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HSN Maths Quiz - HSN forum

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HSN Maths Quiz


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#1 Dave

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Posted 13 January 2005 - 05:29 PM

This pinned thread will be the home of the maths quiz

The thread will remain closed so as to allow others to chance to find the answers for themselves

users can PM me there answers if they wish or ask for help

Questions will be made up be me so cannot be found in textbooks

Qustion will follow this weekend

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#2 Dave

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Posted 14 January 2005 - 06:26 PM

1) If f(x) = (8x-5)1/2 find f'(5) to 3 sig fig

2) Solve {limits: 0-3} integral.gif x3 + sqrt.gif 3x dx

3) Fred keeps fish in his pond, initially he has 20 fish. fred assumes that each year he will lose 25% of his fish stock and so each year he adds 4 fish to his pond.

a) how many fish will Fred have after 3 years

b) for the recurrence relation to have a limit what condition must be met

c) how many fish will be in the pond in the long term

4) The graphs of 2cos x and 4 sin 2x are plotted on one axis. find the points of intersection

5) If tan theta.gif = 4/5 find cos 2 theta.gif and sin 2 theta.gif as a fraction

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#3 Dave

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Posted 22 January 2005 - 02:03 PM

answers

1) f'(x) = 1/2(8x-5) -1/2 * 8
= 4* 1/ sqrt.gif 8x-5
f'(5) = 4 * 1/ sqrt.gif 35

2) {limits} 0 to 3 integral.gif x3 + (3x)1/2 dx
[x4/4 + 3x3/2/3/2]
[x4/4 + 6x3/2/3]
[x4/4 + 2x3/2]

34/4 + 63/2 - 0

20.25 + sqrt.gif 54

3)

a)
Un+1 = 0.75 Un + 4
U3 = 17.69

(answer can be found by finding U1 then U2 to finally find U3)

b) -1<a<1

c) L = b/1-a
L = 4/(1-0.75)
L= 4/0.25
L = 16

4)

2cos x = 4 sin 2x
cos x = 2sin2x
cos x = 2(2sinxcosx)
cos x = 4 sinxcosx
4sinxcosx-cosx = 0
cosx(4sinx-1)=0


cosx =0 4sinx-1=0
x = 90, 270 sinx = 1/4
x = 14.5,165.5

2cos 90 = 0 A(90,0)
2cos 270=0 B(270,0)
2cos 14.5 = 2 C(14.5,2)
2cos 165.5 = -1.9 D(-1.9)

5) if you draw the triangle you will find:

cos theta.gif = 5/ sqrt.gif 41
sin theta.gif = 4/ sqrt.gif 41

cos2 theta.gif = cos2 theta.gif - sin2 theta.gif
= (5/ sqrt.gif 41) power2.gif - (4/ sqrt.gif 41) power2.gif
= 25/41- 16/41
= 9/41

sin2 theta.gif = 2sin theta.gif cos theta.gif
2*(4/ sqrt.gif 41) * (5/ sqrt.gif 41)

40/41


i hope you all did ok in this

Dave

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#4 Dave

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Posted 22 January 2005 - 02:25 PM

This weeks questions

1) Express 3-6x - x power2.gif in the form a(x+p) power2.gif + b

2) Prove that sin 2 theta.gif powerslash.gif 1 + cos 2[theta] = tan theta.gif

3) Find k if 2x power3.gif + x power2.gif + kx -8 is divisible by x+2

4) Find P given (P+1)x[^2] -2(P+3)x + 3P = 0 has equal roots

5)Solve the trignometric equation for 0 lessorequal.gif x lessorequal.gif 360

cos 2x - 3cosx + 2 = 0

Happy mathing

dave

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#5 Dave

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Posted 07 February 2005 - 05:24 PM

answers

1) 3 - 6x - x2
= -x2 - 6x + 3
= -(x2 +6x) +3
= -(x2 +6x+9)-9+3
= -(x+3)2 -6

2) sin 2 theta.gif /1 + cos 2[theta]
= 2sin[theta]cos[theta]/2cos2theta.gif
= sin[theta]/cos[theta] = tan[theta]

3) using remander theorem on (x+2)

solve what you get -20-2k = 0, k = -10

4) as equal roots b2-4ac = 0

sub in values you should get after simplifiing

4(p2 + 6P +9) = 0
4(p+3)(P+3) =
P=-3

5)

cos2x -3cosx + 2 =0
2cos2x -1 -3cosx+2 = 0
2cos2x - 3cosx +1=0
(2cosx - 1)(cosx -1)=0
2cosx-1 =0 , cosx - 1= 0

cosx = 1/2 , cosx x = -1
x = 60,300 , x = 0

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#6 calumrd8

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Posted 04 January 2006 - 04:42 PM

Are these questions common in exams?

#7 calumrd8

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Posted 04 January 2006 - 04:44 PM

can you please direct me to which section Q4 in the Jan 14 2005 questions is in.

Thanks

#8 The Wedge Effect

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Posted 04 January 2006 - 05:16 PM

QUOTE(calumrd8 @ Jan 4 2006, 04:44 PM)
can you please direct me to which section Q4 in the Jan 14 2005 questions is in.

Thanks

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Think it's unit 2, something about sin and cos rules and stuff like that.

#9 Dave

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Posted 04 January 2006 - 08:17 PM

yeah it is unit 2 to do with double angle formula and trig graphs

These are all very common type questions and arent particularly easy ones either. Mostly adapted from the kind of homework questions i got as "formal homework exercises" which were very good exam practice for me back then

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#10 dfx

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Posted 05 January 2006 - 07:22 AM

QUOTE(Dave @ Jan 13 2005, 06:29 PM)
The thread will remain closed so as to allow others to chance to find the answers for themselves

View Post



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#11 duncad

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Posted 10 January 2006 - 02:52 PM

Dave,
I know it was almost a year ago but your answer to question 4 on 22 Jan 2005 is wrong. I suggest you try the integration again.


#12 duncad

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Posted 10 January 2006 - 02:53 PM

Sorry, I mean question 2.

#13 Dave

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Posted 10 January 2006 - 03:50 PM

there was nothing wrong with the integration. The problem was in sub in x=3 where on the 2nd last line i put the bracket around the x not the 2x which was to the power 3/2

fixed now

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#14 duncad

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Posted 10 January 2006 - 04:16 PM

Sorry Dave but the integration is incorrect.
Integrating (3x)^1/2 is not the same as integrating 3(x)^1/2 which is what you have done.

#15 Nathan

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Posted 10 January 2006 - 04:22 PM

I think dave's right :S

#16 Ally

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Posted 10 January 2006 - 04:24 PM

QUOTE(nathanm @ Jan 10 2006, 04:22 PM)
I think dave's right :S

View Post


Ditto.
QUOTE(duncad @ Jan 10 2006, 04:16 PM)
Integrating (3x)^1/2 is not the same as integrating 3(x)^1/2 which is what you have done.

View Post


Why is it not?

edit: I was being stupid. rolleyes.gif

#17 duncad

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Posted 10 January 2006 - 04:38 PM

Think 'function of a function'

or

(3x)^1/2
is the same as 3^1/2 * x^1/2
whereas 3(x)^1/2 is 3*x^1/2

#18 Dave

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Posted 10 January 2006 - 07:43 PM

well actually i am correct (as far as i can see anyway)

it is correct what you say, however 31/2 is no more than a constant and isnt effected by the integration

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#19 Steve

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Posted 10 January 2006 - 08:30 PM

Sorry Dave, I think your answer is wrong.

I was confused as to what the integral was though, \int_0^3{x^3+\sqrt{3x}}\,dx or \int_0^3{x^3+\sqrt{3}x}\,dx. This is what I get:



\begin{align*}
\int_0^3{x^3+\sqrt{3x}}\,dx &= \int_0^3{x^3+(3x)^{\frac12}}\,dx \\
&= \left[\frac14x^4 + \frac29(3x)^\frac32\right]^3_0 \\
\end{align*}

and:



\begin{align*}
\int_0^3{x^3+\sqrt{3}x}\,dx &= \left[\frac14x^4 + \frac{\sqrt3}{2}x^2\right]^3_0 \\
\end{align*}

For the first one, you could also do:



\begin{align*}
\int_0^3{x^3+\sqrt{3x}}\,dx &= \int_0^3{x^3+3^{\frac12}x^{\frac12}}\,dx \\
&= \left[\frac14x^4 + \frac{2\sqrt3}{3}x^\frac32\right]^3_0 \\
\end{align*}

which is the same answer.
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#20 duncad

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Posted 11 January 2006 - 06:29 AM

That is correct Steve.

Could someone tell me how I can type in the math symbols please.

Thank you





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