difficult y11 diff qs? (2 Viewers)

eternallyboreduser · Dec 20, 2023

Luukas.2 said:
$\begin{align*} y &= x^k + (c - x)^k \\ \frac{dy}{dx} &= kx^{k - 1} + k(c - x)^{k - 1} \times -1 = k\left[x^{k - 1} - (c - x)^{k - 1}\right] \\ \\ \text{Put $\frac{dy}{dx} = 0$:} \quad k\left[x^{k - 1} - (c - x)^{k - 1}\right] &= 0 \\ \left(\frac{x}{c - x}\right)^{k - 1} &= 1 \qquad \text{as $k \neq 0$} \\ x &= \frac{c}{2} \\ \text{Put $x = \frac{c}{2}$:} \quad y &= \left(\frac{c}{2}\right)^k +\left(c - \frac{c}{2}\right)^k = 2\left(\frac{c}{2}\right)^k \\ \\ \text{Thus, the stationary point is at} \quad &\left(\frac{c}{2},\ 2\left(\frac{c}{2}\right)^k\right) \\ \\ \frac{d^2y}{dx^2} &= k(k - 1)x^{k - 2} - k(k - 1)(c - x)^{k - 2} \times -1 \\ &= k(k - 1)\left[x^{k - 2} + (c - x)^{k - 2}\right] \\ \\ \text{Put $x = \frac{c}{2}$:} \quad \frac{d^2y}{dx^2} &= k(k - 1)\left[\left(\frac{c}{2}\right)^{k - 2} + \left(c - \frac{c}{2}\right)^{k - 2}\right] = 2k(k - 1)\left(\frac{c}{2}\right)^{k - 2} \end{align*}$

$\text{Since $c > 0$, $\left(\frac{c}{2}\right)^{k - 2} > 0$, the sign of $\cfrac{d^2y}{dx^2}$ matches the sign of $k(k - 1)$.} \\ \\ \text{Thus, if $0 < k < 1$, then $k(k - 1) < 0$ and so the stationary point at $\left(\frac{c}{2},\ 2\left(\frac{c}{2}\right)^k\right)$ is a maximum.} \\ \\ \text{And, if $k > 1$, then $k(k - 1) > 0$ and so the stationary point at $\left(\frac{c}{2},\ 2\left(\frac{c}{2}\right)^k\right)$ is a minimum.}$

$\text{Taking $c = a + b$ where $a > b > 0$ and considering the $0 < k < 1$ case, the maximum} \\ \\ \text{on $0 < x < a + b$ will be at $\left(\frac{a + b}{2},\ 2\left(\frac{a + b}{2}\right)^k\right)$. Since this is a local maximum:}$

$\begin{align*} y &< 2\left(\frac{a + b}{2}\right)^k \\ x^k + (a + b - x)^k &< 2\left(\frac{a + b}{2}\right)^k \qquad \text{as $c = a + b$} \\ \\ \text{Take $x = b$:} \quad b^k + (a + b - b)^k &< 2\left(\frac{a + b}{2}\right)^k \\ \frac{a^k + b^k}{2} &< \left(\frac{a + b}{2}\right)^k \qquad \text{as required} \end{align*}$

Im a bit

confused with this part

eternallyboreduser · Dec 20, 2023

Also @Luukas.2 do you think questions like this would appear in selective school exams or would they be too overboard? Like would qs of such difficulty appear often in hsc exams besides like the last q. This was in my year 10 hw btw for tutoring and was one of the challenging (optional) qs.

Luukas.2 · Dec 20, 2023

eternallyboreduser said:
Im a bitView attachment 41882 confused with this part

This is for the case where $0 < k < 1$ , in which case we have shown that there is a maximum turning point at coordinates

$\left(\frac{c}{2},\ 2\left(\frac{c}{2}\right)^k\right)$

There are no other stationary points on the curve as the derivative

$\frac{dy}{dx} = k\left[x^{k-1} - (c - x)^{k-1}\right]$

can only be zero if $k = 0$ (which never happens) or if

$\begin{align*} x^{k-1} - (c - x)^{k-1} &= 0 \\ x^{k-1} &= (c - x)^{k-1} \\ \frac{x^{k-1}}{(c - x)^{k-1}} &= 1 \\ \left(\frac{x}{c - x}\right)^{k-1} &= 1 \\ \frac{x}{c - x} &= \pm1 \end{align*}$

The only solution if the fraction equals +1 is

$\frac{x}{c - x} = 1 \quad \implies \quad x = c - x \quad \implies \quad 2x = c \quad \implies \quad x = \frac{c}{2}$

And -1 is only a solution if $k - 1$ is an even integer (which it can't be in this case) and even then, the equation becomes $x = x - c$ and thus has no solution unless $x = 0$ .

Thus, the function has exactly one stationary point, a maximum which is not only a local maximum, but in fact the global maximum. Hence, the function must satisfy the inequation

$\begin{align*} y &\le 2\left(\frac{c}{2}\right)^k \quad \text{for all $x$} \\ \\ \text{Hence,} \quad x^k + (c - x)^k &\le 2\left(\frac{c}{2}\right)^k \qquad \text{as $y = x^k + (c - x)^k$} \\ \\ \text{Let $c = a + b$:} \qquad x^k + (a + b - x)^k &\le 2\left(\frac{a + b}{2}\right)^k \qquad \text{is true for all $x$ so long as $c > 0$} \\ \\ \text{So, when $x = b$:} \qquad b^k + (a + b - b)^k &\le 2\left(\frac{a + b}{2}\right)^k \\ b^k + a^k &\le 2\left(\frac{a + b}{2}\right)^k \\ \frac{a^k + b^k}{2} &\le \left(\frac{a + b}{2}\right)^k \end{align*}$

Luukas.2 · Dec 20, 2023

eternallyboreduser said:
Also @Luukas.2 do you think questions like this would appear in selective school exams or would they be too overboard? Like would qs of such difficulty appear often in hsc exams besides like the last q. This was in my year 10 hw btw for tutoring and was one of the challenging (optional) qs.

I have seen this exact result in an MX2 assignment, though it was meant to be done by induction - this is actually a nicer approach.

A question like this is certainly possible in an assessment task for a school with a strong candidature. Remember that exams are meant to facilitate ranking the students who take them from first to last, so a question like this isn't helpful if no one will get anywhere with it. Proving the stationary point is a possible (challenging) Advanced question, though I doubt the last part would appear below MX1. Remember also that assessments are controlled by the schools and so can wander outside what might be reasonable in an HSC exam.

For a year 10 student, given that differential calculus is next year and the reasoning from graphical interpretations is novel, this is a very challenging question, IMO.

eternallyboreduser · Dec 20, 2023

Luukas.2 said:
This is for the case where $0 < k < 1$ , in which case we have shown that there is a maximum turning point at coordinates

$\left(\frac{c}{2},\ 2\left(\frac{c}{2}\right)^k\right)$

There are no other stationary points on the curve as the derivative

$\frac{dy}{dx} = k\left[x^{k-1} - (c - x)^{k-1}\right]$

can only be zero if $k = 0$ (which never happens) or if

$\begin{align*} x^{k-1} - (c - x)^{k-1} &= 0 \\ x^{k-1} &= (c - x)^{k-1} \\ \frac{x^{k-1}}{(c - x)^{k-1}} &= 1 \\ \left(\frac{x}{c - x}\right)^{k-1} &= 1 \\ \frac{x}{c - x} &= \pm1 \end{align*}$

The only solution if the fraction equals +1 is

$\frac{x}{c - x} = 1 \quad \implies \quad x = c - x \quad \implies \quad 2x = c \quad \implies \quad x = \frac{c}{2}$

And -1 is only a solution if $k - 1$ is an even integer (which it can't be in this case) and even then, the equation becomes $x = x - c$ and thus has no solution unless $x = 0$ .

Thus, the function has exactly one stationary point, a maximum which is not only a local maximum, but in fact the global maximum. Hence, the function must satisfy the inequation

$\begin{align*} y &\le 2\left(\frac{c}{2}\right)^k \quad \text{for all $x$} \\ \\ \text{Hence,} \quad x^k + (c - x)^k &\le 2\left(\frac{c}{2}\right)^k \qquad \text{as $y = x^k + (c - x)^k$} \\ \\ \text{Let $c = a + b$:} \qquad x^k + (a + b - x)^k &\le 2\left(\frac{a + b}{2}\right)^k \qquad \text{is true for all $x$ so long as $c > 0$} \\ \\ \text{So, when $x = b$:} \qquad b^k + (a + b - b)^k &\le 2\left(\frac{a + b}{2}\right)^k \\ b^k + a^k &\le 2\left(\frac{a + b}{2}\right)^k \\ \frac{a^k + b^k}{2} &\le \left(\frac{a + b}{2}\right)^k \end{align*}$

Yeah ik how to get to x=c/2 since i could solve part a but do u mind explaining how u arrived at y<2(c/2)^k

Luukas.2 · Dec 20, 2023

eternallyboreduser said:
Yeah ik how to get to x=c/2 since i could solve part a but do u mind explaining how u arrived at y<2(c/2)^k

I was showing not only that x = c/2 is a stationary point, but also that there are no other stationary points.

Draw a set of coordinate axes. Mark on them the max at $\left(\frac{c}{2},\ 2\left(\frac{c}{2}\right)^k\right)$ . Now draw a continuous curve that has that point as a maximum and has no other stationary points... So, it must increase for $x < \frac{c}{2}$ and decrease for $x > \frac{c}{2}$ (consistent with the domain, which may be $0 \le x \le c$ .

Look at the picture you've drawn. Is there any possible curve that has those properties and for which the turning point is not also the global maximum (i.e. the point on the curve with the largest y value)? Hence, for all points on the curve, the y value must be less than or equal to the y value of the stationary point.

Does this make sense?

Or, load up desmos, put in (say) $y = x^k + (10 - x)^k$ and try different values of $k \in (0,\ 1)$ . All will have a maximum at $x = 5$ , and many will be restricted to the domain $0 \le x \le 10$ , though for values like $k = 1/3$ the domain will be all reals. In every case, the height of the curve (y value) will always be at or below the value of the maximum turning point.

eternallyboreduser · Dec 20, 2023

Luukas.2 said:
I was showing not only that x = c/2 is a stationary point, but also that there are no other stationary points.

Draw a set of coordinate axes. Mark on them the max at $\left(\frac{c}{2},\ 2\left(\frac{c}{2}\right)^k\right)$ . Now draw a continuous curve that has that point as a maximum and has no other stationary points... So, it must increase for $x < \frac{c}{2}$ and decrease for $x > \frac{c}{2}$ (consistent with the domain, which may be $0 \le x \le c$ .

Look at the picture you've drawn. Is there any possible curve that has those properties and for which the turning point is not also the global maximum (i.e. the point on the curve with the largest y value)? Hence, for all points on the curve, the y value must be less than or equal to the y value of the stationary point.

Does this make sense?

Or, load up desmos, put in (say) $y = x^k + (10 - x)^k$ and try different values of $k \in (0,\ 1)$ . All will have a maximum at $x = 5$ , and many will be restricted to the domain $0 \le x \le 10$ , though for values like $k = 1/3$ the domain will be all reals. In every case, the height of the curve (y value) will always be at or below the value of the maximum turning point.

OHHHH thats what you were trying to say okay it makes sense now

Luukas.2 · Dec 20, 2023

eternallyboreduser said:
OHHHH thats what you were trying to say okay it makes sense now

Yes, as I said, it's a form of reasoning that you might not be very familiar with yet, but you will use it from time to time going forwards

And it is needed for most of the questions you posted, FYI.

eternallyboreduser · Dec 20, 2023

Luukas.2 said:
Yes, as I said, it's a form of reasoning that you might not be very familiar with yet, but you will use it from time to time going forwards

And it is needed for most of the questions you posted, FYI.

Thank you for the help !!

Luukas.2 · Dec 20, 2023

Another question like this would be:

$\text{Use calculus to prove that, for any $a \in \mathbb{R}^+$, $a + \frac{1}{a} \geqslant 2$.}$

eternallyboreduser · Dec 20, 2023

Luukas.2 said:
Another question like this would be:

$\text{Use calculus to prove that, for any $a \in \mathbb{R}^+$, $a + \frac{1}{a} \geqslant 2$.}$

You would just find the y value and nature of the stationary point of y= a + 1/a right for the given domain

Luukas.2 · Dec 20, 2023

eternallyboreduser said:
You would just find the y value and nature of the stationary point of y= a + 1/a right for the given domain

Pretty much, though this uses the reasoning mentioned above and would actually be at Advanced level. This is a result that appears at the start of plenty of MX2 questions, and which can be done without calculus... so I think it is useful to know. It can also be extended into other problems, like the minimum of $a^n + a^{-n}$ , or finding $a^5 + a^{-5}$ if $a + a^{-1}$ is given.

Luukas.2 · Dec 21, 2023

Once you've got calculus of exponentials, you can explore ideas like:

$x^0 = 1$
$0^x = 0$

$\text{So, $0^0 = 0$ or $1$ or $\frac{1}{2}$ (on average)?}$

Or, what happens to a function like $x.2^x$ as $x$ becomes increasingly negative, or as $x\ln{x}$ gets close to zero...

difficult y11 diff qs? (2 Viewers)

eternallyboreduser

Well-Known Member

eternallyboreduser

Well-Known Member

Luukas.2

Well-Known Member

Luukas.2

Well-Known Member

eternallyboreduser

Well-Known Member

Luukas.2

Well-Known Member

eternallyboreduser

Well-Known Member

Luukas.2

Well-Known Member

eternallyboreduser

Well-Known Member

Luukas.2

Well-Known Member

eternallyboreduser

Well-Known Member

Luukas.2

Well-Known Member

Luukas.2

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 2)