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http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_12.pdf
Someone please help me with question no. 10(a) iii..
Someone please help me with question no. 10(a) iii..
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u just find till wich term are negative...take their sum n add it to 525..as their sum will b negative will be subtracted frm original in the start..http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_12.pdf
Someone please help me with question no. 10(a) iii..
MustafaMotani , A star , daredevil , PANDA-
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_11.pdf
Question No. 7.. whole question, I will try and solve it but you guys solve it as well.. so that I can tally my working.. since I don't exactly understand the question over here...
a)
an = a1 + (n-1)d
cos^2x = 1 +d
d= cos^2x -1
Sn = 10/2[2(1) + (10-1)(cos^2x - 1)]
Sn = 10/2[2(1) + 9(-sin^2x)]
Sn = 5[ 2 - 9sin^2x]
Sn = 10 - 45sin^2x
a = 10 ... b = 45
Thank you very much..tan inverse of gradient .. is the angle the line makes with x-axis , so when ever u need to find the angle of two lines intersecting ! just find their gradients and do tan^-1(gradient)
the one with larger angle minus the smaller angle so u get the angle ... this angle can be acute or obtuse ..
some time cie makes it thougher for us .. wen we get acute angle ! they might ask for an obtuse angle, so at that time you need to 180-(what ever angle u got) to get the obtuse angle
Yea I solved it.. the exact same way.. what about b part i ? I solved that as well but unsure of the working.. since mark scheme shows literally no working.
MustafaMotani , A star , daredevil , PANDA-
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_11.pdf
8 iv ..
How does one know when taking an inverse whether a + sign or a - sign would come before the √ equation?
like (x-2)^2 = y+4-k
x-2 = ±√(y+4-k)
How'd you know whether we will take + or - here??
Yea I solved it.. the exact same way.. what about b part i ? I solved that as well but unsure of the working.. since mark scheme shows literally no working.
MustafaMotani , A star , daredevil , PANDA-
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_11.pdf
8 iv ..
How does one know when taking an inverse whether a + sign or a - sign would come before the √ equation?
like (x-2)^2 = y+4-k
x-2 = ±√(y+4-k)
How'd you know whether we will take + or - here??
Yup, I had the same doubt, and got stuck there too..
But the ms had it pretty clearly explained, the range of f(x) is the domain of f^-1(x)
In f^-1(x) --- x > k-4
f(x) = (x-2)^2 -4 + k
Take it as if f^-1(x) is y and switch as always
x = (y-2)^2 - 4 + k
x+4-k = (y-2)^2
y-2 = ±√(x+4-k)
f^-1(x) = ±√(x+4-k) + 2
Replace x with anything less than k-4, like k-5
f^-1(x) = ±√(k-5+4-k) + 2
y - 2 = ±√(k-5+4-k)
The two k's cancel each other (FINALLY)
y-2 = ±√(5+4)
y-2 = ±√(9)
y - 2 = ±3
y = 5 or y = -1
The range of f^-1(x) is... f >= 2
So y = 5, so +.
f^-1(x) = +√(x+4-k) + 2
Why did ya take k-5 as x in f^-1 when x is supposed to be greater than k-4 in f^-1 ? That part I didn't understand.. everything else is clear cut.
volume of rotation of this graph will be (pie) x integration of the square of the equation in terms of yHi... can someone please help me with this problem.. Would really appreciate it.
I only want the explanation of the 2nd and third part. i dont get how the volume is being calculated.. please explain what cylinder does the MS talk about?
Its from w08.
Thanks
well as an aid roughly sketch the graph and you will find graph varies between -1 and 3 which is the rangehttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_12.pdf
Question 9 i)a)
I always get lost when I see questions like these :c
Help please?
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