2 Unit Revising Marathon HSC '11 (2 Viewers)

[Riproot]

Yes I'm starting this thingy again because yeah... the old one is gone.
I'll start with a simple one on maxima/minima:

 

[hscishard]

AC = 2root(100-x^2)
Therefore A triangle = 1/2 x 2root(100-x^2) x (10+x)

Differentiate it using the product rule. Equate to 0 and solve for x. Determine max point

Use shift cos to prove the two 60 degrees and then use angle sum.

someone post next question

 

[Trebla]

Find the equation of the tangent to the curve y = ln x such that the tangent passes through the origin.

Hence show that

 

[Bored Of Fail]

another question

solve : [ cot (x/2) ]^2 =sqrt(2) for -2pi<=x<=3pi,

give answers in radians correct to 2 decimal places.

 

[Pyrobooby]

Find the equation of the tangent to the curve y = ln x such that the tangent passes through the origin.

Hence show that

y=ln x, dy/dx= 1/x
as the tangent passes through the origin, the tangent has an equation in the form
y=mx. But m=dy/dx,
. ' . solving simultaneously
ln x=1/x * x
ln x=1
x=e

when x=e, y=1
using (e,1) , m=1/e

y=x/e

I was wondering if a proof such as this is valid:
x^e is less than or equal to e^x
when x^e = e^x,
by inspection, x=e
when x approaches positive infinity, e^x>x^e
when x approaches 0, e^x>x^e
hence x^e is less than or equal to e^x?

 

[Bored Of Fail]

y=ln x, dy/dx= 1/x
as the tangent passes through the origin, the tangent has an equation in the form
y=mx. But m=dy/dx,
. ' . solving simultaneously
ln x=1/x * x
ln x=1
x=e

when x=e, y=1
using (e,1) , m=1/e

y=x/e

I was wondering if a proof such as this is valid:
x^e is less than or equal to e^x
when x^e = e^x,
by inspection, x=e
when x approaches positive infinity, e^x>x^e
when x approaches 0, e^x>x^e
hence x^e is less than or equal to e^x?

it is true for x=0

then f(x) = e^(x) -> f ' (x) = e^(x)
g(x) = x^e -> g ' (x) = e x^(e-1)

now we need to show that f ' (x) > g ' (x)

f ' (x) / g ' (x) = [ e^(x) / [e x^(e-1) ] = e^(x-1) / x^(e-1) which is positive for all positive x ( x not equal to zero off course )

so f '(x) > g'(x)

hmm not 100% sure on the last step how we got x not equal to zero, but the statement is true for x=0

but kinda like that

EDIT: lol I think I know why, it is because they ( f(x) and g(x) ) are equal at x=0, and g' (0) =0 , we cannot divide by zero , such a dummy lol

EDIT: ahh not quite, need to show that the ratio is greater than one, that shouldnt be too hard, have just went round in a circle and reduced the powers by one, disregard this post

 

[Pyrobooby]

EDIT: My bad!

 

[Bored Of Fail]

at x=0,
(0)^e = e^0
0=1
???

remember its an inequality

also, I think I have screwed it up and just went around in a circle.

i think im gona stop now XD

 

[cutemouse]

Yes I'm starting this thingy again because yeah... the old one is gone.
I'll start with a simple one on maxima/minima:
View attachment 22464

I've seen that Q before. Probably the 2009 CSSA.

Here's an interesting Q: Prove 3/4+5/36+7/144+9/400+11/900+...= 1

EDIT: Noticed previous Q not solved yet.

As before, eq'n of tangent is y=x/e

Now due to geometry of figure (ie. tangent lies above curve), (x/e) >= ln x (with equality occuring at x=e)

So e^(x/e) >= x (as e^x is an increasing function)

e^x >= x^e
ie. x^e <= e^x (as required)

 

[Bored Of Fail]

I've seen that Q before. Probably the 2009 CSSA.

Here's an interesting Q: Prove 3/4+5/36+7/144+9/400+11/900+...= 1

thats been put up before by trebla i think

 

[NewiJapper]

Find the equation of the tangent to the curve y = ln x such that the tangent passes through the origin.

Hence show that...

That was a Q9 HSC question a couple of years ago. Completely stupid and I hated doing it for revision lol.

 

[Riproot]

AC = 2root(100-x^2)
Therefore A triangle = 1/2 x 2root(100-x^2) x (10+x)

Differentiate it using the product rule. Equate to 0 and solve for x. Determine max point

Use shift cos to prove the two 60 degrees and then use angle sum.

someone post next question

Get out. *points at door*

 

[hscishard]

Get out. *points at door*

2U thread? What am I supposed to say? Inverse cosine function/Inverse cos? People wont know what you're on about, except you and everyone who has done inverse functions of course

 

[Riproot]

I learned to call it inverse cos when i was in year 9. If they don't know, they're retarded.

 

[hscishard]

I learned to call it inverse cos when i was in year 9. If they don't know, they're retarded.

I doubt most students in year 9 would've learnt this term. You must have a good teacher.
Your second statement will be offensive to a lot of people. So please, get off your high horse

 

[Alkanes]

ye some will u start to learn this term in basic trig like soh cah toa and stuff lol =)

 

[Riproot]

I doubt most students in year 9 would've learnt this term. You must have a good teacher.
Your second statement will be offensive to a lot of people. So please, get off your high horse

Your last statement was also offensive, so, I'd say we're about even.