Mathematics: Post your doubts here!

[jrahmed]

link?

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_33.pdf

 

[Thought blocker]

Thanks so much, that was very helpful.

I also need help in the same paper, question 3.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_1.pdf

(1-(sin^2x/cos^2x))
--------------------------- =
(1+(sin^2x/cos^2x))

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

Now,
Using the identity

, if we subtract

from both sides, we obtain

so, 1-2sin^2x / 1 = 1-2sin^2x

 

[kitkat <3 :P]

Question 1 doesn't have a part 2 and here's question 7:
View attachment 40702 View attachment 40703

sorry i meant the functiins question
thanks alot btw

 

[Hadi Murtaza]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_12.pdf
Q1 ii) and Q7 please
m not good with vectors

Q.7
i) OA = (1 0 2)
OB = (k -k 2k)

when k = 2
OB = (2 -2 4)

Cos(Θ) = (a.b) / (|a|*|b|)
Cos(Θ) = [ (2*1) + (0*-2) + (2*4) ] / [ √(1² + 0² + 2²) * √(2² + -2² + 4²) ]
Cos(Θ) = (2 + 0 + 8) / [ √5 * √ 24 ]
Cos(Θ) = 10 / 2√30
Cos(Θ) = 5/√30
[ Θ ≈ 24.1° ]

AB = OB - AB
AB = (k -k 2k) - (1 0 2)
AB = (k-1 -k 2k-2)

|AB|² = (k-1)² + (-k)² + (2k-2)²
1² = k² - 2k + 1 + k² + 4k² - 8k + 4
1 = 6k² - 10k + 5
6k² - 10k + 4 = 0
6k² - 6k - 4k + 4 = 0
6k(k-1) - 4(k-1) = 0
(6k - 4)(k-1) = 0
[ k = 1 , 2/3 ]

 

[Thought blocker]

sorry i meant the functiins question
thanks alot btw

part i)
To find f inverse, convert f(X) to x and x to y
so, x = (root of y + 3 / 2 ) + 1
= ( x - 1 ) => (root of y + 3 / 2 )
= 2(( x - 1 )^2) -3 => y
Now convert y = f inverse.
when you open the bracket you get is 2x^2 - 4x - 1

part ii) (NOTE >= means greater or equal to)
Range of F(X) is Domain of f inverse,
so range of f(X) is x >= 1 hence domain of inverse is >= 1

 

[Hadi Murtaza]

sorry i meant the functiins question
thanks alot btw

y = √[ (x+3)/2 ] + 1
x = √[ (y+3)/2 ] + 1
x-1 = √[ (y+3)/2 ]
(x-1)² = (y+3)/2
y+3 = 2(x-1)²
y = 2(x-1)² - 3
y = 2(x²-2x+1)-3
y = 2x²-4x+2-3
y = 2x²-4x-1

ii) Domain of function = Range of inverse
Range: y ≥ -3
2x²-4x-1 ≤ -3
2x²-4x+2 ≤ 0
x²-2x+1 ≤ 0
(x-1)² ≤ 0
[ x ≥ 1 ]

 

[M Haseeb Javed]

Guys I need help with the (b) parts where we have to find minimum modulus, whats the procedure for that?

 

[sitooon]

Sorry about the x=y .
|z-3|=|z-3i|
now removing modulus sign we get sq rt of ((real component)^2+(imaginary component)^2) on both sides
suppose z as x+yi you get sq rt of ((x-3^2)+y^2)=(x^2+(y-3)^2)
remove sq rt by squaring both sides now solve to get x=y

yeah vector parallel to x axis is always (1,0,0) for y axis (0,1,0) and z axis (0,0,1) remember this
as (1,0,0) lies in x-axis and line which lies on a bigger line is parallel to it. These are the most simple vectors parallel to respective axis, there are others as well but use this !

Sorry for disturbing you again
Got other doubt , when you square
right hand side you get x^2 + ( iy-3i)^2
which dont simplify to x=y when solve
and on what basis did you square root ,
Is this also a rule in complex no.

 

[jrahmed]

can u please do part a for me

 

[Faaiz Haque]

Question 4, 8 i, ii
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s07_qp_1.pdf

Please help, my exam is in 6 days and i'm freaking out.

 

[kitkat <3 :P]

part i)
To find f inverse, convert f(X) to x and x to y
so, x = (root of y + 3 / 2 ) + 1
= ( x - 1 ) => (root of y + 3 / 2 )
= 2(( x - 1 )^2) -3 => y
Now convert y = f inverse.
when you open the bracket you get is 2x^2 - 4x - 1

part ii) (NOTE >= means greater or equal to)
Range of F(X) is Domain of f inverse,
so range of f(X) is x >= 1 hence domain of inverse is >= 1

y = √[ (x+3)/2 ] + 1
x = √[ (y+3)/2 ] + 1
x-1 = √[ (y+3)/2 ]
(x-1)² = (y+3)/2
y+3 = 2(x-1)²
y = 2(x-1)² - 3
y = 2(x²-2x+1)-3
y = 2x²-4x+2-3
y = 2x²-4x-1

ii) Domain of function = Range of inverse
Range: y ≥ -3
2x²-4x-1 ≤ -3
2x²-4x+2 ≤ 0
x²-2x+1 ≤ 0
(x-1)² ≤ 0
[ x ≥ 1 ]

thank u so much

 

[Thought blocker]

Question 4, 8 i, ii
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_1.pdf

Please help, my exam is in 6 days and i'm freaking out.

Q4)

Q8)
i)
given, f(0)=-1 so, a + bcos2(0) = -1 ; a + b(1) = -1
given f(pie/2) = 7, a + bcos2pi/2 = 7 ; a + b(-1) = 7
solve siml. eqn, you get a = 3 and b = -4.
ii)
Put the values : 3 - 4cos2x = 0
cos2x = 3/4
cos inverse 3/4 = 0.722
2x = 0.722
x = 0.36
and see the domain hence, pi - 0.36 you get x = 2.78
hence x = 0.36 , 2.78

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s13_qp_12.pdf
Question 2ii) , 4i),iii) , 5i), 6i), 7, 8i,ii) Please.
Some questions I am just stuck on formula like :¬
in question 4 i) dont we use : theta = cos inverse a^2 + b^2 - c^2 / 2(ab) ?
question 5i) how we got R.H.S as 10 ?
question 7) I got (3,9) what next ?

 

[Rutzaba]

i guess il comcentrate on p3 students lol

 

[Thought blocker]

i guess il comcentrate on p3 students lol

 

[Rutzaba]

i solve the whole question and umbody else posts it lol

 

[Talha Khatri]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_12.pdf
Question 2ii) , 4i),iii) , 5i), 6i), 7, 8i,ii) Please.
Some questions I am just stuck on formula like :¬
in question 4 i) dont we use : theta = cos inverse a^2 + b^2 - c^2 / 2(ab) ?
question 5i) how we got R.H.S as 10 ?
question 7) I got (3,9) what next ?

2ii. There are two possible ways to get x^2. One when the first term of the first bracket multiplies with the a term of the power x^2 from the second bracket and another when the second term of the first bracket multiplies with a term of the power x^0 of the second bracket.
Hence, (1) (60x^2) + (x^2) (6C3*(2x)^3 *(-1/2)^3)
= 60 x^2 - 20 x^2 = 40 x^2

 

[Talha Khatri]

i solve the whole question and umbody else posts it lol

AHAHAHAHA! I so feel you on this.

 

[Thought blocker]

i solve the whole question and umbody else posts it lol

You think I'll do the same, then no need. I'll get it solved by others. Indeed.
Suchal Riaz
ZaqZainab
sitooon
midha.ch
Adeena Shamshir

 

[Thought blocker]

2ii. There are two possible ways to get x^2. One when the first term of the first bracket multiplies with the a term of the power x^2 from the second bracket and another when the second term of the first bracket multiplies with a term of the power x^0 of the second bracket.
Hence, (1) (60x^2) + (x^2) (6C3*(2x)^3 *(-1/2)^3)
= 60 x^2 - 20 x^2 = 40 x^2

Was it that easy ? Ty