HSC 2015 MX1 Marathon (archive) (3 Viewers)

davidgoes4wce · Jul 12, 2015

Re: HSC 2015 3U Marathon

There is an alternate answer which is

davidgoes4wce · Jul 12, 2015

Re: HSC 2015 3U Marathon

I think I may have figured it out, I put the

a sin theta + b cos theta= A sin (theta +alpha)

the book used:

a cos theta + b sin theta = A cos (theta-alpha)

Both from what I gather give you the same answer

davidgoes4wce · Jul 12, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce · Jul 12, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
There is an alternate answer which is

Could someone confirm if my answer of x=sqrt(3) sin (2t + pi/6) is also the right answer?

leehuan · Jul 12, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:

$\\ x=\sin { \left( 2t+\frac { \pi }{ 2 } \right) } +\cos { \left( 2t+\frac { \pi }{ 3 } \right) } \\ v=2\cos { \left( 2t+\frac { \pi }{ 2 } \right) } -2\sin { \left( 2t+\frac { \pi }{ 3 } \right) } \\ \ddot { x } =-4\sin { \left( 2t+\frac { \pi }{ 2 } \right) } -4\cos { \left( 2t+\frac { \pi }{ 3 } \right) } \\ \ddot { x } =-4\left( \sin { \left( 2t+\frac { \pi }{ 2 } \right) } +\cos { \left( 2t+\frac { \pi }{ 3 } \right) } \right) \\ \ddot { x } =-4x \\ $Hence the particle moves in S.H.M. $$

leehuan · Jul 12, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Could someone confirm if my answer of x=sqrt(3) sin (2t + pi/6) is also the right answer?

Not sure why you wrote sin(2t+pi/6) when you had sin(2t+pi/3) in your working out.

As for whether it is true or not...
$\sqrt { 3 } \cos { \left( 2t-\frac { \pi }{ 6 } \right) } \\ =\sqrt { 3 } \cos { \left( \frac { \pi }{ 6 } -2t \right) } \\ =\sqrt { 3 } \sin { \left( \frac { \pi }{ 2 } -\left( \frac { \pi }{ 6 } -2t \right) \right) } \\ =\sqrt { 3 } \sin { \left( 2t+\frac { \pi }{ 3 } \right) } \\ $Hence you are right$$

Properties used:
f(x)=cos(x) is an even function
The sine and cosine ratios are complementary.

leehuan · Jul 12, 2015

Re: HSC 2015 3U Marathon

NEXT QUESTION FOR THE MARATHON:

$\\ $1. Explain why $\int _{ -1 }^{ 1 }{ \sin ^{ -1 }{ x } dx } =0$ but $\int _{ -1 }^{ 1 }{ \cos ^{ -1 }{ x } dx } \neq 0 \\ $2. Find $\frac { d }{ dx } \left( \sqrt { 1-{ x }^{ 2 } } +x\sin ^{ -1 }{ x } \right) \\ $3. Hence, calculate the area bound by the curve $ y=\sin ^{ -1 }{ x } $ and $x=1\\ $4. Use an alternate approach to calculate the area.$$

Hint for 4:

Sketch the curve y=arcsin(x) and observe different regions.

1 is unrelated to the other three parts.

rand_althor · Jul 12, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
Edit: found the error — the limits of integration weren't changed.

leehuan said:
Check whether or not you changed your integration boundaries from the change of variable.

Okay thanks guys! Is this the correct method then?:
$\begin{align*}\textup{When }x=1, \, u&=0\textup{ and when }x=0\,,u=1 \\\int^1_0\frac{\ln{(1+x)}}{1+x^2}\,\mathrm{d}x&=\int^0_1\frac{\ln{(1+\frac{1-u}{1+u}})}{1+(\frac{1-u}{1+u})^2}\times\frac{-2}{(1+u)^2}\,\mathrm{d}u \\&\vdots \\&= -\int^1_0\frac{\ln(1+x)}{1+x^2}\,\mathrm{d}x + \ln2\int^1_0\frac{1}{1+u^2}\,\mathrm{d}u \\2\int^1_0\frac{\ln{(1+x)}}{1+x^2}\,\mathrm{d}x&=\ln2\int^1_0\frac{1}{1+u^2}\,\mathrm{d}u \\\int^1_0\frac{\ln{(1+x)}}{1+x^2}\,\mathrm{d}x&=\frac{\ln2}{2}\int^1_0\frac{1}{1+u^2}\,\mathrm{d}u \\ &=\frac{\ln2}{2}\left [\tan^{-1}x \right ]^1_0 \\ &= \frac{\ln2}{2}\left [ \tan^{-1}1-\tan^{-1}0 \right ] \\&= \frac{\ln2}{2} \times\frac{\pi}{4} \\&= \frac{\pi\ln2}{8}\end{align*}$

davidgoes4wce · Jul 12, 2015

Re: HSC 2015 3U Marathon

Find the general solution

I know the general solution for sin for this example is the following:

my question is with the alpha value. Can we use an alpha value of -Pi/4 instead of the one suggested in the text book of 5Pi/4?

InteGrand · Jul 12, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Find the general solution

I know the general solution for sin for this example is the following:

my question is with the alpha value. Can we use an alpha value of -Pi/4 instead of the one suggested in the text book of 5Pi/4?

Yes you can use any $\alpha$ in that formula such that $2\theta =\alpha$ would satisfy the equation.

Drsoccerball · Jul 12, 2015

Re: HSC 2015 3U Marathon

leehuan said:
NEXT QUESTION FOR THE MARATHON:

$\\ $1. Explain why $\int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \sin ^{ -1 }{ x } dx } =0$ but $\int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \cos ^{ -1 }{ x } dx } \neq 0 \\ $2. Find $\frac { d }{ dx } \left( \sqrt { 1-{ x }^{ 2 } } +x\sin ^{ -1 }{ x } \right) \\ $3. Hence, calculate the area bound by the curve $ y=\sin ^{ -1 }{ x } $ and $x=\frac { \pi }{ 2 } \\ $4. Use an alternate approach to calculate the area.$$

Hint for 4:

Sketch the curve y=arcsin(x) and observe different regions.

1 is unrelated to the other three parts.

1) Sin inverse is an odd function therefore an integral between the two boundaries would cause a result of 0 while cos inverse isn't an odd function the result would not yield a 0 value
2) sin^-1x
3) $\int{sin^{-1}x}dx = [\sqrt{1-x^2}+xsin^{-1}x]_0^{\frac{pi}{2}}$
4) use areas $\int_0^1{sinx}dx$

Sorry if i made mistakes im in a rush

Ekman · Jul 12, 2015

Re: HSC 2015 3U Marathon

leehuan said:
NEXT QUESTION FOR THE MARATHON:

$\\ $1. Explain why $\int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \sin ^{ -1 }{ x } dx } =0$ but $\int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \cos ^{ -1 }{ x } dx } \neq 0 \\ $2. Find $\frac { d }{ dx } \left( \sqrt { 1-{ x }^{ 2 } } +x\sin ^{ -1 }{ x } \right) \\ $3. Hence, calculate the area bound by the curve $ y=\sin ^{ -1 }{ x } $ and $x=\frac { \pi }{ 2 } \\ $4. Use an alternate approach to calculate the area.$$

Hint for 4:

Sketch the curve y=arcsin(x) and observe different regions.

1 is unrelated to the other three parts.

This question doesn't make sense, because the domain of sin^-1x is $-1\leq x \leq 1$

Im assuming you meant to say between y=sin^-1x and x=1

leehuan · Jul 12, 2015

Re: HSC 2015 3U Marathon

Yes I was. Fixing it now, thanks.

I must have been too busy thinking about the sine function, not the inverse sine.

davidgoes4wce · Jul 15, 2015

Re: HSC 2015 3U Marathon

I was just going through Binomial Expansion and this was the question

In the expansion of (1+x)^n, three consecutive coefficients are in the ratio of 6:3:1. I understand how to do this question but will the term have to be always be ' (1+x)^n ?

davidgoes4wce · Jul 15, 2015

Re: HSC 2015 3U Marathon

Disregard my last question, just saw a question in my textbook asking for the binomial expansion of (2+3x)^n in the ratio of 8:15

Ambility · Jul 16, 2015

Re: HSC 2015 3U Marathon

I stumped my class with this one today:

$\text{Make y the subject of the formula}\\2\sqrt{x}\sqrt{y}-y+x^2=x$

VBN2470 · Jul 16, 2015

Re: HSC 2015 3U Marathon

Ambility said:
I stumped my class with this one today:

$\text{Make y the subject of the formula}\\2\sqrt{x}\sqrt{y}-y+x^2=x$

Re-arranging the equation gives,

$x^2=(\sqrt{x}-\sqrt{y})^2 \Rightarrow y=(\sqrt{x}\pm{x})^2$

davidgoes4wce · Jul 19, 2015

Re: HSC 2015 3U Marathon

Not sure how to draw 1 of these out.

davidgoes4wce · Jul 19, 2015

Re: HSC 2015 3U Marathon

This was my attempt. (Some of you guys might find it funny)

InteGrand · Jul 19, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
This was my attempt. (Some of you guys might find it funny)

That diagram is correct, but there isn't any need to draw the circumcircle of triangle ABC for this question. Also, you may want to label sides BD and DC as equal. And it might help you visualise things better if you draw the triangle so that BC is horizontal (if it doesn't, then don't worry).

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