Mathematics: Post your doubts here!

[monoho]

its logarithm question
substitute 5^x by a
yu shall get
a/5 = a - 5 (if yu dont get this step .. tell me)
a = 5a - 25
a =25/4

now
5^x = 25 /4 (apply logs to both sides)

x log (5) = log (25/4)
x = 1.14

i got it thanks

 

[TaffsAsLevel]

Please help me solve the second part, I always struggle finding distances in vectors

 

[Esme]

I am retiring for today ... its 1 am here and my brain has given up ..

Will annoy you tomorrow

 

[D0cEngi]

as far as i can comprehend you would be having difficult in integrating
cos(2x)/sin(2x) dx
u see differential of sin(2x) is 2cos(2x) ...
to get 2 cos(2x) in numerator just multiply the integral with 2/2
u shall get 1/2 S 2 cos(2x)/sin(2x) ( S denotes integral sign)
now u can integrate using integration by recognition.. it will become
1/2 ln (sin(2x))

Thanks a lot. Got it now.

 

[D0cEngi]

Plz plz help me in one more question ryt now. it's:
y^4 + y^2 - 6 = 0.
how to solve this equation to get ( 3^(1/2) - 2^(1/2) i ).

 

[Esme]

Plz plz help me in one more question ryt now. it's:
y^4 + y^2 - 6 = 0.
how to solve this equation to get ( 3^(1/2) - 2^(1/2) i ).

Which paper, please ?

 

[D0cEngi]

Which paper, please ?

Q.10 of this paper. http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_32.pdf

 

[Esme]

Q.10 of this paper. http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_32.pdf

That's not how I did this question. Here's how to do it:
a+bi = √ ( 1 - 2√6i )
(a+bi)^2 = 1-2√6i
a^2 + 2abi - b^2 = 1 - 2√6i
a^2 - b^2 = 1 (equating the real parts)
2ab = 2√6 (equating the imaginary parts)

Now that you have two equations, you can find a and b.

Plz plz help me in one more question ryt now. it's:
y^4 + y^2 - 6 = 0.
how to solve this equation to get ( 3^(1/2) - 2^(1/2) i ).

If you want to do it this way:
Take x=y^2
Your equation will become
x^2 + x - 6 =0
Then you'll get x and take it's √ to get y.

 

[D0cEngi]

Esme Thanks a lot. I got it now. Was doing a mistake. You're awesome.

 

[surinr]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_31.pdf

Please help with the entire Q3

 

[Khalid Mohammed Okiely]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_31.pdf

Please help with the entire Q3

For part (i) you:

Find the normal to the plane. Given that the plane passes through point B and that AB is perpendicular to the plane you try figuring out AB position vector as the normal i.e.

(2, -2, 11) - (-1, 2, 5) = (3, -4, 6)

Now we have the euqation of the plane p without k i.e. 3x - 4y + 6z = k

To find k substitute the point it passes through i.e. point B

So you get: 3(2) -4(-2) +6(11) = k

6+8+66 = k

k = 80; Hence the equation of the plane is: 3x-4y+6z = 80

for part (ii) you:

determine the direction vector which is given as y-axis i.e. (0,1,0)
And we have got the normal as: (3, -4, 6),

Then you use the following formula:
cos x = q(direction vector).n(normal to the plane)/|q|x|n|

Hence we obtain:

cos x = (0)(3) + (1)(-4) + (o)(6)/ sqrt(1) x sqrt(3^2 + (-4)^2 + 6^2)

You then calculate cos x = -4/sqrt(61)

then find x by:
x = cos-1 (-4/sqrt(61))
x = 30.8 degress or equivalent in radians
Hope this helps Good luck!

 

[Zsiddiqui]

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s12_qp_32.pdf
Q8(ii) and Q10 (iii) help please

 

[iKhaled]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_32.pdf

Q8(ii) and Q10 (iii) help please

for 8(ii) ever heard of partial fractions?? if so then apply it here and then integrate!

 

[iKhaled]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_32.pdf

Q8(ii) and Q10 (iii) help please

alright first of all i will use t instead of lamda for all steps in question 10(iii)

they told u that the prependicular distance is equal from point p for both planes n and m and they gave u the formula to find the distance so
we have x = 1+2t y = 1+t z= -1+2t
plane m : x+2y-2z= 1 plane n: 2x-2y+z = 7

now find the perpendicular distance of the point p and the plane m using the formula
dm = |1+2t+2(1+t)-2(-1+2t)-1|/√(1^2+2^2+2^2) now do the math and expand dat u will get dm = |4|/3

now we will find the prepen distance of the point p and plane n
dn = |2(1+2t)-2(1+t)-1+2t-7|/√(2^2+2^2+1^2) do the math and expand it u will get dn = |-8+4t|/3

since dn = dm then |4|/3 = |-8+4t|/3 so..

16/3 = (-8+4t)^2/3 do cross multiplication and u will get t = 3 and t = 1 substitute both of them to find x,y and z of the 2 position vectors and once u get the 2 position vectors substract them from each other and find the distance using the formula u learnt in AS

 

[Zsiddiqui]

for 8(ii) ever heard of partial fractions?? if so then apply it here and then integrate!

I know we have to solve it through partial fractions but still I'm not able to do it I get stuck in middle if possible so can you show the working and thanks for helping

 

[surinr]

For part (i) you:

Find the normal to the plane. Given that the plane passes through point B and that AB is perpendicular to the plane you try figuring out AB position vector as the normal i.e.

(2, -2, 11) - (-1, 2, 5) = (3, -4, 6)

Now we have the euqation of the plane p without k i.e. 3x - 4y + 6z = k

To find k substitute the point it passes through i.e. point B

So you get: 3(2) -4(-2) +6(11) = k

6+8+66 = k

k = 80; Hence the equation of the plane is: 3x-4y+6z = 80

for part (ii) you:

determine the direction vector which is given as y-axis i.e. (0,1,0)
And we have got the normal as: (3, -4, 6),

Then you use the following formula:
cos x = q(direction vector).n(normal to the plane)/|q|x|n|

Hence we obtain:

cos x = (0)(3) + (1)(-4) + (o)(6)/ sqrt(1) x sqrt(3^2 + (-4)^2 + 6^2)

You then calculate cos x = -4/sqrt(61)

then find x by:
x = cos-1 (-4/sqrt(61))
x = 30.8 degress or equivalent in radians
Hope this helps Good luck!

thanks very much i got through with it

 

[iKhaled]

I know we have to solve it through partial fractions but still I'm not able to do it I get stuck in middle if possible so can you show the working and thanks for helping

tomorrow then i am headin to bed now nightttt tag me in the question again tomorrow and i will solve it when i wake up inshAllah

 

[Ashique]

I know we have to solve it through partial fractions but still I'm not able to do it I get stuck in middle if possible so can you show the working and thanks for helping

Here you go:

 

[surinr]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_31.pdf
please help with 8 (ii)

 

[WayneRooney10]

Can anyone solve this identity for me? cos4x-4cos2x+3=8sin^4x