onur369's Methods Question Thread :)

[onur369]

Hi guys, I need some clearing up to do...
No matter how many times I 'attempt or read' these types of questions I do not know what to do.
For example:
For the simultaneous linear equations
mx + 12y = 12
3x + my = m
find the value(s) of m for which the equations have
i. a unique solution
ii. infinitely many solutions.
ii. no solution

I do not know what to do for each, any help will be appreciated.

[luken93]

Rearrange to let y = for both.

From there;
unique solution = intersect once = different gradient, any c value.
infinitely many = always intersect = same gradient, same c value (same line)
no solution = never intersect = same gradient, different c (parallel)

[onur369]

Thanks for that, I think I understood the first two, just having trouble with the last one.

i.e: The simultaneous linear equations (m-2) x + 3y = 6 and 2x + (m - 3) y = m - 1 have no solution for

[luken93]

(m-2) x + 3y = 6               =>    y = -(m - 2)x/3 + 2
2x + (m - 3) y = m - 1       =>    y = -2x/(m - 3) + (m - 1)/(m - 3)

For it to have no solution, they must be parallel lines, ie the gradient must be the same:
-(m - 2)/3 = -2/(m - 3)
-(m - 2)(m - 3) = -6
-m^2 + 5m - 6 + 6 = 0
-m(m - 5) = 0
m = 0, m = 5

Now, we must also check to see that they are NOT the same line, so we must ensure that the 2 c values are not the same.
2 =/= (m - 1)/(m - 3)
2m - 6 =/= m - 1
m =/= 5

Therefore, we can't let it equal 5, because that would make it the same line. Hence sub in 0 to check the eq:
=>    y = 2x/3 + 2
=>    y = 2x/3 + 3

The two lines are parallel, but not the same line, hence they will never intersect.

[onur369]

Hi, Im just stuck with a probability question, having not done probability in yr11 is a b%#%$ch
Q: A kindergarten teacher has found over the years that 25% of children can tie their shoelaces, and 30% can use a pair of scissors, and 18% can do both. Find the probability that a randomly selected child:
a) can neither ties their shoelaces nor use scissors
b) can use scissors, but cannot  tie their shoelaces

for a) i tried doing 0.7x0.75= 0.525, but the solution is 0.63.

[luken93]

Pr(Shoe) = 0.25
Pr(Scissors) = 0.30
Pr(Shoe n Scissors) = 0.18

Since Pr(Scissors u Shoes) = 0.30 + 0.25 - 0.18 = 0.37
Neither = 1 - Pr(Scissors u Shoes) = 1 - 0.37 = 0.63

b) Pr(Scissors only) = Pr(Scissors) - Pr(Scissors n Shoes) = 0.30 - 0.18 = 0.12

[kaushik]

Hint try a karnaugh map. It should help you to understand the associated probabilities. 

[onur369]

Let g: R → R, g(x) = x2.
Show that g(u + v) + g(u – v) = 2(g(u) + g(v)).

How do i do this ?

[tony3272]

and

Let LHS=

[onur369]

cheers
this question is really weird for me :s:
Show that the graph of h(x)= x^n/e^x , where n is a positive integer, has a local maximum at x=n.

[b^3]

First differentiate using quotient rule.

Take an e^x out

Now let this equal zero to find maximum and minimums

so


Then show it is a maximum by subbing in terms to the left and right, i.e. n-1 and n+1, as n is postive, you can should that the left side will have a positive gradient and right side has a negative gradient, hence a maximum.

The gradient on the left at n-1 is
As n is a postive interger, n-1 is also positive and n/n-1 is greater than 1 so the gradient on this side is positive
Now on the right at n+1 the gradient is
As n+1 is greater than n, n/n+1 is a fraction smaller than one and greater than zero, so if you take the one away, you get a negative. The other terms are still positive so overalll it multiplies to a neagtive.

There is proabably a more mathematical way to write this out, so if anyone wants to add to it (and well do it the proper way) feel free to.

EDIT: added last bit of working.

[onur369]

wouldnt that last step take wayy to long for a 3mark question

[b^3]

Yeh probably, and I just editted the post to show that bit of working, looks longer now. I think there is a more mathemical way of showing it but I'm not entirely sure.

[onur369]

Hi guys, got stuck with a reallly basic related rates question

[b^3]

NOTE: change x to h (sorry)

a)

Now you need to use the ratios of the similar triangles.


Sub that in

b)
=



,h=x=5




That seems odd, let me have a check of it.