# Thread: VCE Maths questions help

1. ## Re: maths questions help

Originally Posted by InteGrand
.
Problem is i'm still not understanding (for whatever reason it's just not making sense to me) the bisection methods even after reading my textbook and the summary you linked me

2. ## Re: maths questions help

I'm going to assume bisection method is the same as halving the interval:

You have two points, one positive, one negative, e.g. (0,3) and (4, -4).
Then you bisect this interval and your new x value is x = (0+4)/2 = 2

For example if you sub in x = 2 you get y = 1.5, that means x = 2 is a closer approximation to the root than x = 0 because y = 1.5 is closer than y = 3 to the x axis (the root). So your new interval becomes 2 < x < 4

Hope this helped, don't know how else to explain it.

3. ## Re: maths questions help

How do you efficiently study for a maths assessment/test?

4. ## Re: maths questions help

Originally Posted by boredsatan
How do you efficiently study for a maths assessment/test?
Very open-ended question. Not easy to provide a useful and meaningful answer.

5. ## Re: maths questions help

Originally Posted by Drongoski
Very open-ended question. Not easy to provide a useful and meaningful answer.
Does it really matter if I asked the question?

6. ## Re: maths questions help

Different people will provide different advice. Advice will depend on where you are at currently.

In general, make sure you are on top of the topics you are going to be examined on. Depending on how much time you have before the test, make sure you get on top of areas you may be weak in currently. Then do as many of the basic exercises (if you have not already done this) and then progress to more challenging questions after that - these harder questions can be from your books and preferably from your school's past, but more recent, test questions. Also don't stay up too late. When you know your stuff, you are more confident and less stressed and conversely. When you are less stressed, you tend to perform better.

Good luck.

7. ## Re: maths questions help

Originally Posted by Drongoski
Different people will provide different advice. Advice will depend on where you are at currently.

In general, make sure you are on top of the topics you are going to be examined on. Depending on how much time you have before the test, make sure you get on top of areas you may be weak in currently. Then do as many of the basic exercises (if you have not already done this) and then progress to more challenging questions after that - these harder questions can be from your books and preferably from your school's past, but more recent, test questions. Also don't stay up too late. When you know your stuff, you are more confident and less stressed and conversely. When you are less stressed, you tend to perform better.

Good luck.
thanks

8. ## Re: maths questions help

What's the stationery point of inflection for y = 8(x+1)^3 - 1

9. ## Re: maths questions help

Originally Posted by boredsatan
What's the stationery point of inflection for y = 8(x+1)^3 - 1
y' = 24(x+1)^2
When y'=0,
x = -1

y'' = 48(x+1)
When y''=0,
x = -1

So we can see x = -1 is both a stationary point and a point of inflexion (horizontal p.o.i.)

When x = -1, y = -1

Therefore, the stationary point of inflexion occurs at (-1,-1)

10. ## Re: maths questions help

Originally Posted by boredofstudiesuser1
y' = 24(x+1)^2
When y'=0,
x = -1

y'' = 48(x+1)
When y''=0,
x = -1

So we can see x = -1 is both a stationary point and a point of inflexion (horizontal p.o.i.)

When x = -1, y = -1

Therefore, the stationary point of inflexion occurs at (-1,-1)
Note that in general, y' = 0 and y" = 0 at some point is not sufficient for that point to be an inflection (for example, consider y = x^4 at x = 0).

In this example, the third derivative is non-zero at x = -1, which implies it is an inflection. Or you can note that the second derivative changes sign about x = -1.

11. ## Re: maths questions help

Originally Posted by InteGrand
Note that in general, y' = 0 and y" = 0 at some point is not sufficient for that point to be an inflection (for example, consider y = x^4 at x = 0).

In this example, the third derivative is non-zero at x = -1, which implies it is an inflection. Or you can note that the second derivative changes sign about x = -1.
Yeah, sorry, forgot to put that step in.

12. ## Re: maths questions help

Originally Posted by boredofstudiesuser1
y' = 24(x+1)^2
When y'=0,
x = -1

y'' = 48(x+1)
When y''=0,
x = -1

So we can see x = -1 is both a stationary point and a point of inflexion (horizontal p.o.i.)

When x = -1, y = -1

Therefore, the stationary point of inflexion occurs at (-1,-1)
Thanks for helping

13. ## Re: maths questions help

1. The line with equation y = mx is tangent to the circle with centre (10,0) and radius 5 at the point P(x,y)
a. find the equation of the circle
b. show that the x-coordinate of the point P satisfies the equation (1+m^2)x^2 - 20x + 75=0
c. use the discriminant for this equation to find the exact value of m
d. find the coordinates of P (there are two such points)
e. find the distance of P from the origin

3. A circle has centre at the origin and radius 5. The point P(3,4) lies on the circle
a. Find the gradient of the line segment OP, where O is the origin
b. Find the gradient of the tangent to the circle at P
c. Find the equation of the tangent at P
d. If the tangent crosses the x-axis at A and the y-axis at B, find the length of line segment AB

5. An equilateral triangle ABC circumscribes the circle with equation x^2 + y^2 = a^2. The side BC of the triangle has equation x = -a
a. Find the equations of AB and AC
b. Find the equation of the circle circumscribing triangle ABC

7. For the curve with equation y = square root (x) - 1 and the straight line with equation y = kx, find the values of k such that:
a. the line meets the curve twice
b. the line meets the curve once

Thanks!!

14. ## Re: maths questions help

Originally Posted by boredsatan
1. The line with equation y = mx is tangent to the circle with centre (10,0) and radius 5 at the point P(x,y)
a. find the equation of the circle
b. show that the x-coordinate of the point P satisfies the equation (1+m^2)x^2 - 20x + 75=0
c. use the discriminant for this equation to find the exact value of m
d. find the coordinates of P (there are two such points)
e. find the distance of P from the origin

3. A circle has centre at the origin and radius 5. The point P(3,4) lies on the circle
a. Find the gradient of the line segment OP, where O is the origin
b. Find the gradient of the tangent to the circle at P
c. Find the equation of the tangent at P
d. If the tangent crosses the x-axis at A and the y-axis at B, find the length of line segment AB

5. An equilateral triangle ABC circumscribes the circle with equation x^2 + y^2 = a^2. The side BC of the triangle has equation x = -a
a. Find the equations of AB and AC
b. Find the equation of the circle circumscribing triangle ABC

7. For the curve with equation y = square root (x) - 1 and the straight line with equation y = kx, find the values of k such that:
a. the line meets the curve twice
b. the line meets the curve once

Thanks!!
Bump!!

15. ## Re: maths questions help

Q1. Eqn of circle: (x-10)^2 + y^2 = 5^2; where line meets this circle:
(x-10)^2 + (mx)^2 = 5^2 ==> (1+m^2)x^2 -20x + 75 = 0
(this is a quadratic equation with a repeated roots; discriminant = 0)

.: discriminant = (-20)^2 - 4 x (1+m^2) x 75 = 0 ==> m = +- (1/sqrt(3))

.: lines are: y = +- (1/sqrt(3)) x

.: the original quadratic equation becomes: (4/3)x^2 - 20x + 75 = 0 or 4x^2 - 60x + 225 = 0

i.e. (2x - 15)^2 = 0 ==> x = 15/2 ==> y = +- (1/sqrt(3)) times 15/2

.: you get co-ords of a couple of points, the points of contact of the tangents to the circle.

Sorry - I'll let others do the rest.

16. ## Re: maths questions help

Originally Posted by boredsatan
Bump!!
$\noindent \textbf{Q3.} O(0,0) and P(3,4) and so m_{OP} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4}{3}$

$\noindent Since the point P lies on the circle, OP is a radius and the tangent to the circle at P is perpendicular to OP. So m_{OP} \cdot m_t = -1 \implies m_t = -\frac{3}{4}$

$\noindent Well m_t = -\frac{3}{4} and P(3,4) so the equation of the tangent is y - 4 = -\frac{3}{4}(x - 3). This becomes 4y - 16 = -3x + 9 and so the tangent is 3x + 4y - 25 = 0.$

$\noindent Let y = 0 \implies x_A = \frac{25}{3}. Let x = 0 \implies y_B = \frac{25}{4}. By pythagoras, AB = \sqrt{\left(\frac{25}{3}\right)^2+\left(\frac{25}{ 4}\right)^2} = 10\frac{5}{12} units.$

$\noindent \textbf{Q5.} y = \sqrt{a^2-x^2} \implies \frac{dy}{dx} = \frac{-x}{\sqrt{a^2-x^2}}. Now consider the side AB as the side which touches the circle on its upper half. AB makes an angle of 150 degrees to the positive x-axis and so m_{AB} = \tan{(150)} = -\frac{1}{\sqrt{3}}. Let \frac{dy}{dx} = -\frac{1}{\sqrt{3}} \implies -\sqrt{3}x = -\sqrt{a^2-x^2}. Squaring both sides, 4x^2 = a^2. This means that the sides AB and AC touch the circle at x = \frac{a}{2} \implies y = \pm \frac{\sqrt{3}a}{2} respectively. Now, m_{AB} = -\frac{1}{\sqrt{3}} and touches circle at \left(\frac{a}{2}, \frac{\sqrt{3}a}{2} \right). This means AC: y - \frac{\sqrt{3}a}{2} = \frac{-1}{\sqrt{3}}\left(x-\frac{a}{2}\right) and thus;$

$\noindent AB: 2x + 2\sqrt{3}y - 4a = 0 (\star) \\ AC: 2x - 2\sqrt{3}y - 4a = 0 (\star \star)$

$\noindent The circle circumscribing triangle ABC has centre O(0,0) and passes through A, which lies on the x-axis. Now (\star) + (\star \star): 4x - 8a = 0 \implies x = 2a. Thus, A(2a, 0) and the required circle has radius 2a. And so it's equation is x^2 + y^2 = 4a^2$

$\noindent \textbf{Q7.} If the curve and the line meet twice, then the equation \sqrt{x} - 1 = kx has two, distinct real roots. Squaring both sides gives x = (kx + 1)^2 \implies x = k^2x^2 + 2kx + 1 \implies k^2x^2 + (2k - 1)x + 1 = 0. Two, distinct real roots occur when \Delta > 0. That is, (2k - 1)^2 - 4(k^2)(1) > 0. This is 4k^2 - 4k + 1 - 4k^2 > 0 \implies -4k > - 1, and thus k < \frac{1}{4}. But by squaring both sides, we assumed the lower part of the parabola exists. And so the curve and line meet twice only for 0 < k < \frac{1}{4}.$

$\noindent By graphical inspection, the curve and line meet once when the line has non-positive gradient and when the line is a tangent to the circle. Thus, the curve and line meet once if and only if k \leq 0 or k = \frac{1}{4}$

17. ## Re: maths questions help

Originally Posted by 1729
$\noindent \textbf{Q3.} O(0,0) and P(3,4) and so m_{OP} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4}{3}$

$\noindent Since the point P lies on the circle, OP is a radius and the tangent to the circle at P is perpendicular to OP. So m_{OP} \cdot m_t = -1 \implies m_t = -\frac{3}{4}$

$\noindent Well m_t = -\frac{3}{4} and P(3,4) so the equation of the tangent is y - 4 = -\frac{3}{4}(x - 3). This becomes 4y - 16 = -3x + 9 and so the tangent is 3x + 4y - 25 = 0.$

$\noindent Let y = 0 \implies x_A = \frac{25}{3}. Let x = 0 \implies y_B = \frac{25}{4}. By pythagoras, AB = \sqrt{\left(\frac{25}{3}\right)^2+\left(\frac{25}{ 4}\right)^2} = 10\frac{5}{12} units.$

$\noindent \textbf{Q5.} y = \sqrt{a^2-x^2} \implies \frac{dy}{dx} = \frac{-x}{\sqrt{a^2-x^2}}. Now consider the side AB as the side which touches the circle on its upper half. AB makes an angle of 150 degrees to the positive x-axis and so m_{AB} = \tan{(150)} = -\frac{1}{\sqrt{3}}. Let \frac{dy}{dx} = -\frac{1}{\sqrt{3}} \implies -\sqrt{3}x = -\sqrt{a^2-x^2}. Squaring both sides, 4x^2 = a^2. This means that the sides AB and AC touch the circle at x = \frac{a}{2} \implies y = \pm \frac{\sqrt{3}a}{2} respectively. Now, m_{AB} = -\frac{1}{\sqrt{3}} and touches circle at \left(\frac{a}{2}, \frac{\sqrt{3}a}{2} \right). This means AC: y - \frac{\sqrt{3}a}{2} = \frac{-1}{\sqrt{3}}\left(x-\frac{a}{2}\right) and thus;$

$\noindent AB: 2x + 2\sqrt{3}y - 4a = 0 (\star) \\ AC: 2x - 2\sqrt{3}y - 4a = 0 (\star \star)$

$\noindent The circle circumscribing triangle ABC has centre O(0,0) and passes through A, which lies on the x-axis. Now (\star) + (\star \star): 4x - 8a = 0 \implies x = 2a. Thus, A(2a, 0) and the required circle has radius 2a. And so it's equation is x^2 + y^2 = 4a^2$

$\noindent \textbf{Q7.} If the curve and the line meet twice, then the equation \sqrt{x} - 1 = kx has two, distinct real roots. Squaring both sides gives x = (kx + 1)^2 \implies x = k^2x^2 + 2kx + 1 \implies k^2x^2 + (2k - 1)x + 1 = 0. Two, distinct real roots occur when \Delta > 0. That is, (2k - 1)^2 - 4(k^2)(1) > 0. This is 4k^2 - 4k + 1 - 4k^2 > 0 \implies -4k > - 1, and thus k < \frac{1}{4}. But by squaring both sides, we assumed the lower part of the parabola exists. And so the curve and line meet twice only for 0 < k < \frac{1}{4}.$

$\noindent By graphical inspection, the curve and line meet once when the line has non-positive gradient and when the line is a tangent to the circle. Thus, the curve and line meet once if and only if k \leq 0 or k = \frac{1}{4}$
You're a legend

18. ## Re: maths questions help

Originally Posted by Drongoski
Very few here are familiar with the VCE. But I find the VCE Maths curriculum modern and progressive.
Would you by any chance have any idea of vce maths methods difficulty in getting 40 raw

Sorry, No.

20. ## Re: maths questions help

If you don't know how to do a question, then consider the event that you don't understand question; you know what to do in this case.

Otherwise, you need to figure out, precisely, why you can't do a question. So, if this habit is left unchecked, then you may become more dependent on people helping you to learn new things; clearly a detriment to your cognitive growth and independence.

21. ## Re: maths questions help

Originally Posted by He-Mann
If you don't know how to do a question, then consider the event that you don't understand question; you know what to do in this case.

Otherwise, you need to figure out, precisely, why you can't do a question. So, if this habit is left unchecked, then you may become more dependent on people helping you to learn new things; clearly a detriment to your cognitive growth and independence.
Are you trying to degrade me?

22. ## Re: maths questions help

Originally Posted by He-Mann
If you don't know how to do a question, then consider the event that you don't understand question; you know what to do in this case.

Otherwise, you need to figure out, precisely, why you can't do a question. So, if this habit is left unchecked, then you may become more dependent on people helping you to learn new things; clearly a detriment to your cognitive growth and independence.
What would you do if you found a maths question hard?

23. ## Re: maths questions help

Originally Posted by boredsatan
What would you do if you found a maths question hard?
Keep trying or check the answers (then redo it).

24. ## Re: maths questions help

Originally Posted by boredsatan
Are you trying to degrade me?
No, I apologize for coming off like that, I didn't mean to, but I'm just telling you what I think straight up.

25. ## Re: maths questions help

Originally Posted by boredsatan
What would you do if you found a maths question hard?
In my experience, it's either a lack of knowledge in a topic or it requires a whole new perspective on a concept.

Assume that I've understood the question and have enough knowledge in the relevant topics. Then I usually employ a systematic approach described by George Polya in his book How to Solve It:

1. First principle: Understand the problem

2. Second principle: Devise a plan

3. Third principle: Carry out the plan

4. Fourth principle: Review/extend

Works well in my experience when you don't have solutions. But I used to have a habit of checking solutions (given that I had solutions) before giving 100% on it; missed out on mental development here :P

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