WEBVTT

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Hi, my name's Richard.

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Here we have TMUA 2021 paper two,
question 8.

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Let's read it through.

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So it says,
consider the following statement about

57bbe934-cc0a-433f-83f4-3fef1985c99b-1
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the polynomial P of X where A&
B are real numbers with a less than B.

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So the statement star is that there
exists a number C in between A&B,

fbc75d8b-0804-4369-b1e4-c0cc46436396-1
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so a less than C less than B such that
the derivative of P at C = 0.

fa8152d5-da00-4638-9c44-d3489d6f214a-0
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So we're being asked to consider these
statements.

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Which of them is true?

56cce66e-6eb2-4b30-91c6-fac3c5b41faa-0
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They all involve this condition that P of
A = P of B.

25553ec1-1889-477f-a7f1-6196bfcd70da-0
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So what we're investigating here is the
relationship logically between the

25553ec1-1889-477f-a7f1-6196bfcd70da-1
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statement that P of A = P of B and the
statement that there is a Point C in

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between A&B where P dash of C = 0.

8626270e-720f-4195-9595-8bac2f9fb394-0
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We have to decide whether one is
necessary and sufficient for the other,

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and necessary not sufficient.

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We've got all four possibilities of
combinations of necessary and sufficient

cbe6ed63-cc3c-4a33-ab05-e135cb5271df-1
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to consider.

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OK,
so let's start by considering whether P

9a46736e-270c-430b-b377-fb7b9bce6e77-1
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of A = P of B implies this statement star.

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So I'm going to write that down.

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So we've got P of A = P of B and we are
asking the question whether this implies

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the statement star,
which is that there is this Point C in

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between A&
B where the derivative of P at C = 0.

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So when we are considering this,
we are asking the question is P of A

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equal P of B sufficient for star when we
are considering the implication in this

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direction,
so asking whether this is true or not.

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OK,
so we have to wonder whether P of A

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equaling P of B will guarantee the
existence of this C where the derivative

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is zero.

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OK,
let's imagine what our graph might look

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like.

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Let's put A&B here.

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We have a less than B.

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We are starting by imagining that P of A
= P of B.

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So what that means is that our graph,
our polynomial, agrees our A&B,

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so P of A = P of B.

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So they would be at the same height on
the graph, these two points above A&

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B respectively.

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OK,
so let's wonder what our graph might look

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like.

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It has to pass through here and here.

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If we start at A,
we can imagine that if our graph starts

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to go up,
it's going to have to come back down

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again in order to go through this point.

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And that would create a Point C where the
gradient was 0.

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Equally, if it started to go down,
it's going to have to come back up.

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And again,
that would create a point at least one

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where the gradient was zero.

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If our graph neither goes down nor up the
whole time,

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it must be constant and and we can see
that all points in between A&

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B would satisfy P dash of C = 0.

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So this is true if P of A = P of B.

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It is forced that there is a value in
between A&B where the gradient is 0.

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So I think the answer to this question is
yes, this is true,

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and that means that this is sufficient
for this.

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OK,
so we'll now ask the question whether P

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of A = P of B is necessary for star.

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What this means is we are asking the
question whether star implies that P of A

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= P of B.

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So this is what we mean when we say that
P of A = P of B is necessary for star.

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So we are asking whether star implies
that P of A = P of B.

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So let's imagine star being true.

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This means that we have a polynomial.

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We have an A and AB.

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We've got AC between them,
where the gradient is 0 the the gradient

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of that tangent is 0.

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So we might imagine that in a picture,
perhaps A&B are here,

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perhaps C is here,
and what we're being told is to imagine

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that the gradient of our polynomial is 0
at C,

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so it's got a horizontal tangent there.

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Does this force P of A to be equal to P
of B?

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Does this force P of A to be equal to P
of B?

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Well,
I don't think it does because I can

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imagine drawing a picture here.

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Perhaps I can extend this curve to here
so that we've got P of A taking this

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value.

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And then I'll extend this down to here so
that we've got P of B taking this value.

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And you can see that P of A is not equal
to P of B.

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So the the fact that our function has a
Point C where it's graded to 0 does not

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force P of A to be equal to P of BI.

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Can see this from an example like this.

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So the answer to this question is no,
star does not imply that P of A = P of B.

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And this means that P of A = P of B is
not necessary for star.

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So in summary,
P of A = P of B is sufficient for star

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but not necessary for star.

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So we need to look for which of our
options expresses that we can see that

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it's C so we can go over and put a tip
next to C and that's the solution to our

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problem.