Exam 2: Solutions

[keltingmeith]

Hi

Can you just double check Q9 MC, I'm pretty sure it's A) because A describes the line y=x which does cut through the circle twice

Actually, that describes the line y=-x:

[keltingmeith]

Question 2iii) should be z=root(3) + 3i and z = -2root(3) as cos(pi) = -1

It's funny, because I did this question two times and got it right the first time... Fixed, thanks for that!

[chuck981996]

Question 2iii) should be z=root(3) + 3i and z = -2root(3) as cos(pi) = -1
|z - i| = |z + 1| describes the line y = -x, not y=x

Ahhh you're right.. I think I thought it was |z - i| = |z-1|

[jazzycab]

A couple of minor errors that others have already picked up:

is an asymptote of the graph in Q1b) as is defined over its maximal domain in this part of the question;

for Q2c) as is not the x-value of the curve at the tangential point;

is required in the Vector question.

The third root of the complex cubic is

My solutions for Q4 (note, there may be errors here):
4a)
Similar triangles gives:, giving
Substituting into the Volume of a cone,
Gives:


4b)
and







At :





4c)














4d)




























And my solution to Q2c):


Tangent Line:









Because the gradient of the tangent is 1:





Solving simultaneously: or

Graphically, because ,

Substituting back into the tangent line equation:

[keltingmeith]

A couple of minor errors that others have already picked up:

is an asymptote of the graph in Q1b) as is defined over its maximal domain in this part of the question;

for Q2c) as is not the x-value of the curve at the tangential point;

is required in the Vector question.

The third root of the complex cubic is

My solutions for Q4 (note, there may be errors here):
4a)
Similar triangles gives:, giving
Substituting into the Volume of a cone,
Gives:


4b)
and







At :





4c)














4d)




























And my solution to Q2c):


Tangent Line:









Because the gradient of the tangent is 1:





Solving simultaneously: or

Graphically, because ,

Substituting back into the tangent line equation:

Well thank you, jazzycab! This means I can actually focus on my probability exam tomorrow. Adding to the first post.

[DanielJ]

I don't even get what I was thinking tackling 4c with a perpendicular bisector method. Ah well, 3 marks there if I don't get any method marks, and 2 in mc.