Prelim 2016 Maths Help Thread (1 Viewer)

[InteGrand]

would u then just sub b as (3a+3b) and c as 9ab?
to get x^2-(3a+3b)+9ab? or do we need numerical results?

 

[Green Yoda]

I see,
x^2-12x+18?

 

[Green Yoda]

I made a silly mistake in making 3a*3b = 3ab fml but I was on the right track.

 

[Orwell]

 

[InteGrand]

For the first one, write 4^x as (2^2)^x (as 4 = 2^2) and use index laws.

For the second one, write 2^(2x +1) as 2*(2^(2x)), then let u = 2^x and you'll get a quadratic equation in u, which you can solve via the quadratic formula (or whatever method you like). Once you find the values of u, the values of x are found by recalling that u was 2^x, so x = log₂ u.

 

[Orwell]

Just sat my exam, confident in my receiving of 95%+

Thanks to InteGrand, Rathin, Paradoxica, Drongoski, Jathu and others.

 

[jathu123]

Just sat my exam, confident in my receiving of 95%+

Thanks to InteGrand, Rathin, Paradoxica, Drongoski, Jathu and others.

woah congratulations mate! get that top 5 ranking haha

 

[eyeseeyou]

How do you do this question via differentiation first principles:

Find the derivative of y=x^3

Whenever I do it, I keep getting confused

 

[InteGrand]

How do you do this question via differentiation first principles:

Find the derivative of y=x^3

Whenever I do it, I keep getting confused

 

[pikachu975]

How do you do this question via differentiation first principles:

Find the derivative of y=x^3

Whenever I do it, I keep getting confused

f(x) = x^3
f(x+h) = (x+h)^3
= x^3 + 3hx^2 + 3xh^2 + h^3
so basically:
lim x->0 (x^3+3hx^2+3xh^2 + h^3 - x^3) / h
= lim x->0 ( 3hx^2 + 3xh^2 + h^3 ) / h
= lim x->0 3x^2 + 3xh + h^2
sub 0 into h
= 3x^2

EDIT: h-> 0 not x-> 0 for the limit

 

[eyeseeyou]

f(x) = x^3
f(x+h) = (x+h)^3
= x^3 + 3hx^2 + 3xh^2 + h^3
so basically:
lim x->0 (x^3+3hx^2+3xh^2 + h^3 - x^3) / h
= lim x->0 ( 3hx^2 + 3xh^2 + h^3 ) / h
= lim x->0 3x^2 + 3xh + h^2
sub 0 into h
= 3x^2

I'm confused on this???

 

[DatAtarLyfe]

@pikachu975 the limit is actually as h->0. I know thats what you meant since yo let h = 0 in the final line, but make sure you include it in working as its quite critical for first principles

 

[InteGrand]

I'm confused on this???

That emboldened part was the " – f(x)".

 

[eyeseeyou]

That emboldened part was the " – f(x)".

Oh right right

Thanks

 

[Green Yoda]

If (x+1)^2+(y-2)^2=25
What is the length of the intercept cut off by the circle and the x-axis.

I have no idea what to do here..so far I have found the x intercept of the circle at -1+-sqrt(21)

 

[InteGrand]

If (x+1)^2+(y-2)^2=25
What is the length of the intercept cut off by the circle and the x-axis.

I have no idea what to do here..so far I have found the x intercept of the circle at -1+-sqrt(21)

 

[Paradoxica]

If (x+1)^2+(y-2)^2=25
What is the length of the intercept cut off by the circle and the x-axis.

I have no idea what to do here..so far I have found the x intercept of the circle at -1±√21

The length is the greater ordinate minus the lesser ordinate.

that is 2√21, by inspection.

 

[Paradoxica]

is that what that means? I'm confused.

 

[Paradoxica]

If (x+1)^2+(y-2)^2=25
What is the length of the intercept cut off by the circle and the x-axis.

I have no idea what to do here..so far I have found the x intercept of the circle at -1±√21

If Integrand's interpretation is correct, then using the cosine rule on the angle at the centre,

5² + 5² - (2√21)² = 2(5)(5)cosθ

-17 = 25cosθ

cosθ = -17/25

l = 5cos⁻¹(-17/25)

 

[Green Yoda]

The length is the greater ordinate minus the lesser ordinate.

that is 2√21, by inspection.

Thats the correct answer, I got it anyways thanks