Prelim 2016 Maths Help Thread (2 Viewers)

Status
Not open for further replies.

Paradoxica

-insert title here-
Joined
Jun 19, 2014
Messages
2,556
Location
Outside reality
Gender
Male
HSC
2016
Now, for a more serious solution.....

make the substitution x = y-3

The function, in terms of y

3 - y + √y

Complete the square:

13⁄4 - (1⁄2-√y)²

back substitute to return x

13⁄4 - (1⁄2-√(x+3))²

The maximum value of the expression occurs when the negative term is as small as possible (in this case, 0)

So the maximum value is 13⁄4

The minimum value occurs when the negative term is as large as possible. Since there is no upper bound on this term, the minimum does not exist.

So the function range goes from -∞ to 13⁄4
 

drsabz101

Member
Joined
Dec 28, 2015
Messages
429
Gender
Undisclosed
HSC
N/A
Can someone draw the situation, I know how to draw the two given lines, but not the 'M' and 'n'? snipping.PNG
 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
how do u find the locus of p?
It's the pair of intersecting lines that bisect the angles made by the two given lines (they go through the point of intersection of those two lines).

If you want the equations of the lines, the way they'd expect you to do it is to use the "perpendicular distance formula" and then equate the distances of P to each of the two given lines.
 

drsabz101

Member
Joined
Dec 28, 2015
Messages
429
Gender
Undisclosed
HSC
N/A
why do we need to consider a positive and negative case?
 

drsabz101

Member
Joined
Dec 28, 2015
Messages
429
Gender
Undisclosed
HSC
N/A
can someone draw a diagram labeling the locus, I am not getting this ( i am self-teaching)
 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
can someone draw a diagram labeling the locus, I am not getting this ( i am self-teaching)
Draw the two lines going through the point of intersection of the given lines and bisecting the angles made by these two given lines. This is the locus.
 
Status
Not open for further replies.

Users Who Are Viewing This Thread (Users: 0, Guests: 2)

Top