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Mathematics: Post your doubts here!

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Faaiz Haque

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Thought blocker

Thought blocker

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Faaiz Haque said:
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
Click to expand...
(1-(sin^2x/cos^2x))
--------------------------- =
(1+(sin^2x/cos^2x))

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

Now,
Using the identity
plot-formula.mpl
, if we subtract
plot-formula.mpl
from both sides, we obtain
plot-formula.mpl


so, 1-2sin^2x / 1 = 1-2sin^2x ;)
 
Hadi Murtaza

Hadi Murtaza

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kitkat <3 :p said:
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 :(
Click to expand...
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

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kitkat <3 :p said:
sorry i meant the functiins question :oops:
thanks alot btw :)
Click to expand...
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

Hadi Murtaza

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kitkat <3 :p said:
sorry i meant the functiins question :oops:
thanks alot btw :)
Click to expand...
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 ]
 
S

sitooon

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sudeep1097 said:
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 !
Click to expand...
Sorry for disturbing you again :rolleyes:
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. :eek:
 
K

kitkat <3 :P

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Thought blocker said:
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 :)
Click to expand...

Hadi Murtaza said:
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 ]
Click to expand...
thank u so much :)
 
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Faaiz Haque said:
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.
Click to expand...
Q4)
img_20140430_005906-jpg.40570

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
:)
 
Talha Khatri

Talha Khatri

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Thought blocker said:
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 ?
Click to expand...

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
 
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