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Mathematics: Post your doubts here!

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i m doing Cambridge advanced level pure mathematics 2 & 3 by hugh neil douglas quading....... the problem is i lost the answers of chapter 15-Rational function......plz help me!
 
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From the previous part, you have obtained a simplified form of the expression as p(x) = (x-2)(x+2)(x-4)
Since the question now calls for evaluating p(x squared), you just substitute the x in the expression for x squared.
Your expression would become somewhat like this
(I'll be using x2 for x squared)
p(x2) = (x2 -2)(x2 +2)(x2 -4)
Now, you equate them to zero, each of them. You'll obtain six roots, two of which will be imaginary.
 
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From the previous part, you have obtained a simplified form of the expression as p(x) = (x-2)(x+2)(x-4)
Since the question now calls for evaluating p(x squared), you just substitute the x in the expression for x squared.
Your expression would become somewhat like this
(I'll be using x2 for x squared)
p(x2) = (x2 -2)(x2 +2)(x2 -4)
Now, you equate them to zero, each of them. You'll obtain six roots, two of which will be imaginary.
Thanks! : )
 
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can you please help me in how to answer questions such as exam 9709/32/O/N10 the question 8 part b

its basically after making partial fractions...expanding it to ascending powers upto x^3 etc

how do you do these? plz help :(
 
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See, can u help me in (a) , i cant use the r-a.n equation , sorry.

for finding angle,
u take normal,
2i−3j+6k
and direction vector,
i − 2j + 2k

take scalar product
2+6+12 = 20

now divide with the mod of both vectors
and u get
20/21

20/21 normally =cos (angle)
but when plane and line
it is=sin (angle)
so sin^-1 (20/21) = 72.2
 
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i sorry about that....actually i have tried to solve it....but in Q8(i),my answer is C=-1/3,but the ms is 1/3...and i can‘t find any way to get “-”
I am afraid i am confused.

See, i quoted a post of different member, not u.

And what question are you referring to?
 
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See, can u help me in (a) , i cant use the r-a.n equation , sorry.

for finding angle,
u take normal,
2i−3j+6k
and direction vector,
i − 2j + 2k

take scalar product
2+6+12 = 20

now divide with the mod of both vectors
and u get
20/21

20/21 normally =cos (angle)
but when plane and line
it is=sin (angle)
so sin^-1 (20/21) = 72.2
SURE :D
part a right ?
 
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