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C3 and C4 DOUBTS HERE!!!

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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:

--- Quote from: sweetest angel on June 11, 2010, 09:43:08 am ---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 :/

--- End quote ---

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 )

 :)

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