VCE Methods Question Thread!

[luffy]

But y/where did you know that you had > and <

Couldn't I just divide both sides by (4x^2-1)^5 to get

(4x^2-1) > 1

You can, but you have to realize that might be <0. If that is the case, the sign is reversed.

Hence, 4x^2- 1 > 1 WHEN (4x^2 - 1)^5 > 0 (Solve for both and find the intersection).

OR 4x^2 - 1 < 1 WHEN (4x^2 - 1)^5 < 0 (Solve for both and find the intersection).

That is essentially what xZero did.

[ICECOLD]

Nice. Thanks.

[xZero]

and if (4x^2-1)=0, dividing both side by it will get 0/0 which is undefined

[ashoni]

10. If the graphs y=x-6 and y=x^2-mx+10 intersect at one point only, then m must equal :
A 7 or -9
B -7 or -9
C -7 or 9
D 7 or 9
E 9

7. For what values of k does the following quadratic equation have just one solution?
   5kx^2-6kx+k+4=0

[pi]

10. If the graphs y=x-6 and y=x^2-mx+10 intersect at one point only, then m must equal :
A 7 or -9
B -7 or -9
C -7 or 9
D 7 or 9
E 9

7. For what values of k does the following quadratic equation have just one solution?
   5kx^2-6kx+k+4=0

10. If they intersect once, the discriminant of x^2 - mx + 10 = x-6, must equal 0. Solve from there
(m+1)^2 - 4(16)(1) = 0
m = -9 or m = 7
A


7.  Same principle, discriminant = 0
(6k)^2 - 4(5k)(k+4) = 0
k = 0 or k = 5

[ashoni]

ohhhh okay! thanks heaps!
one more question....
10a) What is the smallest positive integer value k can take for which the equation x^2+(k-2)x-4k=0 had two rational solutions?
    b) For the value of k calculated in part (a), solve the equation.
    c) Solve the equation when k=8

thanks!

[pi]

Just hints this time

10. a) Same principle as earlier, except this time, find when the discriminant > 0. Find the smallest INTEGER solution from there (using that domain, maybe trial and error might help here)
      b) Just solve it
      c) Just solve it with k=8, use quadratic eq. if you ceebs alternate routes

[ashoni]

hmmm, i still dont get it D:
could you explain in more depth..?

[tiluu]

Hey, this might be easy but I've just forgotten the rule..

I'm solving the inverse of f:R->R, where f(x)=4e^(x-3) , which means I swap the values. Then it becomes x=4e^(y-3).. how do I make y the subject? I'm not sure about the log laws.. ehehe.

Sorry! But thanks. 

[Phy124]

y = 4e^(x-3)

for inverse swap x and y

x = 4e^(y-3)

(x/4)=e^(y-3)

ln(x/4) = y - 3

y = ln(x/4) + 3

Using a = e^(b)

b = ln(a)

[tiluu]

y = 4e^(x-3)

for inverse swap x and y

x = 4e^(y-3)

(x/4)=e^(y-3)

ln(x/4) = y - 3

y = ln(x/4) + 3

Using a = e^(b)

b = ln(a)

I got it right after I posted that. Lol, I must've been tired. ~_~
Thanks btw!

[Kanon]

Afternoon guisez!1!
Firstly, thank you all for helping everyone with this, it's a hugely helpful thing for me personally, and I'm sure other people feel the same way

Anyway, two theory based questions.
What's the point in finding the Inverse of a Matrix when solving eqns with Matricies?  Is it because the determinant of a Matrix can only be found in a square matrix?  Therefore, anything that's not in a square matrix we have to use the Inverse of a Matrix?

When not allowed to use calculus, will the mid-point formula work if you have two points with the same y value?

[brightsky]

err..not quite. the inverse of a matrix A, denoted by A^(-1), can only be found for regular square matrices (i.e. square matrices whose determinant does not equal 0).

when you have a matrix equation such as AX = B. in order to get X = some junk, we need to be able to move the matrix A over to the right hand side. now, in normal arithmetic, you'll recall that we can simply divide both sides by A. however, this is not a valid move in matrix arithmetic. there is however an equivalent. by pre-multiplying both sides by A^(-1), you are effectively dividing everything by A, since now you get A^(-1)AX = A^(-1)B, and by definition A^(-1)A = I, which is the matrix equivalent of 1, and so X = A^(-1)B.

you are right in saying that the determinant of a matrix can only be found in a square matrix. not sure what you mean by "anything that's not in a square matrix".

don't quite understand your second question...

[Phy124]

When not allowed to use calculus, will the mid-point formula work if you have two points with the same y value?

You mean something like (2,6) and (4,6)?

The midpoint will have the same y-value as the two points - (6 + 6)/(2) = 6 and of course the x value will be 3 ((2 + 4)/(2) = 3)

I assume by midpoint formula you meant ((x₂ - x₁)/2, (y₂ - y₁)/2)

tl;dr - yes

[Mr. Study]

Just about this graph.

If we had the equation y=1/(|x|+2), Why exactly is it shaped like this. (Picture below)

Because if it was y=1/|x| it looks like a truncus and I understand why that is but since a +2 is added, it changes shapes into the one below.