HSC 2013-14 MX1 Marathon (archive) (4 Viewers)

SpiralFlex · Nov 29, 2012

Thought I would start this early. Please post questions that are reasonable for Extension 1 level.

Post cool questions that you have encountered, elegant results, answers, help each other, explain, enjoy! Don't be shy boys and girls.

HeroicPandas · Nov 29, 2012

Re: HSC 2013 3U Marathon Thread

$Car A and car B are traveling along a straight level road at constant speeds $V_A $ and $V_B$ $respectively.$ Car A is behind car B, but is traveling faster. \\ \\ $ $When car A is exactly D metres behind car B, car A applies its brakes, producing a constant deceleration of $k m/s^2$

$$ i) Using calculus, find the speed of car A after it has travelled a distance x metres under braking.$
$$ ii) Prove that the cars will collide if V_A -V_B >\sqrt{2kD}$

jyu · Nov 29, 2012

Re: HSC 2013 3U Marathon Thread

Kurosaki · Nov 30, 2012

Re: HSC 2013 3U Marathon Thread

Ok guys.
Given that the polynomial P(x) when divided by (x^2 + 4) gives the remainder (x + 8) and when divided by x gives by the remainder Of -4 find P(x) which is a monic cubic polynomial expressing in in the form ax^3 + bx^2 + cx + d

epsilon- · Nov 30, 2012

Re: HSC 2013 3U Marathon Thread

Kurosaki said:
Ok guys.
Given that the polynomial P(x) when divided by (x^2 + 4) gives the remainder (x + 8) and when divided by x gives by the remainder Of -4 find P(x) which is a monic cubic polynomial expressing in in the form ax^3 + bx^2 + cx + d

Use "The Form".

Kurosaki · Nov 30, 2012

Re: HSC 2013 3U Marathon Thread

epsilon- said:
Use "The Form".

What exactly do you mean by that?

Sy123 · Nov 30, 2012

Re: HSC 2013 3U Marathon Thread

$$Find a general quadratic function such that if $ f(x) \\ $ is the quadratic function then $ \\ \\ f(f'(x))=f'(f(x))$

I feel like it belongs here more than it does in the 4U marathon.

Genero · Nov 30, 2012

Re: HSC 2013 3U Marathon Thread

Hopefully I don't make a typo (the c's and x's are too close to each other on the keyboard). What 3u topic is Sy123's question? Here's my 2 cents:

f(x) = ax^2 + bx + c
f'(x) = 2ax + b

f[f'(x)] = f(2ax + b)
= a(2ax+b)^2 + b(2ax + b) + c
= 4a^3.x^2 + 4a^2.xb + ab^2 + 2abx + b^2 + c

f'[f(x)] = f'(ax^2 + bx + c)
= 2a(ax^2 + bx + c) + b
= 2a^2.x^2+ 2abx + 2ac + b

f[f'(x)] = f'[f(x)]

therefore,
4a^3.bx^2 + 4a^2.xb + ab^2 + b^2 + c - 2a^2.x^2 - 2ac - b = 0, which doesn't look right lol

Sy123 · Dec 1, 2012

Re: HSC 2013 3U Marathon Thread

Genero said:
Hopefully I don't make a typo (the c's and x's are too close to each other on the keyboard). What 3u topic is Sy123's question? Here's my 2 cents:

f(x) = ax^2 + bx + c
f'(x) = 2ax + b

f[f'(x)] = f(2ax + b)
= a(2ax+b)^2 + b(2ax + b) + c
= 4a^3.x^2 + 4a^2.xb + ab^2 + 2abx + b^2 + c

f'[f(x)] = f'(ax^2 + bx + c)
= 2a(ax^2 + bx + c) + b
= 2a^2.x^2+ 2abx + 2ac + b

f[f'(x)] = f'[f(x)]

therefore,
4a^3.bx^2 + 4a^2.xb + ab^2 + b^2 + c - 2a^2.x^2 - 2ac - b = 0, which doesn't look right lol

If I were to give this a topic it would be 'Harder 2U'
However you need to use a technique that you probably would not have seen if you haven't done polynomials.

However you do not need it, if you can notice something special about f'(x), and notice how it can only take one function then you can do that too.

epsilon- · Dec 1, 2012

Re: HSC 2013 3U Marathon Thread

Kurosaki said:
What exactly do you mean by that?

I don't know... I asked my friend and he said use "The Form".

Kurosaki · Dec 1, 2012

Re: HSC 2013 3U Marathon Thread

epsilon- said:
I don't know... I asked my friend and he said use "The Form".

Ok......I'll post the solution if noone goes around to answering it by 7pm

Kurosaki · Dec 2, 2012

Re: HSC 2013 3U Marathon Thread

Kurosaki said:
Ok guys.
Given that the polynomial P(x) when divided by (x^2 + 4) gives the remainder (x + 8) and when divided by x gives by the remainder Of -4 find P(x) which is a monic cubic polynomial expressing in in the form ax^3 + bx^2 + cx + d

Solution. we all know that we can write P(x) as P(x)= A(x).Q(x) + R(x) (dot is multiplication)
So we write P(x)= (x^2 +4)(x+k) +(x+8), Why (x+k), with k being a constant? Its monic, so the leading term must have a coeffiecient of zero.
Then we know that when divided by x, the remainder is minus 4. but isnt (x)= (x-0)? From the remainder theorem we know that when divided by (x-a), P(a)= the remainder when you divide by P(a).
so. We sub in x=0
therefore P(0)= 4k +8.
we know that P(0)=-4
therefore
4k +8=-4
k=-3
sub k in and you get the polynomial after expanding

.
epsilon was this what your friend meant?

epsilon- · Dec 2, 2012

Re: HSC 2013 3U Marathon Thread

Kurosaki said:
Solution. we all know that we can write P(x) as P(x)= A(x).Q(x) + R(x) (dot is multiplication)
So we write P(x)= (x^2 +4)(x+k) +(x+8), Why (x+k), with k being a constant? Its monic, so the leading term must have a coeffiecient of zero.
Then we know that when divided by x, the remainder is minus 4. but isnt (x)= (x-0)? From the remainder theorem we know that when divided by (x-a), P(a)= the remainder when you divide by P(a).
so. We sub in x=0
therefore P(0)= 4k +8.
we know that P(0)=-4
therefore
4k +8=-4
k=-3
sub k in and you get the polynomial after expanding .
epsilon was this what your friend meant?

Yeah, I think so

So I guess "The Form" is P(x)= A(x).Q(x) + R(x), forms of polynomials

Syque · Jan 2, 2013

Re: HSC 2013 3U Marathon Thread

Keep the marathon going guys and make sure the questions are accessible to the 3 unit level

All 4 unit questions should be posted here in the link below

http://community.boredofstudies.org/showthread.php?t=294872

Sy123 · Jan 2, 2013

Re: HSC 2013 3U Marathon Thread

$P(2ap, ap^2) \ \ \ $lies on the parabola with the equation$ \ \ \ x^2=4ay$

$$A tangent to the parabola is constucted at P, and the tangent intersects the x-axis at R, and the y-axis at S$

$$Prove that $ \ \ \ SR = RP$

RealiseNothing · Jan 2, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$P(2ap, ap^2) \ \ \ $lies on the parabola with the equation$ \ \ \ x^2=4ay$

$$A tangent to the parabola is constucted at P, and the tangent intersects the x-axis at R, and the y-axis at S$

$$Prove that $ \ \ \ SR = RP$

The equation of the tangent is $y=px-ap^2$

Let y=0, then the 'x' co-ordinate of R is $ap$

The 'x' co-ordinate of P is $2ap$ and of S is $0$

Therefore since P,R,S are collinear, SR=RP.

Deliriously · Jan 2, 2013

Re: HSC 2013 3U Marathon Thread

$\\*\text{If } x+y+z=1 \text{ where } x,y,z \in \mathbb{R}, \\ \text{Prove that: } \\ \\ x^2+y^2+z^2 \geq \frac{1}{3}$

Sy123 · Jan 2, 2013

Re: HSC 2013 3U Marathon Thread

RealiseNothing said:
The equation of the tangent is $y=px-ap^2$

Let y=0, then the 'x' co-ordinate of R is $ap$

The 'x' co-ordinate of P is $2ap$ and of S is $0$

Therefore since P,R,S are collinear, SR=RP.

Correct - this question was more about finding the quickest solution, and you had the solution that I had in mind.

RealiseNothing · Jan 2, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$P(2ap, ap^2) \ \ \ $lies on the parabola with the equation$ \ \ \ x^2=4ay$

$$A tangent to the parabola is constucted at P, and the tangent intersects the x-axis at R, and the y-axis at S$

$$Prove that $ \ \ \ SR = RP$

Suppose a point Q has co-ordinates $(0,ap^2)$

Show that:

$QP^2 + QS^2 = 4QR^2$

SpiralFlex · Jan 2, 2013

Re: HSC 2013 3U Marathon Thread

Deliriously said:
$\\*\text{If } x+y+z=1 \text{ where } x,y,z \in \mathbb{R}, \\ \text{Prove that: } \\ \\ x^2+y^2+z^2 \geq \frac{1}{3}$

I'm sure this goes under harder 3U: Inequalities

$Equivalent of proving$

$3(x^2+y^2+z^2) \geq (x+y+z)^2$

$3x^2+3y^2+3z^2 - z^2- 2zx - x^2 - 2xy - y^2 - 2zy \geq 0$

$$LHS$\;=2x^2+2y^2+2z^2-2zx - 2zy-2xy$

$=(x-y)^2+(z-x)^2+(y-z)^2$

$\therefore (x-y)^2+(z-x)^2+(y-z)^2 \geq 0$

$$Hence,$\; x^2+y^2+z^2 \geq \frac{1}{3}$ #

HSC 2013-14 MX1 Marathon (archive) (4 Viewers)

Well-Known Member

Heroic!

Member

True Fail Kid

?

True Fail Kid

This too shall pass

New Member

This too shall pass

?

True Fail Kid

True Fail Kid

?

New Member

This too shall pass

what is that?It is Cowpea

Member

This too shall pass

what is that?It is Cowpea

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 4)