~~Mathematics Revising Game~~ (1 Viewer)

[bored of sc]

Find the values of k for which:
y = x<SUP>2</SUP> + 2kx + (k+2) = 0
has two real, distinct roots.

Discriminant > 0

4k²-4k-8 > 0 *

k²-k-2 = 0

(k-2)(k+1) = 0
k = -1, 2

Sub in k = 0 into *

-8 > 0 False

Therefore k < -1, k > 2

 

[Js^-1]

Yepo, perfect.

 

[bored of sc]

If f(x) = 4x⁹-3x⁸+9x⁷-5x⁶+x⁵-3x⁴+6x³-3x²+10x-13 find f''''''''(x) (the 8th derivate).

 

[Timothy.Siu]

If f(x) = 4x⁹-3x⁸+9x⁷-5x⁶+x⁵-3x⁴+6x³-3x²+10x-13 find f''''''''(x) (the 8th derivate).

umm
(4x9x8x7x6x5x4x3x2)x-(3x8x7x6x5x4x3x2x1)
= 1451520x- 120960
is that right??
umm

find a primitive of tan x

 

[homijoe]

umm
(4x9x8x7x6x5x4x3x2)x-(3x8x7x6x5x4x3x2x1)
= 1451520x- 120960
is that right??
umm

find a primitive of tan x

the intergral of sinx/cosx .dx = -lncosx + c

find the derivative of y=ln(e^x sin^2 x)

 

[Js^-1]

y' = 1 + 2 cot x ?

I'll post working out when i get home if its right. Out to watch Saw 5

 

[homijoe]

y' = 1 + 2 cot x ?

I'll post working out when i get home if its right. Out to watch Saw 5

if u mean for the question :find the derivative of y=ln(e^x sin^2 x)

thats incorrect...u should start off with y'=u'v+v'u
---------- f(x)

 

[Timothy.Siu]

if u mean for the question :find the derivative of y=ln(e^x sin^2 x)

thats incorrect...u should start off with y'=u'v+v'u
---------- f(x)

lol y=e^xsin^2x y'=e^x.2cos^2x+sin^2x.e^x=e^x(2cos^2x+2in^2x)=e^x(cos^2x+1)

soo derivative of that is

[e^x(cos^2x+1)]/(e^x sin^2x)=cos^2x+1/sin^2x=cot^2x + cosec^2x

 

[Azreil]

y= ln(e^x[1/2{cos2x+1}])
y'=f'(x)/f(x)
f(x)=e^x[1/2{cos2x+1}]
f'(x)=uv'+vu'
u=e^x u'=e^x v=1/2cos2x+1/2 v'=-1/4sin2x
f'(x)=(e^x[cos2x+1])/2 - (e^x sin2x)/4
= e^x/2([cos2x+1] - sin2x/2)
y'= {e^x/2([cos2x+1] - sin2x/2)}/(e^x[1/2cos2x+1])
= 2cos2x + 2 - sin2x / cos2x + 1
= 2cos2x+2/cos2x+1 - sin2x/cos2x+1
= 2 - sin2x/cos2x+1
I have a feeling I've mucked up somewhere in there. Please point it out if you can see the flaw in my working.

Solve:
2e^2x - e^x = 0

 

[homijoe]

y= ln(e^x[1/2{cos2x+1}])
y'=f'(x)/f(x)
f(x)=e^x[1/2{cos2x+1}]
f'(x)=uv'+vu'
u=e^x u'=e^x v=1/2cos2x+1/2 v'=-1/4sin2x
f'(x)=(e^x[cos2x+1])/2 - (e^x sin2x)/4
= e^x/2([cos2x+1] - sin2x/2)
y'= {e^x/2([cos2x+1] - sin2x/2)}/(e^x[1/2cos2x+1])
= 2cos2x + 2 - sin2x / cos2x + 1
= 2cos2x+2/cos2x+1 - sin2x/cos2x+1
= 2 - sin2x/cos2x+1
I have a feeling I've mucked up somewhere in there. Please point it out if you can see the flaw in my working.

Solve:
2e^2x - e^x = 0

ive attached the solution to the derivative question

2e^2x - e^x = 0

e^x(2e^X-1)=0

e^x=o has no solution

2e^x=1
e^x=0.5
x=ln0.5

 

[kangms]

i couldn't post the image up so i attached a question from Syd Boys' 08 trials.
its QUESTION 3 by the way. Took me 20min to get it...farout

 

[Timothy.Siu]

lol i remember doing that question, i've done it b4 so i wont do it

 

[reynoldson]

Hi could someone please have a look at at q9 b) ii) of the Baulkham Hills 2004 Trial, Namu posted it up earlier in the thread, anyways, ive used the change of base rule and convereted the √a base to powers of a half, resulting in the denominator of the change of base = 1/2, so they are times by two, except when i collect them using the log rules, and even get it to one expression, im still not getting the right answer, thanks.

 

[old.skool.kid]

i couldn't post the image up so i attached a question from Syd Boys' 08 trials.
its QUESTION 3 by the way. Took me 20min to get it...farout

LoL man thats ridiculous for q3. Whats the point of even putting that in an exam when its not gunna be in the HSC?

That took me a while and im not sure if its right.

But i got CD+DE+EF+FG = 2sin@+ root3sin@+1.5sin@+3root3/4sin@

=3.5+7root3/4sin@


Oh and new question IF i got it right (which i probably didnt)

One model for the number of mobile phones in use worldwide is the exponential growth model,

N= Ae^kt

Where N is the estimate for the number of mobile phones in use (in millions), and t is the time in years after 1 January 2008.

i)It is estimated that at the start of 2009, when t=1, there will be 1600 million mobile phones in use, while at the start of 2010, when t=2, there will be 2600 million. Find A and k.

 

[Timothy.Siu]

LoL man thats ridiculous for q3. Whats the point of even putting that in an exam when its not gunna be in the HSC?

That took me a while and im not sure if its right.

But i got CD+DE+EF+FG = 2sin@+ root3sin@+1.5sin@+3root3/4sin@

=3.5+7root3/4sin@

dont think so

 

[old.skool.kid]

LOL damn, what way off?

 

[Timothy.Siu]

CD=2sin A
DE=2sin^2 A

soo
a=2sin A
r=sin A
limitin sun = 2sin A/(1-sin A)
i think thats it, not too hard

 

[Js^-1]

lol y=e^xsin^2x y'=e^x.2cos^2x+sin^2x.e^x=e^x(2cos^2x+2in^2x)=e^x(cos^2x+1)

soo derivative of that is

[e^x(cos^2x+1)]/(e^x sin^2x)=cos^2x+1/sin^2x=cot^2x + cosec^2x

This is wrong.

The part in bold is the part where you made the mistake.
d/dx (sin²x) =/= cos²x

The answer is what Homijoe posted, i.e.
y' = [e^x sinx (sinx + 2cosx)] / [e^xsin²x]

This simplifies to:
y' = [e^x sinx (sinx + 2cosx)] / [e^xsinx sinx]
y' = [sinx + 2 cosx] / sinx
y' = sinx/sinx +2cosx/sinx
y' = 1 + 2 cotx

 

[Timothy.Siu]

lol, ok, i seee, not concentrating.

 

[reynoldson]

LoL man thats ridiculous for q3. Whats the point of even putting that in an exam when its not gunna be in the HSC?

That took me a while and im not sure if its right.

But i got CD+DE+EF+FG = 2sin@+ root3sin@+1.5sin@+3root3/4sin@

=3.5+7root3/4sin@


Oh and new question IF i got it right (which i probably didnt)

One model for the number of mobile phones in use worldwide is the exponential growth model,

N= Ae^kt

Where N is the estimate for the number of mobile phones in use (in millions), and t is the time in years after 1 January 2008.

i)It is estimated that at the start of 2009, when t=1, there will be 1600 million mobile phones in use, while at the start of 2010, when t=2, there will be 2600 million. Find A and k.

I remember this one, because the first term is t= 1 not t= 0, you have to divide 2600/1600 with the other values to find k, you cant sub it in like normal