Video: Continuity of a Function Video Lecture | Mathematics (Maths) Class 12 - JEE

hi everyone I am Nina from
mathematically inclined and today we are
beginning with part 5 of continuity well
if you haven't seen the previous four
parts then you will find the link in the
description box below you will also find
the links to the playlist of other
chapters which are already on my channel
in the description box please note all
these videos have been crafted not only
to suit your CBSE needs but to give you
a head start for competitive exams as
well so let's get started today we are
going to discuss that the composition of
two continuous functions is again a
continuous function now before we begin
with this in case you do not know about
composition of two functions please
check out the short video on composition
of functions which I will be linking at
the end of this video to apply the
above-mentioned results please have a
look at the first question
it's a discuss the continuity of the
function sine mod X now how would you
identify that which questions we need to
apply this results the given function
would always be a mixture of two
functions which cannot be directly
defined for instance your final mod X
has been made out of two functions
flying X and mod X the composition for
these has been taken to create your
initial function so from a previous
video you would see sine X being the
trigonometric function is everywhere
continuous and so is modulus of X now in
order to create the composition we will
take the function f of G which is f of
mod X which is sine of mod X we will see
according to the given reserve that
since FX and GX are already continuous
therefore their composition which is the
sine products is continuous as
then please know if in case it's a 102
mark on this kind of proof is sufficient
however prefer drastic smoker you would
also have to show the proof or sign it
and more xD continuous everywhere
the in question is again very similar
once again if I split the existing
function if you do cos x ln g XB x
square once again cos x being a
trigonometric function is continuous
everywhere
and likewise x square comes under the
category of a polynomial function which
is continuous everywhere
so for the composition F of 3 X by which
we mean f of X square which gives us cos
of X square is also continuous now with
the third function here you have to show
that this function is continuous
everywhere
so let's first try to split this
function into two this would give us
let's say the first function be 1 minus
X plus mod X and let's see the second
function BG X which is more days now
once again from your previous video you
can see 1 minus X is a polynomial
function position which is continuous
everywhere and modulus is once again
continuous everywhere
therefore their sum would be continuous
everywhere
therefore each X is continuous moving on
once again G X which is mod of X is
continuous everywhere
please note G of H X gives us G of bleep
substitute HX and then you know GX is
mod of X therefore this is the output
which is your function FX and thus will
be continuous everywhere
now for the fourth question you are
already given that FA
this ocean and you have to find the
points of discontinuity of the
composition of the function f of F X so
first of all FX is given by 1 upon X
minus 1 which is not defined at X equal
to 1 thus one becomes the point of
discontinuity or doe mcafee i f of f of
X this means F of 1 upon X minus 1 which
gives us 1 upon X minus 1 on simplifying
this please note here this one acts as
your age so when we substitute it your f
of capital X will give us 1 upon capital
X minus 1 which is this I am taking the
LCM this is the function which we ended
up getting nearly by having a look at
this new function you know that it's not
defined at X equal to 2 thus two becomes
the second point of its discontinuity
but still to be doubly sure you take
limit X tending to 2 minus of f of FX
which would be limited tending to 0 to
minus H minus 1 upon 2 minus 2 minus H
on substituting the value for H of 0 you
end up getting get yourselves to
infinity similarly for you will extend
into 2 plus we take Lee with H tending
to 0 2 plus h minus 1 upon 2 minus P
plus h which tends towards minus
infinity both of them are not defined
therefore the function f of s f is
discontinuous at 2 now to compute our
question we would say f of FX is
discontinuous at X equal to 1 as well as
2 so this is the end of continuity I
hope all the concepts made sense
please keep practicing
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