VCE Methods Question Thread!

[chansena]

I'm stuck with these questions, could some please explain the steps on solving it


 



 

[Floatzel98]

I'm stuck with these questions, could some please explain the steps on solving it

Sometimes with these questions i like to imagine them as triangles (at least for the first 2 questions). If you make a right angled triangle with angle x and 1 as the adjacent value and 2 as the hypotenuse, then you can find out the final side length using Pythagorean theorem. In this case it is . You can evaluate sin now which is O/H = . Cos is negative in the 2nd quadrant and sin is positive so it works out. try the same for the second one. For the last one you could plot sin and cos on a graph together over 2pi and then find values for sin where cos = 0. Or you could do the same with the values on the unit circle.

[cosine]

I'm stuck with these questions, could some please explain the steps on solving it

1). Find given that and x is in the second quadrant.







We have two solutions for sin, but our given info says that x is in the second quadrant, which we know that sin is positive there.




2). Find given that and x is in quadrant 3







Now we have two solutions, but our given info says that x is in quadrant 3, where cos is also negative so we take the negative option.




3). If cos(x) = 0, find two possible values of sin(x)

Well ask yourself at what angles is cos (x-value) equal to zero? That's right, at the angle of and the angle of . So we can say that and can be two values of sin(x).

[chansena]

 Why do I let it = 1

[cosine]

Why do I let it = 1

Have a look at the dark red line. cos(x) is defined as the x-axis and sin(x) is define as the y-axis. So, using Pythagorus' theorem of we have      hypotenuse = 1 and and .

[Alter]

Why do I let it = 1

This is an equivalence relation and true for all values of x in the unit circle. Just think of it as a mathematical rule.
edit: lol ninja'd with a way better explanation.

[knightrider]

If a questions says that the two new graphs required have local minimums  at the end points of a bridge .

Does this mean that we use the points on the end of the domain where the domain starts and ends for the bridge?

Say the bridge  has  a restricted domain of [0,30].

Does this mean that the new graphs will have minimum turning points at 0 and 30?

So they will be of form y=a(x-0)^2+0 and y=a(x-30)^2+0?

[qwerty101]

for addition of ordinate questions, i know dom of the two is the intersection of the two? but is there any other conditions, ie. if one graph as a y asymptote at y = 0, does the overall graph still exists on the otherside? say y = e^x + y=loge(x), is there any point it is undefined, (as dom is (0,inf)

and if one graph has x asy at x = 0, does the overall addition graph have the same condition? (i.e. do we do dotted lines)

[keltingmeith]

for addition of ordinate questions, i know dom of the two is the intersection of the two? but is there any other conditions, ie. if one graph as a y asymptote at y = 0, does the overall graph still exists on the otherside? say y = e^x + y=loge(x), is there any point it is undefined, (as dom is (0,inf)

and if one graph has x asy at x = 0, does the overall addition graph have the same condition? (i.e. do we do dotted lines)

I can't think of an example where a y-asymptote is in one graph but not the sum, so my gut says that yes, the y-asymptote stays (if someone can come up with a counter example, by all means bring it). However, I know for a fact that the x-asymptote won't always stay - consider y=x+1/x. 1/x has an x-asymptote at y=0, however y=x+1/x doesn't, and instead has an asymptote at y=x.

[knightrider]

If a questions says that the two new graphs required have local minimums  at the end points of a bridge .

Does this mean that we use the points on the end of the domain where the domain starts and ends for the bridge?

Say the bridge  has  a restricted domain of [0,30].

Does this mean that the new graphs will have minimum turning points at 0 and 30?

So they will be of form y=a(x-0)^2+0 and y=a(x-30)^2+0?

Anyone?

[kinslayer]

I can't think of an example where a y-asymptote is in one graph but not the sum, so my gut says that yes, the y-asymptote stays (if someone can come up with a counter example, by all means bring it). However, I know for a fact that the x-asymptote won't always stay - consider y=x+1/x. 1/x has an x-asymptote at y=0, however y=x+1/x doesn't, and instead has an asymptote at y=x.

y = 1/x and y = -1/x ?

[keltingmeith]

y = 1/x and y = -1/x ?

I'm actually disappointed with myself for not thinking this simply.

So yes, just because a function has an asymptote, this doesn't mean that a sum of it will have an asymptote.

[Shinkaze]

Find two quadratic functions f and g such that f(1)=0,g(1)=0 and f(0)=10, g(0) = 10 and both have a maximum value of 18

Can anyone help me? Thank you so much ^_^

[qwerty101]

I'm actually disappointed with myself for not thinking this simply.

So yes, just because a function has an asymptote, this doesn't mean that a sum of it will have an asymptote.

ok thanks!!

also for this, doesnt x > 1? either way x = 2, but just in case? since x > 0 in this case will not always be the case cause we want to avoid "0" within the log?

[kinslayer]

ok thanks!!

also for this, doesnt x > 1? either way x = 2, but just in case? since x > 0 in this case will not always be the case cause we want to avoid "0" within the log?

Sure, x > 1, but this implies that x > 0 automatically, so writing x > 0 isn't wrong.