Author Topic: Mathematics Question Thread (Read 1626079 times) Tweet Share

inescelic · « **Reply #2775 on:** October 13, 2017, 01:22:19 pm »

Please help with these questions from 2004 HSC, no solutions from nesa

itssona · « **Reply #2776 on:** October 13, 2017, 02:00:39 pm »

pls help<br>
a parabola with vertical axis has its vertex at origin. if the line 8x-y-4=0 is a tangent to the parabola fine yhe value of k

RuiAce · « **Reply #2777 on:** October 13, 2017, 02:17:35 pm »

Quote from: inescelic on October 13, 2017, 01:22:19 pm

Please help with these questions from 2004 HSC, no solutions from nesa

Some of these were previously answered.
more

Quote from: sssona09 on October 13, 2017, 02:00:39 pm

pls help<br>
a parabola with vertical axis has its vertex at origin. if the line 8x-y-4=0 is a tangent to the parabola fine yhe value of k

What's k?

Natasha.97 · « **Reply #2778 on:** October 13, 2017, 02:33:49 pm »

Quote from: inescelic on October 13, 2017, 01:22:19 pm

Please help with these questions from 2004 HSC, no solutions from nesa
2004 6bii)

Hi!

$\text{part i)} \\ \text{YM is common} \\ AM = \text{CM (given)} \\ \angle AMY = \angle CMY \text{ (AC is perpendicular to YM)} \\ \text{Therefore } \triangle AYM \equiv \triangle \text {CYM (S.A.S)}$

$\text{part ii)} \\ \angle YAB = \angle YAM \text{(AY bisects Angle BAC)} \\ \angle YAM = \angle YCM \text{(corresponding angles in congruent triangles)} \\ ∴ \angle YAB = \angle YAM = \angle YCM$
$\text {In } \triangle \text {ABC, } \angle ABC + \angle BAC + \angle YCM = 180^{\circ} \\ ∴ 90^{\circ} + \angle YAB + \angle YAM + \angle YCM = 180^{\circ} \\ ∴ 90^{\circ} + 3(\angle YCM) = 180^{\circ} \\ ∴ 3\angle YCM = 90 ^{\circ} \\ ∴ \angle YCM = 30^{\circ}$
$\text{tan } \angle YCM = \frac{MY}{MC} \\ \text {tan }30^{\circ} = \frac {1}{\sqrt{3}} \\ ∴ \frac{MY}{MC} = \frac{1}{2\sqrt{3}}$

$\text{but AC } = 2 \text{MC} \\ ∴ \frac{MY}{AC} = \frac{MY}{2MC} = \frac{1}{2\sqrt{3}} \\ ∴ MY:AC = 1:2\sqrt{3}$

Edit: Added LaTeX

av-angie-er · « **Reply #2779 on:** October 13, 2017, 11:47:34 pm »

Hi! Can I get some help for part (ii)? I've got the rationalised denominators down, but I'm not quite sure how to approach applying AP/GP sums. Thanks!

RuiAce · « **Reply #2780 on:** October 14, 2017, 12:12:08 am »

Quote from: av-angie-er on October 13, 2017, 11:47:34 pm

Hi! Can I get some help for part (ii)? I've got the rationalised denominators down, but I'm not quite sure how to approach applying AP/GP sums. Thanks!

Already addressed in the compilation. It is not related to arithmetic or geometric sums.

georgiia · « **Reply #2781 on:** October 14, 2017, 12:33:15 pm »

Could I please have help with this? Thanks!

RuiAce · « **Reply #2782 on:** October 14, 2017, 04:41:05 pm »

Quote from: georgiia on October 14, 2017, 12:33:15 pm

(Image removed from quote.)

Could I please have help with this? Thanks!

$\text{One possible approach to show that }x_A\neq x_B\text{ for all }t\ge 0 \\ \text{is to show that either }x_A < x_B\text{ for all }t \ge 0\\ \text{or }x_A > x_B\text{ for all }t \ge 0.\\ \text{To do this, we will consider the displacement }\textbf{between the particles}.$
Inspiration: If you try graphing $x_A$ and $x_B$, you will find that the graph of $x_B$ will always be above $x_A$. Because one graph is always above the other, there will not be any points of intersection, and hence the particles will not meet. But we will pretend we didn't know that.
_________________________________________
$\text{Let }f(t) = x_A - x_B\\ \text{i.e. }\boxed{f(t)=\frac{1}{2}t+1 - \ln (t+1)}\\ \text{Then, }f^\prime(t) = \frac{1}{2}-\frac{1}{t+1}$
$\text{It can be shown that the stationary point of }f(t)\text{ occurs at }t=1\\ \text{Since }f^{\prime\prime}(t) = \frac{1}{(t+1)^2} \ge 0\text{, we conclude that this stationary point is a local minimum}\\ \text{Further checking shows that this is indeed the global minimum}.$
$\text{For convenience, we graph }f(t).$

$\text{From the graph, it appears that }\textbf{for all}\, t\ge 0, \quad f(t) \ge 0$
$\text{Hence }\frac{x}{2} + 1 - \ln (x+1) \ge 0\\ \text{which rearranges to }\boxed{\frac{x}{2} + 1 \ge \ln (x+1)}$
$\text{Subbing back in, we have }\boxed{x_A > x_B}\\ \text{so the particles never intersect}\\ \text{because the displacement of }A\text{ will always be}\\ \text{greater than that of }B$

Mathew587 · « **Reply #2783 on:** October 14, 2017, 07:50:18 pm »

Quote from: RuiAce on October 14, 2017, 04:41:05 pm

$\text{One possible approach to show that }x_A\neq x_B\text{ for all }t\ge 0 \\ \text{is to show that either }x_A < x_B\text{ for all }t \ge 0\\ \text{or }x_A > x_B\text{ for all }t \ge 0.\\ \text{To do this, we will consider the displacement }\textbf{between the particles}.$
Inspiration: If you try graphing $x_A$ and $x_B$, you will find that the graph of $x_B$ will always be above $x_A$. Because one graph is always above the other, there will not be any points of intersection, and hence the particles will not meet. But we will pretend we didn't know that.
_________________________________________
$\text{Let }f(t) = x_A - x_B\\ \text{i.e. }\boxed{f(t)=\frac{1}{2}t+1 - \ln (t+1)}\\ \text{Then, }f^\prime(t) = \frac{1}{2}-\frac{1}{t+1}$
$\text{It can be shown that the stationary point of }f(t)\text{ occurs at }t=1\\ \text{Since }f^{\prime\prime}(t) = \frac{1}{(t+1)^2} \ge 0\text{, we conclude that this stationary point is a local minimum}\\ \text{Further checking shows that this is indeed the global minimum}.$
$\text{For convenience, we graph }f(t).$
(Image removed from quote.)
$\text{From the graph, it appears that }\textbf{for all}\, t\ge 0, \quad f(t) \ge 0$
$\text{Hence }\frac{x}{2} + 1 - \ln (x+1) \ge 0\\ \text{which rearranges to }\boxed{\frac{x}{2} + 1 \ge \ln (x+1)}$
$\text{Subbing back in, we have }\boxed{x_A > x_B}\\ \text{so the particles never intersect}\\ \text{because the displacement of }A\text{ will always be}\\ \text{greater than that of }B$

Couldn't we have just graphed both the equations of the same graph and show that they never meet?

RuiAce · « **Reply #2784 on:** October 14, 2017, 07:54:21 pm »

Quote from: Mathew587 on October 14, 2017, 07:50:18 pm

Couldn't we have just graphed both the equations of the same graph and show that they never meet?

How would you have known that the graphs don't intersect?

I've seen students not introduce points of intersection when they should be there. I've also seen students who introduce points of intersection when they should not have as well.

bdobrin · « **Reply #2785 on:** October 15, 2017, 11:47:42 am »

Hi there,

I was just wondering if someone could explain for the answer to this question, why they did the 'democracy rule' where they share the denominator instead of integrating it normally? Dont you just check if the derivative of the bottom is equal to the numerator (which it is in this case).

Thanks,
Ben

Natasha.97 · « **Reply #2786 on:** October 15, 2017, 12:12:15 pm »

Quote from: bdobrin on October 15, 2017, 11:47:42 am

Hi there,

I was just wondering if someone could explain for the answer to this question, why they did the 'democracy rule' where they share the denominator instead of integrating it normally? Dont you just check if the derivative of the bottom is equal to the numerator (which it is in this case).

Thanks,
Ben

Hi Ben!

In this case, the "democracy rule" would have to be used, as the derivative of the denominator is $2x$. If the denominator was $x^2+x$, then the answer would be $ln(x^2+x) + C$.

Hope this helps

caitlinlddouglas · « **Reply #2787 on:** October 17, 2017, 04:42:28 pm »

Hey i was just wondering how to work out which side of the triangle went with which for this question? I can't work it out

Thanks!:)

pokemonlv10 · « **Reply #2788 on:** October 17, 2017, 06:29:20 pm »

Hey, was wondering if c and d are both the correct answer in the 2nd screenshot? apparently the answer is c but i think d is plausible as well. Also, is my drawing for 1st picture correct? (the point of inflection is poorly drawn sorry)

RuiAce · « **Reply #2789 on:** October 17, 2017, 07:56:51 pm »

Quote from: caitlinlddouglas on October 17, 2017, 04:42:28 pm

Hey i was just wondering how to work out which side of the triangle went with which for this question? I can't work it out

Thanks!:)

Because the HSC always does things correctly (as opposed to some certain textbooks), the ordering goes with the order they are listed in.
$\text{So since we needed to prove that }\triangle DBA|||\triangle ABC\\ DB\text{ corresponds to }AB\\ BA\text{ corresponds to }BC\\DA\text{corresponds to }AC$

Quote from: pokemonlv10 on October 17, 2017, 06:29:20 pm

Hey, was wondering if c and d are both the correct answer in the 2nd screenshot? apparently the answer is c but i think d is plausible as well. Also, is my drawing for 1st picture correct? (the point of inflection is poorly drawn sorry)

I don't know what you mean by c and d. Your second screenshot involves a graph.

I assume that in your first screenshot you were given the black curve and needed to draw the blue one, which was its primitive. If that were the case, the only problem I have with it is that as you go further to the left $ (x < -4) $, your graph becomes horizontal. That definitely should not happen.

Other than that it looks fine.

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inescelic

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