Prelim 2016 Maths Help Thread (1 Viewer)

Status
Not open for further replies.
I

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
eyeseeyou said:
Without actually solving the simultaneous equations, state whether the following pairs of lines intersect are parrallel or coincide

a) x+y-7=0 and x+y-8=0
Click to expand...
They are clearly parallel and don't coincide or meet anywhere.
 
E

eyeseeyou

Well-Known Member
Joined
Nov 20, 2015
Messages
4,125
Location
Space
Gender
Undisclosed
HSC
N/A
What about 6x-5y-24=0 and 9x-4y-22=0. Do you so these things by inspection?
 
E

eyeseeyou

Well-Known Member
Joined
Nov 20, 2015
Messages
4,125
Location
Space
Gender
Undisclosed
HSC
N/A
Consider the curve given by y=x^3-6x+4

FInd the coordinates of the stationary points and determine their nature

Find the coordinates of any points of inflection

Sketch the curve for the domain -3=<x=<3

What is the maximum value of x^3-6x+4 in the domain -3=<x=<3
 
E

eyeseeyou

Well-Known Member
Joined
Nov 20, 2015
Messages
4,125
Location
Space
Gender
Undisclosed
HSC
N/A
If a:b:c is 1:2:3 and a^2+b^2+c^2=504 find a, where a is a positive number
 
I

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
eyeseeyou said:
Consider the curve given by y=x^3-6x+4

FInd the coordinates of the stationary points and determine their nature

Find the coordinates of any points of inflection

Sketch the curve for the domain -3=<x=<3

What is the maximum value of x^3-6x+4 in the domain -3=<x=<3
Click to expand...
Differentiate and find where the derivative is 0 for stationary points. Determine their nature using for instance the second derivative test. Sub. in the x-values of the stationary points to the f(x) expression to get the coordinates.

Find where f"(x) = 0 for the inflection point.

Last part can be done by inspection after doing the previous parts.
 
E

eyeseeyou

Well-Known Member
Joined
Nov 20, 2015
Messages
4,125
Location
Space
Gender
Undisclosed
HSC
N/A
InteGrand said:
Differentiate and find where the derivative is 0 for stationary points. Determine their nature using for instance the second derivative test. Sub. in the x-values of the stationary points to the f(x) expression to get the coordinates.

Find where f"(x) = 0 for the inflection point.

Last part can be done by inspection after doing the previous parts.
Click to expand...
IS this the highest part of the graph?
 
I

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
Kreisler said:
Click to expand...
The answer is just the perpendicular bisector of A and B. We can find the equation of this line by finding the midpoint of A and B, and then finding the slope of the line (which will be the negative reciprocal of the slope of AB, since it's perpendicular to AB). Once we've worked out the slope and the midpoint, we can use point-gradient formula to get the equation.
 
pikachu975

pikachu975

Premium Member
Joined
May 31, 2015
Messages
2,739
Location
NSW
Gender
Male
HSC
2017
Orwell said:
Click to expand...
i) (k-1)^2 - 4(2)(2)
= k^2 - 2k + 1 - 16
= k^2 - 2k - 15

ii) Intersects when y = y
2x^2 - x + 7 = 5 - kx
2x^2 + kx - x + 2 = 0
2x^2 + (k-1)x + 2 = 0
So this is the equation from part i, so we use that discriminant. Intersects at more than one point when discriminant > 0

k^2 - 2k - 15 > 0
(k - 5)(k+3) > 0

Test k = 0, which is less than 0

Therefore k > 5, k< -3
 
Status
Not open for further replies.

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

Top