HSC 2012 MX1 Marathon #1 (archive) (1 Viewer)

[nightweaver066]

Re: 2012 HSC MX1 Marathon

I'm not sure if I follow :/

You had the discriminant, but you let it = 0 and solved for values of k.

When the discriminant = 0, you find a quadratic equation with 1 real root.

If discriminant is > 0, you look for values of k such that it has real roots.

So you are solving the equation:
(k + 5)(k - 3) > 0

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

You had the discriminant, but you let it = 0 and solved for values of k.

When the discriminant = 0, you find a quadratic equation with 1 real root.

If discriminant is > 0, you look for values of k such that it has real roots.

So you are solving the equation:
(k + 5)(k - 3) > 0

I'm quite sure that there should be equality associated with that inequality sign.

Even one root is still a root.

 

[Examine]

Re: 2012 HSC MX1 Marathon

So it's 3>k, -5<k?

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

So it's 3>, -5<

Your notation is a bit strange, but x > 3 or x < -5.

 

[nightweaver066]

Re: 2012 HSC MX1 Marathon

I'm quite sure that there should be equality associated with that inequality sign.

Even one root is still a root.

Yeah my mistake lol.

Examine forgot to state the condition the question requires..

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

Yeah my mistake lol.

Examine forgot to state the condition the question requires..

The only condition was 'quadratic is equal to zero'.

 

[CrackerMo]

Re: 2012 HSC MX1 Marathon

Solve sinx + cosx = cos2x using the auxiliary method!

Please give me a hint ^ Thanks in advance

 

[Nira123]

Re: 2012 HSC MX1 Marathon

Hope attached solutions will be helpfull for you.




[/U]'Spirit of Maths & Physics' across wollongong region

 

[CrackerMo]

Re: 2012 HSC MX1 Marathon

Hope attached solutions will be helpfull for you.

View attachment 24397


[/U]'Spirit of Maths & Physics' across wollongong region

Thanks so much! + Rep

 

[Examine]

Re: 2012 HSC MX1 Marathon

kx^2+3x+2=0

eqn has no real roots eqn<0
3^2-4(k)(2)<0
9-8k<0
9<8k
9/8<k
k>9/8

Right/wrong?

 

[Timske]

Re: 2012 HSC MX1 Marathon

right

for no real root its discriminant < 0, discriminant = b^2 - 4ac

kx^2+3x+2=0

9-(4)(k)(2) < =
9 - 8k < 0
- 8k < -9
k > 9/8

so its correct

 

[RivalryofTroll]

Re: 2012 HSC MX1 Marathon

kx^2+3x+2=0

eqn has no real roots eqn<0
3^2-4(k)(2)<0
9-8k<0
9<8k
9/8<K
k>9/8

Right/wrong?

b^2-4ac = (3)^2-4(k)(2)
= 9 - 8k < 0 (if equation has no real roots then discriminant is less than zero)
9 < 8k
k > 9/8
Therefore, you are CORRECT?!

 

[Examine]

Re: 2012 HSC MX1 Marathon

Not sure how to tackle these question...

Solve for values of x between 0 and 360

sin(2x+15)=0.6

2cos2y=-1

Cos(1/2y+71)=-0.3420

I don't want the answers though a steer to the right direction would be nice.

 

[Timske]

Re: 2012 HSC MX1 Marathon

im pretty sure you use auxilliary method i havent really nailed those questions
Rsin(x+a)
sin(2x + 15)=0.6
2x + 15 = inverse sin(0.6)

 

[nightweaver066]

Re: 2012 HSC MX1 Marathon

Not sure how to tackle these question...

Solve for values of x between 0 and 360

sin(2x+15)=0.6

2cos2y=-1

Cos(1/2y+71)=-0.3420

I don't want the answers though a steer to the right direction would be nice.

1. You need a calculator.

You end up with 2x + 15 = ???
x =

Find the solutions between 0 to 360

2. Rearrange and solve.





3. Same approach for 1.

 

[RivalryofTroll]

Re: 2012 HSC MX1 Marathon

Not sure how to tackle these question...

Solve for values of x between 0 and 360

sin(2x+15)=0.6

2cos2y=-1

Cos(1/2y+71)=-0.3420

I don't want the answers though a steer to the right direction would be nice.

2cos2y = -1
cos2y = -1/2
Let 2y = x
cosx = -1/2
2nd Quad 180 - 60
3rd Quad 180 + 60
x = 120, 240
2y = 120, 240
2y = 120, 240, 120 + 360, 240 + 360
y = 60, 120, 240, 300

As for the others, have you tried using the expansions for Sin(A+B)= SinAcosB + SinBcosA and Cos(A+B)= CosAcosB - SinAsinB?

EDIT: Nvm, about what I said above ^ (Silly me)

 

[kingkong123]

Re: 2012 HSC MX1 Marathon

Find the term independent of x in the expansion without expanding the whole bracket.

 

[bleakarcher]

Re: 2012 HSC MX1 Marathon

15

 

[Timske]

Re: 2012 HSC MX1 Marathon

inverse triggg help
Evaluate the following;
sin(inverse cos (t/4)

let A = inverse cos t/4
therefore, cosA=t/4
therefore sin(inverse cos (t/4) = sinA

A^2 - t^2 = x^2??
i get stuck on this part

 

[AAEldar]

Re: 2012 HSC MX1 Marathon