WEBVTT

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Hi, my name's Richard.

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We're looking at Tamua 2021 Paper two,
question 17.

05668523-2525-48cf-9191-d088917b0c86-0
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Consider the following functions defined
for X bigger than 1F of X is log base 2

05668523-2525-48cf-9191-d088917b0c86-1
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of log base two of the square root of
X&

05668523-2525-48cf-9191-d088917b0c86-2
00:00:19.446 --> 00:00:23.800
G of X is log base two of the square root
of the square of X.

26913e14-2473-48df-afa5-5d3efec94d9b-0
00:00:24.160 --> 00:00:28.800
We're asked which of the following is
true for all values of X greater than one.

9b46cc73-8cae-46ea-83f9-b6b6fab96ec8-0
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So you can see the options here A to F
They all involve the relationship between

9b46cc73-8cae-46ea-83f9-b6b6fab96ec8-1
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F&G.

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Is F less than or equal to G?

777bbc53-9e0d-4a13-b06c-d878cc753490-0
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Is G less than or equal to F?

f350b96b-c164-42db-9fe1-d32a225ffa7c-0
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And also their relation to some
particular constants like 0, a, half,

f350b96b-c164-42db-9fe1-d32a225ffa7c-1
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and 1.

b96d5d88-5f66-40f3-9e40-66496f8e8c53-0
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So let's move these to the side for now.

e00dd558-a345-4dcd-8dae-b0ce627c293f-0
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So the approach we're going to take to
this is to think about what the graphs of

e00dd558-a345-4dcd-8dae-b0ce627c293f-1
00:00:54.486 --> 00:00:56.320
F of X&G of X might look like.

7eb66e1e-670b-4c1c-8c30-5660257c7992-0
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Where do they meet?

752e50ce-43ff-4c52-a2ba-a31a247bc447-0
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Where might F be bigger than G?

b03bb29b-150f-4e0f-9571-fabfb0eb2e9e-0
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Where might G be less than FA?

51e1fbdf-78cd-4283-bc34-e28e61dbf561-0
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Couple of things to note.

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We're told that X is bigger than one here.

bd07a3a3-ce89-4a3d-8978-0bb12aa6fe0f-0
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This is really important because these
functions will only be defined for X

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bigger than one.

8d123f1b-0446-486d-8690-786d12b59c62-0
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For example, if X is bigger than one,
then the square root of X is bigger than

8d123f1b-0446-486d-8690-786d12b59c62-1
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one,
which means that log base two of the

8d123f1b-0446-486d-8690-786d12b59c62-2
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square root of X is bigger than 0.

ee1dff01-4ec9-4647-8385-80cd5227dcc6-0
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So we can then feed that into log base 2
again.

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Another thing to notice about these two
functions is that they are both

e97196ce-4a8d-4cdd-ad17-b9a12159d677-1
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increasing.

8f4bf165-84d2-4bb8-b4c1-3e5eede5b0a8-0
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If we consider F of X,
we start with the square root of X which

8f4bf165-84d2-4bb8-b4c1-3e5eede5b0a8-1
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is itself an increasing function.

b012c0b5-838d-4705-abd4-ace2c45549ef-0
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We apply log base 2 to that which is
another increase increasing function,

b012c0b5-838d-4705-abd4-ace2c45549ef-1
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and then we apply log base 2 to that
another increasing function.

accb52d6-7cf3-435a-8f62-9ce6c7b975cd-0
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Because all those functions were
increasing,

accb52d6-7cf3-435a-8f62-9ce6c7b975cd-1
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F will be an increasing function.

64fbd030-8678-40a9-97fc-3475a62625cc-0
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Similarly with G,
we start with log base 2 of X,

64fbd030-8678-40a9-97fc-3475a62625cc-1
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which is an increasing function.

1cddfd7a-6a4b-44ff-81af-176e0a1c1509-0
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We then do the square root function to
that which itself is increasing and

1cddfd7a-6a4b-44ff-81af-176e0a1c1509-1
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follow up with log base 2, again,
another increasing function.

d9888a95-a358-4f85-90ca-9e0750dd924a-0
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So we know the graphs of F&
G will increase,

d9888a95-a358-4f85-90ca-9e0750dd924a-1
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they'll go up as X increases.

a0d1df39-1ed0-4213-b3df-7fe3df752f04-0
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So that's two important things to note
before we start.

47552aeb-330a-49db-aa8b-adff389e37c9-0
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So the first thing we're going to
consider is,

47552aeb-330a-49db-aa8b-adff389e37c9-1
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are there any values of X for which F of
X = g of X?

6e7534f6-40db-4419-a10d-ebb3b07a2a6e-0
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So I can just use the expressions for F
of X&G of X.

4a79560c-0645-4458-95e7-79c7ab4dfda7-0
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And I can say this happens if and only if
log base 2 of log base two of the square

4a79560c-0645-4458-95e7-79c7ab4dfda7-1
00:02:24.370 --> 00:02:30.360
root of X is the same as log base two of
the square root of X.

63e12e79-4515-4b12-a14c-4c4b6dd84e1c-0
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I can say that this will happen if and
only if these two values are the same.

a6417254-ebde-45a5-b389-b93f2a0246f4-0
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There are lots of different ways of
thinking about that.

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One of them is you.

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Imagine taking 2 to the power of this and
2 to the power of this.

75bc8076-7818-4e1a-bf68-80404ca37a33-0
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This will have the effect of effectively
cancelling this logbase 2 of X.

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Now on the right hand side we have log
base 2 of X.

b7c55f28-adf7-4f98-bdbb-827276c9528f-0
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On the left hand side we have log base
two of the square root of X.

d82ccdcc-32dc-4cbf-b954-cad4697f9509-0
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But using the familiar property of the
logarithm function,

d82ccdcc-32dc-4cbf-b954-cad4697f9509-1
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we can write this in terms of log base 2
of X.

37ef1e55-557a-4df4-8e40-e44a663f7851-0
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So imagine this as log base 2 of X to the
half.

1bb2707c-025c-42b2-97f2-1f21647bafc4-0
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We can bring that half to the front and
we can say that this is the same as.

a0e0ad88-2aa6-4aea-b5eb-94f8b9451672-0
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The next thing I'm going to do is square
both sides of this.

db3a739d-cfc8-4070-976a-d65166948a70-0
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I can actually include this as an if and
only if step because I know that I'm

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squaring 2 positive values here.

174a2fba-36b1-4e7f-9adb-e9246b5e2d3d-0
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And next I can use the fact that since X
is bigger than one log base,

174a2fba-36b1-4e7f-9adb-e9246b5e2d3d-1
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2 of X is not equal to 0,
so I can cancel a log base 2 of X from

174a2fba-36b1-4e7f-9adb-e9246b5e2d3d-2
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both sides of this equation.

ca849755-155b-439b-8499-72de46e3e7fb-0
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So at this point I will have 1/4 of log
base 2 of X = 1.

a1770dcc-b787-4bc1-b98c-e2abef2559d6-0
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So this is the same as saying that log
base 2 of X = 4.

40ba8c63-4b7b-420e-b8c0-f2603ff06df3-0
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And from this I can see that X must be 16.

8d3e2b3a-aeea-4922-b134-cd79e172c700-0
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So log base 2 of 16 is equal to 4.

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So this is if and only if X = 16.

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So I've managed to show that the only
value of X bigger than one where F of X =

841c2131-1dba-4c60-a4df-39c35a1bee7a-1
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g of X is when X = 16.

c78a8263-7505-4d3d-bde8-f159062665fd-0
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I want to include this point on my graph.

c1ed9c90-a3e8-45e0-b20c-58c04d1ffc70-0
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So now I'm going to evaluate the common
value F of 16 and G of 16.

eb57b372-b3f1-4914-bc69-9e3a4b49ef2d-0
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So let's consider F of 16.

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So we imagine this X is 16.

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We take its square root, we get 4.

61ddb62b-5d27-4c77-813b-0cef50287df0-0
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Log base two of four is the #2,
and then log base two of two is the

61ddb62b-5d27-4c77-813b-0cef50287df0-1
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number one.

4529ea55-839c-477b-aec5-15d17eb98885-0
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So we can see that F of 16 = 1.

4553abf8-4b4e-45ef-aa57-81fb0e079e67-0
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And we know because of our work here that
G of 16 will also be 1.

39d840a3-ca98-4309-b943-39dccccd2b3c-0
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OK,
so now we can start to think about a

39d840a3-ca98-4309-b943-39dccccd2b3c-1
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diagram which includes the graph of F of
X and the graph of G of X on the same

39d840a3-ca98-4309-b943-39dccccd2b3c-2
00:05:04.016 --> 00:05:04.280
axis.

49101068-edda-4452-84c5-572280bef511-0
00:05:05.120 --> 00:05:10.960
We've calculated that at 16 the two
functions agree and they take the value 1.

35b7e564-c63e-489a-b20f-a5af4243fca7-0
00:05:11.520 --> 00:05:16.139
So this point here is 16, one,
and that point is on both the graph of F

35b7e564-c63e-489a-b20f-a5af4243fca7-1
00:05:16.139 --> 00:05:18.000
of X and the graph of G of X.

cdbd74a3-fd28-485d-9bcd-5ecb799486c1-0
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We also saw from our algebra that this
was the only value of X greater than one

cdbd74a3-fd28-485d-9bcd-5ecb799486c1-1
00:05:23.047 --> 00:05:26.160
where the two functions agree,
so they will cross here.

95b8c49c-2cfe-4159-9324-56f92d6e52a5-0
00:05:27.040 --> 00:05:31.798
So in order to determine whether F of X
is greater than or equal to G of X or

95b8c49c-2cfe-4159-9324-56f92d6e52a5-1
00:05:31.798 --> 00:05:36.190
less than or equal to G of X,
we just need to test the point in each of

95b8c49c-2cfe-4159-9324-56f92d6e52a5-2
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the regions to the first being the one to
the right of X = 16 and the other one to

95b8c49c-2cfe-4159-9324-56f92d6e52a5-3
00:05:41.253 --> 00:05:45.523
the left of 16 and bigger than one to see
which way round F of X&

95b8c49c-2cfe-4159-9324-56f92d6e52a5-4
00:05:45.523 --> 00:05:47.720
G of X are in each of those regions.

2e586f08-9a98-4c78-a7a5-268604165381-0
00:05:48.880 --> 00:05:52.981
Now, because F of X&
G of X both involve square roots and

2e586f08-9a98-4c78-a7a5-268604165381-1
00:05:52.981 --> 00:05:57.016
taking log base two,
it's really good to choose X as a power

2e586f08-9a98-4c78-a7a5-268604165381-2
00:05:57.016 --> 00:05:57.480
of two.

b700cd90-c18d-4f79-869e-a0dc058a9884-0
00:05:57.480 --> 00:06:03.076
To make your calculation a little bit
easier, I'm going to choose X = 64,

b700cd90-c18d-4f79-869e-a0dc058a9884-1
00:06:03.076 --> 00:06:06.480
and I'm going to compare F of 64 to G of
64.

d9afd51d-1100-45b4-9d1a-fca96eec8050-0
00:06:07.320 --> 00:06:10.240
So let's start by calculating F of 64.

2af9f558-decf-47a5-9881-3555351a5209-0
00:06:11.640 --> 00:06:13.760
So I'm taking X to be 64 here.

56e06728-6621-48ec-ae46-f13c158c06b8-0
00:06:14.280 --> 00:06:18.520
The square root will be 8,
so this will be log base two of eight.

bc791d62-351b-49fc-9d7a-b947069eba3e-0
00:06:19.120 --> 00:06:22.920
Now log base two of eight is 3 because 2
^3 is 8.

dd511404-c2c8-4f7e-abd2-7615f72f997d-0
00:06:23.440 --> 00:06:29.560
So F of 64 is actually log base two of
three.

c11a9a65-877d-4ffc-bfc9-e3d2cc0af8e1-0
00:06:30.960 --> 00:06:33.680
Now let's have a look at G of 64.

e80e1a95-3873-4a8b-a28d-a3c0445ef5d0-0
00:06:34.680 --> 00:06:41.640
So this time I start with log base 2 of
64 and 64 is 2 to the power of 6.

413b7ecc-c29d-41cc-9ac6-f05d68049b99-0
00:06:42.240 --> 00:06:48.127
So this will give me 6 here,
and then I'll take sqrt 6 and log base 2

413b7ecc-c29d-41cc-9ac6-f05d68049b99-1
00:06:48.127 --> 00:06:48.800
of that.

20077b9d-be0a-49d1-ab2b-76a65d8b334b-0
00:06:48.800 --> 00:06:53.560
So this is log base 2 of sqrt 6.

01b9e2cf-ff62-4772-9b82-5cd7793d593a-0
00:06:54.080 --> 00:06:56.880
So now I need to ask myself which of
these is larger.

32a12ddc-9f2b-4ada-bfd9-a54363165a85-0
00:06:57.800 --> 00:07:01.087
Well,
I know that 3 is bigger than root 6

32a12ddc-9f2b-4ada-bfd9-a54363165a85-1
00:07:01.087 --> 00:07:05.080
because 3 ^2 is 9 and root 6 ^2 is 6,
for example.

9fa36377-21cd-4df2-9c80-5350f67d9e26-0
00:07:05.880 --> 00:07:11.199
And because log base 2 is an increasing
function and this 3 is bigger than this

9fa36377-21cd-4df2-9c80-5350f67d9e26-1
00:07:11.199 --> 00:07:14.523
root 6,
log base two of three will be bigger than

9fa36377-21cd-4df2-9c80-5350f67d9e26-2
00:07:14.523 --> 00:07:15.920
log base 2 of root 6.

27f9ce39-087c-4301-81c7-fbc69457a03a-0
00:07:16.120 --> 00:07:20.520
So we can see that F of 64 is larger than
G of 64.

d424b252-a8d0-4f63-ae82-0083859528ec-0
00:07:20.680 --> 00:07:23.821
And we also know that they're both larger
than one,

d424b252-a8d0-4f63-ae82-0083859528ec-1
00:07:23.821 --> 00:07:26.480
which is the value they took when X was
16.

d33755da-a951-4ed5-90cc-804237d10006-0
00:07:26.960 --> 00:07:34.818
So I might include F of 64 on the Y axis
here and G of 64 on the Y axis slightly

d33755da-a951-4ed5-90cc-804237d10006-1
00:07:34.818 --> 00:07:36.080
further down.

00b62503-ea7f-4c43-ade5-601494e2d2b4-0
00:07:37.160 --> 00:07:39.640
And then we can say that we have two
points on the graph.

8d2494af-0808-4a0f-96c2-219a59c0407c-0
00:07:39.720 --> 00:07:43.760
This one is on the graph of G and this
one is on the graph of F.

10c1a7be-607d-4dea-afce-fbcbff1d61bf-0
00:07:44.000 --> 00:07:48.798
So now we can see that if X is greater
than 16, so we're in this region,

10c1a7be-607d-4dea-afce-fbcbff1d61bf-1
00:07:48.798 --> 00:07:53.333
we've got F greater than or equal to G,
and we've got both of F&

10c1a7be-607d-4dea-afce-fbcbff1d61bf-2
00:07:53.333 --> 00:07:55.240
G greater than or equal to 1.

26f21971-3ae3-40be-b630-a6b89ce1f8ba-0
00:07:56.960 --> 00:07:59.880
Now let's test the value between 1 and 16.

b4c7ba63-4bf9-48dd-bbf5-e17bbf44a9b5-0
00:08:00.360 --> 00:08:03.280
Again, taking a power of 2 is a good idea.

0c11e305-33f3-4d38-b9c4-3969445cacbe-0
00:08:03.280 --> 00:08:04.560
I'm going to choose 4.

7587706e-c146-4823-85df-2a5ee50e8d0e-0
00:08:05.360 --> 00:08:08.040
So let's now consider F of four.

06488707-acee-43b6-9327-9c2e7c41f35a-0
00:08:08.640 --> 00:08:10.840
So we need to imagine that this X is 4.

0fa8febe-eb4d-4d44-b38c-092708de565f-0
00:08:11.240 --> 00:08:13.040
We take its square root, we get 2.

f08c8253-59ec-4a62-8363-7b08d7a59eae-0
00:08:13.400 --> 00:08:20.160
Log base two of two is 1,
and log base two of one is 0.

f6fa2f19-511b-4932-90cf-d3637724d1a1-0
00:08:20.520 --> 00:08:22.920
So F of four is actually zero.

08995be6-5993-48f2-87be-f4d887a8915a-0
00:08:23.640 --> 00:08:26.000
And now let's have a look at G of four.

e89b9d42-8c5d-43c9-b2a1-636c7d492682-0
00:08:26.520 --> 00:08:30.240
So this time we take log base two of four,
which is 2.

44907b46-a202-4cf6-b1e1-4111aa0560f0-0
00:08:30.560 --> 00:08:37.960
We take its square root root 2 and we
have log base 2 of sqrt 2.

750a765d-77dd-4e67-a7c7-d630b52d700e-0
00:08:39.600 --> 00:08:44.881
This is quite a nice log to calculate
because we know that this is just two to

750a765d-77dd-4e67-a7c7-d630b52d700e-1
00:08:44.881 --> 00:08:49.160
the half and so this will be equal to 1/2
log base 2 of root 2.

091043f1-2c5f-420b-a686-e8a52ef3fbe4-0
00:08:50.520 --> 00:08:54.778
Another way to think about that is that
you bring the half down and you have this

091043f1-2c5f-420b-a686-e8a52ef3fbe4-1
00:08:54.778 --> 00:08:56.960
as a half log base two of two, so 1/2 * 1.

dcae8206-5a66-494d-ba49-546c21adae24-0
00:08:57.560 --> 00:08:58.560
So this one is 1/2.

a51c28f9-a763-4ac9-8823-b2d9ebde333d-0
00:08:59.040 --> 00:09:07.080
So F of four is actually 0 and G of four,
if I can squeeze that in, is equal to 1/2.

b388450a-6021-4f35-b71d-2361d14034ec-0
00:09:08.760 --> 00:09:16.192
So what we have now is that at 4G is
bigger than F and both F&

b388450a-6021-4f35-b71d-2361d14034ec-1
00:09:16.192 --> 00:09:19.520
G are less than or equal to 1.

e988e973-9c0e-4d24-97cd-c7844f8dd2a1-0
00:09:20.640 --> 00:09:23.560
In summary,
we've got 2 possible situations.

2a5f0e00-9b0b-49d3-954d-1eb2da1e5ae1-0
00:09:23.960 --> 00:09:29.840
We've got either F is less than or equal
to G and both are less than or equal to 1.

3457047c-1972-423d-bccd-d89b83988c8c-0
00:09:30.160 --> 00:09:36.240
Or we have F is greater than or equal to
G and both are greater than or equal to 1.

7e6ee2ab-39d2-40b6-b5c6-261522e108e9-0
00:09:38.400 --> 00:09:46.240
So if we return to our options,
we can see that this is option F.

59a731e0-4c2b-4f21-be14-b68e60d2a985-0
00:09:47.440 --> 00:09:48.120
We've got this.

ab8b2685-3e8c-43cb-af52-348e1b3cbb83-0
00:09:48.200 --> 00:09:51.345
This is the situation for X bigger than
16,

ab8b2685-3e8c-43cb-af52-348e1b3cbb83-1
00:09:51.345 --> 00:09:54.920
and this is the situation for X between 1
and 16.

b1723464-ff69-4190-91ad-2f14cdb0c1a9-0
00:09:56.600 --> 00:09:57.720
So we'll give that one a tick.

8c74d34c-71a3-4402-8839-a6c5afaaf8ad-0
00:09:58.600 --> 00:10:02.200
Looking back on this question,
in our calculation,

8c74d34c-71a3-4402-8839-a6c5afaaf8ad-1
00:10:02.200 --> 00:10:06.858
we identified that F&
G both agreed when X = 16 and they both

8c74d34c-71a3-4402-8839-a6c5afaaf8ad-2
00:10:06.858 --> 00:10:08.200
took the value one.

6f2a6cf8-36a0-4927-bce3-10757a8f74e6-0
00:10:08.560 --> 00:10:14.002
At this point we knew that the answer
must have been one of E or F We then went

6f2a6cf8-36a0-4927-bce3-10757a8f74e6-1
00:10:14.002 --> 00:10:18.832
on to test F of 64 and G of 64,
and we saw that the two functions were

6f2a6cf8-36a0-4927-bce3-10757a8f74e6-2
00:10:18.832 --> 00:10:23.799
both greater than or equal to 1,
and G of 64 was less than or equal to F

6f2a6cf8-36a0-4927-bce3-10757a8f74e6-3
00:10:23.799 --> 00:10:25.160
of 64 at this point.

e3adbfb8-03d4-4ab0-807a-6d91773ce3c9-0
00:10:25.480 --> 00:10:28.240
So actually we didn't really need to do
any further work.

1a940819-cdff-4e06-a7ed-98d6f70c5c26-0
00:10:28.240 --> 00:10:32.446
We could have identified the answer as
being F at that point by process of

1a940819-cdff-4e06-a7ed-98d6f70c5c26-1
00:10:32.446 --> 00:10:33.120
elimination.