~~Mathematics Revising Game~~ (1 Viewer)

[veloc1ty]

Someone put up a new question.

(Do you just make them up yourself or pull them from obscure sources?)

 

[xclo]

given pi is irrational, prove pi^-1 + pi^-2 + pi^-3 + ... is irrational

 

[Aerath]

 

[Graceofgod]

given pi is irrational, prove pi^-1 + pi^-2 + pi^-3 + ... is irrational

a = 1/pi, r = 1/pi

S(infinity) = (1/pi)/(1-1/pi)
S(infinity) = (1/pi)/[(pi-1)/pi]
S(infinity) = 1/(pi-1)

And that is where I give up

 

[Js^-1]

That's a nice question. What is the limiting sum to:

1/2-1/6+1/18-...

 

[Cathy Sander]

1/2-1/6+1/18...=1/2[1-1/3+1/9-1/27...]

Limiting sum: a=1, r=-1/3

Hence:
S(infinity)=1/[1-(-1/3)]
S(infinity)=3/4

Therefore, limiting sum is:
1/2*3/4=3/8

[Q.E.D? Nah...]

 

[charlesdinio]

(i) 0.15ⁿ
(ii) 3 seedlings

Next question:
The region by the curve y = ln x, the co-ordinate axes, and the line y = ln 2 is rotated about the y-axis. Find the volume of the solid so formed.

Please explain 3 seedlings bit..

Not sure why im not getting it!

 

[Hatta]

Can I join?
As no one's posted a new question, I'll put an easy one

Find the equation of the normal to the curve y=e^(-x) at the point (1, e^(-1))

 

[Js^-1]

y = e^-x
dy/dx = - e^-x
m_t = - e^-1
.: m_n = e
y - 1/e = e (x-1)
y= ex - (e²+1)/e

 

[Hatta]

y = e^-x
dy/dx = - e^-x
m_t = - e^-1
.: m_n = e
y - 1/e = e (x-1)
y= ex - (e²+1)/e

Correct, but it can be simplified.

y= ex- ((e^2/e)+(1/e))
= ex-e+(1/e)

 

[Trebla]

Find the values of a and b such that the integral I = ∫_a^b(2 + x – x²) dx is a maximum. Justify your answer.

 

[Timothy.Siu]

b=2 and a=-1?

umm
differentiate: x^3sin(cos(sin x^2)

 

[Hatta]

b=2 and a=-1?

umm
differentiate: x^3sin(cos(sin x^2)

This is as far as I got before admitting defeat:
y'= 3sin.(cos.2cos(x^2)+sin(x^2).-sin)+((cos.sin(x^2)).-3cos).x^((3sin.cos.sin(x^2))-1)

=3sin.(2(cos^2)(x^2)-(sin^2)(x^2)+(cos.sin(x^2).-3cos).x^((3sin.cos.sin(x^2))-1)

= extra simplifying steps that I'm too lazy to type up

=((x^2)(2(cos^2)-(sin^2))-3(cos^2).(sin^2).3sinx^((3sin.cos.sin(x^2))-1)

:rofl: too much english study, my 75% from the maths trial has just vanished.

Also, badly done paint image of what the last line is meant to look like.


I just looked at it again, I don't think the -1 needs to be at the end of the x power... help?

EDIT: *facepalm*
Did it again and got y'=6(cos^2)x.-sinx^(3(sin^2)(x^2).cos)

 

[Timothy.Siu]

lol i dont know the answer , i just made it up and too lazy to do it

 

[Trebla]

b=2 and a=-1?

Justify your answer.

 

[Timothy.Siu]

because 2+x-x^2 is the derivative of the function and to hav the highest value u wud want b to have the highest value which is when b=2 (max turning point) and a= -1 (minimum)
basically...doesn't really make sense but yeah

 

[Js^-1]

Find the values of k for which:
y = x² + 2kx + (k+2) = 0
has two real, distinct roots.

 

[-::Sanni::-]

Find the values of k for which:
y = x<SUP>2</SUP> + 2kx + (k+2) = 0
has two real, distinct roots.

I got k=2 and k=-1

Discriminant formula --> inserted info into formula --> whatever it is equals zero as it has real roots --> and solved it!

I think that's it... um..I don't have any questions at the moment. I'll post one up when I find one!

 

[Timothy.Siu]

I got k=2 and k=-1

Discriminant formula --> inserted info into formula --> whatever it is equals zero as it has real roots --> and solved it!

I think that's it... um..I don't have any questions at the moment. I'll post one up when I find one!

wrong, thats one double root....

 

[Js^-1]

I got k=2 and k=-1

Discriminant formula --> inserted info into formula --> whatever it is equals zero as it has real roots --> and solved it!

I think that's it... um..I don't have any questions at the moment. I'll post one up when I find one!

Not quite correct...You're on the way though.

Discriminant = 0 for a double root.
Discriminant > 0 for two distinct roots
Discriminant < 0 for no real roots.