsize="21600,21600"> ,</v:shapetype> oh well<?xml:namespace prefix = o ns = "urn:schemas-microsoft-comfficeffice" /><o></o>
<o></o>
here goes,<o></o>
y=x⁴ + 2x³-3x²+4x-1. Solve finding roots, turning points and inflection points. Also, graph the eqn (graph is optional)<o></o>
<o></o>
have fun!!

 

[SoulSearcher]

Just one thing, don't forget to use the spoiler tags when answering the question, so that other people may try it too , i.e. these tags [spoiler ][ /spoiler], without the spaces.

 

[Riviet]

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

 

[pLuvia]

Spoiler

(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]

Spoiler

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

 

[Sober]

You forgot that √(x²) = |x|

 

[Riviet]

Also don't forget to post up a new question, Templar.

 

[Templar]

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

 

[angmor]

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]

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

Spoiler

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