Mathematics: Post your doubts here!

[19islandprincess96]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_1.pdf
Please help me solve qn no. 8 (i & ii).. an explanation will be appreciated.

Thanks!

f(x) = a+bcos 2x
f(0)=−1
=> a + bcos 2(0) = -1

Solve cos 2(0) in radian mode.
=> a + b = -1


and f(π/2) = 7
=> a + bcos 2(π/2) = 7
=> a - b =7

Solve them as simultaneous equations.

a + b = -1
a - b =7

a will be 3 and b -4.


Part ii)

y would be 0 when the curve intersects the x axis.
Putting the values of a and b found in i, the equation becomes
=> 3 - 4cos2x

And because y=0

3 - 4cos2x = o

I guess u can solve the rest.
Hope I helped!

 

[TERMINATOR]

A PROBABILITY QUESTION

Helen has some black tiles, some white tiles and some grey tiles. She places a single row of
8 tiles above her washbasin. Each tile she places is equally likely to be black, white or grey.
Find the probability that there are no tiles of the same colour next to each other

The ans is 128/2187

Any help will be highly appreciated.

 

[abcde]

I am jumping into you conversation out of nowhere but I had the same problem... My question is, how'd you know that max (sin^2 x) = 1 and min(cos^2 x) = 0?

The range of both the sine and cosine functions is -1 to 1. The range of their squares would be from 0 to 1 [ (-1)^2 = 1 ].

 

[Sandhya Mahat]

Plz help me with these questions guys..
1) Solve: 2^(2x-1) < 3^(3x-2)
2) Solve: (3* Mod(x)) / (x-1) < 2 , where Mod() means modulus or absolute value.
I would be very grateful.

 

[princess787]

A PROBABILITY QUESTION

Helen has some black tiles, some white tiles and some grey tiles. She places a single row of
8 tiles above her washbasin. Each tile she places is equally likely to be black, white or grey.
Find the probability that there are no tiles of the same colour next to each other

The ans is 128/2187

Any help will be highly appreciated.

yo terminator! so c now ther r 8 tiles! so 1st calculate the probability of getting d same color tiles! tht wld be (1/3)^8.... now its 1/3 cause therz an equal probability of al three colors... so (1/3)^8 and multiplied by 3 cause ther r 3 different colors! so u get 1/6561 into 3 = 3/6561 which reduces to 1/2187.... now the question asks tht no same color tile shld b aside so u subtract tht frm 1.... 1-1/2187.... which gives 2187-1/2187 giving 2186/2187....... thou i dunno hw they get 127.... well thts til where i cld reach..

 

[ahmed abdulla]

can any one help me bro's ???
q no. 1 q5 (ii)


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

i need also the same help >>> can any one help us please ?
WAITING

 

[Dug]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_33.pdf
Can someone help me?
Q10) i)
I made an equation in u but I don't understand how to integrate it.

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)

 

[Farahhassan]

I need help with S2 please :'(
Hypothesis testing !!

 

[ahmed abdulla]

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)

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

 

[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)