VCE Methods Question Thread!

[gameboy99]

I have been stuck on two log questions 9 and 10 for a long time. Can somebody please show working out to get the answer?
Here are the two questions , http://imgur.com/a/wPjmf
Here are the answers to the two questions, http://imgur.com/a/KuFfD
I would appreciate it a lot if someone can help.
Thanks 

[clarke54321]

Hi everyone,

 Could I please have some help with the following?

For question 2, I only require help for part b.

For question 3, I don't know why there is a square root on the top also.

For question 4, I require help for b-d.

Thanks so much!!! 

[zhen]

Hi everyone,

 Could I please have some help with the following?

For question 2, I only require help for part b.

For question 3, I don't know why there is a square root on the top also.

For question 4, I require help for b-d.

Thanks so much!!! 

Here's what I got for 1, 2 and 3. Someone else should check my answers, since I might not be right.

[deStudent]

For this http://m.imgur.com/hxoZceZ I'm getting C but the answer says D?

Thanks

[zhen]

For this http://m.imgur.com/hxoZceZ I'm getting C but the answer says D?

Thanks

I'm getting C too. I don't know what's wrong with it.

[Shadowxo]

For this http://m.imgur.com/hxoZceZ I'm getting C but the answer says D?

Thanks

The correct answer is C

Hi everyone,

 Could I please have some help with the following?

For question 2, I only require help for part b.

For question 3, I don't know why there is a square root on the top also.

For question 4, I require help for b-d.

Thanks so much!!! 

For 4, first you need to find the initial mass (mass at t=0) which is 2*e⁰=2
b) you want to find when it's half of M0 (initial mass), = half of 2 = 1
1=2e^-0.2t and solve
same method for c
d) rate of decay is change in mass over change in time - dm/dt
find the derivative:
dm/dt = -0.2*2e^-0.2t
m = 2e^-0.2t therefore
dm/dt = -0.2m

[Shadowxo]

I have been stuck on two log questions 9 and 10 for a long time. Can somebody please show working out to get the answer?
Here are the two questions , http://imgur.com/a/wPjmf
Here are the answers to the two questions, http://imgur.com/a/KuFfD
I would appreciate it a lot if someone can help.
Thanks 

Hi, question 9 looks a little bit hard - do you have the relevant formula?
Formula is log_ab=1/log_ba
Derivation: Change of base rule
log_ab=log_rb/log_ra
if r=b
log_ab=1/log_ba

so log₅x=16*log_x5
1/log_x5=16*log_x5
1=16(log_x5)²
1/16=(log_x5)²
log_x5=±1/4
x^1/4=5            or    x^-1/4=5
x=^{5^4}=625            x=5^-1/4
                              x=1/5⁴=1/625

10. just taking the log of both sides
log₅(q^p)=log₅25
p*log₅q=2
log₅q=2/p

[deStudent]

http://m.imgur.com/a/O7Icq

For f)I) the answer has nsin(Pi/n)cos(Pi/n). I'm not sure what to do but I'm abit confused on where their cos(pi/n) came from?

Thanks

[MattBro]

Can some one help me with this:
Sketch the graph of f: R \ {0} -> R, f(x)=1/x+1/(x2) using addition of ordinates
Thanks

[Shadowxo]

http://m.imgur.com/a/O7Icq

For f)I) the answer has nsin(Pi/n)cos(Pi/n). I'm not sure what to do but I'm abit confused on where their cos(pi/n) came from?

Thanks

The area of a triangle is 1/2 b h
For the half-triangle, you know the angle is π/n, and the radius/hypotenuse is 1
base = rsinx = 1*sin(π/n)
height = rcosx = 1*cos(π/n)
So the area = 1/2 *sin(π/n)*cos(π/n)
Continuing on from there, area of full triangle = sin(π/n)*cos(π/n) and area of full polygon = n*sin(π/n)*cos(π/n)

I think your mistake may have been using the radius (1) instead of the height (1cos(π/n))

Can some one help me with this:
Sketch the graph of f: R \ {0} -> R, f(x)=1/x+1/(x2) using addition of ordinates
Thanks

For this, graph 1/x and also graph 1/x², then for y=1/x+1/x², add the graphs to produce this graph (you add the two y values from each of the graphs to find the final y values)

[clarke54321]

Could someone please explain to me why x=0 isn't also a stationary point?

For the second question, I just need some steps shown, as to how I can start it off. It's getting pretty messy for me.

Thanks very much 

[QueenSmarty]

Can I please get some help with these questions?
1. Show that (k+1)x^2-2x-k=0 has a solution of all values of k. (In the answers, it says we need to show that the discriminant is greater than 0. I'm having trouble understanding that. Why do we have to show that the discriminant is greater than 0 instead of showing that it's equal to 0?)

2. Factorise (a-b)^3+(a+b)^3.

3. Find 'm' for which y=mx and y=1/x +5 touch, and give coordinates. (I've found 'm' when the discriminant equals 0, not sure where to go from there)

4. Given that y=2x^3-6x^2+18x, find dy/dx, hence show that dy/dx>0 for all x.

5. a) Show that f:R-->R, f(x)=x^3 is a strictly increasing function for R by showing that f'(x)>0 for all non-zero x, and showing that, if b>0, then f(b)>f(0) and if 0>b, then f(0)>f(b)
b) Show that f:R-->R, f(x)=-x^3 is a strictly decreasing function for R

Thank you!

[Samuel1130]

Can someone please help me with this question!!
If A(4,0) and B(0,-8) the equation of a perpendicular bisector is AB is??

[Shadowxo]

Could someone please explain to me why x=0 isn't also a stationary point?

For the second question, I just need some steps shown, as to how I can start it off. It's getting pretty messy for me.

Thanks very much 

For 10, first you find the derivative


I don't think you posted the full question for the other one.

Can someone please help me with this question!!
If A(4,0) and B(0,-8) the equation of a perpendicular bisector is AB is??

You know it's a line - what you need in order to find the line is the gradient, and a point on the line
The gradient of the line joining AB is (0+8)/(4-0)=2
So the gradient of the perpendicular bisector is -1/2 (the lines are perpendicular, m1*m2=-1)
The line passes through the midpoint of the two points (bisector - splits the line connecting them into two equal parts therefore it passes through the midpoint)
The midpoint is ( (4+0)/2 , (0-8)/2) = (2, -4)
From here you can use y=mx+c or y-y1=m(x-x1) (I prefer this one)
y-(-4)= -1/2 *(x-2)
y=-x/2-3

[Shadowxo]

Can I please get some help with these questions?
1. Show that (k+1)x^2-2x-k=0 has a solution of all values of k. (In the answers, it says we need to show that the discriminant is greater than 0. I'm having trouble understanding that. Why do we have to show that the discriminant is greater than 0 instead of showing that it's equal to 0?)

2. Factorise (a-b)^3+(a+b)^3.

3. Find 'm' for which y=mx and y=1/x +5 touch, and give coordinates. (I've found 'm' when the discriminant equals 0, not sure where to go from there)

4. Given that y=2x^3-6x^2+18x, find dy/dx, hence show that dy/dx>0 for all x.

5. a) Show that f:R-->R, f(x)=x^3 is a strictly increasing function for R by showing that f'(x)>0 for all non-zero x, and showing that, if b>0, then f(b)>f(0) and if 0>b, then f(0)>f(b)
b) Show that f:R-->R, f(x)=-x^3 is a strictly decreasing function for R

Thank you!

1. It'll have a solution if the discriminant =0 also. It may have just been saying that in this case, the discriminant will always be greater than 0 therefore it has a solution for all values of k.
2. Should be able to use standard rules for factorising an expression like this


3. Point of intersection: mx=1/x+5
mx²=1+5x
mx²-5x-1=0
If they touch then there's only one solution, one point of intersection as it only touches the other line, doesn't pass through
discriminant =0
25+4m=0
m= -25/4
Then just plug it in to find the point of intersection - quadratic formula
x=(5±√0)/2m = 5/(-25/2)=-2/5
y=mx = -25/4 * -2/5 = 5/2

4. standard derivative,
dy/dx = 6x²-12x+18 = 6(x²-2x+3)
Discriminant = 4-4*3*1 = -8
Therefore no x intercepts, and as it's a positive parabola that means dy/dx>0 for all x.

5.a) f'(x)=3x² which is always ≥0, and is > 0 when x≠0
f(0)=0, and is strictly increasing therefore if b>0, f(b)>0 therefore f(b)>0
If b<0, f(b)<0 as it's strictly increasing, therefore f(b)<f(0)
b) Same thing, f'(x)=-3x² which is ≤0 for all x, and <0 when x≠0. Therefore it is strictly decreasing.