[CIE] All Pure Mathematics (P1, P2 and P3) doubts here !

[elemis]

Question 3:
Complete the square of f(x) to get (x-2)^2+3 to H(x-2)=(x-2)^2+3----> so H(x) = x^2+3.....
Question 4:
length of PQ is the (length of radius of the arc - height of triangle ) x 2
----> (5-5cos(0.6rad))x2


@ Ari: do you know how marks are distributed on sketches of graphs? (not drawings)

You get marks for the general shape, intersection points with coordinate axis and drawing in any asymptotes.

[SkyPilotage]

You get marks for the general shape, intersection points with coordinate axis and drawing in any asymptotes.

ok so if its a curve like sinx, i can draw it free hand and just show where it intersects with the axis right?
-If they ask us to sketch a graph and its domain in x>0 , i should only draw the graph on the positive x-axis; should I show him the negative axis and leave it blank?
-when they ask us abt solving eqns do we need to show the quadratic formula or shall we just write the factors then the answer??
-which is better? to write the final answer as fractions or decimals? and if there is a square root of irration number do we leave it tht way or convert to decimals?
THankS

[Tohru Kyo Sohma]

Question 3:
Complete the square of f(x) to get (x-2)^2+3 to H(x-2)=(x-2)^2+3----> so H(x) = x^2+3.....
Question 4:
length of PQ is the (length of radius of the arc - height of triangle ) x 2
----> (5-5cos(0.6rad))x2

i understood the 4th qns.thanks ...but im still stuck at the 3rd qns i got till H(x-2)=(x-2)^2+3 myself too. but i cant get the final answer as H(x) = x^2+3
can you show me the steps??
plus do you understand may/june 2010 p11 qns 1??because im so lost in that qns

[SkyPilotage]

i understood the 4th qns.thanks ...but im still stuck at the 3rd qns i got till H(x-2)=(x-2)^2+3 myself too. but i cant get the final answer as H(x) = x^2+3
can you show me the steps??
plus do you understand may/june 2010 p11 qns 1??because im so lost in that qns

Actually you have to be able to see tht what gives u f(x) when u substitute x by x-2....
-June 2010...Yh it was answered before in this topic..Just go back some pages..

[$H00t!N& $t@r]

question

Q10 (ii) and (iii) m/j 2009
 

[SkyPilotage]

question

Q10 (ii) and (iii) m/j 2009
 

II)
the axis of symmetry is at x=3 so the distance from both sides of the axis must be equal so A=6
III)
to find the range in the domain given, the minimum point is at the vertex(x=3) and y=-5
the largest y is when x=0 or x=6---->y=13
----->the range is -5<y<13

[$H00t!N& $t@r]

II)
the axis of symmetry is at x=3 so the distance from both sides of the axis must be equal so A=6
III)
to find the range in the domain given, the minimum point is at the vertex(x=3) and y=-5
the largest y is when x=0 or x=6---->y=13
----->the range is -5<y<13

oh! thanks  
for the question about range... i got a bit confused because i was told that whenever you have a(x + b)^2 + c ... c is always the range. I didn't consider the part where they said "When A has this value..."
So what if they only said find the range of f. Would it just be x>= -5 ?

[........]

If you have a given domain and a equation just substitute x values into the equation to get the range.

Correct me if i am wrong.

[elemis]

If you have a given domain and a equation just substitute x values into the equation to get the range.

Correct me if i am wrong.

True.

[$H00t!N& $t@r]

yeah I know that...
 I just wanted to know if the statement "whenever you have a(x + b)^2 + c ... c is always the range" is right or not...

[Arthur Bon Zavi]

yeah I know that...
 I just wanted to know if the statement "whenever you have a(x + b)^2 + c ... c is always the range" is right or not...

Almost all cases. Look for the domain first by the way.

[$H00t!N& $t@r]

Almost all cases. Look for the domain first by the way.

ok thanks

[$H00t!N& $t@r]

Q6 ii)

y = kx² +1   &  y = kx

equate the two to get k = 4 (follow up from the previous- but k² - 4k = 0 )
equate k=4 in kx² - kx +1 = 0 to get x = 1/2
equate x = 1/2 in one of the equations to get the y coordinate

Q 10

Expressing h in terms of x

Vol = 4 cm3 (given)

vol of object = x * 0.5x * h
4 = 0.5 * h * x^2
Change the sub. of formula to h to get h = 8/x²

Total Surface area (consider each side and add them up to make things easy)

= 2xh + 2*0.5*xh + 0.5 * x* x + 4/5 x * 5/4 x

solve, and substitute h = 8/x² to get the desired answer.

ii)

dA/ dx = 1.5 * 2 * x + 24 * -1 * * x ^-1-1

= 3x - 24/x^2

equate this to zero to obtain the stationary point as 2

d²A/ dx² = 3 - 24 * -2 * x ^-2-1

= 3 + 48/x^3

equate x =2 to get 9
since 9 > 0, point of minima

Thanks a lot Vin  I thought no one replied.

[Vin]

No problem

Okay, i have a stupid doubt which i cant waste my time on thinking on much.

http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9709%20-%20Mathematics/&file=9709_s09_qp_1.pdf

Q4 part i

-.-

[SkyPilotage]

No problem

Okay, i have a stupid doubt which i cant waste my time on thinking on much.

http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9709%20-%20Mathematics/&file=9709_s09_qp_1.pdf

Q4 part i

-.-

Ahaha it happens....Actually its not stupid
-For c:
-----Substitute x=0 to get c=3
-For a:
Remember max value of sinx is 1 and min is -1.. so substitute sin(bx)=1 --->9=a(1)+3---->a=6
-For b:
The graph oh sinx has a whole wave from 0 to 2pi but here there are 2 curves in the domain 0 to pi...Therefore B=2...