VCE Maths questions help (1 Viewer)

1729 · May 4, 2017

Re: maths questions help

boredsatan said:
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 $k = \frac{1}{4}$

boredsatan · May 4, 2017

Re: maths questions help

1729 said:
$\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

boredsatan · May 5, 2017

Re: maths questions help

Drongoski said:
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 · May 5, 2017

Re: maths questions help

Sorry, No.

He-Mann · May 5, 2017

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 · May 5, 2017

Re: maths questions help

He-Mann said:
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 · May 5, 2017

Re: maths questions help

He-Mann said:
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 · May 5, 2017

Re: maths questions help

boredsatan said:
What would you do if you found a maths question hard?

Keep trying or check the answers (then redo it).

He-Mann · May 5, 2017

Re: maths questions help

boredsatan said:
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 · May 5, 2017

Re: maths questions help

boredsatan said:
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

boredsatan · May 5, 2017

Re: maths questions help

He-Mann said:
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 · May 5, 2017

Re: maths questions help

boredsatan said:
would you check worked solutions or non worked solutions?

Worked solutions.

boredsatan · May 8, 2017

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

boredsatan · May 9, 2017

Bump!!

boredsatan · May 9, 2017

What does the domain R- mean?

InteGrand · May 9, 2017

boredsatan said:
What does the domain R- mean?

Probably the negative reals. What was the context?

boredsatan · May 9, 2017

InteGrand said:
Probably the negative reals. What was the context?

to sketch the graph of y = 2-x^2, x belongs to R-

InteGrand · May 9, 2017

boredsatan said:
to sketch the graph of y = 2-x^2, x belongs to R-

Yeah, so it means sketch it for x < 0.

boredsatan · May 9, 2017

InteGrand said:
Yeah, so it means sketch it for x < 0.

so basically sketch just the left hand side of the parabola?

boredsatan · May 9, 2017

InteGrand said:
Yeah, so it means sketch it for x < 0.

So R- means <0?

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VCE Maths questions help (1 Viewer)

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