Prelim 2016 Maths Help Thread (5 Viewers)

[InteGrand]

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

They are clearly parallel and don't coincide or meet anywhere.

 

[eyeseeyou]

What about 6x-5y-24=0 and 9x-4y-22=0. Do you so these things by inspection?

 

[eyeseeyou]

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

 

[eyeseeyou]

If a:b:c is 1:2:3 and a^2+b^2+c^2=504 find a, where a is a positive number

 

[InteGrand]

What about 6x-5y-24=0 and 9x-4y-22=0. Do you so these things by inspection?

Pretty much

 

[InteGrand]

If a:b:c is 1:2:3 and a^2+b^2+c^2=504 find a, where a is a positive number

Write b = 2a, c = 3a, then substitute into the given equation and solve for a.

 

[InteGrand]

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

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.

 

[eyeseeyou]

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.

IS this the highest part of the graph?

 

[InteGrand]

IS this the highest part of the graph?

Yeah the max. value is the y-value of the highest point on the graph.

 

[Orwell]

 

[Kreisler]

Please help with this :O

 

[Kreisler]

 

[InteGrand]

 

[InteGrand]

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.

 

[Orwell]

I was your 2222 like <3

I feel honoured.

 

[Orwell]

If someone could hook me up with an answer to this, much appreciated.

 

[Kreisler]

Please help me thanks...

 

[InteGrand]

Please help me thanks...

Here are some hints/answers.

 

[Orwell]

 

[pikachu975]

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