Mathematics: Post your doubts here!

[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)
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[leadingguy]

http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_s08_qp_3.pdf

How to solve Q#2?

e^x + e^2x = e^3x take e^x as p

now substitute for p

p + p^2 = p^3

p( 1+ p) = p^3

1 + p = p^2

p^2 - p - 1 = 0

solve quadraticaly, u wil get two values of of p one wil be 1.62 and other wil be -0.62

now as we use p = e^x so again reform the eq. as


e^x = p
e^x = 1.62 take ln on both sides

xlne = ln1.62 xlne = ln(-0.62)

x = ln 1.62 x = not posible

x= 0.48 Ans

 

[2pac]

(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}

okay i understood till the part where u cancel 2 but after that its just too vague.Could you please write it on a piece of paper,the signs above are really confusing.I know its too much to ask but would really be grateful if you could.

 

[anonymous123]

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)
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Q3
when x=0 , y=b
when y=0 , x=a
p(a,0) and q(0,b)

PQ = √45
√45 = √a^2 + b^2
sq both sides
a^2 + b^2 = 45

grad = -1/2
plug pts p and q
we get 2 more eqs

c = b
c = a/2
set them equal u get a=2b
put this in eq a^2 + b^2 = 45
u get b = 3
and a=6

 

[anonymous123]

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

Q11
y=x^2 - 2
x^2 = y+2
x = +-√(y+2)

in the q it says for the domain x =<0
sketch the parabola for x=<0 and reflect it in y=x u get the sign frm the graph

 

[Hammad Siddiqi]

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

Q11 (iv), tou are supposed to find inverse of f and substitute the fuction g(x) in it
inverse of function f is (x-1)/2
after substituting g(x) in it it becomes
(x^2 -2-1)/2
i.e ((x^2) -3)/2

 

[sea_princess]

can somone please explain the answers of question 10 : (ii) and (v)
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s10_qp_62.pdf

 

[Hammad Siddiqi]

can somone please explain the answers of question 10 : (ii) and (v)
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_62.pdf

there isnt any Q10,
and if its Q7 u are asking then
(ii) the question says that pink and green card must not be next to each other,
the total random cards are 7
for such question, always draw some spaces
we need 15 spaces here => _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
now fill the spaces with numbers for random cards starting from second space. you must leave a space in between two numbers
it will look something like this
_ x 7 x _ x 6 x _ x 5 x _ x 4 x _ x 3 x _ x 2 x _ x 1 x _
now we have 2 cards remaining and 8 spaces left
and the cards wont be next to each other
now multiply 8 x 7 to 7!
you will get 362580
(v)
find the arrangements for pink and green cards to be together and minus it from total arrangements p(iii)
which will be 7 x 2 x 2!
as there are 3 spaces (consider it 2 as pink and green will be together) and 2 are already taken by pink and green card
so only one card will make it from 7
now as pink and green cards have been assured spaces, it will be written as 7x2
as pink and green cards can change places, it will be written as 7x2x2!
which is 28
minus it from 504..
Cheers!!

 

[sea_princess]

there isnt any Q10,
and if its Q7 u are asking then
(ii) the question says that pink and green card must not be next to each other,
the total random cards are 7
for such question, always draw some spaces
we need 15 spaces here => _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
now fill the spaces with numbers for random cards starting from second space. you must leave a space in between two numbers
it will look something like this
_ x 7 x _ x 6 x _ x 5 x _ x 4 x _ x 3 x _ x 2 x _ x 1 x _
now we have 2 cards remaining and 8 spaces left
and the cards wont be next to each other
now multiply 8 x 7 to 7!
you will get 362580
(v)
find the arrangements for pink and green cards to be together and minus it from total arrangements p(iii)
which will be 7 x 2 x 2!
as there are 3 spaces (consider it 2 as pink and green will be together) and 2 are already taken by pink and green card
so only one card will make it from 7
now as pink and green cards have been assured spaces, it will be written as 7x2
as pink and green cards can change places, it will be written as 7x2x2!
which is 28
minus it from 504..
Cheers!!

got it now , thanx a lot

 

[anonymous123]

AoA wr wb
9709_w04_qp_1
Q9 iv
explanation plz

 

[Hammad Siddiqi]

AoA wr wb
9709_w04_qp_1
Q9 iv
explanation plz

The formula used here is called completing the square method. The formula is,
x^2 + nx = (x + n/2)^2 - (n/2)^2
in this case, the value of n is -6,
so it will become
x^2 - 6x = (x - 6/2)^2 - (-6/2)^2
=> (x-3)^2 - 9

 

[anonymous123]

The formula used here is called completing the square method. The formula is,
x^2 + nx = (x + n/2)^2 - (n/2)^2
in this case, the value of n is -6,
so it will become
x^2 - 6x = (x - 6/2)^2 - (-6/2)^2
=> (x-3)^2 - 9

thx but i was asking abt part iv..

 

[Nikesh]

AoA wr wb
9709_w04_qp_1
Q9 iv
explanation plz

you have h: x → x2 − 6x, change to complete square form
which will be,
h(x) = x^2 - 6x
---->= x^2 - 2.x.3 + 3^2 - 3^2
---->=(x-3)^2 - 9
note how is it formed??
h:x --> subtract 3 ---> square ----> subtract 9
now its inverse will be jus opposite
h-1: x ---> add 9 ----> square root -----> add 3
i.e. h-1(x) = (x+9)^1/2 + 3
for domain of it find the range of h(x) as the function will be one to one onto function

 

[imanmalik]

Q11 (iv), tou are supposed to find inverse of f and substitute the fuction g(x) in it
inverse of function f is (x-1)/2
after substituting g(x) in it it becomes
(x^2 -2-1)/2
i.e ((x^2) -3)/2

But the thing is i got that answer. And according to the markscheme it should be f-1g(x) = 1/((2(x^2 -3))

 

[Hammad Siddiqi]

But the thing is i got that answer. And according to the markscheme it should be f-1g(x) = 1/((2(x^2 -3))

no. u r misreading the mark scheme. mark scheme says (1/2) is multiplied by (x^2 -3)
(x^2 -3) isnt in denominator
(1/2)(x^2 -3) and ((x^2) -3)/2 is same thing..

 

[Rampag3r]

QUESTION 5(iii). If someone could do it, please.
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_61.pdf

 

[Rampag3r]

Sorry for the double post, but I got it. Misread the question :/
For anyone else who's wondering, the question asks for only the probability of the letter weighing MORE than 12 grams above the mean. Therefore we shall use a p value of 0.97 (the z value of 12/6.38) and use it in a binomial distribution.

 

[CaptainDanger]

Q- 7. Last part??

9709_w10_qp_12.pdf

9709_w10_ms_12.pdf

 

[anonymous123]

you have h: x → x2 − 6x, change to complete square form
which will be,
h(x) = x^2 - 6x
---->= x^2 - 2.x.3 + 3^2 - 3^2
---->=(x-3)^2 - 9
note how is it formed??
h:x --> subtract 3 ---> square ----> subtract 9
now its inverse will be jus opposite
h-1: x ---> add 9 ----> square root -----> add 3
i.e. h-1(x) = (x+9)^1/2 + 3
for domain of it find the range of h(x) as the function will be one to one onto function

i dont knw how to decide which sign to put wid inverse whn i get smthing lyk f-1 = +- √asd

 

[Esme]

Q- 7. Last part??

9709_w10_qp_12.pdf

9709_w10_ms_12.pdf

if u see the completed square form of f(x) is (x-2)^2+3.
now as g(x) is x-2. h(x) will be x^2+3. becuz then we put the equation of gx in hx n then we get fx.