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Mathematics: Post your doubts here!

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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. :)
 
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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. :)
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.
 
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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

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

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

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
 
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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.
 
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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
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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