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A/L math p6 chem help
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උඩයි යටයි මාතය සමාන උනාම නේද නියතයක් එන්න පුලුවන් වෙනම ?ඔව් බ්ං ගාන වැරදි නේද
ඔය එක එකාගේ පේපර් හදන්න එපා බන් .
exm පේපර් හදපන් .
නැත්නම් හරි හමන් එකෙක්ගේ tute එකක ගන්න හදන්න .
------ Post added on Aug 30, 2024 at 8:29 AM
It shldn't be k.... but in the form of f(x) which is linear. In fact f(x) = x + a +b + c
The complete partial fraction expansion is
x^4/{(x-a)(x-b)(x-c)} = a^4/{(a-b)(a-c)(x-a)} - b^4/{(a-b)(b-c)(x-b)} - c^4/{(a-c)(c-b)(x-c)} + a + b + c + x
Because Sin2θ = 2SinθCosθ
Sinθ = x and hence Cosθ = √(1-x^2)
How did you workout No#46 ?
What's meant by repeated use of the remainder theoem? Can anyone explain?
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Maybe there's some trick. I am not aware of a repeated use of the reminder theorem. My current knowledge of maths is limited mostly to what I taught my son when he sat for A/L , but it's a quite while now.
If P(x) is a polynomial with a degree of m, then the remainder when divided by (x-a)^n (when m >= n) will be another polynomial of degree n-1. Let's say it's R(x)
Then P(x) = (x-a)^n Q(x) + R(x)
When n = 1, R(x) will have a degree of 0 (a constant) , let's say r0
Then P(x) = (x-a) Q(x) + r0, with x = a this gives us that P(a) = r0
When n = 2, R(x) will have a degree of 1 which is of the form R(x) = r1x + r0
Then P(x) = (x-a)^2 Q(x) +(r1x +r0), with x=a this gives us P(a) = r1a + r0
Differentiation gives us P'(x) = 2(x-a)Q(x) + (x-a)^2 Q'(x) + r1 This gives us P'(a) = r1. So these set of equations will give you r1 and r0
Note: This is also quite evident from the Taylor series expansion of P(x) if you know about it.
P(x) = P(a) + P'(a) (x-a) + P"(a) (x-a)^2 / 2 +........ etc ( can't type more terms in simple text)
So the given expression 2x^3+7x^2+13x-3 when divied by x+1 will have a remainder of -11
when divided by (x+1)^2 will have a remainder of 5x-6 (calculate r1 and r0 as mentioned above)
when divided ny (x+1)^3 will have a remainder of x^2+7x-5
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@dilann උබ තරහ වෙන්න එපා බං ගානක් ඒම අහගන්න ඉන්නෙ උබ විතරයි ම්ං ක්ලාස් ඒමත් යනව 
------ Post added on Sep 8, 2024 at 11:32 PM
Thx
------ Post added on Sep 8, 2024 at 11:32 PM
AFAIK, there's isn't any other way. I have outlined the general approach when P(x) is divided by (x-a)^n. Hope you understood.@dilann උබ තරහ වෙන්න එපා බං ගානක් ඒම අහගන්න ඉන්නෙ උබ විතරයි ම්ං ක්ලාස් ඒමත් යනව
Thx
------ Post added on Sep 8, 2024 at 11:32 PM
Did you at least take the time to try the Google translation of what I wrote above?ම්ං ආතල් එකක් ගන්නව නෙමෙයි
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Says a lot about yourself and your attitudes....ස්තුති බ්ං ඒ ටයිම් එකේ වෙන එකක් අරන් හැදුවෑකිනෙ
ඔය ගාන රූල් 3න් හ්දන්න නෙ තියෙන්නේ
හරි හරි බ්ං ඕව ගනන් ගන්න එපා ස්තුති මචන්Says a lot about yourself and your attitudes....
විවරනේ නෙමේ method එක ඕනNo need to thank me. It's the generalized method to solve any of these problems had you to cared to understand it. There isn't any විවරනේ in it. Tht's up to you.හරි හරි බ්ං ඕව ගනන් ගන්න එපා ස්තුති මචන්
