MATHS study thread. (1 Viewer)

[YBK]

Horizontal asymptote is at 0

i think that's because bottom power is one higher than the top and has coefficient 1

so it's 0/1 = 0

 

[insert-username]

Horizontal asymptote is at 0

From the limit as x -> infinity, the curve approaches 0 as it goes to infinity, yes, but the curve is actually 0 when x is -1 or -3, so it's not an asymptote per se. Unless you want to start debating about the definition of "asymptote" (stupid maths teacher...)


I_F

 

[YBK]

Haha same here, didn't have time to find the pof's, we only had 45 mins and this question was 1 of 2... the other was a hard parametrics question.

u could simply sketch f'x = 0

 

[insert-username]

He could, but that usually means finding f''(x) anyway to sketch f'(x)


I_F

 

[YBK]

When above x-axis then gradient is +ve and below gradient is -ve at that particular point.

 

[Riviet]

Haha nah, i wouldn't have had enough time for it but thx for the idea.

 

[kido_1]

who is doing integration of implicit and parametric functions?

 

[Riviet]

Definitely not me lol.

 

[pLuvia]

who is doing integration of implicit and parametric functions?

Me neither, having even touched on parametrics

Slow down kido

 

[Riviet]

For anyone that's interested:

Find the gradient of the tangent to the curve y=x²⁴-98x¹³ at x=2

[Yes i know the numbers are a little exaggerated. ]

 

[YBK]

For anyone that's interested:

Find the gradient of the tangent to the curve y=x²⁴-98x¹³ at x=2

[Yes i know the numbers are a little exaggerated. ]

haha

dy/dx=24x²³-98*13x¹²

replace x=2 into derrivative ...

and ... tada!!!

 

[Boxxxhead]

Me neither, having even touched on parametrics

Slow down kido

Did 'em for the prelims.

 

[YBK]

here, i have a harder one (harder... not too hard though )... from our end of yr MX 1 assessment

The equation x^3 + kx + 2 = 0 has 2 equal roots. Find all three roots and the value of k.

 

[Riviet]

If no one wants to do it, then i will.

I have little experience with this type of question, but i think i've got it.

Let the equal root be "a" and the other root be "b".

Then (x-a)²(x-b) = x³+kx+2

Expanding and then factorising LHS into the form px³+qx²+rx+s we obtain the equality:

x³-(b+2a)x²+(2ab+a²)x-a²b = x³+0x²+kx+2

Equating coefficients of q, r, and s (x³'s cancel), we obtain 3 equations:

-(b+2a) = 0 *

2ab+a² = k **

-a²b = 2 ***

From *,
b = -2a
sub -> ***
=> 2a³ = 2
a³ = 1
a = 1

.: b = -2

.: k = 2ab+a²
= -4+1
= -3

Therfore, the equal root is 1, the other root is -2, and k = -3

 

[Riviet]

Next question:

Find

/
| x(x²+1)³ dx
/

 

[YBK]

Next question:

Find

/
| x(x²+1)³ dx
/

what do the /'s mean?

 

[YBK]

If no one wants to do it, then i will.

I have little experience with this type of question, but i think i've got it.

Let the equal root be "a" and the other root be "b".

Then (x-a)²(x-b) = x³+kx+2

Expanding and then factorising LHS into the form px³+qx²+rx+s we obtain the equality:

x³-(b+2a)x²+(2ab+a²)x-a²b = x³+0x²+kx+2

Equating coefficients of q, r, and s (x³'s cancel), we obtain 3 equations:

-(b+2a) = 0 *

2ab+a² = k **

-a²b = 2 ***

From *,
b = -2a
sub -> ***
=> 2a³ = 2
a³ = 1
a = 1

.: b = -2

.: k = 2ab+a²
= -4+1
= -3

Therfore, the equal root is 1, the other root is -2, and k = -3

Correct! btw, in my exam, i did it in a different way to yours but it was still right.

What I did was find the sum of roots, and product of roots. Made one of them an equation in alpha and then substituted in the other...

I think it's shorter than your method since there really is no need to expand and factorise...

 

[insert-username]

what do the /'s mean?

Integral sign. By the way, the Windows Character Map (Start Menu, All Programs, Accessories, System Tools) has a nice integral sign - ∫


I_F

 

[YBK]

what do the /'s mean?

Integral sign. By the way, the Windows Character Map (Start Menu, All Programs, Accessories, System Tools) has a nice integral sign - ∫


I_F

ahh, cool thanks.

We haven't done integration yet .

 

[HamuTarou]

if u expand it....

∫x(x^2+1)^3 dx

=∫x(x^6+3x^4+3x^2+1)dx

=∫(x^7+3x^5+3x^3+x)dx

=1/8(x^8)+3/6(x^6)+3/4(x^4)+1/2(x^2)+c

=1/8(x^8)+1/2(x^6)+3/4(x^4)+1/2(x^2)+c

i think there is a quicker and more easier method...but this is one way =P