2u Mathematics Marathon v1.0 (1 Viewer)

[Forbidden.]

I didn't use calculus at all

But the notations dy/dx, f'(x), y' are from calculus ...

 

[Riviet]

Haha, you know that he wasn't really differentiating anything, but I do admit that's the first time I've seen y' being used to refer to a gradient.

 

[Riviet]

Speaking of calculus...

Next Question:

Find any critical point(s) on the curve y=xe^-2x and determine their nature.

 

[madwoman]

(-1/2, -e1/2)
max
how does that sound?

 

[SoulSearcher]

Speaking of calculus...

Next Question:

Find any critical point(s) on the curve y=xe^-2x and determine their nature.

Spoiler

y=xe^-2x
y'=e^-2x-2xe^-2x
=e^-2x(1-2x)
y'=0 for critical points
0=e^-2x(1-2x)
Therefore e^-2x = 0 or 1-2x = 0
but e^-2x cannot equal 0, therefore
1-2x = 0
3x = 1
x = 1/2
Therefore y = 1/2*e^-1
= 1/(2e)
Nature of stationary point, test points x = 0 and x = 1
x = 0, y' = e⁰(1-0) = 1*1 = 1
x = 1, y' = e^-2(1-2) = -e^-2
Since the gradient is positive when x = 0 and the gradient is negative when x = 1, (1/2,1/(2e)) is a maximum stationary point

 

[SoulSearcher]

Later, a bit busy at the moment

 

[SoulSearcher]

Solve 2cos²theta - 1 = 0, where 0 <= theta <= 360

Spoiler

2cos²@ - 1 = 0
2cos²@ = 1
cos²@ = 1/2
cos @ = +1/rt2
@ = 45, 135, 225, 315

Next question:
i) Show that sin(x+y)sin(x-y) = sin²x-sin²y
ii) Hence solve the equation sin²3@-sin²@ = sin2@, for 0 < @ < pi

 

[Riviet]

Bonus Question:

i) Show that

d/dx{[(tan³x)/3] + tanx} = sec⁴x

ii) Find the equation of the normal to the curve y=(tan³x)/3 + tanx at x=pi/3.

* See previous page for latest question by SoulSearcher.

 

[Riviet]

It's not a fraction, rather two separate bits; (tan³x)/3 being the first bit and tanx being the other separate part. You should give it another go.

 

[SoulSearcher]

Bonus Question:

i) Show that

d/dx{[(tan³x)/3] + tanx} = sec⁴x

ii) Find the equation of the normal to the curve y=(tan³x)/3 + tanx at x=pi/3.

* See previous page for latest question by SoulSearcher.

Spoiler

a) d/dx{[(tan³x)/3] + tanx} = 3tan²xsec²x/3 + sec²x
= tan²xsec²x + sec²x
=(sec²x-1)sec²x + sec²x
= sec⁴x- sec²x + sec²x
= sec⁴x

b)x = pi/3,
y' = (1/2)⁴
= 1/16
Therefore gradient of normal is -16
x = pi/3, y = 3rt3/3 + rt3
= rt3 + rt3
= 2rt3
Therefore equation of normal is
y - 2rt3 = -16(x-pi/3)
y = 2rt3 + 16pi/3 - 16x

My question still stands, Exphate you didn't get it right.

 

[B35tY]

Hmm i got stuck halfway through the second one... any hints?

Spoiler

1.

LHS = sin(x+y)sin(x - y)

= (sinxcosy + cosxsiny)(sinxcosy - cosxsiny)

= sin^2(x)cos^2(y) - cos^2(x)sin^2(y) (Difference of 2 squares)

= sin^2(x) (1 - sin^2(y)) - (1 - sin^2(x)) sin^2(y)

= sin^2(x) - sin^2(x)sin^2(y) -sin^2(y) + sin^2(x)sin^2(y)

= sin^2(x) - sin^2(y)

2.

Prove:

sin^2(3@) - sin^2(@) = sin2@, for 0 <= @ <= pi

LHS = sin(3@ + @)sin(3@ - @) (From result proven in Q1)

= sin(2@ + 2@) * sin2@

???

 

[SoulSearcher]

It's solving the equation sin²3@ - sin²@ = sin2@ for the range 0 < x < pi, not proving it

 

[zeek]

Prove: 1+1=2

 

[Forbidden.]

Prove: 1+1=2

That's not in the 2 Unit Mathematics Preliminary Course or HSC Course ...

One and a one gives two, happy ?

 

[zeek]

no... prove it

 

[JhK89]

let x=1

x+x=2x

as x=1

=2(1)
therefore = 2

 

[SoulSearcher]

Prove: 1+1=2

http://mathforum.org/library/drmath/view/51551.html

Don't ask me how he does it

 

[Riviet]

i) Show that sin(x+y)sin(x-y) = sin²x-sin²y
ii) Hence solve the equation sin²3@-sin²@ = sin2@, for 0 < @ < pi

Spoiler

(ii) sin4x.sin2x=sin2x, using (i)
sin2x(sin4x-1)=0
sin2x=0 or sin4x=1
2x=0, pi, 2pi; 4x=pi/2, 5pi/2
.'. x=0, pi/2, pi, pi/8, 5pi/8.

 

[SoulSearcher]

Question please

 

[Riviet]

Next Question:

In a complex pulley system, the length of rope L needed for operation and the distance x m from a specific pulley to the centre of the system is defined by:

L=(x²-20x+136)^1/2 + (x²+64)^1/2

Find the distance x m that the pulley should be from the centre of the system that minimises the length of rope required.