Prelim 2016 Maths Help Thread (2 Viewers)

[InteGrand]

Help lads. Need help with both.

 

[HeroicPandas]

Here is another way to do Q21

Factorise 3 and complete the square,



The result is clearly positive for all values of x

 

[Kreisler]

help.

 

[Orwell]

Anyone?

 

[si2136]

help.

http://i.imgur.com/V5h5F9D.jpg

 

[si2136]

Anyone?

Use trig identities and simplify.

 

[jathu123]

http://i.imgur.com/V5h5F9D.jpg

You made a mistake when solving k^2+k-2>0, it should be k>1 or k<-2

 

[Green Yoda]

help.

Δ=4k^2-4(-1)(k-2)>0
4k^2+4k-8>0
k^2+k-2>0
(k+2)(k-1)>0
k<-2 or k>1

 

[Green Yoda]

Anyone?

sec^2(x)=tan^2(x)+1
so
tan^2x+(tan^2(x)+1)=0
2tan^2(x)=8
tan^2(x)=4
tan(x)=+-2

 

[Kreisler]

im this close to crying, more help ples

 

[Drongoski]

Q49

This is simply a quadratic equation in (x + 1/x). Were it a quadratic equation in m, say, you'd have no problem solving it as it'd simply be:

m² - 7m + = 0

so: (m-6)(m-1) = 0 ==> m = 1, 6

For the original equation, this means: x + 1/x = 1 or x + 1/x = 6

Each will result in a quadratic equation in x. So for 1st case: x² -x + 1 = 0

Solving will give you complex roots, outside the scope of 2U Maths.

For other case: x² - 6x + 1 = 0; for this you can use the quadratic formula to obtain 2 real roots.

 

[Green Yoda]

im this close to crying, more help ples

Q50
Let 2x^2+x+1=Ax(x+1)+B(x-1)+C
RHS=Ax^2-Ax+Bx-B+c
=Ax^2+(-A+B)+(-B+C)
Therefore A=2 - (1)
-A+B=1 - (2)
Sub (1) into (2)
so -2+B=1
B=3 - (3)
-B+C=1 -(4)
sub (3) into (4)
so -3+C=1
C=3

So
A=2 B=3 C=4
Sub it into RHS in the original equation
=2x(x+1)+3(x-1)+4

 

[Green Yoda]

im this close to crying, more help ples

Q51
Vertex at (1,3)
Note: it is a negative y^2=4ax graph
Focal length is 3units
Diretrix is at x=4

 

[jathu123]

Q51
Vertex at (3,1)
Note: it is a negative y^2=4ax graph
Focal length is 3units
Diretrix is at x=0

the vertex is actually (1,3). So the focal length is 3 units so you move 3 units right of the vertex (since negative y^2=4ax) to get the directrix which is x=4

 

[Green Yoda]

the vertex is actually (1,3). So the focal length is 3 units so you move 3 units right of the vertex (since negative y^2=4ax) to get the directrix which is x=4

My mistake sorry, very silly mistake.
It is indeed (1,3) so the directrix is +3 units left which is as you said x=4

 

[Orwell]

Could someone please explain and show their working out?

 

[InteGrand]

Could someone please explain and show their working out?

By the geometric definition of the parabola, it's a parabola with focus S(2,3) and directrix y = 5. So the vertex is halfway between these at the point (2,4) and the focal length is 1. The parabola is downward facing. This is now enough information to write down the equation of the parabola.

 

[Green Yoda]

Could someone please explain and show their working out?

S(2:3)
Directrix: y=5
So we can conclude it is a vertical negative parabola
Since in parabola PS=PD, eccentricity=1
Thus the vetex is at half way between focus and directrix at (2,4)
Therefore the a or focal length is 1units
Thus the equation: (x-2)^2=-4(y-4)

 

[Orwell]

Parts 2 and 3 are what I'm stumped on.

 

[si2136]

Parts 2 and 3 are what I'm stumped on.

Part 2- Add the fractions and once finding the sum and product of roots, the answer will be apparent to you.

Part 3- Unsure tbh