CSSA 2019 Maths Ext.1 Question 11e (1 Viewer)

[Jacobagel]

Hey guys, I know that there was a big fuss about question 11e of the CSSA 2019 trial paper, and that some schools only allowed an answer of two solutions, while others accepted both two solutions or four solutions for the second part. I was curious as to whether if we were given something like this in the HSC, would they consider 4 solutions correct at all?

The question is below:

 

[Trebla]

I don’t see how the first two solutions satisfy the original equation?

The identity in part (i) is only true if . Therefore you should apply that restriction when solving the equation in (ii).

FYI solving



is not exactly the same as solving



because the former has a slightly more restricted domain than the latter.

 

[DrEuler]

I don’t see how the first two solutions satisfy the original equation?

The identity in part (i) is only true if . Therefore you should apply that restriction when solving the equation in (ii).

FYI solving



is not exactly the same as solving



because the former has a slightly more restricted domain than the latter.

Yep, I didn't even read it properly ;-;

 

[HeroWise]

Yeah i remember this question most of the kids fell for it in my class.

Only 2 solutions exists because of the tantheta

 

[Jacobagel]

See but the thing was that even in the official CSSA solutions they included all four solutions, and didn't disregard those that didn't satisfy the domain of tan

 

[HeroWise]

2017 CSSA paper had many wrong solutions.

Universally this is wrong. IF you can prove tanpi/2 exists (no dont say infinity) then it is true.

My teacher is marking with 4 solutions too which I respect but font agree with.

 

[quthoang]

is f(x)= (2tan(x))/(1+tan^(2)(x)) a continuous function?

 

[blyatman]

is f(x)= (2tan(x))/(1+tan^(2)(x)) a continuous function?

limx->0 of that function is 0.

So i guess all 4 solutions are acceptable. Even though you cant actually compute tan(90) in the calculator.

I don't think that's classified as a continuous function. It's only continuous if you define f(0) = 0. It's similar to the sinc(x) function, which is sin(x)/x. The function itself is only continuous because sinc(0) is defined to be the limiting value as x->0, but otherwise it's discontinuous there. Another example is y = 3x/x, which is equal to 3 at all non-zero values of x, and is discontinuous at x=0 even though the limiting value is 3.

 

[fan96]

is f(x)= (2tan(x))/(1+tan^(2)(x)) a continuous function?

It depends on what exactly is meant by "continuous".

When we say a function is continuous we usually mean either of two things:

1. the function is continuous at every point in its domain.
In this sense, functions like are considered continuous, and so is .
Because is not in the domain of , we don't need to worry about what happens at that point.
(I believe this is the definition that's formally used.)

2. the function is continuous at every point in .
In this sense, is not continous because it is not defined at .

 

[quthoang]

Looks like the online graphing programs do not recognise the discontinuity for these kinds of graphs. I tried desmos, geogebra and wolfram alpha. They all converted the function to sin2x before graphing them.

Is there a better graphing program out there that shows the discontinuity?

 

[blyatman]

Looks like the online graphing programs do not recognise the discontinuity for these kinds of graphs. I tried desmos, geogebra and wolfram alpha. They all converted the function to sin2x before graphing them.

Is there a better graphing program out there that shows the discontinuity?

View attachment 27120

It shows sin2x because the function is equal to sin2x EXCEPT at the points where tan(x) is undefined.

In regards to the graphing programs not showing the discontinuity - it's because the program is simply plotting a large number of points where the function is defined. The function only APPEARS to be continuous since the program is simply plotting plots closer and closer to the discontinuity - it's not going to the plot the discontinuity itself in the same way that it's not going to plot 1/x at x=0: it doesn't exist there, so it can't plot it.

As far as I'm aware, that's how all graphing calculators work - by plotting points where the function exists. It's up to the user to know it's limitations and understand the results.

 

[quthoang]

That clears things up a bit.
Thank you