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Mathematics: Post your doubts here!

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Jimmy Russels

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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.

Thank you. :)
 
Nikesh

Nikesh

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Jimmy Russels said:
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.

Thank you. :)
Click to expand...
Substitute the value of y in the given equation of curve and you will get a new equation in the form of ax^2 + bx + c = 0 if not change the equation to this form. And question says that the line cuts exactly once that means the discriminant of the equation is equal to 0. i.e. b^2 - 4ac = 0 use this u will get 2 values of k.
 
Mehroz

Mehroz

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imanmalik said:
two questions.
Question 3. I'm getting the values for b and a the wrong way around. Why? (s11 paper13)
Question 11 (iv) Isnt the inverse of f y= (x-1)/2 ? The mark scheme says something completely different. (s11 paper11)
View attachment 7768
View attachment 7769
Click to expand...

This is the answer for q3. If you find it useful then just like it:)
 

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V

vita199

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can u please help in these questions..they're from p3..may/2011/31/q10a and q9a and q6b
 
aaakhtar19

aaakhtar19

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2pac said:
Could you please explain me q9 of may 2010 p31?
I have done half of the procedure but then can't figure what to do after the 2 gets cancelled out.
Thanks
Click to expand...
I have uploaded the picture its the gradient of curve at (x,y)
the gradient of the normal to the curve at (x,y)
would be its reciprocal with opposite sign as m1*m2=-1
so ans is
(1+x)*(1-x^2)^0.5

Hope u got it :D
 

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leadingguy

leadingguy

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vita199 said:
can u please help in these questions..they're from p3..may/2011/31/q10a and q9a and q6b
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qstn 6 ) part b

u have the frmula θ = 2/5π + sin θ

u are given an initial value of θ whixh is 2.1 put this value and obtain another value of θ ;


θ = 2/5π + sin (2.1) = 1.29328..
put this new value in the same eq. in place of θ

θ = 2/5π + sin (1.29328..) = 1.2792..

now U again ahave a new value put it in eq again and solve it.
repeat the procedure until U get a value whch is not changin(root value) ..(2.105)

that wil be the exact value of theta (θ)

now we know that the total angle sum of a triangle is 180 degri means, π.

U have one angle, 2.105 now subtract ths frm π U wil get 1.04
divide this by two to get the angle of each side ABO , BAO (its an isoseles triangle )

now U have the angle ABO = 0.52rad. u know the length o of the side BO =10
use sin(0.52) = p /10 to calculate p . this p is the height of triangle.


there are 2 methods for finding area of triangle equate them lyk this

1/2absinθ = 1/2 h * AB ........(we already calculated h, the height )

1/2 (10)^2 sin(2.105) = 1/2 * h * AB
find Ab by this:) a ltle long but not complicated itx eaSy
 
leadingguy

leadingguy

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vita199 said:
can u please help in these questions..they're from p3..may/2011/31/q10a and q9a and q6b
Click to expand...

qstn 9 iia) we have proved that cos4θ 4 cos2θ = 8cos^4θ - 3

so for solving cos4θ 4 cos2θ =1 we can solve 8cos^4θ - 3 =1

8cos^4θ - 3 =1
8cos^4θ = 1+3
8cos^4θ = 4
cos^4θ = 4/8
cos^4θ = 1/2 take square root on both sides

cos^2θ = 0.707 take square root again :)

cosθ = 0.84 now θ = cos-1(0.841 = 0.572 Ans
 
leadingguy

leadingguy

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vita199 said:
here is the link to the paper http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s11_qp_31.pdf
please answer this urgently!!!1
Click to expand...


qstn 10 part a)
dn/dt = N (1800 -N)/3600

multiply bth sides by "dt"
dn = N (1800 -N)/3600.dt

now arrange the eq. by arranging the N containing terms to the left hand side.

1/(N (1800 -N)).dn = 1/3600.dt now we have 2 integrate both sides, but before doing so remember that eq. on left hand side cannot be integrated directly it should be broken by partioal fractions


so 1/(N (1800 -N)) = A/N + B/( 1800 -N)
solve them to get values of A and B as we do normaly fr partial fractions.

u wil get value sof A and B equal to "2" substitute them in eq.

2/N + 2/ ( 1800 -N) dn = 1/3600.dt now we can integrate

2lnN + (-2ln(1800 -N) ) = (1/3600)t + c

now find "c" by substituting N = 300 and t = o (as given in qstn )

2ln300 -2ln(1800 - 300) = 0 +c
2ln (300) -2ln (1500) = o +c
2ln(300/1500) = c
2ln(1/5) = c or -2ln5 now substitute the value of C and obtain eq.
 
2

2pac

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aaakhtar19 said:
I have uploaded the picture its the gradient of curve at (x,y)
the gradient of the normal to the curve at (x,y)
would be its reciprocal with opposite sign as m1*m2=-1
so ans is
(1+x)*(1-x^2)^0.5

Hope u got it :D
Click to expand...
Thank you so much.Really like your approach of answering by uploading files instead of typing it.The signs really confuse me and hence why I really appreciate your method.
Can you help me with q8 (ii) b as well?
 
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