How do you do these questions?? (1 Viewer)

mmmm.

Active Member
Joined
Dec 18, 2019
Messages
219
Gender
Male
HSC
2020
Could someone help/explain q 11b, 13 (is there a proper way, i was only taught trial and error, and don't rlly understand the working from document) and question 28c

Although there are answers on the document, i don't really understand how they did it.
 

Attachments

cossine

New Member
Joined
Jul 24, 2020
Messages
29
Gender
Male
HSC
2017
Using the result of 11a) you can find sin A

cos A = 23/32.

This means 23 is the adjacent side and hypotenuse is 32 units long.

Using Pythagoras theorem you can then find the value of the opposite side.

Alternatively, you could use sin^theta + cos^2 theta = 1.


I am assuming you have done 13a)

For 13b) just consider each case.

you can write y = |ax + b| as piecewise function

y = { ax + b, if x>= 2.5
{ - ax - b, if x < 2.5

That means -ax-b >= 3 for x< 2.5 or ax+b >= 3 for x>= 2.5.
Then just solve.

For cumulative distributive function, the answer is just

cdf(8) - cdf(5)

or integral pdf with lower bound 5 and upper bound 8.

You just need to know what cdf is and how to use it to perform calculations

In this case, it looks like they use complementary probability to make the calculations easier.
 
Last edited:

CM_Tutor

Well-Known Member
Joined
Mar 11, 2004
Messages
1,469
Q13... You have the function



(a) From the given graph, you know at least two points that satisfy the function, and . Using the first point:



and using the second point,



So, we have two possible solutions:

Possibility 1: which leads to

Possibility 2: which leads to

But these are the same, as

So, both solutions are valid. The solutions on the exam are flawed, however, in that they find by starting with without first establishing the value of .


Part (b) requires you to solve . The solution will be the same whichever of the previous solutions you choose as they both result in the same and so have the same equation to solve, . This can be solved by cases:

Case 1: , in which case and so



and this solution is valid throughout the restricted domain .

Case 2: , in which case as and so



and this solution is valid throughout the restricted domain .


So, the complete solution of is or . This can be expressed in interval notation as .

Part (b) can also be solved graphically by adding the line to the given diagram, noting that there are intersections at and , and then recognising that when the given graph lies above the line that you have added. The solution immediately follows.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top