Prelim 2016 Maths Help Thread (4 Viewers)

davidgoes4wce · Jun 12, 2016

$$ Find the point of intersection in which case $ x=\frac{-1}{12}, $ and knowing that the coordinate (x,y) = $ (\frac{-1}{12}, 1)$

$$ By rearranging the formula, we can express $ y=\frac{-12x}{5}+\frac{4}{5}$

$Use the perpendicular rule $ m_1 \times m_2 = -1 $ to solve for the gradient that is perpendicular to that equation$

$\frac{-12}{5} \times m_2 = -1$

$m_2=\frac{5}{12}$

$y-1=\frac{5}{12} (x+\frac{1}{12})$

$y=\frac{5x}{12} +\frac{149}{144}$

Here is my graph below ( Someone may have to correct any errors of mine). I'm aware that what Integrand said there is also a second line:

InteGrand · Jun 12, 2016

davidgoes4wce said:
$$ Find the point of intersection in which case $ x=\frac{-1}{12}, $ and knowing that the coordinate (x,y) = $ (\frac{-1}{12}, 1)$

$$ By rearranging the formula, we can express $ y=\frac{-12x}{5}+\frac{4}{5}$

$Use the perpendicular rule $ m_1 \times m_2 = -1 $ to solve for the gradient that is perpendicular to that equation$

$\frac{-12}{5} \times m_2 = -1$

$m_2=\frac{5}{12}$

$y-1=\frac{5}{12} (x+\frac{1}{12})$

$y=\frac{5x}{12} +\frac{149}{144}$

Here is my graph below ( Someone may have to correct any errors of mine). I'm aware that what Integrand said there is also a second line:

You have found a line that is perpendicular to one of the given lines, but this isn't what we want. The lines we want are essentially the lines containing all those points that are equidistant from the two given lines (they're equidistant because the lengths of the perpendiculars from P to the two given lines are equal, and these lengths are what is meant by distance of P to a line, since we generally mean perpendicular distance when we say that.).

drsabz101 · Jun 12, 2016

thanks!where is the other line from?

davidgoes4wce · Jun 12, 2016

$$ By using the rule $ tan \ \theta=|\frac{m_1-m_2}{1+m_1 \times m_2}|$

$$ Given that the $ \theta $ must be $ 45^\circ , \ m_1=\frac{5}{12}$

$tan \ 45^\circ=|\frac{\frac{5}{12}-m_2}{1+\frac{5}{12} \times m_2}|$

$1= |\frac{\frac{5}{12}-m_2}{1+\frac{5}{12} \times m_2}|$

$1+\frac{5}{12} \times m_2={\frac{5}{12}-m_2}$

$m_2 = \frac{-7}{17}$

$$ Substituting the coordinates $ (\frac{-1}{12},1) $ into y=mx+b gives y $ =\frac{-7x}{24}+\frac{211}{204}$

Again, someone correct me if there are any errors.

eyeseeyou · Jun 13, 2016

Okay someone please correct what I am doing wrong:

Find the point of intersections of the curves y=f(x)and y=f^(-1(x) ) if:
a.f(x)=3x(2-x),x≤ 1

so what I did for this equation was solve simultaneously for the equations y=3x(2-x) and y=x
So then I got the equation 3x(2-x)=x
6x-3x^2=x
x(5-3x)=0 then we have x=0 and x=5/3 and then we have y=0 y=5/3
and my answers were (0,0)and (5/3,5/3)and apparently the answers were just (0,0)
What am I doing wrong?

eyeseeyou · Jun 13, 2016

b.f(x)=1- √(x+2)

so again for this question I solved simultaneously for the equations y=1-√(x+2) and y=x
x=1-√((x+2) )
x^2=1-(x+2)
x^2=1-x-2
x^2=-x-1
x^2+x+1=0
x=(-1±√(1-4(1)(1)))/(2(1))
What am I doing wrong here???

eyeseeyou · Jun 13, 2016

c.f(x)=x^3

What I did for this was solve simulataneously
y=x^3 and y=x
solve simultaneously and you get x^3=x
then I did x^3-x=0
x(x^2-1)=0
x(x+1)(x-1)=0
Then I got the following values for x: 0, -1, 1
And for the y values I got y=0,0,0
And apparently the answers for this are (0,0), (-1,-1) and (1,1)
What the hell is wrong???

Shuuya · Jun 13, 2016

eyeseeyou said:
Okay someone please correct what I am doing wrong:

Find the point of intersections of the curves y=f(x)and y=f^(-1(x) ) if:
a.f(x)=3x(2-x),x≤ 1

so what I did for this equation was solve simultaneously for the equations y=3x(2-x) and y=x
So then I got the equation 3x(2-x)=x
6x-3x^2=x
x(5-3x)=0 then we have x=0 and x=5/3 and then we have y=0 y=5/3
and my answers were (0,0)and (5/3,5/3)and apparently the answers were just (0,0)
What am I doing wrong?

Note that x≤1, so 5/3 is not a solution.

Drongoski · Jun 13, 2016

eyeseeyou said:
b.f(x)=1- √(x+2)

so again for this question I solved simultaneously for the equations y=1-√(x+2) and y=x
x=1-√((x+2) )
x^2=1-(x+2)
x^2=1-x-2
x^2=-x-1
x^2+x+1=0
x=(-1±√(1-4(1)(1)))/(2(1))
What am I doing wrong here???

This is wrong. You are in fact saying here: (a - b)² = a² - b²

Better to proceed like this:

$\\ \therefore x - 1 = -\sqrt{x+2}\\ \\ \therefore (x-1)^2 = (-\sqrt {x+2})^2\\ \\ \therefore x^2 - 2x + 1 = x+2\\ \\ \cdots \cdots$

eyeseeyou · Jun 13, 2016

Drongoski said:
This is wrong. You are in fact saying here: (a - b)² = a² - b²

Better to proceed like this:

$\\ \therefore x - 1 = -\sqrt{x+2}\\ \\ \therefore (x-1)^2 = (-\sqrt {x+2})^2\\ \\ \therefore x^2 - 2x + 1 = x+2\\ \\ \cdots \cdots$

Oh I see now thanks

davidgoes4wce · Jun 13, 2016

eyeseeyou said:
b.f(x)=1- √(x+2)

so again for this question I solved simultaneously for the equations y=1-√(x+2) and y=x
x=1-√((x+2) )
x^2=1-(x+2)
x^2=1-x-2
x^2=-x-1
x^2+x+1=0
x=(-1±√(1-4(1)(1)))/(2(1))
What am I doing wrong here???

$y=1-\sqrt{x+2}$

$$ Domain of this function $ \{x \in \mathbb{R}: x \geq -2 \}$

$$ Range of this function $ \{y \in \mathbb{R}: y\leq -1 \}$

$To find the inverses of each other , interchange the x and y terms.$

$x=1-\sqrt{y+2}$

$\sqrt{y+2}=1-x$

$y+2=(1-x)^2$

$y=(1-x)^2-2$

$$ Domain of the inverse function $ f^{-1} (x) : \{x \in \mathbb{R}: x \leq -1 \}$

$$ Range of the inverse function $ f^{-1} (x) : \{y \in \mathbb{R}: y \geq -2 \}$

drsabz101 · Jun 13, 2016

can someone help solve this q. outlining the steps... thank

s

InteGrand · Jun 13, 2016

drsabz101 said:
can someone help solve this q. outlining the steps... thankView attachment 33269s

$\noindent The parabola is defined as the set of all points for which the distance to the focus equals the distance to the directrix (or equivalently the squared-distances are equal). Let $P=(x,y)$, then its squared-distance to the focus is $(x-2)^2 + (y-4)^2$. Also, its distance to the line $y=-2$ is $\left(y-(-2)\right)^2=\left(y+2\right)^2$. So the point $P$ lies on the parabola if and only if $(x-2)^2 + (y-4)^2=\left(y+2\right)^2$, from which we can simplify things to get the equation of the parabola.$

Drongoski · Jun 13, 2016

drsabz101 said:
can someone help solve this q. outlining the steps... thankView attachment 33269s

I suggested to you once, several months ago, after seeing a number of your posts that you may need help, and quickly. You said you were ok. You were in denial as can be seen from your currents posts.

Solution:

From definition:

Let P(x,y) be any point on this parabola. Therefore:

distance of P from S = distance of P from the given line y = -2 (the directrix)

.: (LHS) squared = (RHS) squared

.: (x-2)² + (y-4)² = (y - [-2])²

.: (x-2)² = (y+2)² - (y-4)² = (y+2 + y-4)(y+2 - y+4) = 12(y-1)

i.e (x-2)² = 4 x 3 x (y-1)

the equation of an upright parabola with vertex V(2,1) and focal length 3.

eyeseeyou · Jun 13, 2016

drsabz101 said:
can someone help solve this q. outlining the steps... thankView attachment 33269s

Is this locus and parabola or parametrics?

eyeseeyou · Jun 13, 2016

Shuuya said:
Note that x≤1, so 5/3 is not a solution.

Thanks Shuuya

eyeseeyou · Jun 13, 2016

eyeseeyou said:
c.f(x)=x^3

What I did for this was solve simulataneously
y=x^3 and y=x
solve simultaneously and you get x^3=x
then I did x^3-x=0
x(x^2-1)=0
x(x+1)(x-1)=0
Then I got the following values for x: 0, -1, 1
And for the y values I got y=0,0,0
And apparently the answers for this are (0,0), (-1,-1) and (1,1)
What the hell is wrong???

Someone please help me with this???

Shuuya · Jun 13, 2016

eyeseeyou said:
Someone please help me with this???

How did you get those y values? Subbing x=0,-1,1 into either y=x³ or y=x gives you the coordinates (0,0), (1,1) and (-1,-1)

eyeseeyou · Jun 13, 2016

I subbed all the x values back into the cubic equation

Green Yoda · Jun 13, 2016

drsabz101 said:
can someone help solve this q. outlining the steps... thankView attachment 33269s

Notice that the focal length is 3 so the vertex is (2,1). We can notice its a shifted positive x^2=4ay parabola with a=3
Thus equation of the parabola can be written in the form of (x-h)^2=4a(y-k) where (h,k) is the vertex and a being the focal length.
So the equation of the parabola is (x-2)^2=12(y-1)

EDIT MADE A SILLY MISTAKE

Prelim 2016 Maths Help Thread (4 Viewers)

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