Past HSC problems... (1 Viewer)

[nick1048]

HSC 2U 2001
Question 6: c)iii) For what values of k has the equation x^3 + x^2 - x + 2 = k three real solutions?

Question 10: b) Both these questions are quite dense so if you know of these and have the worked solutions could you please post em up.

HSC 2002 2U
Question 8: a)ii) A drug is used to control a medical condition. It is known that the quantity ! of drug remaining in the body after t hours satisfies an equation of form

Q = Ae^(-k)t

The initial dose is 6 milligrams and after 15 hours the amount remaining in the body is half the initial dose.

When will one-eighth of the initial dose remain.

(This seems really easy to calculate, however the actual answer, which is 45 hours does not agree with my answer. [despite having the first part of the question correct] I'm not sure what I'm doing wrong, if anyone cares to solve thankyou.)

HSC 2003 2U
Question 8: d)i) Show that for all values of m, the line y = mx - 3m^2 touches the parabola x^2 = 12y (I have no idea how to prove this or indicate this to be honest)

d)iii) Hence determine the equations of the two tangents to the parabola x^2 = 12y from the point (5,2) (I believe this correlates to my lack of understanding for the first part).

I'm especially concerned about the last problem and the first one because they follow the same sort of trend with a function with a pronumeral. I'm not sure if I'm just having a complete mental blank or if I haven't done enough practise on these sorts of things. I am using the fitzpatrick 2U mathematics book, so if any of this work falls under an exercise or something could you please direct me to these areas. Thanks alot, and goodluck for the 2u exam tomorrow everyone!

 

[Trev]

6ciii)
Find stationary points, k lies between thes y values of these stationary points (not equal to).

 

[Trev]

Q=Ae^-kt
t=0;Q=6 so A=6.
Q=6e^-kt
t=15; Q=3
ln(1/2)=-15k
k=(ln2)/15
For 1/8th dose:
1/8=e^(ln2)/15t
[15ln(1/8)]/(ln2)=t
t=45 hours.
Mistake there somewhere because the answer comes out negative, but yeah doesn't matter

 

[Trev]

8di)
y=mx-3m² (1)
x²=12y (2)
(1) into (2) gives:
x²=12mx-36m²
x²-12mx+36m²=0
For it to be a tangent it must touch the parabola, so discriminant must equal zero.
= (12m)²-4.1.36m² = 144m²-144m² = 0. So regardless of what m equals it will always touch the parabola and be a tangent.

 

[Jexi]

HSC 2002 Question 8

I got 45.

the equation should be 1/8 x 6 = 6 e ^ (-0.046209812 t)

Then you cancel the six off both sides. then -0.046209812t = ln 0.125

And you type in calculator.

HSC 2001 Question 6

Okay this question, you need to rely on the graph. See x^3+x^2-x+2 = k means the intersection of y = x^3+x^2-x+2 and y = k. And notice how y is a stright line? therefore if you look at the graph, K must be between the Max and Min to have three values.

I hope that makes sense.

The rest are really long. The keyboard is sooo not maths friendly.

 

[Trev]

8diii)
y=mx-3m², from point (5,2); sub in to find values of m when line passes through this point.
2=5m-3m²
3m²-5m+2=0
Solve to find values of m, which will give you equations of two tangents.

 

[nick1048]

Thanks guys I can see where all my mistakes are (sigh). Your time to answer my questions is greatly appreciated I'd rep you, but that died ages ago