Mathematics: Post your doubts here!

[Konstantino Nikolas]

I dont find them that difficult, its just thinking a bit and that's all, but my marks goes only in this parts Hence I am poor at it.. Anyways...

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.

 

[Saad the Paki]

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

 

[Rizwan Javed]

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.

 

[Saad the Paki]

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

 

[Saad the Paki]

Part (ii) for this please .

 

[Rizwan Javed]

Part (ii) for this please .View attachment 59564

What's the answer? I'll post the solution if my answer will be correct.

 

[qwertypoiu]

Part (ii) for this please .View attachment 59564

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

 

[Rizwan Javed]

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.

 

[Saad the Paki]

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

Wow thanks. Did u make that?

 

[Saad the Paki]

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

 

[Anum96]

Oh thanks! I spotted out my mistake.

Good to know You're welcome

 

[Rizwan Javed]

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

I've marked the point F. The region CADF is coloured red.

And yes that was a triangle area that I subtracted. the triangle was CBD.

 

[qwertypoiu]

Wow thanks. Did u make that?

You're welcome.
I didn't make it. I screenshotted the thingy and edited with paint.

 

[Saad the Paki]

View attachment 59566
I've marked the point F. The region CADF is coloured red.

And yes that was a triangle area that I subtracted. the triangle was CBD.

Ohk thanks a lot.

You're welcome.
I didn't make it. I screenshotted the thingy and edited with paint.

Yeah that's what I meant to ask. Thank you again

 

[Raj21999]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s15_qp_13.pdf
please help me with question 8 ( iii)
i know how to differentiate it but don't know how to find the value of x in this case.
i would appreciate your help

 

[Anum96]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s15_qp_13.pdf
please help me with question 8 ( iii)
i know how to differentiate it but don't know how to find the value of x in this case.
i would appreciate your help

 

[Copy Cat]

(ii) & (iii) part

 

[Raj21999]

View attachment 59573

Thanks a lot for your help.

 

[Raj21999]

http://papers.xtremepapers.com/CIE/...S 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.

 

[The Sarcastic Retard]

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