HSC Mathematics Marathon (7 Viewers)

[jyu]

 

[deterministic]

Well, technically it is. But still it's boring.

Only because the question asked for induction. Its really just a repeated application of triangle's inequality.

 

[jyu]

Not worry about 'hence'


At
for

 

[deterministic]

"Hence" makes it easier



Note that the integration step only works for x>=0

 

[jyu]

The locus of point P(x,y) is (x/2)^2+(y/b)^2=1, b>0. Find, in terms of b, the maximum value of x^2+2y.

 

[Trebla]

Not worry about 'hence'


At
for

There is a flaw in this approach. 2^x > 2x is not true when 1 < x < 2

 

[jyu]

There is a flaw in this approach. 2^x > 2x is not true when 1 < x < 2

Thanks for that.
In that case, just drop 2^x in the inequality.

To justify e^x >2x: x=0, 1>0. d/dx of e^x = d/dx of 2x when x=ln2. When x=ln2, 2>2ln2. Since both are increasing, .: e^x>2x in [0,ln2]
For x>ln2, d/dx of e^x > d/dx of 2x, .: e^x>2x.

 

[deterministic]

To justify e^x >2x: x=0, 1>0. d/dx of e^x = d/dx of 2x when x=ln2. When x=ln2, 2>2ln2. Since both are increasing, .: e^x>2x in [0,ln2]
For x>ln2, d/dx of e^x > d/dx of 2x, .: e^x>2x.

The result is correct, but this reasoning is rather incomplete.

 

[deterministic]

Try this approach:

Let f(x)=e^x-2x and show that f(x)>0 using simple calculus.

 

[jyu]

The result is correct, but this reasoning is rather incomplete.

Why is it incomplete?

 

[jyu]

Try this approach:

Let f(x)=e^x-2x and show that f(x)>0 using simple calculus.

How do you show that?

 

[deterministic]

Why is it incomplete?

It is not immediate that this:
"d/dx of e^x = d/dx of 2x when x=ln2.
When x=ln2, 2>2ln2."

justifies this:
"Since both are increasing, .: e^x>2x in [0,ln2]"

 

[deterministic]

How do you show that?

Similar to what you have tried to do but clearer. Just show the minimum of f(x) takes a positive value.

 

[jyu]

It is not immediate that this:
"d/dx of e^x = d/dx of 2x when x=ln2.
When x=ln2, 2>2ln2."

justifies this:
"Since both are increasing, .: e^x>2x in [0,ln2]"

I thought I did that by pointing out at x=0, e^0>2(0).

 

[deterministic]

The fact the relationship between the derivatives of the 2 functions don't change as x moves from 0 to ln2 is actually pretty vital in your argument.

 

[jyu]

The fact the relationship between the derivatives of the 2 functions don't change as x moves from 0 to ln2 is actually pretty vital in your argument.

The relationship does change. At x=0, d(e^x)/dx < d(2x)/dx. At x=ln2, they are the same.
So I pointed out x=0, e^0>2(0) AND x=ln2, 2>2ln2.

 

[deterministic]

The relationship does change. At x=0, d(e^x)/dx < d(2x)/dx. At x=ln2, they are the same.
So I pointed out x=0, e^0>2(0) AND x=ln2, 2>2ln2.

I mean in between 0 and ln2 (not inclusive of ln2), d(e^x)/dx < d(2x)/dx which is why your logic still works.

In general, the existence of changes in the relationship of derivatives of 2 functions can mean while they are still increasing functions, it is entirely possible for relationship between 2 functions to change as well, while still satisfying the other facts you have stated. Try constructing piecewise functions (still continuous though) which demonstrates this.

 

[jyu]

Thanks.

The relationship does change for x>=0. For x>ln2, d(e^x)/dx > d(2x)/dx. So I looked at each interval.

 

[deterministic]

The locus of point P(x,y) is (x/2)^2+(y/b)^2=1, b>0. Find, in terms of b, the maximum value of x^2+2y.

New question

 

[hscishard]

Is it b^2 / 4 by any chance?