# VCE Maths questions help (1 Viewer)

#### 1729

##### Active Member
Re: maths questions help

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

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

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

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

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

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

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

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

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

#### boredsatan

##### Member
Re: maths questions help

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

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

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

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

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

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

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

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

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

#### boredsatan

##### Member
Re: maths questions help

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

#### Drongoski

##### Well-Known Member
Re: maths questions help

Sorry, No.

#### He-Mann

##### Vexed?
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.

#### boredsatan

##### Member
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.
Are you trying to degrade me?

#### boredsatan

##### Member
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.
What would you do if you found a maths question hard?

#### pikachu975

##### I love trials
Moderator
Re: maths questions help

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

#### He-Mann

##### Vexed?
Re: maths questions help

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.

#### He-Mann

##### Vexed?
Re: maths questions help

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

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#### boredsatan

##### Member
Re: maths questions help

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
would you check worked solutions or non worked solutions?

#### He-Mann

##### Vexed?
Re: maths questions help

would you check worked solutions or non worked solutions?
Worked solutions.

#### boredsatan

##### Member
Is it concerning if I need to use worked solutions to do maths homework? (I have tutor as well)

Bump!!

#### boredsatan

##### Member
What does the domain R- mean?

#### InteGrand

##### Well-Known Member
What does the domain R- mean?
Probably the negative reals. What was the context?

#### boredsatan

##### Member
Probably the negative reals. What was the context?
to sketch the graph of y = 2-x^2, x belongs to R-

#### InteGrand

##### Well-Known Member
to sketch the graph of y = 2-x^2, x belongs to R-
Yeah, so it means sketch it for x < 0.

#### boredsatan

##### Member
Yeah, so it means sketch it for x < 0.
so basically sketch just the left hand side of the parabola?