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Mathematics: Post your doubts here!

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I dont find them that difficult, its just thinking a bit and that's all, but my marks goes only in this parts :p Hence I am poor at it.. :p Anyways... :D
Hahaha ... That's what I'm talking about. Practice more and you will hopefully get it. Complex numbers and vectors also require a bit of imagination, or rather, visualisation.
 
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Part (ii) I used probability as 0.8 however the mark scheme has used probability of 0.9... can somebody explain why? This is M/J 2005 p6
20160304_171756-1.jpg
 
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Part (ii) I used probability as 0.8 however the mark scheme has used probability of 0.9... can somebody explain why? This is M/J 2005 p6
View attachment 59563
P(1.9 - b < X < 1.9 + b) = 0.80

P(X < 1.9 + b) - P(X < 1.9 - b) = 0.80

Standardizing X, using Z = X - μ / σ

P( Z < 1.9 + b - 1.9 / 0.15 ) - P(Z < 1.9 - b -1.9) = 0.80

P(Z< b/0.15) - P(Z <-b/0.15) = 0.80
Ф(b/0.15) - Ф(-b/0.15) = 0.80
2Ф (b/0.15) - 1 = 0.80
2Ф (b/0.15) = 1.80
Ф (b/0.15) = 0.90

Using Normal Distribution tables,

b/0.15 = 1.281
b = 0.192 (3sf)

You can now use this value of b to find the safety limits. :)
 
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P(1.9 - b < X < 1.9 + b) = 0.80

P(X < 1.9 + b) - P(X < 1.9 - b) = 0.80

Standardizing X, using Z = X - μ / σ

P( Z < 1.9 + b - 1.9 / 0.15 ) - P(Z < 1.9 - b -1.9) = 0.80

P(Z< b/0.15) - P(Z <-b/0.15) = 0.80
Ф(b/0.15) - Ф(-b/0.15) = 0.80
2Ф (b/0.15) - 1 = 0.80
2Ф (b/0.15) = 1.80
Ф (b/0.15) = 0.90

Using Normal Distribution tables,

b/0.15 = 1.281
b = 0.192 (3sf)

You can now use this value of b to find the safety limits. :)
Damn mashallah! Thanks
 
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Part (ii) for this please :).View attachment 59564
Untitled.png

I think the diagram above will help.
You can find the pink area using sector area formula.
You can find yellow area by first finding area of sector CAB, then subtracting the right angle triangle CAB. (multiply two since there are two yellow parts)
You can find shaded area by first finding area of circle CBED, then subtracting yellow part then subtracting pink part.

Sorry I didn't do the actual calculation :D
 
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Part (ii) for this please :).View attachment 59564
Find the angle CBD.
θ = 2 sin^-1 (1/√2)
= pi/2
Then,
find the area of the sector, CBD.
A = 1/2 (2√r)^2 * pi/2
= (pi r^2 )/2

Now, you know the area of sector CBD, so you can find the area of the region, CADF. (I have taken F a point on the arc to the left of E)
Area of region CADF = (pi r^2 )/2 - (1/2 * 2r * r)
= (pi r^2) /2 - r^2

Find the area of the hemisphere of the smaller circle and subtract the area of the region CADF from it.
Hence, area of shaded region will be:

A = (pi r^2) /2 - ((pi r^2) /2 - r^2)
= r^2 Ans.
 
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View attachment 59565

I think the diagram above will help.
You can find the pink area using sector area formula.
You can find yellow area by first finding area of sector CAB, then subtracting the right angle triangle CAB. (multiply two since there are two yellow parts)
You can find shaded area by first finding area of circle CBED, then subtracting yellow part then subtracting pink part.

Sorry I didn't do the actual calculation :D
Wow thanks. Did u make that?
 
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Find the angle CBD.
θ = 2 sin^-1 (1/√2)
= pi/2
Then,
find the area of the sector, CBD.
A = 1/2 (2√r)^2 * pi/2
= (pi r^2 )/2

Now, you know the area of sector CBD, so you can find the area of the region, CADF. (I have taken F a point on the arc to the left of E)
Area of region CADF = (pi r^2 )/2 - (1/2 * 2r * r)
= (pi r^2) /2 - r^2

Find the area of the hemisphere of the smaller circle and subtract the area of the region CADF from it.
Hence, area of shaded region will be:

A = (pi r^2) /2 - ((pi r^2) /2 - r^2)
= r^2 Ans.
Thanks I got most of it except the CADF part. What is (1/2*2r*r)? Is it a triangle? And can you be more specific about the location of point F. Appreciate it
 
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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s14_qp_11.pdf

Please help me with question 10 (iii)
Also, in this question they say "set of values for which each expression is valid." Do they mean find domain of the function or the range?
I'm really confused.Thanks in advance.
f(x) = 3x - 2
replace x with y and f(x) with x
x = 3y - 2
make y subject.
y = 1/3(x + 2)
f^-1(x) = 1/3(x + 2)
Range of f(x) = domain of f^-1(x)
range of f(x) :domain of f^-1(x) = -5 <= x <= 1
f(x) = 4/(5-x)
x = 4/(5 - y)
y = 5 - (4/x)
f^-1(x) = 5 - (4/x)
domain of f^-1(x) : range of f(x) = 1 < x <= 4
 
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