C3 and C4 DOUBTS HERE!!!

[break freak]

I need answers for these question s pleaaase ( x2 means x square )?
1) The value of cos(sin-1x) is: A. 1 B. x C. 1/x D. 1/?1-x2 E. ?1-x2

2) The domain of f(x)= (3?x)/(1-x) is: A. (1,?) B. (0,1) U (1,?) C. (0,?) D. none of these
3)C is the circle (x-1)2 + (y-2)2 =144 and the point P(10,10), then: A. P is the center of C B. P is inside C but not the center C. P is outside C D. P is on C E. the location of P relative to C cannot be determined

[break freak]

I need answers for these questions pleaaaaase today ( x2 means x square )?
1) The value of cos(sin-1x) is: A. 1 B. x C. 1/x D. 1/?1-x2 E. ?1-x2

2) The domain of f(x)= (3?x)/(1-x) is: A. (1,?) B. (0,1) U (1,?) C. (0,?) D. none of these
3)C is the circle (x-1)2 + (y-2)2 =144 and the point P(10,10), then: A. P is the center of C B. P is inside C but not the center C. P is outside C D. P is on C E. the location of P relative to C cannot be determined

[astarmathsandphysics]

1)Put y=sin^-1 x then we need cos y
y=sin^-1 x so siny =x so cosy =sqrt(1-sin^2y) =sqrt(1-x^2)2) x=1 is not in domain f(x) since f(1) not defined so domain could be B
3)centre is (1,2)
distance from (1,2) to P os sqrt((1-10)^2 +(2-10)^2)=sqrt(81+64)=sqrt(145)>sqrt(144) so P is outside C,
Answer is C.

[qooqo111]

Hey
I got this formula for integration by parts
i used it in loads of questions and they all seem to come true(even the past paper questions) but am still curious if it is correct if i used it in my C4 exam the formula goes like this :
u?vdx-?[du/dx.?vdx]dx
P.S

 for some reason i cant put the integral mark it just comes as a question mark ..... so yeah the question mark is integral
the second integral is for all the bracket.
thank you

[mr.incredible]

Jan '06 4-b :show did they do the last step??

i got till 1/8cos(arcsin(x/4)) but the next step is magic to me :/

Since that the domain of sin(2y+6) is not stated, therefore you will have to take +ve/-ve values of cos(2y+6). You can obtain this value by either drawing a right-angled triangle and using the pythagoras theorem to get the unknown side and then cos(2y+6)=adj./hyp. OR by using the identity sin^2(2y+6) + cos^2(2y+6) = 1 and given that sin(2y+6)= x/4 you will be able to get the value of cos(2y+6).

The final answer is 1/(±2(16-x^2 )^2 )

 

[Pias]

sin A= 3/5; find cos A;
after finding the value of cos A by formula, why is it that the negative value is taken of the root of the square of cos A.
Please could someone help me with that.

[Darkstar3000]

How do they get 45 and 135 in question 8? I managed to get 22.5 and 112.5 but I have no idea how they got the other two values

Here is the question paper: http://www.scribd.com/doc/26846381/Edexcel-GCE-January-2010-Core-Mathematics-C3-QP (Jan 2010)
Here is the mark scheme :  http://www.yourschoolmaster.com/mathematics/ks5/answers/2010web/C3_jan_2010ms.pdf


And also in question 3, why did they take their "alpha" to 4 decimal places, I took mine to 2 decimal places and my final answer has a 0.1 difference from theirs. Does this matter since they didn't state to what number of decimal places we should write alpha in ?

[astarmathsandphysics]

cosec^2 2x - cot 2x =1
1+cot^2 2x  -cot 2x =1
cot^2 2x -cot 2x =0
cot 2x(cot 2x -1)=0
cot 2x =0  so tan 2x =infinity so 2x =90, 90+180=270 so x=22.5,135
or cot 2x -1=0 so cot 2x =1 so tan 2x =1 so 2x=45, 45+180=225 so x=22.5,112.5

[astarmathsandphysics]

3 dp because in part b they want the answer to 2 dp

[Darkstar3000]

cosec^2 2x - cot 2x =1
1+cot^2 2x  -cot 2x =1
cot^2 2x -cot 2x =0
cot 2x(cot 2x -1)=0
cot 2x =0  so tan 2x =infinity so 2x =90, 90+180=270 so x=22.5,135
or cot 2x -1=0 so cot 2x =1 so tan 2x =1 so 2x=45, 45+180=225 so x=22.5,112.5

how is 2x=90? That's the part I don't understand

[astarmathsandphysics]

cos tan^{-1} infinity =90

[hmh]

Q: Express as a single fraction in its simplest form

f(x)= ( 2x + 2 / x^2 - 2x - 3 ) - ( x + 1 / x - 3 )

[astarmathsandphysics]

f(x)= ( 2x + 2 / x^2 - 2x - 3 ) - ( x + 1 / x - 3 )

[hmh]

Thanks Astar

[astarmathsandphysics]

No prob