Mathematics: Post your doubts here!

[reina81]

for Q5 part i, I was able to simplyfy it but i dont unerstand what we are supposed to do next and i'm not able to solve the expansion part.

Also in Q6,ii, i got the answer by trying different number, is there any other way?

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

 

[sea_princess]

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?

yes , but take care, sometimes you add a half and at others you subtract a half

 

[sea_princess]

for Q5 part i, I was able to simplyfy it but i dont unerstand what we are supposed to do next and i'm not able to solve the expansion part.

Also in Q6,ii, i got the answer by trying different number, is there any other way?

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

in question 6 , you equate dy/dx to 0 ,you'll get x^2 =y substitute in the equation of the curve
it will be x^3 + 2 x^6= 3x^3
2x^6 =2x^3 divide by 2x^3
x^3= 1 >>>>> x= 1 , put this back in the equation of the curve and get y

in question 5 :
you already have that 1/ square root( 1+x) +square root (1-x) is equal to ( square root(1+x)+square root(1-x))/ 2x
separate the fraction to be square root(1+x)/2x - square root( 1-x)/ 2x
now u can expand it:
1/2x X( 1+x)^0.5 - 1/2x X(1-x)^0.5 and complete the expansion like normal

 

[CaptainDanger]

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!

(i) y = kx^2 + 1

y = kx

so

kx^2 + 1 = kx

kx^2 -kx + 1 = 0

As the lines have no common points so discriminant is less than zero.

(-k)^2 - 4 (k) (1) < 0
k^2 - 4k <0
k ( k - 4) < 0

k < 0 and k < 4

I hope you know how to make the range from it now...

(ii) Discriminant is equal to zero.
k^2 - 4k = 0
k^2 = 4k
k = 4

Put k = 4 in the given equations and solve them simultaneously to get the coordinates...

 

[Aahliya]

lnz = ln(y+2) - ln(y^2)
lnz = ln [(y+2) / (y^2)]
z = (y+2) / (y^2)

Thanks a lot for ur kind help .. n btw if u don't mind then can u pls tell me how do u DIFFERENTIATE : (1+tanx) ??

 

[unique840]

Thanks a lot for ur kind help .. n btw if u don't mind then can u pls tell me how do u DIFFERENTIATE : (1+tanx) ??

is it AS or A2?

 

[2pac]

ur welcum

Hi there its me again.Was doing 9709 may 2010 31 and couldn't solve q9.I differentiated u but couldn't complete the quesition using chain rule.second part also an enigma.
Could you please solve them?
thanks

 

[anooshraja]

May June 2008 question 10/8/20
May June 2006 question 22 AND 23
Oct/nov 2006 question 5/13
Oct/nov 2007 question 21/16
May June 2009 question 3/6and 8/11 part b

 

[gary221]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_12.pdf
Ques 2, 3, 4

 

[unique840]

Hi there its me again.Was doing 9709 may 2010 31 and couldn't solve q9.I differentiated u but couldn't complete the quesition using chain rule.second part also an enigma.
Could you please solve them?
thanks

(u'v - v'u)/v^2
{(1-x)' (1+x) - (1+x)' (1-x)} / (1+x)^2
{(-1)(1+x) - (1)(1-x)} / (1+x)^2
{-1-x - 1+x} / (1+x)^2
-2/(1+x)^2

y = {(1-x)/(1+x)} ^0.5
y' = [0.5 {(1-x)/(1+x)}^-0.5] * {(1-x)/(1+x)}'
we found the derivative of {(1-x)/(1+x)} above so substitute it here
y' = [1/2 {(1+x)/(1-x)}^0.5] * [-2/(1+x)^2]
2 cancels out
y' = {-(1+x)^0.5} / {[(1-x)^0.5] * [(1+x)^2]}
bring (1+x)^0.5 in the denominator
y' = -1/{[(1-x)^0.5] [(1+x)^2-0.5]}
y' = -1/{[(1-x)^0.5] [(1+x)^3/2]}
y' = -1 / {[(1-x)^0.5] (1+x) [ (1+x)^0.5]}
(1+x)(1-x) will be changed to (1)^2 - (x)^2 inside the sqr root due to power 0.5
y' = -1/{(1+x) (1-(x^2))^0.5}
this was gradient of tangent. for normal it will be
y' = {(1+x) (1-(x^2))^0.5}

 

[unique840]

Hi there its me again.Was doing 9709 may 2010 31 and couldn't solve q9.I differentiated u but couldn't complete the quesition using chain rule.second part also an enigma.
Could you please solve them?
thanks

in part 2 u will diffrenciate the gradient of normal found in part 1. and equate it to zero to find the value of x

 

[unique840]

Thanks a lot for ur kind help .. n btw if u don't mind then can u pls tell me how do u DIFFERENTIATE : (1+tanx) ??

if its for A2 then the derivative of tan is sec^2

 

[gary221]

if its for A2 then the derivative of tan is sec^2

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_12.pdf
Ques 2, 3, 4

cud u help me wth this??

 

[redapple20]

Yes and this is called s continuity correction!

 

[Nikesh]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_12.pdf
Ques 2, 3, 4

2)i) Use this formula ( 1- 3/2x)^6 = 6C0*1 + 6C1 * 1 * (-3/2x) + 6C2 * 1 * (-3/2x)^2 + 6C3 * 1 * (-3/2x)^3 +................
evaluate it u will get the terms including x^2 and x^3
ii) here, use above expanded equation to find terms with x^3
i.e. (k + 2x)( 6C0*1 + 6C1 * 1 * (-3/2x) + 6C2 * 1 * (-3/2x)^2 + 6C3 * 1 * (-3/2x)^3 )
u will find, terms including x^3 will be given by (k* 6C3 * 1 * (-3/2x)^3 + 2x * 6C2 * 1 * (-3/2x)^2 )
find the coefficient of x^3 and then solve it by making right hand side = 0 as question says there is no terms with x^3. u will then get k = 1.

3)i) Given that -3 and 5 are roots
i.e. (x+3)(x-5) = 0
--> x^2 -5x + 3x - 15 = 0 (expand above eqn.)
--> x^2 -2x -15 = 0 compare this with given equation x^2 + px + q = 0
u will get ur value
ii) when u have ur values substitute it and then new equation including r is formed...
given that there are equal roots, i.e. b^2 - 4ac = 0
solve it u will get ur ans.

4)i) Use differentiation formula and find dy/dx which will give u the slope of tangent when x=2
and when u have slope and (x1,y1) = (2,2) find equation of line
ii) u know slope= tanA for angle
i.e. tanA= (-3)
i.e. A = (-71.57○)
i.e. A=108.43○

 

[Aahliya]

is it AS or A2?

Its A2 question..

The parametric equations of a curve are
x = 1 + tanθ , y = secθ ,
for −1π/2 < θ < 1π/2.

Show that dy/dx = sin θ.

 

[imanmalik]

two questions.
Question 3. I'm getting the values for b and a the wrong way around. Why? (s11 paper13)
Question 11 (iv) Isnt the inverse of f y= (x-1)/2 ? The mark scheme says something completely different. (s11 paper11)

 

[unique840]

Its A2 question..

The parametric equations of a curve are
x = 1 + tanθ , y = secθ ,
for −1π/2 < θ < 1π/2.

Show that dy/dx = sin θ.

y = 1/cosθ
y= cosθ inverse
dy/dθ = -1(cosθ^-2)(-sinθ)
dy/dθ = {sinθ/(cos^2)θ}


dx/dθ = 0 + (sec^2)θ
dx/dθ = (sec^2)θ
dx/dθ = 1/(cos^2)θ

dy/dx = {sinθ/(cos^2)θ} / {1/(cos^2)θ}
dy/dx = sinθ

 

[imanmalik]

two questions.
Question 3. I'm getting the values for b and a the wrong way around. Why? (s11 paper13)
Question 11 (iv) Isnt the inverse of f y= (x-1)/2 ? The mark scheme says something completely different. (s11 paper11)
View attachment 7768
View attachment 7769

 

[junaidaftab]

http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s08_qp_3.pdf

How to solve Q#2?