Mathematics: Post your doubts here!

[A star]

q no. 1 q5 (ii)


http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_11.pdf can u help bro

Q1 is easy
part i) tan (180 - x) = -tan x since pie = 180 we can see
tan (pi- x) = -k
part ii) tan(90- x)= 1/tanx so use it to get 1/k
for last part u know tan x = perp/base therefore
perpendicular = k
base = 1
hypotenuse = underroot(k^2 +1) hense sin x =perp/hypotenuse ull get the anser

 

[A star]

q no. 1 q5 (ii)


http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_11.pdf can u help bro

ans of 5 part ii) u can see a+bcos^2 x is a curve and each surve has max min points these are the greatest and least values of f(x)

 

[A star]

P(X<6)=0.105 Let (6-mean)/sd be a
P(Z<a)=0.105
a is negative (im not sure how you've been taught to determine whether a is -ve or +ve. What we do is that if both the sign is less than, i.e <, and the probability is less than 0.5, or if both are more than, then a is -ve)
1- Fi(a)= 0.105
Fi(a)= 0.895
a= -1.253

thanks alot aleezay

 

[Dug]

q no. 1 q5 (ii)


http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_11.pdf can u help bro

0 ≤ cos² x ≤ 1

f(x) = 2 - 5cos² x

When x = 0,
f(x) = 2 - 5[cos(0)]² = -3 (least value)

When x = 1,
f(x) = 2 - 5[cos(1)]² = 2 (greatest value)

 

[A star]

0 ≤ cos² x ≤ 1

f(x) = 2 - 5cos² x

When x = 0,
f(x) = 2 - 5[cos(0)]² = -3 (least value)

When x = 1,
f(x) = 2 - 5[cos(1)]² = 2 (greatest value)

hes asking part two i think

 

[abruzzi]

Hello,

I'm new to this form

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s07_qp_1.pdf
In qn no. 11(ii), why is the domain 0 < x ≤ 2 and not just x ≤ 2?

Thanks

 

[A star]

Hello,

I'm new to this form

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_1.pdf
In qn no. 11(ii), why is the domain 0 < x ≤ 2 and not just x ≤ 2?

Thanks

because function is defined for values of 0 ≤ x

 

[abruzzi]

because function is defined for values of 0 ≤ x

Thank you for your reply..

Yeah, but that's the domain for the inverse.. not the function. I'm confused

 

[tanmaydube]

HELP ME with the (ii) part,
i got (i) some what could you tell me whether i am right!? dx/dt= -3acos^2 t *sin^2 t and dy/dt = 3a sin^2 t* cos t ! now what to do?

 

[19islandprincess96]

I = ⌡(tanⁿ⁺² x + tanⁿ x) dx
I = ⌡[tanⁿ x(tan² x + 1)] dx
u = tan x
dx = du/sec² x

I = ⌡[tanⁿ x(tan² x + 1)] du/sec² x
I = ⌡uⁿ du
I = uⁿ⁺¹/(n + 1)

u = tan x
tan(π/4) = 1
tan(0) = 0

I = 1/(n + 1)

Thank you SO much! May you get loads of A* and great success in this life and in the hereafter!

 

[luniaz226]

need help! JazakAllah.

3. The first three terms in the expansion of (1 − 2x)^2 (1 + ax)^6, in ascending powers of x, are 1 − x + bx2. Find the values of the constants a and b. (9709/13, May/June 2012)

 

[A star]

Thank you for your reply..

Yeah, but that's the domain for the inverse.. not the function. I'm confused

when we inverse a function its domain becomes the range and range the domain of the inverse BUT the defining range remains constant . take it like this that only and only for value of x greater then 0 is the curve following the formulae given so it will only have a valid inverse function after zero

 

[Tkp]

need help! JazakAllah.

3. The first three terms in the expansion of (1 − 2x)^2 (1 + ax)^6, in ascending powers of x, are 1 − x + bx2. Find the values of the constants a and b. (9709/13, May/June 2012)

(1-4x+4x2)(1+6c1*ax+6c2*(ax)2)
(1-4x+4 x2)(1+6ax+15a2x2)
For a u need the coefficient of x
6a-4=-1
A=.5
For b
4+15a2-24a=b(put a =.5)
B=-17/4

 

[Khunkar]

help please!! Question number: 5
Question paper: http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_31.pdf
Marking Scheme: http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_ms_31.pdf

 

[Yousif Mukkhtar]

When a function as such is written, the 'a' is the amplitude of the wave i.e. its maximum height/displacement from the mean position. Hence, a = 9 -3 =6.
For a normal sine curve, b = 1. However, if we compare a normal curve with the given, we find that this one completes two wave cycles in a time period of 2 pi. Therefore, b = 2.
'c' is the value of how high or low the mean position of the wave is when compared to the line y = 0. Here the graph is raised +3 respective of the line y = 0, hence c = 3.

Thank you for your help as it really was useful.

 

[tanmaydube]

The answer to the first part is 1/x + 10/x^2 + 1/(10-x)


Cannot do part (ii) please help!

 

[Yousif Mukkhtar]

Can anyone explain Q1 i) ii) iii)
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_11.pdf

 

[parthrocks]

Hey can anyone help me with this
A(a,-3); B(-3,1);C(2,-9) are collinear,Find A....can someone plz show me how to solve this sum plz asap!!

 

[A star]

Can anyone explain Q1 i) ii) iii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_11.pdf

i solved it for someone u can go check that post

 

[abruzzi]

when we inverse a function its domain becomes the range and range the domain of the inverse BUT the defining range remains constant . take it like this that only and only for value of x greater then 0 is the curve following the formulae given so it will only have a valid inverse function after zero

Oh.. alright.. thanks