Mathematics Marathon HSC 09 (1 Viewer)

[ForrestGump]

You may have seen similar threads in other forums. In past years students have found it a great way to quiz each other on various areas of the syllabus and share their knowledge in preparation for the HSC Exams so I thought I would start one for the 2009 Mathematics students.

How To Play:
The first person asks a Mathmatics question (maybe one you've had trouble with). The next person answers it then posts another question for the next person then so on.

Tip:
If you feel an area is your weakness then try asking a question on this area to see others responses to it which may help you in your understanding.

1st Question: (Worth 3 marks in 2000 HSC)

• Find the equation of the tangent to the curve y=2logex at (1,0)

 

[ninetypercent]

Find the equation of the tangent to the curve y=2logex at (1,0)

dy/dx = 2/x
when x =1, m = 2

equation of tangent
2 = y - 0/x-1
2x -2 = y

next question:

differentiate 4cos2x + 6x

 

[ForrestGump]

differentiate 4cos2x + 6x

Well dy/dx of cos 2x = - 2 sin 2x

So if y = 4cos2x + 6x, then dy/dx = -8 sin 2x + 6

Okay how about:

If (4,-3) is the centre of a circle with radius 5 units.
a) Find the equation of the circle and b) Check if it passes through pt (4,2)

 

[boxhunter91]

(x-4)^2 + (y+3)^2 = 25

at (4,2)
(0)^2 + (5)^2= 25 therefore satisfies.

Solve tanx=2 for 0 < x < 2pi

 

[ForrestGump]

Solve tanx=2 for 0 < x < 2pi

The tanx curve doesnt reach 2 does it?

If you meant 2tanx=2, tanx=1
base angle = pi/4, tan is positive in first + third quadrant
therefore x=pi/4 and x=5pi/4

 

[boxhunter91]

You need to solve for x
Go to Q 5 2000 paper. you will see it there.

 

[ForrestGump]

You need to solve for x
Go to Q 5 2000 paper. you will see it there.

hahah im such an idiot i did that question yesterday.

okay base angle = 62.4349... degrees
to change to rads x pi/180
therefore x = 1.107148718 rads
tan is positive in 1st + 3rd quadrants
so x = 1.11 (2dp) and x = pi + 1.107... = 4.25 (2dp)

Okay next question
Solve the pair of simultaneous equations:
x + y = 1 and 2x - y = 5

 

[boxhunter91]

y=1-x
.: 2x-(1-x) = 5
.: 2x-1 + x = 5
.:3x+6
.:x=2
sub back into y=1-x
.:y=-1
.: (2,-1)

The third term of an arithmetic series is 32 and the sixth term is 17.
a)find the common difference.
b)find the sum of the first 10 terms.

 

[ForrestGump]

The third term of an arithmetic series is 32 and the sixth term is 17.
a)find the common difference.
b)find the sum of the first 10 terms.

Booyah series is my fave.
a) okay well: t3 = a + 2d = 32...(1) and t6 = a + 5d = 17...(2)
(2) - (1) = 3d = -15 so d = -5
sub d = -5 into (1) to get a
a - 10 = 32, so a = 42

b) sum to 10 terms = 10/2 (42 + (9 x -5))
S10 = 5(-3) = -15 (i think)

Okay try:
factorise x^3 - 27

 

[Zak Ambrose]

Sn = n/2 [2a + (n-1)d]

so S10 = 5 [84 - 45]
= 195


x^3 - 27 = (x-3)(x^2 +3x + 9)


solve for x.
(2e^2x) - e^x = 0

 

[iMAN2]

- Isn't this already factorised? :S



Question: Show that

 

[Iruka]

^ I think you need some limits of integration.

 

[iMAN2]

Solve for x:
Let


or
OR
OR
ln 0 is an invalid solution, does not exist.




Use the LaTeX Equation Editor for equations

 

[iMAN2]

^ I think you need some limits of integration.

my bad, forgot to put them in. Thanks.

 

[Zak Ambrose]

Solve for x:
Let


or

Use the LaTeX Equation Editor for equations

you've got to finish it off.

 

[untouchablecuz]

- Isn't this already factorised? :S



Question: Show that

if sinx=u, find (cos²x)^-1 in terms of u

 

[ForrestGump]

Okay so we're up to this question:

Question: Show that

 

[boxhunter91]

sin^2x = u^2
.:cos^2x = 1-sin^2x
=(1-U^2)^-1?
Maybe i need sec^2x in there.

 

[iMAN2]

if sinx=u, find (cos²x)^-1 in terms of u

(cos²x)^-1
=(1-sin²x)^-1
=(1-u²)^-1
=1/1-u²

 

[iMAN2]

The curve secx is rotated around the x-axis between x=0 and x=pi/3 to form a solid. Find the volume of this solid.