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Mathematics: Post your doubts here!

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:eek: That's a hard one!
Maybe it has to do with parametric equations, don't u think?
points lieing on a circle means the distance from the centre to these points will be same.
check that out prove that out.(y)
If the three points lie on the circumference of a circle there distances from the centre (2,0) must be same [RADIUS ]

Find the distance b/w (7,12) and (2,0)

Find the distance b/w (-3,-12) and (2,0)

Find the distance b/w (14,-5) and (2,0)

These three distances must be same!

tnx 4 ur help guys :) i solved it myself, smzimran yh they r all radius
 
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can anyone plzzz help me out with question 2 and question 3.....when i tried and did it.....my answer is nt matching with the textbook answers help
 
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Please could you sketch and explain them : 1) 4sin(3x) ,0 (<or =) x ( <or=) 360
2) 9-5cos(2x) , 0 (<or =) x ( <or=) 360
3) 3+4sin(2x) , 0 (<or =) x ( <or=) 180
4)Tan(2x) , 0 (<or =) x ( <or=) 180

Thanks alot ! :)
 
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Please could you sketch and explain them : 1) 4sin(3x) ,0 (<or =) x ( <or=) 360
2) 9-5cos(2x) , 0 (<or =) x ( <or=) 360
3) 3+4sin(2x) , 0 (<or =) x ( <or=) 180
4)Tan(2x) , 0 (<or =) x ( <or=) 180

Thanks alot ! :)

Aoa! Why don't you go to http://www.wolframalpha.com/ and check out the graphs for yourself? But if you're unsuccessful, your welcome back to carry this query forward! :)
 
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"Integration A2"

So there's this question on how to integrate cos^3 x. I did it by breaking it into cos^2 x and cos x, you know, the by-parts method. But i seem to be getting the wrong answer. Could anyone do it for me, so i could compare my working ?? Thankyou! :)
 
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"Integration A2"

So there's this question on how to integrate cos^3 x. I did it by breaking it into cos^2 x and cos x, you know, the by-parts method. But i seem to be getting the wrong answer. Could anyone do it for me, so i could compare my working ?? Thankyou! :)
AoA,
This is best done using substitution, and not by parts:
dfs.png
 
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"Integration A2"

So there's this question on how to integrate cos^3 x. I did it by breaking it into cos^2 x and cos x, you know, the by-parts method. But i seem to be getting the wrong answer. Could anyone do it for me, so i could compare my working ?? Thankyou! :)

cos^3 x
= (cos^2 x)(cos x)
= (1-sin^2 x)(cos x)
= cos x - (sin^2 x)(cos x)

after integration,
it become:
= -sin x - (1/3)(sin^3 x) + c
 
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Can anyone help me in this question?

Thanks in advance. :)

View attachment 14038

Let f(x) = (x^2 - 1)Q(x) + ax + b
f(x) = (x-1)(x+1)Q(x) + ax + b

we know that (x+1) is a factor,
f(-1) = 0,
b-a = 0, this is equation 1,

(x-1) gives remainder of 4,
f(1) = 4,
a+b = 4 , this is equation 2,

solve simultaneous equation 1 and 2,
then you can find a and b.
 
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hye b
Let f(x) = (x^2 - 1)Q(x) + ax + b
f(x) = (x-1)(x+1)Q(x) + ax + b

we know that (x+1) is a factor,
f(-1) = 0,
b-a = 0, this is equation 1,

(x-1) gives remainder of 4,
f(1) = 4,
a+b = 4 , this is equation 2,

solve simultaneous equation 1 and 2,
then you can find a and b.
hey bro but why did u substitute f(-1) in the remainder part it is supposed to be in the original equation thats f(x) and hey also thats what we call remainder theorem right?
 
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hye b

hey bro but why did u substitute f(-1) in the remainder part it is supposed to be in the original equation thats f(x) and hey also thats what we call remainder theorem right?

f(-1) means we substitute the x with -1 in here:
f(x) = (x-1)(x+1)Q(x) + ax + b
every x above will turn into -1
 
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Let f(x) = (x^2 - 1)Q(x) + ax + b
f(x) = (x-1)(x+1)Q(x) + ax + b

we know that (x+1) is a factor,
f(-1) = 0,
b-a = 0, this is equation 1,

(x-1) gives remainder of 4,
f(1) = 4,
a+b = 4 , this is equation 2,

solve simultaneous equation 1 and 2,
then you can find a and b.

Thank you! :D
 
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