2U Maths Help (1 Viewer)

[bmn]

Q1-3, and 5 have been solved by tommykins.

Found out 4 myself.

 

[tommykins]

回复: 2U Maths Help

1) f(a) = 4a-5, f(b) = 4b-5
f(a)-f(b) / a-b = [ 4a-5 - (4b-5)]/a-b
= 4a - 4b / a-b = 4(a-b)/a-b = 4.

2) 16/[2^3x * 8^(1-x)] = 16/[2^3x * 2^3(1-x) ] = 16/[2^3x*2^3-3x] = 16/[2^3] = 16/8 = 2.

3) f(x) = 1-x^2/x
f(-x) = 1-x^2/-x = x^2-1/x
-f(x) = - [1-x^2/x] = x^2 - 1/x = f(-x)

.'. f(x) is odd as f(-x) = -f(x).

4) cbf.

 

[bmn]

Thanks, I understand 1, 2 and 3 now. And with 2, because its 2 x 2 you dont times them together, only add the power like in algerbra? Thats where i went wrong... i done 4^3 instead of 2. Thanks for that too....

Anybody help with 4?

and also, need help with a fifth one.

5) Prove that tanX - (sin^3X)/cosX = sinXcosX (was originally the 0 like thing, but X works...) And its sine to the power of 3 then X, not power of 3X. Usually im good with ones like that, but not that one.

 

[tommykins]

回复: Re: 2U Maths Help

index rules.
x^a*x^b = x^a+b.
For example, 2^1*2^3 = 2*8 = 16 = 2^4.

5)tanx - sin^3x/cos = sinxcosx
LHS = (sinx-sin^3 x)/cosx = sinx(1-sin^2 x)/cosx = sinxcos^2x/cosx = sinxcosx = RHS.

 

[bmn]

I see, didnt see that common factor, and thanks for explaining 2.

Thanks overall for you very fast replys.