VCE Methods Question Thread!

[Phy124]

Hey guys, just to clarify, if X is normally distributed, and Pr (X>a)= 0.025, it means that a is 2 standard deviations from the mean right? Sorry for the stupid question haha

Approximately, yes.

[fleet street]

Hi everyone,

I've been doing the 2006 Sample Exam 2 for Methods (CAS) and I am confused as to how you would do Q1biv.

Consider the function

Find the real values of for which the equation , where , has exactly one solution.

According to itute (VCAA did not make an examiners' report that I could see), the answer is , but they only offer a CAS explanation and the question is worth 2 marks.

Can anyone help? Thanks for reading this question.

[SocialRhubarb]

I would just sketch the graph and see the regions in which the graph is one-to-one, and adjust p so that the answer falls within the region.

I'm assuming that's the CAS method they used?

[Daenerys Targaryen]

If you draw f(x), and find local max's and local min's, end points

Then f(x)=p is just a horizonal line anywhere

We want to know for what x values will a horizontal line cut f(x) only once

May be a tangent at a tp, or maybe just an end point

[Alwin]

Consider the function

Find the real values of for which the equation , where , has exactly one solution.

According to itute (VCAA did not make an examiners' report that I could see), the answer is , but they only offer a CAS explanation and the question is worth 2 marks.

Graph of f(x):


We require f(x) = p , wherever a horizontal line y=p cuts the graph of y=f(x) once only in the domain [0, 2].

From the graph, we can see that there are turning point at (1,1) [local max] and (5/3, 23/27) [local min]
Also, the y-intercept is (0, -1)

As this is only a 2 mark question, i will forgo the algebraic explanation.

As the domain is [0, 1] then the range is [-1, 1]. So p is in [-1, 1]
Clearly, for 23/27≤y≤1 any horizontal line will intersect the graph twice. So, for only one intersection p cannot be in [23/27, 1]

Thus, it follows that p must be from -1 (inclusive)  to 23/27 (exclusive) , ie -1 ≤ p < 23/27 as required

[abcdqd]

I'm hopeless at probability, thanks in advance .

[Jaswinder]

Im not sure if im right but is it E? 

[Phy124]

(Image removed from quote.)
I'm hopeless at probability, thanks in advance .

is the same as

So we want to find the pink area below.



We know that which are the blue areas in the two images below.



If the area of the blue is given by then the area of the green is given by

Looking back at the first picture to find the pink area, , we need to subtract 2 of the green area from the entire area (1), so:



Or mathematically:



Alternatively you could just simplify it and say that the pink area is equal to the green area (that is a single green area, not both sides) subtracted from the blue area, so

Or mathematically:



edit: restructured response

[darklight]

A random variable X has prob. density function f with rule:

f(x)= 0, for x <0
f(x) = 2x*e^(-x^2), for x greater than or equal to 0

Show that e^-(x^2) is an antiderivative of -2x *e^(-x^2) and hence find the interquartile range of X.

--> I've shown what I needed to prove. However, after I sub in the limits a, equal it to 0.25, then solve for a, CAS says false... :/

[Jaswinder]

my cas says its between ~0.69 and ~0.99

[darklight]

my cas says its between ~0.69 and ~0.99

The value for a is 0.536 according to solns, value for b is 1.1774, hence interquartile range = 0.6411

When I put this equation into the CAS:
e^(-a^2) - e^0 = 0.25, it returns 'false'...

Similarly for b, and the 75% quartile...

[Phy124]

The value for a is 0.536 according to solns, value for b is 1.1774, hence interquartile range = 0.6411

When I put this equation into the CAS:
e^(-a^2) - e^0 = 0.25, it returns 'false'...

Similarly for b, and the 75% quartile...

It's saying that it's false, because it is false

You're typing in which is equal to , not 0.25. It is equal to 0.25 for a certain value of a, not a in general. You should be solving the expression for a.

From the beginning, assuming you have given the correct equations in the original part of the question, your expression should be something along the lines of solve(int(2*x*e^(-x^2),x,0,a)=1/4,a)

You could quite easily do the integration by hand and end having to solve , which gives:



This is approximately 0.536, the value you were looking for.

[Nato]

Hey guys, quick question

when graphing a trig function such as 

am i required to find the x-intercepts, or do i just draw the graph cutting through the x axis (not at a particular point).

thanks

[McFleurry]

I always do, unless the question states not to (which is highly unlikely).

[EspoirTron]

Hey guys, quick question

when graphing a trig function such as 

am i required to find the x-intercepts, or do i just draw the graph cutting through the x axis (not at a particular point).

thanks

When the questions asks for a 'sketch', for example, sketch  , it is implied that you accurately label intercepts and usually I'd do that for one cycle depending on how many cycles they have requested. However, if it specifies to sketch within a restricted domain, I would label all  - x and y - intercepts, label key points such as turning points, and also label endpoints of the graph.