Mathematics: Post your doubts here!

[anonymous123]

http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s09_qp_1.pdf
Q4ii)
im stuck at sin2x=-1/2

AoA wr wb
sin2x = -0.5
to make it easier
take 2x = y and make the sign +
sin y = 0.5
y=(1/6)pi
now if sin y = -0.5 use the quadrant method...sin is negative in 3rd nd 4th quadrnt
so for 3rd quadrant it wud be = pi + (1/6)pi = (7/6)pi
and for the fourth quadrant its = 2pi - (1/6)pi = (11/6)pi
as we took y = 2x
2x = (7/6)pi
x=(7/12)pi
do the same for 11/6 pi and the value u get for x will be greater thn dis one...we need the smallest value so its 7/12pi
u dnt need to do it for the fourth quadrant its obvious i did it just to xplain thngs clearly

 

[alphabravocharlie]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_31.pdf

Q 7 part 3 and Q 9 part 1

 

[bamteck]

Please help me out :

1. Solve the equation ln(1+x) = 1 + lnx , giving your answer correct to 2 s.f.
2.Solve the equation 2^x + 1/2^x - 1 = 5 , giving your answer correct to 3 s.f
3. Given that 2log2 x = y and log2 (2x) = y + 4. find the value of x.

 

[smzimran]

Please help me out :

1. Solve the equation ln(1+x) = 1 + lnx , giving your answer correct to 2 s.f.
2.Solve the equation 2^x + 1/2^x - 1 = 5 , giving your answer correct to 3 s.f
3. Given that 2log2 x = y and log2 (2x) = y + 4. find the value of x.

AoA
1.
ln(1+x) - lnx = 1
ln[ (1+x) / x] = 1
[ (1+x) / x] = e^1
(1+x) / x = e
1 + x = xe
1 = xe - x
1 = x(e - 1)
x = 1 / (e-1)


2.
let 2^x = y
The equation becomes
y + 1/ (y-1) = 5
y -5 = - 1/ (y-1)
(y - 1)(y - 5) = -1
y^2 - 6y + 5 = -1
y^2 - 6y + 6 = 0
Using completing the square method
(y - 3)^2 = -6 + (-3)^2
(y - 3)^2 = 3
(y - 3) = +/- √3
y = 3 +/- √3

2^x = 3 +/- √3
x = ln(3 +/- √3) / ln2
Dont have a calculator, solve the equation to get ans to 3 s.f

3.
2log2 x = y and log2 (2x) = y + 4
2log2 x = y
log2 (x^2) =y
2 ^ y = x^2 ----eq(1)

log2 (2x) = y + 4
log2 2 + log2 x = y + 4
1 + log2 x = y + 4
log2 x = y + 3
2^ (y+3) = x
(2^y) * (2^3)= x
(2^y) * 8 = x
(2^y) = x/8 ---eq(2)

Substitute eq(1) in eq(2)
x^2 = x/8
8x^2 = x
8x^2 - x = 0
x(8x - 1) = 0
x = 0
OR
(8x - 1) = 0
x = 1/8

 

[leadingguy]

qstn 7 part 3
I have sketched the diagram but will be good if some one sketch it! a rough sketch is acceptble jxt need an idea

qstn 10 part 3
a detailed solution plz alitle weak at it


http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_w05_qp_3.pdf
http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_w05_ms_3.pdf

 

[Mehroz]

Assalamoalaikum.Can anyone help me out in qs 9 part1 os this paper....
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s10_qp_31.pdf

 

[reina81]

Q3 part i, i know we have to differentiate and i got the correct derevative, but everytime i equate it to zero, i get x =0
http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_w06_qp_3.pdf

http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_w06_ms_3.pdf

 

[redapple20]

Ho gaya

 

[redapple20]

we got this in our mock!

The probability of 3cm is 4/5 and of 5cm is 1/5. Hence Expecance will be (3 x 4/5)+(5 x 1/5). similarly variance will be deduced using the formula Var(X)= E(X^2)-E(X)^2. Do you know the ans ??[/quote]
Yes thanku

 

[leadingguy]

Q3 part i, i know we have to differentiate and i got the correct derevative, but everytime i equate it to zero, i get x =0
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w06_qp_3.pdf

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w06_ms_3.pdf

6e^x - 3e^3x = 0

6e^x = 3e^3x

xlne + ln6 = 3xlne + ln3 .......... lne = 1

x + ln6 = 3x +ln3

ln6 - ln3 = 3x - x

ln(6/3) = 2x

0.692 =2x

x = 0.692/2

x= 0.346 ANs

 

[redapple20]

Here: sin^2 (2x) (cosec^2 (x)-sec^2 (x))= 4cos2x
Sin^2 (2x) è4sin^2x cos^2 x
cosec^2 (x)-sec^2 (x) è 1/sin^2 x – 1/ cos^2 x = (cos^2 x – sin^2 x ) / (sin^2 x cos^2 X)
cos^2x – sin^2 x = cos 2x

now just simplify...
P.S. kinda difficult to put each and every step...that’s why did this way..

Thank you so much

 

[reina81]

Whenusing normal distribution and a question asks us to find the probability of being less than for eg, 99, we sometimes take 98.5. Why do we do this in some places and not others?

 

[Amy Bloom]

Can somebody help me with number 4
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_31.pdf

 

[bamteck]

AoA
1.
ln(1+x) - lnx = 1
ln[ (1+x) / x] = 1
[ (1+x) / x] = e^1
(1+x) / x = e
1 + x = xe
1 = xe - x
1 = x(e - 1)
x = 1 / (e-1)


2.
let 2^x = y
The equation becomes
y + 1/ (y-1) = 5
y -5 = - 1/ (y-1)
(y - 1)(y - 5) = -1
y^2 - 6y + 5 = -1
y^2 - 6y + 6 = 0
Using completing the square method
(y - 3)^2 = -6 + (-3)^2
(y - 3)^2 = 3
(y - 3) = +/- √3
y = 3 +/- √3

2^x = 3 +/- √3
x = ln(3 +/- √3) / ln2
Dont have a calculator, solve the equation to get ans to 3 s.f

3.
2log2 x = y and log2 (2x) = y + 4
2log2 x = y
log2 (x^2) =y
2 ^ y = x^2 ----eq(1)

log2 (2x) = y + 4
log2 2 + log2 x = y + 4
1 + log2 x = y + 4
log2 x = y + 3
2^ (y+3) = x
(2^y) * (2^3)= x
(2^y) * 8 = x
(2^y) = x/8 ---eq(2)

Substitute eq(1) in eq(2)
x^2 = x/8
8x^2 = x
8x^2 - x = 0
x(8x - 1) = 0
x = 0
OR
(8x - 1) = 0
x = 1/8

Thank You very much !

Would you mind helping me with this one?
State the exact value of the constants k such that 3^x = e^kx , for all x.

 

[ousamah112]

plz someone help me with o/n 11 41 q6 ii
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_41.pdf

 

[allysaleemally]

Could anyone please explain question 6 in P1 oct/nov 2010. I really have almost no idea how to solve such questions with k in them.

http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w10_qp_12.pdf
heres the marksheet
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w10_ms_12.pdf

Thanks, really it means alot!

 

[anonymous123]

AoA wr wb
9709_s10_qp_41
Q 6 part ii
plz explain the first solution

9709_w06_qp_1
Q 10 part v

 

[iFuz]

Could anyone please explain question 6 in P1 oct/nov 2010. I really have almost no idea how to solve such questions with k in them.

http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w10_qp_12.pdf
heres the marksheet
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w10_ms_12.pdf

Thanks, really it means alot!

For part 1 you have to find points on which the curve and line are NOT meeting each other...
so, for that you'll substitute eq.2 in eq1 which will result in following equation

kx^2 - kx + 1 = 0

Now we will find its discriminant and will put that < 0 as these never meet...

so, B^2 - 4AC < 0

where B = -k
A = k and C = 1

The Range is answer to its part i..

for part ii )
as its tangent so discriminant should be = 0

B^2 - 4AC = 0
where B = -k , A = k and C = 1
Solving it will give you value of K i.e 4
and putting it into equations and solving it accordingly will lead you to required answers!

 

[sea_princess]

Whenusing normal distribution and a question asks us to find the probability of being less than for eg, 99, we sometimes take 98.5. Why do we do this in some places and not others?

you use the halves , when this normal distribution was originall a binomial distribution ( meaning when you use noraml approximation) but if the question stated that the variable was modelled by a normal distribution , then you don't put the half as in the example u gave

 

[reina81]

you use the halves , when this normal distribution was originall a binomial distribution ( meaning when you use noraml approximation) but if the question stated that the variable was modelled by a normal distribution , then you don't put the half as in the example u gave

ohhh thanks, so this means that any time binomibnal distribution is changed to normal distribution and they ask us to find the probability of something being greater or less than a certain value we use this method?