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HSC 2013 MX2 Marathon (archive) (5 Viewers)

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study1234

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Re: HSC 2013 4U Marathon

Suppose that three points on the parabola y=x^2 have the property that their normal lines intersect at a common point. Show that the sum of their x-coordinates is 0.
 

Sy123

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Re: HSC 2013 4U Marathon

Suppose that three points on the parabola y=x^2 have the property that their normal lines intersect at a common point. Show that the sum of their x-coordinates is 0.
Let the points be A(a,a^2), B(b,b^2), C(c,c^2)
Find through differentiating that nromal to curve at A is:

x+2ay=2a^3+a

Intersection of normals at A and B are

(-2ab(a+b), a^2+ab+b^2+1/2)

Intersections of normals at A and C are

(-2ac(a+c), a^2+ac+c^2+1/2)

Equating x-y coordinates we get that a+b+c=0, hence proof is complete.

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RealiseNothing

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Re: HSC 2013 4U Marathon

The first inequality isn't always true for all positive integers n.
It works for integers

The cases of 1, 2, 3 can just be manually tested I guess. At the time I originally had which works for all integers, until I realised it's n-1 not n+1, and thought it still worked without actually testing it lol.
 

Trebla

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Re: HSC 2013 4U Marathon

Hint: first prove that for x > 1

x - 1 > ln x

then go from there :p
 

RealiseNothing

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Re: HSC 2013 4U Marathon

Hint: first prove that for x > 1

x - 1 > ln x

then go from there :p
When LHS=RHS

differentiating both sides gives LHS'=1 and RHS'=1/x so for x>1 LHS>RHS

then apply what nightweaver did, and the solution is done
 

Trebla

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Re: HSC 2013 4U Marathon

When LHS=RHS

differentiating both sides gives LHS'=1 and RHS'=1/x so for x>1 LHS>RHS

then apply what nightweaver did, and the solution is done
lol you can't differentiate both sides of an inequality
 

RealiseNothing

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Re: HSC 2013 4U Marathon

lol you can't differentiate both sides of an inequality
no not like that

like, if you differentiate x-1 you get 1, and if you differentiate ln(x) you get 1/x, so x-1 increases faster than ln(x) for x>1, but at x=1 they are equal, thus for x>1, x-1>ln(x)
 

Trebla

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Re: HSC 2013 4U Marathon

no not like that

like, if you differentiate x-1 you get 1, and if you differentiate ln(x) you get 1/x, so x-1 increases faster than ln(x) for x>1, but at x=1 they are equal, thus for x>1, x-1>ln(x)
Oh right. I would normally just show that the maximum value of y = ln x - x is -1.
 

Sy123

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Re: HSC 2013 4U Marathon

It works for integers

The cases of 1, 2, 3 can just be manually tested I guess. At the time I originally had which works for all integers, until I realised it's n-1 not n+1, and thought it still worked without actually testing it lol.
ah ok, well done.


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Sy123

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Re: HSC 2013 4U Marathon











b) Resolve all forces with respect to the perpendicular axes of the Tension and tangential acceleration vectors, we find that



Where a is the acceleration, the decomposed gravity vector opposes the tangential acceleration



But from part (a),

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








 

study1234

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Re: HSC 2013 4U Marathon

Consider the ellipse x^2/100 + y^2/25 = 1, and let P be the point (6,4) on the ellipse. If the normal at P meets the major axis at G, and OH is the perpendicular drawn from the origin to the tangent at P, show that PG x OH = 25.
 

Sy123

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Re: HSC 2013 4U Marathon

Very similar to last year's BOS trial question 15 by carrot.
Well it is given that E_n approaches a finite limit, and

H_n - ln(n) = E_n + 1/n, so the result is practically given, for my one it is not given.
 
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