Examples : Absolute maxima and minima Video Lecture | Mathematics (Maths) Class 12 - JEE

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before watching this video please make
sure that you have watched part 1
Department focus take for example here
we have to say that the way to find the
point interval at what points in this
interval does this function sine 2x
attains its maximum value so you see in
this question the interval is been
defined so it is an absolute global
Maxima question so if I told I told
earlier the critical points are our
turning points right not differentiable
and end points correct so the turning
point is nothing but where my f dash x
is 0 that does nothing but have less
many mud cause 2 X into 2 pi because my
FX is sine 2x so my f dash X will be cos
2x into 2 and this is equal to 0 if this
is required to 0 my 2x will be nothing
but I can say PI by 2 also 3 PI by 2
also 5 PI by 2 also and 7 PI by 2 also I
am taking in this range because I want X
to be of the form 0 to 2 pi or I can say
X is nothing but PI by 4 3 PI by 4 5 PI
by 4 and 7 I can't take more because
these are the only values in 0 to 2 pi
where my sine 2x or cos 2x is 0 there
the only turning points correct there
are other turning points also but since
my domain is fixed now in this fix to
mean the only
now this is done not definition point
not a mixture because sine 2x is a
differential function end points yes you
have to take 0 and 2pi also so in this
will take critical points all right here
tickle points I'll add 0 to n pi also so
at 0 and then PI by 4 3 by 4 PI 5 by 4 5
7 5 X 4 and PI 2 pi so is added in the
last because I wanted to maintain the
consistency gonna end anywhere so these
are my critical points these two are end
points and these twos are these are
turning reversible now I have to find
the value at all these points
let's find f of 0 is nothing but sine of
C f of X so let me write here to make
just show you clearly f of X is 9 so f
of 0 is so I know you mean to 0 that is
finds out and that is 0 next thing F of
next value is PI by 4 which is nothing
but sine of 2 into PI by 4
there is sine PI by 2 and that is
nothing but 1 this guy's also done next
is F of 3 by 4 by 3 by 4 PI and that is
nothing but sine of 2 or 3 by 4 5 and
there is nothing but 3 by 2 pi that is
minus 1 similarly let us take 5 by 4 PI
F of 5 by 4 PI is nothing but sine of 2
into 5 by 4 PI and that is nothing but
sine of 5 by 2 pi that is 1 if you're
having difficulty in this understanding
why sine of Pi by 2 pi is 1 and all
please watch my trigonometry video where
we explain everything clear
the next is this guy 7 by 4 PI so f of 7
by 4 PI that is sign off 2 into 7 by 4
pi that is sign 7 by 2 pi that is again
minus 1 and then 2 pi F of 2 pi is
nothing but sign of 2 of 2 pi that a
sign for pi that is so if you see the
maximum value is 1 and the minimum value
is minus 1 this is my max and this
memory correct so we had to find the
point at which it attains the maximum
value Ashley so the maximum values are
this guy and this guy Correll put a star
here so the points at which this guy
attains the maximum value is X this guy
is X is equal to PI by 4 and PI by
because this is f of X that is what I
was asked and the maximum value is max
value do one question asked me find the
points at which this diet is the maximum
value the points are PI over 4 and 5 by
4 let's take one more example you find
the maximum value of this function in
this interval and we don't even find the
maximum value for this function the same
function in the other in trouble so here
also if you see same thing same logic
the critical points can be either my
turning points my non differentiable
points and my end points
since it is a function it's a continuous
function there is no non definitely the
point so we will talk about turning
points and end points below
but turning points had to find fsx and
that has to be equal to 0 or F dash F is
nothing but 2 X u becomes 2 into 3 X
square 24 X becomes 24 + 1 knots on
become 0 this is my equal to 0 because
this is a blessings or ice all read
further this is nothing but 6 X square
minus 24 and take 6 into common X square
minus 4 is equal to 0
you saw this X is nothing but plus minus
2 so they had two points I get wrong
turning points end points end points if
I assume this guy is one entry so my
critical points I write here
nickel points will be 1 3 2 n minus 2
correct these are my critical points but
are these my critical points no we have
to be little cautious here why because I
am saying my interval is 1 & 3 minus 2
is not part of my interval voluntary all
right
so minus 2 is not very good please note
trick here is X was 2 and minus 2 but
minus 2 is not part of my interval so I
will not consider minus 2 at all because
I am bothered only about these intervals
even if I give it a turning point this
is not in that interval I will not take
that please note here this is a review
critical part here X was minus 2 and
plus 2 we took 2 because 2 comes here
and minus to come somewhere here minus 2
is not in this range so I will not take
- correct so this is my critical point
set 1
so I'll find the value of 1 F of 1 F of
1 will be nothing but if you find Ashley
you into 1 Q because 2 into 1 Q minus 24
into X that is 1 plus 1 not service you
saw all that this becomes 24 minus 1 by
minus 23 plus 1.7 this is minus 23 sorry
do 2 minus 24 is minus 20 to plus 1 not
7 that is 85 similarly if you find F of
2 that will be 2 into Q minus 24 into 2
plus 1 not 7 if you solve this you get
75 and if you find f of 3 that is 2 into
G Q minus 24 into 2 plus 1 not 7 what
you get is anything so if you compare
all this this is my maximum and that is
my answer from the first one in the
second part
let's draw a line here second part my
interval is changing so my critical
point will be I'll take the endpoints
for sure
now I have to add minus 2 out of 2 and
minus 2 you see minus 2 is in this page
you is not so here I'll take - - and
I'll ignore - please note why because 2
is not a member of - 3 n - so I'll
ignore - here in this case minus 2 is
not member of 1/2 so I ignored minus 2
here I ignore correct
so hey don't say anything I have these
critical points you can find the value
of F of minus 3 that will come out to be
125 you find the value of F of minus 2
just work on you own you get 129 you
find the value of alpha minus 1 that
will come out to be 139 and then you see
this is the next one so for the same
function if you see when I do
we changes the maximum value also change
correct
and that is quite true as if my this is
my function right so in this particular
domain if you see this is no Madison
value but in this particular to make my
maximal is let's take my example it does
the little odd away no problem here we
have been told that at X is good to 1
the function attains the maximum value
we have to find a value of e so again I
don't maximum value can be obtained at
endpoints turning point and non
differentiable in this case it's
continuous motion nor in the future is
not in picture X is going to 1 is not
the end point so we'll ignore this that
means X is equal to 1 is a turning point
turning point means F dash X equal to 0
FS X is what X 4 becomes 4x q 62 into X
square becomes 2x e X becomes a 9
becomes 0 that is equal to 0 turning
point so what I get for X cube minus 124
X plus is equal to 0 if you solve this
this is a plastics and I am told that
sorry F does X is this machine I am told
that F dash 1 is equal to 0 why because
at X is equal to 1 my function is
turning it is maximum value so this is 4
into 1 square minus 12 1 24 into 1 plus
is equal to 0 or I get a as 1 and that
is nice correct but after I found the
value of F dash X and then I knew that f
dash of 1 is equal to 0 I found the
value of F X 1 this guy in this is equal
to 0 and I found a is one-to-one friend
thank you
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