Mathematics: Post your doubts here!

[sea_princess]

thanx a lot smzimran and ffaadyy

 

[DANGERBP]

ok this m8 seem a little stupid but seriously as much as i try to solve log questions i never seem to manage solvin
even a silly question like this on >>> 0<x<4 LOG5(4-x) - 2LOG5(x) = 1
gets me confused and doesnt even know how to start !
so someone plz kindly help mee !!
apreciated ~ !

 

[Mustehssun Iqbal]

Q10(i)

r = i −2k +s(2i +j +3k)

x(L) = 1 + 2s
y(L) = s
z(L) = -2 + 3s

r = 6i −5j +4k +t(i −2j +k)

x(M) = 6 + t
y(M) = -5 - 2t
z(M) = 4 + t

Equate x(L) with x(M).

1 + 2s = 6 + t
-5 + 2s = t

Equate y(L) with y(M).

s = -5 - 2t
s = -5 - 2(-5 + 2s)
5s = 5
s=1

Substitute s=1 in '-5 + 2s = t'.

-5 + 2s = t
t = -3

Substitute the value 't=-3' in 'x(L)', 'y(L)' & 'z(L)' and the value of 's=1' in 'x(M)', 'y(M)' & 'z(M)' to show that they intersect.

x(L) = 3
y(L) = 1
z(L) = 1

x(M) = 3
y(M) = 1
z(M) = 1

Therefore, the point of intersection is ( 3, 1, 1).

Q10(ii)

First we'll find the direction of the plane using the common perpendicular rule.

i j k
2 1 3
1 -2 1

i (1+6) + j (3-2) + k (-4-1)
7x + y - 5z = k

To find the value of 'k', substitute the point of intersection of the lines in '7x + y - 5z = k'

7(3) + 1 - 5 = k
17 = k

Therefore, the equation of the plane is 7x + y - 5z = 17.

that's very helpful! thanks a bunch!

 

[ffaadyy]

ok this m8 seem a little stupid but seriously as much as i try to solve log questions i never seem to manage solvin
even a silly question like this on >>> 0<x<4 LOG5(4-x) - 2LOG5(x) = 1
gets me confused and doesnt even know how to start !
so someone plz kindly help mee !!
apreciated ~ !

log5(4-x) - 2log5(x) = 1
log5(4-x) - log5(x)^2 = 1
log5[(4-x)/(x^2)] = 1
(4-x)/(x^2) = 5
5x^2 + x - 4 = 0
5x^2 + 5x - 4x - 4 = 0
(5x - 4) (x + 1) = 0

Keeping in mind the limit 0<x<4, the answer is '4/5'.

 

[DANGERBP]

log5(4-x) - 2log5(x) = 1
log5(4-x) - log5(x)^2 = 1
log5[(4-x)/(x^2)] = 1
(4-x)/(x^2) = 5
5x^2 + x - 4 = 0
5x^2 + 5x - 4x - 4 = 0
(5x - 4) (x + 1) = 0

Keeping in mind the limit 0<x<4, the answer is '4/5'.

thx man !! but how did you go from this step
log5[(4-x)/(x^2)] = 1 >> to that
(4-x)/(x^2) = 5 ???

 

[ffaadyy]

thx man !! but how did you go from this step
log5[(4-x)/(x^2)] = 1 >> to that
(4-x)/(x^2) = 5 ???

If the base of the Log is same (5 in this case) and that when there is a negative sign between two log function's, they are divided. Just like this:

log5(4-x) - log5(x^2) = 1
log5[(4-x)/(x^2)] = 1

And once we've a function in this form 'logA(B) = C', we can write it as 'B = A^C'. This is why
'log5[(4-x)/(x^2)] = 1' can be written as '(4-x)/(x^2) = 5^1'.

 

[DANGERBP]

If the base of the Log is same (5 in this case) and that when there is a negative sign between two log function's, they are divided. Just like this:

log5(4-x) - log5(x^2) = 1
log5[(4-x)/(x^2)] = 1

And once we've a function in this form 'logA(B) = C', we can write it as 'B = A^C'. This is why
'log5[(4-x)/(x^2)] = 1' can be written as '(4-x)/(x^2) = 5^1'.

thank you dude !!! =) lol guess i need to revise all the rules again =p

 

[legion]

as many as possible 8-cm discs are made from a rectangular cardboard of side 170cm by 90cm. how much cardboard will be left after all possible discs are made?

 

[floral123]

kindly do P1 nov 08 q no 7 part i..plz anyone...

 

[ffaadyy]

kindly do P1 nov 08 q no 7 part i..plz anyone...

We know that the length of the wire equals to 80cm therefore the sum of the perimeter of the square and the perimeter of the circle should be equal to 80cm.

4x + 2πr = 80
2x + πr = 40
r = (40 - 2x)/π

Next, we are told that the total area of the circle and that of the square is equal to A cm^2.

A = πr^2 + x^2

Substitute 'r = (40 - 2x)/π' in 'A = πr^2 + x^2'

A = πr^2 + x^2
A = π x [(40 - 2x)/π] + x^2
A = [(1600 - 160x + 4x^2)/π] + x^2
A = (1600 - 160x +4x^2 + πx^2)/π
A = [1600 - 160x + x^2 (4 + π)]/π

Therefore, A = [1600 - 160x + x^2 (4 + π)]/π.

 

[farhan143]

please help me solve this question....

 

[farhan143]

hehehehe

 

[farhan143]

 

[leadingguy]

View attachment 5129

please help me solve this question....

1a) 3-4cos^kx range from (0 to pie)

putting k=2 will make the eq as follows 3-4cos^2x or we can represent it as 3-4(cosx)^2
now enter the values of x in te equation. put x=0 will give u 3-4(cos(o))^2 = -1
put x=pi will give u 3-4(cos(pi))^2 = -1
put x= pi/2 will give u 3-4(cos(pi/2))^2 = 3

now the values of function f u get is varing frm -1 to 3 so the range is frm -1 to 3

1b) 3-4(cosx)^2 = 1
now make x the sbjct -4(cosx)^2 = -1-3
4(cosx)^2 =4
(cosx)^2=1
now taking root on bth sides will simplify cosx= +-(1)
and x=cos^-1(1) and x=cos^-1(-1)
x= 0 x=180(pi)

2) k=1 means the eq will be 3-4cosx=y

use value of x as 0, pi/2 and pi to fuind values of y
and plot the graph it should be a straight line ranging values of y frm -1 to 7.

 

[farhan143]

 

[leadingguy]

whats this?? u gt my point??

 

[xXRUSHXx]

Hey guys, I was just wondering if anyone could upload some notes on the following topics:

Quadratic Equations
Quadratic Functions & Graphs
Co-ordinate Geometry
Functions & Notations

It's a long list but any help would be greatly appreciated. On a side note, this is my first (and hopefully last) year of AS maths.

 

[CaptainDanger]

9709_w10_qp_11.pdf

9709_w10_ms_11.pdf

I need solution for Q7 in detail...?

 

[leadingguy]

9709_w10_qp_11.pdf

9709_w10_ms_11.pdf

I need solution for Q7 in detail...?

1) f(x)=3-2tan(1/2x)........................... having range of x frm (0 to pi)

now put values of x in the equation
when x =0
f(x)=3-2tan(1/2(o)) = 3
when x=pi
f(x)= 3-2tan(1/2(pi)) = 2.95 now the value of the function f(x) lies between 2.95 to 3
we can say that x<3 Ans.

2) f(x)= 3-2tan(1/2x) here put the value of x = (2/3.pi)
= 3-2tan(1/2(2/3.pi)) simplify it u will get,
= 3-2tan(1/3.pi) now look pi= 180, and 1/3rd of pi = 180/3= 60 so,
=3-2tan(60)
=3-2(3)^1/2 Ans.


3) f(x) = y = 3-2tan(1/2x)


on the x-axis plot the values from 0 to pi which should be (o, pi/2, pi)

find the values of y with respect to these values of x

u will have y= 3 at x= 0 and y=1 at x=pi/2 and y=2.95 at x=pi. plot the line Ans.


APart i have to move some where will tell u the rest soon please I apologise for the inconvineance.

 

[farhan143]

whats this?? u gt my point??

it's nothin i got u....
thnks....

i hve anther doubt 2011 may/june Q4. ii


i dont understnd the formula for (ii)