2u Mathematics Marathon v1.0 (2 Viewers)

[Riviet]

Check your answer, darkliight, you forgot the constant at the front.

 

[pLuvia]

y = sqrt(xtan(x))

dy/dx = 1/2(xtanx)^-1/2*(tanx+xsec²x)
= [tanx+xsec²] / [2√xtanx]

Add a question later

 

[kido_1]

Challenge Question

Hey heres a question for use guys:
Y= ||||||x-1|-2|-3|-4|-5|-1|-2|
a) Sketch the following.
b) Find the value of a that the function above has 6 solutions.

*Its supposed to be challenging!!!

 

[Mountain.Dew]

Re: Challenge Question

Hey heres a question for use guys:
Y= ||||||x-1|-2|-3|-4|-5|-1|-2|
a) Sketch the following.
b) Find the value of a that the function above has 6 solutions.
(a)
*Its supposed to be challenging!!!

okay, Y= ||||||x-1|-2|-3|-4|-5|-1|-2| = |x-1||-2||-3||-4||-5||-1||-2|
=|x-1|(2*3*4*5*1*2) = 240|x-1|

so y = 240|x-1|
(b) uhhhh...where's "a" in the equation?

 

[Slidey]

Re: Challenge Question

Mountain.Dew, ||x-a|-b| does not equal |x-a||b|.

 

[Mountain.Dew]

Re: Challenge Question

Mountain.Dew, ||x-a|-b| does not equal |x-a||b|.

mmmmmmm true true...but i wonder, what is the mathematical proof for that?

and i postulate that kido_1's question is not really 2U lvl, is it? better to pose the question in the 4U maths forum.

but of course, anyone who is able to answer this question here will be greatly appreciated!

 

[Nerd Queen]

Re: Challenge Question

mmmmmmm true true...but i wonder, what is the mathematical proof for that?

you shouldn't need mathematical proof to show that

||x-a|-b| doesn't equal |x-a||b|

as they are not asking the same thing ... the first tells you to find the absolute value of b being taken away from the absolute value of x-a

the second is the absolute value of x-a times the absolute value of b ... as u can see two completely different things ...

as for the actual question ... havent seen it but i doubt i can solve it if i do ... if its as hard as all that

 

[Riviet]

Re: Challenge Question

Hey heres a question for use guys:
Y= ||||||x-1|-2|-3|-4|-5|-1|-2|
a) Sketch the following.
b) Find the value of a that the function above has 6 solutions.

*Its supposed to be challenging!!!

I won't be providing a sketch, but I'll verbally sketch the graph for you, please use your imagination. :

Spoiler

Imagine the graph of y=|x-1|. Now shift it down 2 units, then reflect all parts of the curve below the x-axis in the x-axis.
Using this new reflected graph, shift it down by 3 units, then reflect all parts of this new curve below the x-axis in the x-axis.
Using this new reflected graph, shift it down by 4 units, then reflect all parts of this new curve below the x-axis in the x-axis.
Using this new reflected graph, shift it down by 5 units, then reflect all parts of this new curve below the x-axis in the x-axis.
Using this new reflected graph, shift it down by 1 units, then reflect all parts of this new curve below the x-axis in the x-axis.
Using this new reflected graph, shift it down by 2 units, then reflect all parts of this new curve below the x-axis in the x-axis.

I think you get the idea of the graph now. XD
Now get sketching!

 

[SoulSearcher]

Here's a sketch of the graph that you should get.

 

[SoulSearcher]

You have got to love computers at a time like that.

Hmm next question:

Solve the equation 4x³ - 12x² + 9x - 2 = 0 if two if the roots are equal.

3 unit polynomials, but a hint for those at the 2 unit level

Spoiler

for the general cubic equation
ax³ - bx² + cx - d = 0, where the roots of the equation are α, β and γ,
α + β + γ = - b / a
αβ + αγ + βγ = c / a
αβγ = - d / a

 

[Riviet]

Soulsearcher, feel free to post up the next question. ;D

 

[kido_1]

Yes soul searcher your graph is correct for part a, greatly appreciate your effort. Part b is intended to mean, how much does it have to be shifted down again in order to obtain 6 solutions.

 

[Mountain.Dew]

riviet, ur method of reasoning of the graph was brilliant. i bow down before you *pays homage to riviet*

Here's a sketch of the graph that you should get.

wow very nice SoulSearcher! excellent work!

by looking at the graph, i dont think there is a value to shift the graph down (or up) to obtain 6 solutions, providing that solutions = x-intercepts.

 

[pLuvia]

Re: Challenge Question

Nice work Riviet

 

[pLuvia]

Spoiler

P(x)=4x³ - 12x² + 9x - 2 = 0

(x-2) is a factor using factor theorem

Using division transformation

P(x)=(x-2)(4x²-4x+1)
= (x-2)(2x-1)(2x-1)

.: x=2,1/2,1/2

4unit Method:

Since two roots are the same hence P(x) has a double root
P(x)=4x³ - 12x² + 9x - 2 = 0
P'(x)=12x²-24x+9 = 0
(6x-3)(2x-3)=0
x=1/2,3/2

P(1/2)=0
P(3/2)≪0
.: x=1/2 is a double root hence

P(x)=(2x-1)²Q(x)

From observation
Q(x)=(x-2)
Hence roots are x=1/2, 1/2, 2

 

[sando]

what program did u use to draw the graph like that... i need sumthin like that to help me understand trig functions ?

 

[pLuvia]

what program did u use to draw the graph like that... i need sumthin like that to help me understand trig functions ?

http://www8.pair.com/ksoft/

Try this one

 

[SoulSearcher]

I used a program that I bought a few years ago. Studyworks Mathematics its called.

 

[pLuvia]

*Bump*

If the Kth term of an arithmetic series is L, and the Lth term is K:
(i) Show L=a+(K-1)d
(ii) Show another expression for K
(iii) By solving the two equations founded, show d=-1
(iv) Hence, find the first term of L and K

 

[SoulSearcher]

I'm bored, so I'll give it a shot.

Spoiler

(i) T_n = a + (n-1)d
now since the Kth term of an arithmetic series is L, then T_K = a + (K-1)d
which is L = a + (K-1)d ... (1)
(ii) now the Lth term of the arithmetic series is K, so K = a + (L-1)d ... (2)
(iii) you have L = a + (K-1)d ... (1)
and K = a + (L-1)d ... (2)
(1) - (2)
L - K = (K-1)d - (L-1)d
L - K = d[ K - 1 - L + 1 ]
L - K = d[ K - L ]
(L-K) / (K-L) = d
-(K-L) / (K-L) = d
-1 = d
therefore d = -1
(iv) for this part you have to find the first term, otherwise known as a
substitute d = (1)
L = a + (K-1) * -1
L = a - (K-1)
L = a - K + 1
a = L + K - 1
therefore the first term of the arithmetic series is L + K - 1