Warning: Illegal string offset 'html' in /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php on line 909

HSN Maths Quiz - HSN forum

# HSN Maths Quiz

23 replies to this topic

### #1Dave

Ruler (but not owner) of hsn

• Moderators
• 4,252 posts
• Location:kilmarnock(ok kilmaurs)
• Interests:programming, exercising, brass band, using this board
• Gender:Male

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

If i am not here i am somewhere else

### #2Dave

Ruler (but not owner) of hsn

• Moderators
• 4,252 posts
• Location:kilmarnock(ok kilmaurs)
• Interests:programming, exercising, brass band, using this board
• Gender:Male

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} x3 + 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 = 4/5 find cos 2 and sin 2 as a fraction

If i am not here i am somewhere else

### #3Dave

Ruler (but not owner) of hsn

• Moderators
• 4,252 posts
• Location:kilmarnock(ok kilmaurs)
• Interests:programming, exercising, brass band, using this board
• Gender:Male

Posted 22 January 2005 - 02:03 PM

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

2) {limits} 0 to 3 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 + 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 = 5/ 41
sin = 4/ 41

cos2 = cos2 - sin2
= (5/ 41) - (4/ 41)
= 25/41- 16/41
= 9/41

sin2 = 2sin cos
2*(4/ 41) * (5/ 41)

40/41

i hope you all did ok in this

Dave

If i am not here i am somewhere else

### #4Dave

Ruler (but not owner) of hsn

• Moderators
• 4,252 posts
• Location:kilmarnock(ok kilmaurs)
• Interests:programming, exercising, brass band, using this board
• Gender:Male

Posted 22 January 2005 - 02:25 PM

This weeks questions

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

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

3) Find k if 2x + x + 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 x 360

cos 2x - 3cosx + 2 = 0

Happy mathing

dave

If i am not here i am somewhere else

### #5Dave

Ruler (but not owner) of hsn

• Moderators
• 4,252 posts
• Location:kilmarnock(ok kilmaurs)
• Interests:programming, exercising, brass band, using this board
• Gender:Male

Posted 07 February 2005 - 05:24 PM

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

2) sin 2 /1 + cos 2[theta]
= 2sin[theta]cos[theta]/2cos2
= 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

If i am not here i am somewhere else

### #6calumrd8

Showing Improvement

• Members
• 19 posts
• Gender:Male

Posted 04 January 2006 - 04:42 PM

Are these questions common in exams?

### #7calumrd8

Showing Improvement

• Members
• 19 posts
• Gender:Male

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

### #8The Wedge Effect

HSN Legend

• Members
• 2,477 posts
• Location:Paisley
• Gender:Male

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

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

### #9Dave

Ruler (but not owner) of hsn

• Moderators
• 4,252 posts
• Location:kilmarnock(ok kilmaurs)
• Interests:programming, exercising, brass band, using this board
• Gender:Male

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

If i am not here i am somewhere else

### #10dfx

Fully Fledged Genius

• Members
• 1,955 posts
• Gender:Male

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

Showing Improvement

• Members
• 29 posts
• Gender:Male

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.

Showing Improvement

• Members
• 29 posts
• Gender:Male

Posted 10 January 2006 - 02:53 PM

Sorry, I mean question 2.

### #13Dave

Ruler (but not owner) of hsn

• Moderators
• 4,252 posts
• Location:kilmarnock(ok kilmaurs)
• Interests:programming, exercising, brass band, using this board
• Gender:Male

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

If i am not here i am somewhere else

Showing Improvement

• Members
• 29 posts
• Gender:Male

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.

### #15Nathan

Fully Fledged Genius

• Members
• 1,736 posts
• Location:Aberdeen, Scotland
• Gender:Male

Posted 10 January 2006 - 04:22 PM

I think dave's right :S

### #16Ally

HSN Legend

• Moderators
• 2,912 posts
• Interests:Just finished 1st year at uni studying medicine.
• Gender:Male

Posted 10 January 2006 - 04:24 PM

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

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.

Why is it not?

edit: I was being stupid.

Showing Improvement

• Members
• 29 posts
• Gender:Male

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

### #18Dave

Ruler (but not owner) of hsn

• Moderators
• 4,252 posts
• Location:kilmarnock(ok kilmaurs)
• Interests:programming, exercising, brass band, using this board
• Gender:Male

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

If i am not here i am somewhere else

### #19Steve

Top of the Class

• 435 posts
• Location:Edinburgh
• Gender:Male

Posted 10 January 2006 - 08:30 PM

I was confused as to what the integral was though, or . This is what I get:

and:

For the first one, you could also do:

HSN contribute: Help the site grow!

Looking for a Maths tutor in West Lothian? Just PM me!

Showing Improvement

• Members
• 29 posts
• Gender:Male

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

#### 1 user(s) are reading this topic

0 members, 1 guests, 0 anonymous users