luken93's Methods/Spesh Question thread...

[luken93]

Hey guys, I thought I may as well make my own thread, but I'll have both Spesh + Methods in the one thread to save the clutter...

Anyways, first question:




The simultaneous linear equations

(1)...
(2)...

Have No Solution for.....

[francis8nho]

Use matrices
For no solution det=0
(m-2)(m-3)-2*3=0
Simplify to get m= 0 and m = 5

[luken93]

That's what I got, however the answers are:

A. R \ {0,5}
B. R \ {0}
C. R\ {6}
D. m = 5
E. m = 0

I would've said A, but apparently not!

[francis8nho]

you mean the anwer is m=0?
I just tried again using CAS for m=5, x=2-y
for m=0, yes there is no solutions

[sajib_mostofa]

You get no solutions when both lines have different equations but the same gradient. So sub in the value for m that fits that description and you should get the answer, which is m=0 in this case.

[francis8nho]

sajib, your method is quicker

[sajib_mostofa]

Helps when you've finished VCE 

[luken93]

You get no solutions when both lines have different equations but the same gradient. So sub in the value for m that fits that description and you should get the answer, which is m=0 in this case.

So what was your actual method there? I rearranged them to make a linear y = equation, and then let both the gradients equal eachother which gave m = 0 and 5?

[sajib_mostofa]

Now that you've found the values of m, you have to verify which one gives no solutions. If you sub in into both equations, you get:




By rearranging the equations to make y the subject, you can see that both equations have the same gradients but different equations, so they are parrallel lines. Hence they give no solution. However, if we sub in , you get these two equations:




When simplified, these two equations are essentially indicate the same line, hence there is infinite no. of solutions. Therefore is your only answer.

[luken93]

Aaahh, thanks for that

[sajib_mostofa]

Aaahh, thanks for that

Whenever you get these type of questions where you have to determine unique, infinite, no solutions etc, the important thing is to always verify your answers or its very easy to make a mistake. Unless you only get the one answer of course.

[luken93]

SPESH Q
Looking for a nice solution to this, as any of mine get very messy

Let and
Express in modulus - Argument form:

[brightsky]

Modulus is trivial.
For argument, the complex number is in the 4th quadrant.
So tan(x) = (1/cos(t))/(1/sin(t)) = sin(t)/cos(t) = tan(t), pi/2 < t < pi
x = arctan(tan(t)) = t, pi/2 < t < pi
But since x is in the 4th quadrant, then we have 0 < x < -pi/2.

[luken93]

Thanks, I ended up getting it anyway. Essentials has a tendency to keep changing r cis (x) to -r cis (x - pi), but once you get used to it it's not so bad...

[Andiio]

Is modulus-argument form just polar form? O_O