HSC 2013 Maths Marathon (archive) (1 Viewer)

leesh95 · Oct 17, 2013

Re: HSC 2013 2U Marathon

mahmoudali said:
$prove: \frac{cos(2x)}{cos(x)} = \frac{1-tan^2(x)}{sec(x)}$

I can't get this question? How do you do this.

Menomaths · Oct 17, 2013

Re: HSC 2013 2U Marathon

Carrotsticks said:
For purposes of the HSC yes.

That's gay

Carrotsticks · Oct 17, 2013

Re: HSC 2013 2U Marathon

leesh95 said:
I can't get this question? How do you do this.

It's out of the 2U course, so no need to worry.

Menomaths said:
That's gay

No, it's not 'gay'.

The point of the question is for you to practise deducing some expression from another given expression. Accepting the required result and playing with it until you get a known identity is technically valid for that question, but defeats the purpose of it.

Sorry if I'm being a bit snappy, but people saying "Oh.. that's gay" really rustles my jimmies.

leesh95 · Oct 17, 2013

Re: HSC 2013 2U Marathon

Carrotsticks said:
Not quite, the angles are in AP but the actual terms may not necessarily be in AP.

Hint: Consider the identity sin(x) = cos(90-x)

Can you please explain a bit more? I can't understand.

Carrotsticks · Oct 17, 2013

Re: HSC 2013 2U Marathon

leesh95 said:
Can you please explain a bit more? I can't understand.

Just because we have A + B + C + D + ..., doesn't always mean we have an AP.

The series was not an AP. You can easily check using a calculator that T_2 - T_1 is not equal to T_3 - T_2.

Sy123 · Oct 17, 2013

Re: HSC 2013 2U Marathon

leesh95 said:
Can you please explain a bit more? I can't understand.

$\sin^2 89^{\circ} = \cos^2 (90 - 89)^{\circ} = \cos^2 1^{\circ}$

Menomaths · Oct 17, 2013

Re: HSC 2013 2U Marathon

Show working/thought process

mahmoudali · Oct 17, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
Show working/thought process

c

$$working out$ \ \ x^2 +y^2 +6y +9 =16\ \$
$x^2 +(y+3)^2=4^2 \ \ therefore\ \ center (0,-3) radius 4$

Menomaths · Oct 17, 2013

Re: HSC 2013 2U Marathon

mahmoudali said:
c

$$working out$ \ \ x^2 +y^2 +6y +9 =16\ \$
$x^2 +(y+3)^2=4^2 \ \ therefore\ \ center (0,-3) radius 4$

How did you know you had to add 9?

mahmoudali · Oct 17, 2013

Re: HSC 2013 2U Marathon

$\int_0^\frac{\pi}{2} \frac{1}{sin(x)} dx$

mahmoudali · Oct 17, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
How did you know you had to add 9?

because you look at 6y and you need y^2 +6y +c to be a perfect square and only 9 makes it so
so add 9 to both sides since genral equation of a circle is (x-h)^2 +(y-k)^2 = r

mahmoudali · Oct 17, 2013

Re: HSC 2013 2U Marathon

mahmoudali said:
because you look at 6y and you need y^2 +6y +c to be a perfect square and only 9 makes it so
so add 9 to both sides since genral equation of a circle is (x-h)^2 +(y-k)^2 = r

idk what im talking about lol this was the pattern i saw in solutions

Sy123 · Oct 17, 2013

Re: HSC 2013 2U Marathon

mahmoudali said:
$\int_0^\frac{\pi}{2} \frac{1}{sin(x)} dx$

I'm unable to do this integral can you please help me and post your solution right now

-----------

$\int_0^{\pi /4} \tan^2 x \ dx$

Menomaths · Oct 17, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
I'm unable to do this integral can you please help me and post your solution right now

-----------

$\int_0^{\pi /4} \tan^2 x \ dx$

$\int_0^{\pi /4} \tan^2 x \ dx$

$\int_0^{\pi /4} \sec^2 x -1\ dx$

$[tanx-x]$ bounded by pi/4 and 0

$=tan\frac{\pi}{4}-\frac{\pi}{4}$

$=0.21$

RealiseNothing · Oct 17, 2013

Re: HSC 2013 2U Marathon

Sy posted this up I think ages ago, but it's still probably the best 2U integral I've seen:

$\int \frac{dx}{x\ln(x)}$

Sy123 · Oct 17, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
$\int_0^{\pi /4} \tan^2 x \ dx$

$\int_0^{\pi /4} \sec^2 x -1\ dx$

$[tanx-x]$ bounded by pi/4 and 0

$=tan\frac{pi}{4}-\frac{pi}{4}$

$=0.21$

Yea well done

$$i) The point$ \ \ P(a, b) \ \ $lies on the function$ \ \ y=\sqrt{1-x^2}$

$$Find the equation of tangent to the function at$ \ \ P$

$ii) Hence prove that the tangent to a circle is perpendicular to the radius$

Shazer2 · Oct 17, 2013

Re: HSC 2013 2U Marathon

${ \int _{ 0 }^{ \frac { \pi }{ 4 } }{ ({ sec }^{ 2 }x-1)\quad dx } }\\ \indent = { [tanx-x] }_{ 0 }^{ \frac { \pi }{ 4 } }\\ \indent = [1-\frac { \pi }{ 4 } ]\\ \indent = \frac { 4-\pi }{ 4 }$

Shazer2 · Oct 17, 2013

Re: HSC 2013 2U Marathon

RealiseNothing said:
Sy posted this up I think ages ago, but it's still probably the best 2U integral I've seen:

$\int \frac{dx}{x\ln(x)}$

I think this is right?

$\int \frac {1}{xln(x)} dx \\ \\ \indent = x \int \frac {\frac {1}{x}}{xln(x)} dx \\ \\ \indent = xln(xln(x)) + c$

HeroicPandas · Oct 17, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
How did you know you had to add 9?

He 'completed the square'

ColdLipstick · Oct 17, 2013

Re: HSC 2013 2U Marathon

HSC 2013 Maths Marathon (archive) (1 Viewer)

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