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2u Mathematics Marathon v1.0 (1 Viewer)

Riviet

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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. :p
 

Riviet

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Speaking of calculus...

Next Question:

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

SoulSearcher

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Riviet said:
Speaking of calculus...

Next Question:

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

Riviet

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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.
 
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Riviet

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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

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Riviet said:
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.
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

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


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@

???
 

Riviet

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

Riviet

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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.
 

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