HSC Mathematics Marathon (1 Viewer)

Drongoski · Mar 19, 2011

muzeikchun852 said:
$\text{If the roots of the equation}:\ x^4+ax^3+bx^2+abx+1=0:\ \text{are}:\ \alpha,\beta,\gamma,\delta,\\\\\text{show that}(\alpha +\beta +\gamma )(\alpha +\beta +\delta )(\alpha +\gamma +\delta )(\beta +\gamma +\delta )=1$

$\alpha + \beta + \gamma + \delta = -a$

The expression $= (-a-\alpha)(-a-\beta)(-a-\gamma)(-a-\delta)$

$= (a+\alpha)(a+\beta)(a+\gamma)(a+\delta)$

= constant term of the quartic equation:

$(x-a)^4 + a(x-a)^3 + b(x-a)^2 + ab(x-a) + 1$

$= a^4 -a^4 +a^{2}b - a^{2}b + 1$

= 1

hscishard · Mar 19, 2011

ohexploitable said:
Use basic geometry. The locus is a circle, contruct a triangle joining the centre, (0,1) and (0,-1) and make a chocolate cake. The length of the base is obvious (2), the angle subtended at the centre is twice the angle at the circumference (pi/2). Now find the height of the circle to get the coordinates of the centre then find the radius. The cake is a lie.

You need to restrict the range too if you're going to show it algebraically

blackops23 · Mar 20, 2011

ohexploitable said:
Use basic geometry. The locus is a circle, contruct a triangle joining the centre, (0,1) and (0,-1) and make a chocolate cake. The length of the base is obvious (2), the angle subtended at the centre is twice the angle at the circumference (pi/2). Now find the height of the circle to get the coordinates of the centre then find the radius. The cake is a lie.

I see...

muzeikchun852 · Mar 20, 2011

Drongoski said:
$\alpha + \beta + \gamma + \delta = -a$

The expression $= (-a-\alpha)(-a-\beta)(-a-\gamma)(-a-\delta)$

$= (a+\alpha)(a+\beta)(a+\gamma)(a+\delta)$

= constant term of the quartic equation:

$(x-a)^4 + a(x-a)^3 + b(x-a)^2 + ab(x-a) + 1$

$= a^4 -a^4 +a^{2}b - a^{2}b + 1$

= 1

how did you get from this line
$(x-a)^4 + a(x-a)^3 + b(x-a)^2 + ab(x-a) + 1$
to this line ?
$= a^4 -a^4 +a^{2}b - a^{2}b + 1$

if i expand this:
$(x-a)^4 + a(x-a)^3 + b(x-a)^2 + ab(x-a) + 1\\\\\text{i get:}\: x^4 -3ax^3+3ax^2-a^3x+bx^2+abx+1$

deterministic · Mar 20, 2011

muzeikchun852 said:
how did you get from this line
$(x-a)^4 + a(x-a)^3 + b(x-a)^2 + ab(x-a) + 1$
to this line ?
$= a^4 -a^4 +a^{2}b - a^{2}b + 1$

Consider only constant terms in the quartic equation.

Drongoski · Mar 20, 2011

deterministic said:
Consider only constant terms in the quartic equation.

Exactly.

rawrence · Mar 20, 2011

Drongoski · Mar 20, 2011

rawrence said:
$\\a)\text{ Given that }I_n = \int_{0}^{1} (1-x^2)^n\text{ show that }I_n = \frac{2n}{2n+1}I_{n-1} \\\\b)\text{ Hence, show that }I_n = \frac{2^{2n}(n!)^2}{(2n+1)!}$

rawrence - looks like you are ahead of the pack; most have not done this topic yet. Have you yourself already done this question?

a)

$I_n = \int ^1 _0 (1-x^2)^n dx = x(1-x^2)^n]^1_0 - \int ^1 _0 x \cdot n(1-x^2)^{n-1} (-2x)dx $ ( IBP)$$

$= 0 -2n\int _0 ^1[(1-x^2) - 1](1-x^2)^{n-1}dx$

$= -2n\int ^1 _0 [(1-x^2)^n - (1-x^2)^{n-1} ] dx$

$= -2n I_n + 2n I_{n-1}$

$\therefore I_n = \frac {2n}{2n+1} I_{n-1}$

b)

$I_0 = \int _0 ^1 (1-x^2)^0 dx = x]_0 ^1 = 1$

$I_n = \frac {2n}{2n+1} I_{n-1} = \frac {2n}{2n+1} \times \frac {2(n-1)}{2n-1} \times \frac {2(n-2)}{2n-3} \times \cdots \times \frac {2(1)}{2+1} \times I_0$

$= \frac {2n}{2n+1} \times \frac {2(n-1)}{2n-1} \times \cdots \times \frac {2(1)}{2+1} \cdot \frac{2n \times 2(n-1) \times 2(n-2) \times \cdots \times 2(1)}{2n \times 2(n-1) \times 2(n-2) \times \cdots \times2(1)}$

$= \frac {2 ^{2n}(n!)^2}{(2n+1)!}$

rawrence · Mar 21, 2011

Drongoski said:
rawrence - looks like you are ahead of the pack; most have not done this topic yet. Have you already done this question?

a)

$I_n = \int ^1 _0 (1-x^2)^n dx = x(1-x^2)^n]^1_0 - \int ^1 _0 x \cdot n(1-x^2)^{n-1} (-2x)dx $ ( IBP)$$

$= 0 -2n\int _0 ^1[(1-x^2) - 1](1-x^2)^{n-1}dx$

$= -2n\int ^1 _0 [(1-x^2)^n - (1-x^2)^{n-1} ] dx$

$= -2n I_n + 2n I_{n-1}$

$\therefore I_n = \frac {2n}{2n+1} I_{n-1}$

b)

$I_0 = \int _0 ^1 (1-x^2)^0 dx = x]_0 ^1 = 1$

$I_n = \frac {2n}{2n+1} I_{n-1} = \frac {2n}{2n+1} \times \frac {2(n-1)}{2n-1} \times \frac {2(n-2)}{2n-3} \times \cdots \times \frac {2(1)}{2+1} \times I_0$

$= \frac {2n}{2n+1} \times \frac {2(n-1)}{2n-1} \times \cdots \times \frac {2(1)}{2+1} \cdot \frac{2n \times 2(n-1) \times 2(n-2) \times \cdots \times 2(1)}{2n \times 2(n-1) \times 2(n-2) \times \cdots \times2(1)}$

$= \frac {2 ^{2n}(n!)^2}{(2n+1)!}$

Correct!

And yes, I have done it, my school's putting everything except mechanics and probability from harder 3 unit

D :

ohexploitable · Mar 29, 2011

HyperComplexxx · Mar 29, 2011

ohexploitable said:
$\text{Using the property }\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx,$

$\text{Evaluate }\int_0^{\frac{\pi}{2}}\frac{\sin^{69} x}{\cos^{69} x+\sin^{69} x}\,dx$

$\frac{\pi}{4}$

ZachBC_94 · Apr 4, 2011

Slightly unrelated, but ohexploitable, is your avatar from Gossip Girl?

Also, as we have an absence of questions:

If In = (Integral from 1 to 0) x^n tan(inverse)x dx, n = 0, 1, 2, ...
Show that In = 1/(n+1) * π/2 - 1/[n(n+1)] - (n-1)/(n+1) * I(n-2), n = 2, 3, 4, ...

Sorry, don't know latex

ohexploitable · Apr 6, 2011

^ cbf doing that lol

New Q:

$\int\sec^3 2x\,dx$

ohexploitable · Apr 6, 2011

ZachBC_94 said:
Slightly unrelated, but ohexploitable, is your avatar from Gossip Girl?

Yes.

Drongoski · Apr 6, 2011

ohexploitable said:
^ cbf doing that lol

New Q:

$\int\sec^3 2x\,dx$

I'll let others have a go first for the next 24 hrs or so.

ZachBC_94 · Apr 6, 2011

ohexploitable said:
Yes.

Respect...

ZachBC_94 · Apr 6, 2011

ohexploitable said:
New Q:

$\int\sec^3 2x\,dx$

$\int\sec^3 2x\,dx = \int\sec^2 2x sec 2x\,dx$

$= 1/2\int\(1 - tan2x) sec 2x\,d 2x$

$= -1/2(1/2 sec 2x) + 1/2\int\sec 2x\,dx + C$

$= -1/4sec 2x + 1/2(x sec 2x - \int\ x \,d(sec 2x)) + C$

$= -1/4sec 2x + (x sec 2x)/2 - 1/2(sec 2x (x^2)/2) + C$

$= -1/4sec 2x + (x sec 2x)/2 - x^2 sec 2x$

$= sec 2x (-1/4 + x/2 + x^2)$

I think. I did that all on the computer, and I am terrible at Latex, so I might have gone on a crazy tangent there

Drongoski · Apr 6, 2011

Not bad for one terrible in LaTeX.

ZachBC_94 · Apr 6, 2011

why thank you. I'm only new at it, that's why all the fractions are still in x/y form for instance

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