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Mathematics: Post your doubts here!

A

Anoushay

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Zishi said:
Okay, for Model 1, use the Sum formula with n= 40, d and a= 1000. Multiply the sum by 0.05 to get the money which will go to charity. As for Model 2, again use the formula with r=1.1 and n=40. Multiply that sum by 0.05(5%) to get the answer. Do you get it now?
Click to expand...
Yes,I get it. Thanks.
 
Dug

Dug

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AoA wr wb
9709_s03_qp_4
Q7 part ii
why is the KE at A greater thn the PE at h.max
 
Nikesh

Nikesh

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hm12 said:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_11.pdf

in question 11 part (iii) i found out the two values of k i.e k=4 and k=-8 but i dont get how and why does the mark scheme give k<-8 and k>4 the answers can anyone explain please ????
Click to expand...

Solution
--> Substitute value of y= x + k in equation y = 9/2-x
--> You will get a new equation in the form ax2 + bx + c
--> Since question says intersection is at 2 distinct points, its discriminant i.e. b^2 - 4ac > 0
use this and generate a equation involving k
--> you will get (k + 8)(k -4) > 0
--> deduce possible conditions which will be k>-8, k<-8, k>4, k<4
--> merge the conditions you will get k<-8 , -8<k<4 and k>4
--> use above condition in equation involving k, u will find that for k<-8 , k>4 equation is satisfied
Thus you answer is k<-8 , k>4
 
H

hm12

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Nikesh said:
Solution
--> Substitute value of y= x + k in equation y = 9/2-x
--> You will get a new equation in the form ax2 + bx + c
--> Since question says intersection is at 2 distinct points, its discriminant i.e. b^2 - 4ac > 0
use this and generate a equation involving k
--> you will get (k + 8)(k -4) > 0
--> deduce possible conditions which will be k>-8, k<-8, k>4, k<4
--> merge the conditions you will get k<-8 , -8<k<4 and k>4
--> use above condition in equation involving k, u will find that for k<-8 , k>4 equation is satisfied
Thus you answer is k<-8 , k>4
Click to expand...


can you please explain more why did you use b^2-4ac>0 i mean why greator than 0 sign?
 
bamteck

bamteck

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USMAN ALI (MANI) said:
sina/cosa + cosa/sina now take lcm

sin^2a + cos^2a /(cosa.sina)
1/(sina.cosa ) now remember the identity sin2a = 2sina.cosa

so we have to multiply the denominator by 2 to gt 2sina.cosa


when multipling denominator numenator should also be muliplied by;)

1*2/(sina.cosa )*2 = 2/2sina.cosa = 2/sin2a proven
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1. Prove that cosx - cos3x = 4sin^2cosx. Hence, otherwise, find the value of x of the equation cosx - cos3x = tan^2x where 0 <x<360
2. Prove that 1+cos2x +sin2x/1-cos2x +sin2x = cotx
3. Prove that 1+sinA - cosA/1+sinA + cosA = tanA/2 .
4. Prove that sin^4x +cos^4x = 1 -1/2sin^2 2x .
Thank you very much in advance !
 
Nikesh

Nikesh

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hm12 said:
can you please explain more why did you use b^2-4ac>0 i mean why greator than 0 sign?
Click to expand...

it is because equations in the form ax^2 + bx + c = 0
will have exactly one root when
b^2 - 4ac = 0
will have 2 roots when
b^2 - 4ac > 0
and no real roots when
b^2 - 4ac < 0
its a basic rule that is to be known
 
H

hm12

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Nikesh said:
it is because equations in the form ax^2 + bx + c = 0
will have exactly one root when
b^2 - 4ac = 0
will have 2 roots when
b^2 - 4ac > 0
and no real roots when
b^2 - 4ac < 0
its a basic rule that is to be known
Click to expand...

okay thanks a lot for your help:)
 
leadingguy

leadingguy

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bamteck said:
1. Prove that cosx - cos3x = 4sin^2cosx. Hence, otherwise, find the value of x of the equation cosx - cos3x = tan^2x where 0 <x<360
2. Prove that 1+cos2x +sin2x/1-cos2x +sin2x = cotx
3. Prove that 1+sinA - cosA/1+sinA + cosA = tanA/2 .
4. Prove that sin^4x +cos^4x = 1 -1/2sin^2 2x .
Thank you very much in advance !
Click to expand...

2. Prove that 1+cos2x +sin2x/1-cos2x +sin2x = cotx

( cos^2x + sin^2x + cos^2x - sin^2x + sin2x)/ ( cos^2x + sin^2x - cos^2x +sin^2x + sin2x)

(2cos^2x + sin2x)/( 2sin^2x + sin2x) remembr sin2x = 2sinx.cosx

so (2cos^2x + 2sinx.cosx )/( 2sin^2x + 2sinx.cosx ) now take comman

2cosx(cosx + sinx) /2sinx(sinx +cosx) cancel (cosx + sinx ) frm bth sides

2cosx /2sinx cancel 2 frm both sides
cosx/sinx = cotx :)
 
J

Jimmy Russels

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Assalamalaikum. These are probably stupid questions so don't laugh :) Can anyone help me?
Qs.1) Find the values of k such that the straight line y=2x+k meets the curve with equation x^2+2xy+y^2=5 exactly once.

Qs.2) An Open cylindrical wastepaper bin, of radius r cm and capacity Vcm^3 is to have a surface area of 5000 cm^3.
(a) Show that V=1/2r(5000πr^2)
(b) Calculate the maximum possible capacity of the bin.​
 
leadingguy

leadingguy

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bamteck said:
1. Prove that cosx - cos3x = 4sin^2cosx. Hence, otherwise, find the value of x of the equation cosx - cos3x = tan^2x where 0 <x<360
2. Prove that 1+cos2x +sin2x/1-cos2x +sin2x = cotx
3. Prove that 1+sinA - cosA/1+sinA + cosA = tanA/2 .
4. Prove that sin^4x +cos^4x = 1 -1/2sin^2 2x .
Thank you very much in advance !
Click to expand...


1. Prove that cosx - cos3x = 4sin^2cosx.

cosx - cos3x
cosx -(cos(2x+x))
cosx -(cos2x.cosx - sin2x.sinx)
cosx - ( (cos^2x - sin^2x)cosx - 2sinx.cosx.sinx)
cosx - (cos^3x - sin^2x.cosx - 2sin^2x.cosx)
cosx - (cos^3x - 3sin^2x.cosx )

cosx -cos^3x + 3sin^2x.cosx
cosx(1-cos^2x) + 3sin^2x.cosx
cosx(sin^2x) + 3sin^2x.cosx
4sin^2x.cosx proven :)

now value of cosx - cos3x = tan^2x equals to

4sin^2x.cosx =tan^2x
4sin^2x.cosx = sin^2x/cos^2x cos^2x wil be sent to the left side of eq. and mutiplied so

4 sin^2x.cosx.cos^2x = sin^2x
4 sin^2x.cos^3x = sin^2x nw cancel sin^2x frm both sides
4 cos^3x = 1
cos^3x = 1 /4 now taking cube root on both sides

cosx = 0.63
x= cos-1(0.63 = 50.95 :) in frst quadrant
cos is also positive in fourth quadrant so
360 - 50.95 = 309 in 4rth quadrant
 
sasukeuchiha

sasukeuchiha

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hello people,
what does mf9 consist of? all the list of formulae given in the syllabus section 6?? For Maths 9709. Please reply soon :confused:
 
leadingguy

leadingguy

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sasukeuchiha said:
hello people,
what does mf9 consist of? all the list of formulae given in the syllabus section 6?? For Maths 9709. Please reply soon :confused:
Click to expand...
yes it cocnsist of the essential frmulae needed fr calculating
mean, standard deviation, variance of groued and ungrouped data
some other formulae too lyk ... in binomial distribution we use meu = np variance = npq

th is all given in stats portion of mf9 and most importantly the table to calculate the probability is given in mf9
 
leadingguy

leadingguy

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hm12 said:
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w10_qp_11.pdf

question 7 part(i) and part(iii)

i really have a lot of problem finding ranges and domains of function can anyone help me with it? how do we find them?

please someone help me with it I have posted it the 3rd time please seriously need help with it
Click to expand...


here range means the boundries in which the values of "f" lies.
the eq. is given
fx = 3-2tan(1/2x)
jxt put al the values of x (from 0 to π) and find values of "f" at each value of x

f(0) = 3-2tan(1/2(o)) = 3
f( π/2) = 3-2tan( 1/2* π/2) = 1
f( π) = 3-2tan(1/2* π) = not real

so the greatest value of f is 3 and least is 1 range of is 1<f<3.

fr drawing graph it is stated that fx = y
so what ever the vaues of fx are at x = 0 and π/2 use them to plot a graph
i.e
x 0 π/2
y 3 1 sketch I hope m not doing that wrong as nt seen ms yet :p
 
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