Mathematics: Post your doubts here!

[reina81]

ON 07 P3 Q10iii
http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_w07_qp_3.pdf

 

[Esme]

i need help with tan graphs...

sumone PLEASE provide me with notes or any explanation as to how to sketch them.

 

[leadingguy]

qstn 3 part 3 can Any one help me in it . as I have obtained the quadratic expression 2x^2 + x +2


now it cannot be quadraticaly expanded how 2 solve this
http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_w04_ms_3.pdf
http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_w04_qp_3.pdf

 

[Nikesh]

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

if the given equation is too complex then change it to simple form as in complete square form
and when ur simplified equation turns to be like f(x) = -√2x
then ur f-1 (x) will be in the following way
f-1: x ----> divide by 2 -----> square ----> multiply by positive sign
i.e. f-1(x) = (x/2)^2
just oppose the process u will get ur inverse equation

 

[leadingguy]

ON 07 P3 Q10iii
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w07_qp_3.pdf

direction vector of line L is (1 -2 +2)
unit vector of the plane P is also given (2 -3 +6)
now the cross product of these direction vector (1 -2 +2).(2 -3 +6) = (-6 -2 +1)
this wil be the direction vector of the line whose eq. is to be found out.

Now itx stated in the question that the line ahs a point A. we have already found out the position vector of that point "A"
make the q. of the line now. r = (3 +2 +1) +t(-6 -2 1) this is the eq. of the linE

 

[reina81]

direction vector of line L is (1 -2 +2)
unit vector of the plane P is also given (2 -3 +6)
now the cross product of these direction vector (1 -2 +2).(2 -3 +6) = (-6 -2 +1)
this wil be the direction vector of the line whose eq. is to be found out.

Now itx stated in the question that the line ahs a point A. we have already found out the position vector of that point "A"
make the q. of the line now. r = (3 +2 +1) +t(-6 -2 1) this is the eq. of the linE

thanks was just confused about which two vectors we were supposed to take for the cross product. Btw if i change the order for eg. (2-3+6).(1-2+2), the signs of the answer are different, i.e.(6+2-1), is this a big deal?

 

[smzimran]

qstn 3 part 3 can Any one help me in it . as I have obtained the quadratic expression 2x^2 + x +2


now it cannot be quadraticaly expanded how 2 solve this
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w04_ms_3.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w04_qp_3.pdf

p(x) has two roots
when factorised
p(x) = (x - 2) (2x^2 + x +2)
p(x) > 0
(x - 2) (2x^2 + x +2) > 0
Now we have to make a sketch, but the root (2x^2 + x +2) should first be checked if it has real roots
b^2 - 4ac = (1)^2 - 4(2)(2) = 1 - 16 = -15
That means no real roots and the whole curve lies above x-axis, never touches it
So the only point of p(x) that touches x-axis is x =2
(x - 2) (2x^2 + x +2) > 0
(x - 2) > 0
x > 2

 

[anonymous123]

if the given equation is too complex then change it to simple form as in complete square form
and when ur simplified equation turns to be like f(x) = -√2x
then ur f-1 (x) will be in the following way
f-1: x ----> divide by 2 -----> square ----> multiply by positive sign
i.e. f-1(x) = (x/2)^2
just oppose the process u will get ur inverse equation

i knw hw to find an inverse but there is a negative sign for the root as well how do u choose...

 

[anonymous123]

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.

reword this exp plz

 

[Nikesh]

i knw hw to find an inverse but there is a negative sign for the root as well how do u choose...

i didn't get ur problem
will u please elaborate ur problem

 

[anonymous123]

i didn't get ur problem
will u please elaborate ur problem

lets say f(x) = x^2 + 10 for x >= 0
thn f-1 = +- √(x) - 10
but this answr is wrong it can either be +√x or -√x
how to select ??

 

[Nikesh]

lets say f(x) = x^2 + 10 for x >= 0
thn f-1 = +- √(x) - 10
but this answr is wrong it can either be +√x or -√x
how to select ??

i think in this case the domain of f(x) has clearly mentioned that there is not inclusion of negative numbers of x so in inverse to there won't be any negative sign, only √x is accepted
m not pretty much sure about this.....

 

[anonymous123]

Jazakallah nikesh

lets say f(x) = x^2 + 10 for x >= 0
thn f-1 = +- √(x) - 10
but this answr is wrong it can either be +√x or -√x
how to select ??

sm1 clear my doubt plz?

 

[2pac]

y = A (x^n), natural log each side to get
ln y = ln A + n (ln x)
The gradient is n and y-intercept is ln A.
Just draw the line of best fit to find the two values. It should not be hard.

could u please explain q9 of 9709 may 2009 p3.
thanks

 

[sea_princess]

MJ08 P3 Q2
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s08_qp_3.pdf

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s08_ms_3.pdf

e^x +e^2x=e^3x ( divide by e^x)
1 +e^x =e^2x ( substitute for e^x = u)
1 +u = u^2
u^2 - u -1 =0
u = 1 +square root ( -1^2 -4X 1 X -1) /2 X1
u = 1.62 ( the other u would be negative , so it is rejected)
e^x =1.62
x = ln (1.62 )=0.481

 

[reina81]

thanks here's another one

direction vector of line L is (1 -2 +2)​

unit vector of the plane P is also given (2 -3 +6)​

now the cross product of these direction vector (1 -2 +2).(2 -3 +6) = (-6 -2 +1)​

this wil be the direction vector of the line whose eq. is to be found out.​

​

Now itx stated in the question that the line ahs a point A. we have already found out the position vector of that point "A"​

make the q. of the line now. r = (3 +2 +1) +t(-6 -2 1) this is the eq. of the linE ​

thanks was just confused about which two vectors we were supposed to take for the cross product. Btw if i change the order for eg. (2-3+6).(1-2+2), the signs of the answer are different, i.e.(6+2-1), is this a big deal?

 

[sea_princess]

thanks here's another one

no it doesn't matter actually , they can't possibly expect all candidates to use the vectors in the same order. but if you don't mind would you give me the link to the question?

 

[reina81]

no it doesn't matter actually , they can't possibly expect all candidates to use the vectors in the same order. but if you don't mind would you give me the link to the question?

Q10 iii
http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_w07_qp_3.pdf

 

[sea_princess]

Q10 iii
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w07_qp_3.pdf

nope the order doesn't matter, even in markschemes they allow for the difference in signs

 

[gestetner]

Can someone please help me in this question of Oct/Nov 2010 paper 11 Ques 5?