HSC 2013 MX2 Marathon (archive) (1 Viewer)

RealiseNothing · Apr 15, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
bump.

Difficulty: 3.14159/5

That should be enough to keep 2013'ers busy for a while.

This is where I got up to last time, I'll try again though:

Dividing by $x^2$ gives:

$x^2 + ax + b + \frac{a}{x} + \frac{1}{x^2} = 0$

Grouping appropriate terms:

$x^2 + \frac{1}{x^2} + 2 + ax + \frac{a}{x} + (b-2) = 0$

$(x+\frac{1}{x})^2 + a(x+\frac{1}{x}) + (b-2) = 0$

Now using the quadratic formula:

$x + \frac{1}{x} = \frac{-a \pm \sqrt{a^2 - 4b + 8}}{2}$

Now let $x+\frac{1}{x} = U$

We want to find the values of U such that it has atleast one real root:

$x^2 - Ux + 1 = 0$

$U^2 - 4 \geq 0$

$U \geq |2|$

So $x+\frac{1}{x} \geq |2|$

Therefore going back to our quadratic equation and the roots:

$\frac{-a \pm \sqrt{a^2 - 4b + 8}}{2} \geq |2|$

$-a \pm \sqrt{a^2 - 4b + 8} \geq |4|$

RealiseNothing · Apr 15, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
This is where I got up to last time, I'll try again though:

Dividing by $x^2$ gives:

$x^2 + ax + b + \frac{a}{x} + \frac{1}{x^2} = 0$

Grouping appropriate terms:

$x^2 + \frac{1}{x^2} + 2 + ax + \frac{a}{x} + (b-2) = 0$

$(x+\frac{1}{x})^2 + a(x+\frac{1}{x}) + (b-2) = 0$

Now using the quadratic formula:

$x + \frac{1}{x} = \frac{-a \pm \sqrt{a^2 - 4b + 8}}{2}$

Now let $x+\frac{1}{x} = U$

We want to find the values of U such that it has atleast one real root:

$x^2 - Ux + 1 = 0$

$U^2 - 4 \geq 0$

$U \geq |2|$

So $x+\frac{1}{x} \geq |2|$

Therefore going back to our quadratic equation and the roots:

$\frac{-a \pm \sqrt{a^2 - 4b + 8}}{2} \geq |2|$

$-a \pm \sqrt{a^2 - 4b + 8} \geq |4|$

Ok now if we take the positive solution we get:

$|a| + \sqrt{a^2 -4b + 8} \geq 4$

Make the square root the subject and squaring gives:

$a^2 - 4b + 8 \geq 16 + a^2 - 8|a|$

$-4b \geq 8 - 8|a|$

$-b \geq 2 - 2|a|$

Now we want to find the minimum value of $a^2+b^2$ so we will make the $|a|$ the subject:

$|a| \geq \frac{b+2}{2}$

$a^2 \geq \frac{b^2+4b+4}{4}$

Hence we can deduce that:

$a^2+b^2 \geq b^2 + \frac{b^2+4b+4}{4}$

The minimum value of $a^2+b^2$ is thus the minimum value of the RHS, so lets consider that side:

$b^2 + \frac{b^2+4b+4}{4}$

$\frac{5}{4}(b^2+\frac{4}{5}b+\frac{4}{5})$

Upon completing the square:

$\frac{5}{4}[(b+\frac{2}{5})^2 + \frac{16}{25}]$

Hence the minimum value of this is when the term with $b$ goes to 0:

$\frac{5}{4}[(-\frac{2}{5}+\frac{2}{5})^2 + \frac{16}{25}]$

$\frac{5}{4} \times \frac{16}{25}$

$\frac{4}{5}$

Hence the minimum possible value of $a^2+b^2$ for the polynomial to have atleast one real root is $\frac{4}{5}$

shongaponga · Apr 16, 2013

Re: HSC 2013 4U Marathon

$\\ $Find $ \\\\ \int\frac{x^3 e^{x^2}}{(x^2+1)^2}dx$

Sy123 · Apr 17, 2013

Re: HSC 2013 4U Marathon

shongaponga said:
$\\ $Find $ \\\\ \int\frac{x^3 e^{x^2}}{(x^2+1)^2}dx$

$I=\int \frac{x^3e^x}{(x^2+1)^2} \ dx$

$u=x^2$

$\Rightarrow \ \ I = \frac{1}{2}\int \frac{u e^u}{(u+1)^2} \ du = \frac{1}{2}\int e^u \cdot \left(\frac{1}{u+1} - \frac{1}{(u+1)^2} \right ) \ du$

$Note that$

$\frac{d}{dt}f(t) e^{t} = (f(t)+f'(t))e^t$

$$From the above integral, by inspection it comes that$ \ \ f(t) = \frac{1}{t+1}$

$\Rightarrow \ \ I = \frac{1}{2}\frac{e^u}{u+1} + c$

$\therefore \ \ I = \frac{e^{x^2}}{2(x^2+1)} + c$

Sy123 · Apr 17, 2013

Re: HSC 2013 4U Marathon

$$Take the function$ \ \ y=f(x)$

$$continuously differentiable in the interval$ \ \ a < x < b$

$$and continuous in the interval$ \ \ a \leq x \leq b$

$$Let$ \ \ A(a,f(a)) \ \ \ B(b, f(b))$

$$i) By considering integration, prove that the length of the arc$

$$along the function, from points$ \ \ A \ $to$ \ B$

$Is given by$

$\overarc{AB} = \int_{a}^{b} \sqrt{1+\left (\frac{dy}{dx} \right )^2} \ dx$

$$Hence prove that the circumference of a circle with radius$ \ \ r$

$$Is$ \ \ 2\pi r$

iBibah · Apr 17, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$I=\int \frac{x^3e^x}{(x^2+1)^2} \ dx$

$u=x^2$

$\Rightarrow \ \ I = \int \frac{u e^u}{(u+1)^2} \ du = \int e^u \cdot \left(\frac{1}{u+1} - \frac{1}{(u+1)^2} \right ) \ du$

$Note that$

$\frac{d}{dt}f(t) e^{t} = (f(t)+f'(t))e^t$

$$From the above integral, by inspection it comes that$ \ \ f(t) = \frac{1}{t+1}$

$\Rightarrow \ \ I = \frac{e^u}{u+1} + c$

$\therefore \ \ I = \frac{e^{x^2}}{x^2+1} + c$

If u=x^2, why was x^3 on the numerator replaced with u?

You may want to check your answer btw

seanieg89 · Apr 17, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$Take the continuous function$ \ \ y=f(x) \ \$

$$continuous in the interval$ \ \ a \leq x \leq b$

$$Let$ \ \ A(a,f(a)) \ \ \ B(b, f(b))$

$$i) By considering integration, prove that the length of the arc$

$$along the function, from points$ \ \ A \ $to$ \ B$

$Is given by$

$\overarc{AB} = \int_{a}^{b} \sqrt{1+\left (\frac{dy}{dx} \right )^2} \ dx$

$$Hence prove that the circumference of a circle with radius$ \ \ r$

$$Is$ \ \ 2\pi r$

You will need to assume more than continuity if you are going to differentiate your y.

twinklegal19 · Apr 17, 2013

iBibah said:
If u=x^2, why was x^3 on the numerator replaced with u?

You may want to check your answer btw

When you sub u=x^2 you divide by 2x which cancels one of the x"s and you are left with x^2, which becomes u

twinklegal19 · Apr 17, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$I=\int \frac{x^3e^x}{(x^2+1)^2} \ dx$

$u=x^2$

$\Rightarrow \ \ I = \int \frac{u e^u}{(u+1)^2} \ du = \int e^u \cdot \left(\frac{1}{u+1} - \frac{1}{(u+1)^2} \right ) \ du$

$Note that$

$\frac{d}{dt}f(t) e^{t} = (f(t)+f'(t))e^t$

$$From the above integral, by inspection it comes that$ \ \ f(t) = \frac{1}{t+1}$

$\Rightarrow \ \ I = \frac{e^u}{u+1} + c$

$\therefore \ \ I = \frac{e^{x^2}}{x^2+1} + c$

I think you are missing a $\frac{1}{2}$

twinklegal19 · Apr 17, 2013

Re: HSC 2013 4U Marathon

Solve:

$\frac{4x^2}{(1-\sqrt{1+2x})^2} \leq 2x+9$

Sy123 · Apr 17, 2013

Re: HSC 2013 4U Marathon

iBibah said:
If u=x^2, why was x^3 on the numerator replaced with u?

You may want to check your answer btw

$u=x^2$

$du=2x \ dx$

twinklegal19 said:
I think you are missing a $\frac{1}{2}$

yes I am, thank you.

==========

Sean, is saying that in the interval [a,b] it is differentiable sufficient?

Sy123 · Apr 17, 2013

Re: HSC 2013 4U Marathon

twinklegal19 said:
Solve:

$\frac{4x^2}{(1-\sqrt{1+2x})^2} \leq 2x+9$

$x\neq 0$

$x \geq -1/2$

$\frac{2x^2}{(1+x)-\sqrt{1+2x}} \leq 2x+9$

$\frac{2x^2(1+x+\sqrt{1+2x})}{x^2} \leq 2x+9$

$2+2x+\sqrt{1+2x} \leq 2x+7$

$\sqrt{1+2x} \leq \frac{7}{2}$

$x\leq \frac{45}{8}$

Therefore the solution:

$\frac{-1}{2} \leq x \leq \frac{45}{8} \ , \ \ \ x\neq 0$

seanieg89 · Apr 17, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$u=x^2$

$du=2x \ dx$

yes I am, thank you.

==========

Sean, is saying that in the interval [a,b] it is differentiable sufficient?

To be on the safe side I would say continuously differentiable on (a,b) and continuous on [a,b]. Continuity of the derivative implies that the integral exists, which we would not get from differentiability alone.

RealiseNothing · Apr 17, 2013

Re: HSC 2013 4U Marathon

twinklegal19 said:
Solve:

$\frac{4x^2}{(1-\sqrt{1+2x})^2} \leq 2x+9$

Let $u = \sqrt{1+2x}$

Then the inequality simply becomes:

$\frac{(u^2-1)^2}{(1-u)^2} \leq u^2 + 8$

Using difference of two squares on the numerator:

$\frac{(u-1)^2(u+1)^2}{(u-1)^2} \leq u^2 + 8$

$(u+1)^2 \leq u^2+8$

$u^2+2u+1 \leq u^2+8$

$u \leq \frac{7}{2}$

Therefore:

$\sqrt{1+2x} \leq \frac{7}{2}$

$x \leq \frac{45}{8}$

Using the domain:

$\frac{-1}{2} \leq x \leq \frac{45}{8}$

Also $x \neq 0$

HeroicPandas · Apr 17, 2013

Re: HSC 2013 4U Marathon

Lieutenant_21 said:
Nice proof but it can be done in 3 lines

Prove that for any integer n>1, $\frac{1}{2^n} = \cos \frac{2\pi}{2^n-1} \cdot \cos \frac{4\pi}{2^n-1} \cdot \dots \cdot \cos \frac{2^n}{2^n-1}$

^This is the question that u said that it can be done in 3 lines! Can u please show that it can be done in 3 lines?

Sy123 · Apr 19, 2013

Re: HSC 2013 4U Marathon

$Prove the following$

$$i)$ \int_0^{\pi/4} \sin(2x) \ln(\cos x) \ dx = \frac{1}{4}(\ln2-1)$

$$ii)$ \int_0^{\pi/4} \cos(2x) \ln(cos x) \ dx = \frac{1}{8}(\pi - \ln 4 - 2)$

$Hence evaluate$

$$iii)$ \int_{\pi/4}^{\pi/2} (\cos(2x)+\sin(2x)) \ln(\cos x + \sin x) \ dx$

HeroicPandas · Apr 20, 2013

Re: HSC 2013 4U Marathon

$$Find $\int \frac{du}{\sqrt{u} + 1}$

Sy123 · Apr 20, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
$$Find $I = \int \frac{du}{\sqrt{u} + 1}$

$x^2=u$

$du= 2x dx$

$I = 2 \int \frac{x}{x+1}$

$I = 2(x+1) - 2\ln(x+1) + c$

$I = \ln \left(\frac{e^{\sqrt{u}+1}}{\sqrt{u}+1} \right )^2 + c$

==============================

$Prove that$

$\sum_{r=0}^{n} \binom{n}{r} \cos(\alpha+2r\beta) = 2^n \cos^n \beta$

HeroicPandas · Apr 20, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$x^2=u$

$du= 2x dx$

$I = 2 \int \frac{x}{x+1}$

$I = 2(x+1) - 2\ln(x+1) + c$

$I = \ln \left(\frac{e^{\sqrt{u}+1}}{\sqrt{u}+1} \right )^2 + c$

==============================

$Prove that$

$\sum_{r=0}^{n} \binom{n}{r} \cos(\alpha+2r\beta) = 2^n \cos^n \beta$

$\begin{align*}LHS&= \sum_{r=0}^{n} \binom{n}{r} \cos(\alpha+2r\beta)\\&=cos\alpha +cos(\alpha + \beta) + cos(\alpha + 2 \beta) + \dots + cos(\alpha + 2n \beta)\\&= cos\alpha + cos\alpha sin2\beta - cos2\beta sin\alpha + \dots + cos\alpha sin2n\beta - sin\alpha cos2n\beta \\&=cos\alpha (1 + sin2\beta + \dots + sin2n\beta) - sin\alpha(cos 2\beta + cos4\beta + \dots + cos2n\beta) \end{align*}$

$$ Consider the sum: cis^2 \beta + cis^4 \beta + \dots + cis^{2n} \beta = $\frac{cis^2 \beta [1-(cis^2 \beta)^n]}{1-cis^2 \beta}$

$\therefore LHS = cos\alpha (1 + \left Im( \frac{cis^2 \beta [1-(cis^2 \beta)^n]}{1-cis^2 \beta} \right )) - sin\alpha (Re\left ( \frac{cis^2 \beta [1-(cis^2 \beta)^n]}{1-cis^2 \beta} \right ))$

And simplify and should i get RHS?

Sy123 · Apr 20, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
$\begin{align*}LHS&= \sum_{r=0}^{n} \binom{n}{r} \cos(\alpha+2r\beta)\\&=cos\alpha +cos(\alpha + \beta) + cos(\alpha + 2 \beta) + \dots + cos(\alpha + 2n \beta)\\&= cos\alpha + cos\alpha sin2\beta - cos2\beta sin\alpha + \dots + cos\alpha sin2n\beta - sin\alpha cos2n\beta \\&=cos\alpha (1 + sin2\beta + \dots + sin2n\beta) - sin\alpha(cos 2\beta + cos4\beta + \dots + cos2n\beta) \end{align*}$

$$ Consider the sum: cis^2 \beta + cis^4 \beta + \dots + cis^{2n} \beta = $\frac{cis^2 \beta [1-(cis^2 \beta)^n]}{1-cis^2 \beta}$

$\therefore LHS = cos\alpha (1 + \left Im( \frac{cis^2 \beta [1-(cis^2 \beta)^n]}{1-cis^2 \beta} \right )) - sin\alpha (Re\left ( \frac{cis^2 \beta [1-(cis^2 \beta)^n]}{1-cis^2 \beta} \right ))$

And simplify and should i get RHS?

Note the binomial co-efficient at the front =)

HSC 2013 MX2 Marathon (archive) (1 Viewer)

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