Mathematics Challenge Marathon

[RuiAce]

For Q.a did you mean

a^x ln(x) ?

Maybe I'm just doing something wrong?
(Image removed from quote.)

Hint: Whilst it's doable that way, that isn't the best way to apply the identity given

[georgiia]

OMG IT HAS SURFACED!!!

Sorry I'm very messy 

[RuiAce]

Spoiler

Find a contradiction!

[RuiAce]

Hint

You may like to use a graph.

[georgiia]

(Image removed from quote.)

Sorry I'm very messy 

What have I done wrong?

[RuiAce]

Lol, sorry, that was actually my fault. I'll fix that up now.

[RuiAce]

[georgiia]

1.  Max. value at y=1

2. There is no min. as f(x) is a bell curve (i think?)
∴ as x → +∞, y → 0
ans as  x → -∞, y → 0

[RuiAce]

1.  Max. value at y=1

2. There is no min. as f(x) is a bell curve (i think?)
∴ as x → +∞, y → 0
ans as  x → -∞, y → 0

Right idea with the bell curve (it isn't actually one, it just looks like it; a bell curve would be \( y=e^{-\frac12x^2}\) but it's better to make it explicit that there's an asymptote going on.

[RuiAce]

This isn't a computational question but rather just testing your understanding of terminology. Because a lot of 2U students get confused over this.

Explain the difference between a sequence and a series.

[RuiAce]

[Opengangs]

1. From first principles, we get:
lim(h -> 0) (|0 + h| + |0|)/h, which can be broken into two smaller cases: when h tends towards 0+ and 0-
From the right side (0+), we get lim(h -> 0) (h/h) = 1, since h > 0
From the left side (0-),  we get lim(h -> 0) (-h/h) = -1, since h < 0

Since lim(h -> 0+) is not the same as lim(h -> 0-), at the point x = 0, it is not differentiable.

2. Explaining why it's not differentiable at the point, x = 0.
As f(x) tends towards 0 from the positive side, its differential is incrementally tending towards 1, while from the negative side (0-), it is incrementally tending towards -1. This means that a differential cannot exist at the point x = 0, as the change in x (delta x) is not defined at that particular point.

3.

[pikachu975]

1) f(x+h) = |x+h|
f(x) = |x|
f'(x) = lim(h->0) [|x+h| - |x|]/h
at x = 0, f'(x) = lim(h->0) h/h and 0/0 is undefined.

2) There is no tangent at the corner part of the absolute value function hence no gradient/derivative at x = 0.

3) For x>0, f'(x) = 1, for x<0, f'(x) = -1, which can be generalised as f'(x) = x/|x|

[RuiAce]

1) f(x+h) = |x+h|
f(x) = |x|
f'(x) = lim(h->0) [|x+h| - |x|]/h
at x = 0, f'(x) = lim(h->0) h/h and 0/0 is undefined.

2) There is no tangent at the corner part of the absolute value function hence no gradient/derivative at x = 0.

3) For x>0, f'(x) = 1, for x<0, f'(x) = -1, which can be generalised as f'(x) = x/|x|

Just be careful of this. Everything was right except for this part, which Opengangs correctly answered.

If this were the case, because I'm taking a limit I'm allowed to cancel out the h's to just get 1. That would imply f'(0)=1, which is definitely not what we wanted to see.

The problem was in that when x=0, you should've left the numerator as |h|, not take off the absolute values without breaking the cases.

[RuiAce]

(Note: Stddev^2 = Var)

_______________________________