# HSC 2013 Maths Marathon (archive) (1 Viewer)

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#### mahmoudali

##### Member
Re: HSC 2013 2U Marathon

this thread is stressing my out so much dont know how to do half these questions...
half the questions on here are Q12-14 hardness

#### nakruf

##### New Member
Re: HSC 2013 2U Marathon

but i cant because sy's questions are hard
would you like him to spoonfeed you the answer?

#### mahmoudali

##### Member
Re: HSC 2013 2U Marathon

would you like him to spoonfeed you the answer?
as a matter of fact i'd really like that -_-

#### Sy123

##### This too shall pass
Re: HSC 2013 2U Marathon

Bumping for more exposure

#### mahmoudali

##### Member
Re: HSC 2013 2U Marathon

Yea well done

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

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

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

$\bg_white y=\sqrt{1-x^2}$

$\bg_white \frac{dy}{dx} = \frac{-x}{\sqrt{1-x^2}}$

$\bg_white M=\frac{-a}{\sqrt{1-a^2}}\ \ = \therefore \frac{-a}{b} \ \ since \ \ b=\sqrt{1-a^2}$

ii)

$\bg_white tangents gradient = \frac{-a}{b} \ \ and OP is\ \ = radius= \ \ \frac{rise}{run} = \frac{b}{a}$

$\bg_white \frac{-a}{b} \star \frac{b}{a} = -1$

$\bg_white \therefore tangent to a circle is perpendicular to radius$

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#### nakruf

##### New Member
Re: HSC 2013 2U Marathon

$\bg_white (i) find \frac {d}{dx} (xlnx)$

$\bg_white (ii) hence evaluate \int_{1}^{3} lnx dx$

#### mahmoudali

##### Member
Re: HSC 2013 2U Marathon

$\bg_white (i) find \frac {d}{dx} (xlnx)$

$\bg_white (ii) hence evaluate \int_{1}^{3} lnx dx$
I)

$\bg_white u=x\ \ \ \ v=lnx$

$\bg_white u'=1 \ \ \ \ v'= 1/x$

$\bg_white \frac{d}{dx}(xlnx)\ \ = u'v+v'u\ \ (1)(lnx)+1 \ \ = lnx +1$

ii)

$\bg_white \int_{1}^{3} lnx dx$

$\bg_white [{xln(x)-x}]_{1}^{3}$

$\bg_white =(3ln3 -3)-(ln1 -1) \ \ =3ln3 - 2$
----------------------------------------------------------------
$\bg_white Find the volume generated by area of \ \ f(x) = ln(x) bounded by \ x=2 \ \ and \ \ x=3 \ \ being rotated about the\ \ y-axis$

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#### Menomaths

##### Exaı̸̸̸̸̸̸̸̸lted Member
Re: HSC 2013 2U Marathon

I)

$\bg_white u=x\ \ \ \ v=lnx$

$\bg_white u'=1 \ \ \ \ v'= 1/x$

$\bg_white \frac{d}{dx}(xlnx)\ \ = u'v+v'u\ \ (1)(lnx)+1 \ \ = lnx +1$

ii)

$\bg_white \int_{1}^{3} lnx dx$

$\bg_white [{xln(x)-x}]_{1}^{3}$

$\bg_white =(3ln3 -3)-(ln1 -1) \ \ =3ln3 - 2$
How do you integrate the lnx?

#### Sy123

##### This too shall pass
Re: HSC 2013 2U Marathon

How do you integrate the lnx?
The proper answer is:

$\bg_white \frac{d}{dx} x\ln x = \ln x + 1$

Integrate both sides

$\bg_white \therefore \ x\ln x = \int \ln x \ dx + \int 1 \ dx$

$\bg_white \therefore \ \ \int \ln x \ dx = x\ln x- x + c$

Then you just sub in values

#### mahmoudali

##### Member
Re: HSC 2013 2U Marathon

The proper answer is:

$\bg_white \frac{d}{dx} x\ln x = \ln x + 1$

Integrate both sides

$\bg_white \therefore \ x\ln x = \int \ln x \ dx + \int 1 \ dx$

$\bg_white \therefore \ \ \int \ln x \ dx = x\ln x- x + c$

Then you just sub in values
can you calm down i thought the whole state knew that step its not my fault i dont like showing every step Mr sy

#### Sy123

##### This too shall pass
Re: HSC 2013 2U Marathon

can you calm down i thought the whole state knew that step its not my fault i dont like showing every step Mr sy
I know you've memorised the integral of ln x and you just want to take every opportunity to show off your supreme memorisation power

#### mahmoudali

##### Member
Re: HSC 2013 2U Marathon

I know you've memorised the integral of ln x and you just want to take every opportunity to show off your supreme memorisation power
xD MEMORIES GREATER THAN LOGIC ANY DAY OF THE MONTH BUDDY

#### Sy123

##### This too shall pass
Re: HSC 2013 2U Marathon

$\bg_white By making use of differentiation by first principles of \ \sin x$

$\bg_white EVALUATE$

$\bg_white \lim_{h\to 0} \frac{\sin h}{h}$

#### mahmoudali

##### Member
Re: HSC 2013 2U Marathon

$\bg_white By making use of differentiation by first principles of \ \sin x$

$\bg_white EVALUATE$

$\bg_white \lim_{h\to 0} \frac{\sin h}{h}$

$\bg_white was first principal \frac{ f(h+m) - f(x) }{m} ?$

$\bg_white \frac{\frac{sin(h+m)}{h+m} -\frac{sin h}{h}}{m}$

is this rights so far?

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#### Sy123

##### This too shall pass
Re: HSC 2013 2U Marathon

Do I really need to say, provide proof

#### Menomaths

##### Exaı̸̸̸̸̸̸̸̸lted Member
Re: HSC 2013 2U Marathon

The proper answer is:

$\bg_white \frac{d}{dx} x\ln x = \ln x + 1$

Integrate both sides

$\bg_white \therefore \ x\ln x = \int \ln x \ dx + \int 1 \ dx$

$\bg_white \therefore \ \ \int \ln x \ dx = x\ln x- x + c$

Then you just sub in values
Omg so dull I still don't get it =.=...

#### nakruf

##### New Member
Re: HSC 2013 2U Marathon

$\bg_white was first principal \frac{ f(x+h) - f(x) }{h} ?$
forgot i had google

1
top is f(x+h)-f(x)

#### Sy123

##### This too shall pass
Re: HSC 2013 2U Marathon

Omg so dull I still don't get it =.=...
Its like saying

$\bg_white \frac{d}{dx} \sin x = \cos x$

THEREFORE

$\bg_white \int \cos x = \sin x + c$

we are just reversing what we've done, in this case I re-arranged the integrals

$\bg_white \frac{d}{dx} x\ln x = \ln x + 1$

$\bg_white x\ln x = \int (\ln x + 1) \ dx$

$\bg_white x \ln x = \int \ln x \ dx + \int 1 \ dx$

#### Menomaths

##### Exaı̸̸̸̸̸̸̸̸lted Member
Re: HSC 2013 2U Marathon

I understood up until there

#### Menomaths

##### Exaı̸̸̸̸̸̸̸̸lted Member
Re: HSC 2013 2U Marathon

I understood up until there
wtf... Yeah I get it. Just derped like hell...

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