Mathematics: Post your doubts here!

[Raiyan3]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_S08_qp_01.pdf
Guys, how to do 8ii?

Just substitute one of the value of K found in 8.i) in the equation of fg(x)=x. Which was i believe 36/2-x +2k = x. you will end up with a quadratic equation and solve it then!

 

[syed1995]

Put values of k in f[g(x)]=x

then solve f[g(x)]-x=0

Put 5 or -7 in (x^2 + 2kx - 2x + 36 - 4k) and equate to 0.


x^2 + 8x + 16 =0 or x^2 - 16x + 64 =0

Roots come as -4 when k=5 and 8 when k=-7

 

[Fatima18]

Put values of k in f[g(x)]=x

then solve f[g(x)]-x=0

Put 5 or -7 in (x^2 + 2kx - 2x + 36 - 4k) and equate to 0.


x^2 + 8x + 16 =0 or x^2 - 16x + 64 =0

Roots come as -4 when k=5 and 8 when k=-7

Thank you!
Can you please tell me how to do number 8i) in oct paper 2008...I dont know how to find da gradient through differentiation?

 

[syed1995]

Thank you!
Can you please tell me how to do number 8i) in oct paper 2008...I dont know how to find da gradient through differentiation?

Just differentiate the equation and put the value of x in it to get gradient.

y=5-8x^-1
dy/dx=8/x^2

since co-ordinate is (2,1) and x =2

put x = 2 in dy/dx

gradient = 8/2^2
gradient = 2

normal gradient will be found with m1 x m2 = -1

so normal gradient will be -1/2

 

[Fatima18]

Just differentiate the equation and put the value of x in it to get gradient.

y=5-8x^-1
dy/dx=8/x^2

since co-ordinate is (2,1) and x =2

put x = 2 in dy/dx

gradient = 8/2^2
gradient = 2

normal gradient will be found with m1 x m2 = -1

so normal gradient will be -1/2

So why is it that dx over dy is= 8 over x squared?

 

[syed1995]

So why is it that dx over dy is= 8 over x squared?

Basic Differentiation.

-8/x can be written as -8(x)^-1

Differentiating that will be .. -8(x)^-1-1 * -1

which becomes 8(x)^-2 which can also be written as 8/x^2

Like if y= 2x^n then dy/dx = 2x^n-1 * n = 2n(x^n-1)

working with powers while writing on a forum is difficult.

 

[nooshi]

No.4
i) By resolving .. Fcos=12cos30 in x-axis direction(1)
Fsin+12sin30=10 in y-axis direction.(2)
Make Fsin the subject in the second equation .. Fsin=1o-12sin30=4
Do you know that Sin /cos = tan , right ? So to get rid of sin and cos in the equation ..we will divide equation 2 over equation 1 .. F will be cancelled from the equation and sin/cos will be tan ---> Tan=4/12cos30 ... Shift tan you willl find the angle = 21.1 .. and by putting this value in the fist or the second equation your F will be 11.13
ii) 12 is not here anymore . right ? so you have the force f to resolve :
So in x-axis direction : Fcos = 11.1cos21.1 (1)
in y-axis direction : Fsin-10 ---> 11.1sin21.1-10(2)
R = 1square +2square under square root .
and direction : tan (2)/(1)
Hope u understand incha'allah

Thank you so so so much

 

[Fatima18]

Basic Differentiation.

-8/x can be written as -8(x)^-1

Differentiating that will be .. -8(x)^-1-1 * -1

which becomes 8(x)^-2 which can also be written as 8/x^2

Like if y= 2x^n then dy/dx = 2x^n-1 * n = 2n(x^n-1)

working with powers while writing on a forum is difficult.

Thank you!

 

[Fatima18]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s09_qp_1.pdf
Number 10i and ii?

 

[Tkp]

You have maths as well?

yup

 

[leosco1995]

The answer to part (i) is 7.5, I got that.

For part (ii), we use v^2 = u^2 + 2as. But u is taken as 0 instead of 5.2 and s is taken as 7.5 instead of -6.2.. why?

 

[Minato112]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s09_qp_1.pdf
Number 10i and ii?

a) f : x → 2x^2− 12x + 13
f(x) = 2x^2− 12x + 13
= 2 (x^2 - 6x) + 13
= 2 [ (x - 3)^2 - 9 ] + 13
= 2 (x - 3)^2 - 18 + 13
= 2 (x - 3)^2 - 5

b) x = 3

 

[sma786]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w09_qp_12.pdf .. Questn 1, and Question 4 (I) :\

 

[leosco1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_12.pdf .. Questn 1, and Question 4 (I) :\

For Q1, you just integrate the given equation:

3x^-1/2 - x (I shifted the root(x) to the numerator)

Integrating it gives,

y = 6x^1/2 - (x^2 / 2) + c

Because it passes through (4, 6), just replace x and y with 4 and 6 respectively. You'll get c as 2.. so the equation is y = 6x^1/2 - (x^2 / 2) + 2

 

[PhyZac]

The answer to part (i) is 7.5, I got that.

For part (ii), we use v^2 = u^2 + 2as. But u is taken as 0 instead of 5.2 and s is taken as 7.5 instead of -6.2.. why?

For part (ii) you are to find the speed it reaches ground.
So, we can use v^2 = u^2 + 2as, and you should consider from when the particle was in the greatest height, that is the TOP, and then use the formula.
At the TOP distance from ground is 7.5 and u is 0

 

[yousef]

Assalamoalaikum Wr Wb!

Post your doubts here. Make sure you give the link to the question paper when posting your doubts.


May Allah give us all success in this world as well as the HereAfter...Aameen!!

And oh yeah, let me put up some links for A level Maths Notes here.

Check this out! - Nice website, with video tutorials for everything! MUST CHECK

My P1 Notes! - Only few chapters available at the moment!

Maths Notes by destined007

Some Notes for P1 and P3 - shared by hamidali391

A LEVEL MATHS TOPIC WISE NOTES

MATHS A LEVEL LECTURES

compiled pastpapers p3 and p4 - by haseebriaz

Permutations and Combinations (my explanation)- P6

Permutations and Combinations - P6

Vectors - P3

Complex No. max/min IzI and arg(z) - P3

Sketcing Argand Diagrams - P3 (click to download..shared by ffaadyy)

Range of a function. - P1

can any one answer this plz >>>>> a circle with centre O and radius 8 cm. Points A and B lie on the circle. The
tangents at A and B meet at the point T, and AT = BT = 15 cm.
(i) Show that angle AOB is 2.16 radians, correct to 3 significant figures...................P1 may/june2006 ....q7

 

[Iffat]

Can sum1 plz help me with the foll qs:
2007 oct-nov p1:
Q4(ii),(iii)
Q9(i),(ii)
Q11(v)
2008 may-jun p1:
Q8(i)
2008 oct-nov p1:
Q7(i)
Q10(iii),(iv)

thanx in advance

 

[Dug]

can any one answer this plz >>>>> a circle with centre O and radius 8 cm. Points A and B lie on the circle. The
tangents at A and B meet at the point T, and AT = BT = 15 cm.
(i) Show that angle AOB is 2.16 radians, correct to 3 significant figures...................P1 may/june2006 ....q7

OAT and OBT are both congruent right-angled triangles.
Using this fact, find ∠AOT or ∠BOT and multiply it by 2 to get ∠AOC.

∠AOT:
tanΘ = 15/8
Θ = 1.08

∠AOB = 2(1.08) = 2.16 rad

 

[sma786]

Thnks

For Q1, you just integrate the given equation:

3x^-1/2 - x (I shifted the root(x) to the numerator)

Integrating it gives,

y = 6x^1/2 - (x^2 / 2) + c

Because it passes through (4, 6), just replace x and y with 4 and 6 respectively. You'll get c as 2.. so the equation is y = 6x^1/2 - (x^2 / 2) + 2

!!

 

[tanmaydube]

Please solve this!!

Thank you