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LoginA.Maths Qn
15 posts
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Originally posted by wishboy:
need help with this qn
A straight line is in the form Y=mX+chere gradient m = (3-1)/(4-1)=2/3
Hence Y=(2/3)X + c
Sub in X=1, Y=1, we get 1=2/3+c so c=1/3
Therefore, Y=(2/3)X + 1/3
Now Y is actually ln y and X is actually ln x
ln y = (2/3) ln x +1/3
y = exp(2/3 ln x +1/3)
y = x exp(2/3)exp(1/3)
y=x exp(1)
Not 100% sure.
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how to expand (1 + ax + x²)^10 ?
EDIT: ok i got it =D
ok the qn was
Given that the expansion of (1 + ax + x²)^10 up to the x² term is given by (1 + 20x + bx² + ...),
i) find the values of a and b.
ii) hence, without using the calculator, find the value of (1.0404)^10, giving your answer to 3 decimal places.
i found tat a = 2, b = 190 for part (i)
how do i do part (ii)?
Edited by wishboy 27 Apr `08, 7:52PM
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how to expand (1 + ax + x²)^10 ?
EDIT: ok i got it =D
ok the qn was
Given that the expansion of (1 + ax + x²)^10 up to the x² term is given by (1 + 20x + bx² + ...),
i) find the values of a and b.
ii) hence, without using the calculator, find the value of (1.0404)^10, giving your answer to 3 decimal places.
i found tat a = 2, b = 190 for part (i)
how do i do part (ii)?
Given that your answer to part one is right, i.e., a=2,b=190,
then (1 + ax + x²)^10 is also (1 + 2x + x²)^10 ,
so u sub x =0.02 into (1 + 20x + bx² + ...) to get the value of (1.0404)^10... goddit?
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(1 + ax + x²)^10
= ((1 + ax) + x²)^10
= (1 + ax)^10 + 10(1+ax)^9 (x^2) + ....all the rest will be x with powers above 2
= 1 + 10ax + 45a^2x^2 + 10(1 + ...)x^2 + ....
= 1 + 10ax + (45a^2 + 10)x^2 + ...
therefore by comparing of coefficients, 10a=20=>a=2
45a^2+10=190
(ii) By inspection sub in x = 0.02. The answer is an approximated solution so use curvy equality.
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