My Suggested Solutions to SP2008 Exam 2 [Confirmed]

[Synesthetic]

i got E for 2 as well..

If a = ±2, x^2 ±2x + y^2 = 1 => (x±1)^2 + y^2 = 1 => circle defined

That's why I chose C.

[Synesthetic]

and for MC, Q6, doesnt imaginary include i?

No that's a common misconception, Im(z) refers to the value preceding i.

e.g. Im(5i) = 5

[pfftlah]

+ can someone please expain why 12 is A?

[eza]

this is how I did mc2) - not sure if it's right

(x + a/2)^2 - (a^2)/4 + y^2 + 1 = 0
(x + a/2)^2 + y^2 =  (a^2)/4 - 1

For this to be a circle:
(a^2)/4 > 1
a^2 > 4
a<-2, a>2
Hence, E.

[Synesthetic]

+ can someone please expain why 12 is A?

Integral from -3 to 0 = F(0) - F(-3)

Observe the cubic (the antiderivative function), the corresponding points at -3 and 0 are -2 and 7 respectively.

Thus F(0) - F(-3) = -2-7 = -9 [A]

[chid]

this is how I did mc2) - not sure if it's right

(x + a/2)^2 - (a^2)/4 + y^2 + 1 = 0
(x + a/2)^2 + y^2 =  (a^2)/4 - 1

For this to be a circle:
(a^2)/4 > 1
a^2 > 4
a<-2, a>2
Hence, E.

Yes that was my appraoch also.
If a=2 then circle is not defined (radius=0). Therefore C is incorrect.

[csq7]

I got E for MC 12.. Anyone else? maybe just me? but its a definite integral.. and the area is clearly positive.. so E??

[hamtarofreak]

This is what I did for the very last q

Area = Area under the top half of circle

Minus area under the bottom half of the circle

Minus the triangle of area under the point

Triangle area =

Therefore, area within bounds is:

[Synesthetic]

this is how I did mc2) - not sure if it's right

(x + a/2)^2 - (a^2)/4 + y^2 + 1 = 0
(x + a/2)^2 + y^2 =  (a^2)/4 - 1

For this to be a circle:
(a^2)/4 > 1
a^2 > 4
a<-2, a>2
Hence, E.

Yeah I see your logic, I was fixated on the circles with centre (±1,0). I'm still not entirely sure which is correct.

*edit*

this is how I did mc2) - not sure if it's right

(x + a/2)^2 - (a^2)/4 + y^2 + 1 = 0
(x + a/2)^2 + y^2 =  (a^2)/4 - 1

For this to be a circle:
(a^2)/4 > 1
a^2 > 4
a<-2, a>2
Hence, E.

Yes that was my appraoch also.
If a=2 then circle is not defined (radius=0). Therefore C is incorrect.

Point taken I'll change it

[csq7]

*hi five* i got exactly the same answer for that last qn

[Synesthetic]

I got E for MC 12.. Anyone else? maybe just me? but its a definite integral.. and the area is clearly positive.. so E??

It's the area bounded by the parabola and the axes, under the x-axis, so it IS negative.

[negatron]

wtf..... hey how come you dont restrict the ellipse?

isnt it jst tha top bit? not  the bottom because t is greater than zero

also with tcostheta and tsintheta can you do it straight away n obtain tantheta straight away?

[eza]

Ah you have the 1 on the wrong side of the equation.

"If a = ±2, x^2 ±2x + y^2 = 1 => (x±1)^2 + y^2 = 1 => circle defined"

should be
x^2 ± 2x + y^2 + 1 = 0

for example, let us take a=2
x^2 + 2x + y^2 + 1 = 0
(x + 1)^2 + y^2 = 0

[Captain]

(Image removed from quote.)

t=6

To verify this [the easy way]

Get your TI-89 Titanium

Mode --> Graph --> Parametric
y= editor
x(t) =
y(t) =

Window editor
tmin: 0
tmax: 3pi
xmin: -1.2
xmax: 1.2
ymin: -0.6
ymax: 0.6

Graph.

Now, try the following
tmin: 0
tmax: 6pi
xmin: -1.2
xmax: 1.2
ymin: -0.6
ymax: 0.6

QED

[Synesthetic]

Ah you have the 1 on the wrong side of the equation.

"If a = ±2, x^2 ±2x + y^2 = 1 => (x±1)^2 + y^2 = 1 => circle defined"

should be
x^2 ± 2x + y^2 + 1 = 0

for example, let us take a=2
x^2 + 2x + y^2 + 1 = 0
(x + 1)^2 + y^2 = 0

Thanks for that. Definitely E then