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JEE MAIN & ADVANCE 12th PCM Maths -Relation & Function Module-1 Demo Video
Hello, students, so continuing with examples on even and odd functions, next we have this, we need to prove two statements.
Now number one, we will be starting with g of minus x, which is equal to this. Try to see, it is equal to g of x, so g must be an even function. Proceeding in a similar function, second, g of minus x is equal to this which is nothing of minus of g x, so g must be an odd function. Remember f of x plus f of minus x is an even function for any f. And f of x minus f of minus x is an odd function for any f.
Now we have an important remark, try to see, “Every function can be represented as a sum of an even function and an odd function”. How, let us see, try to see. We just proved that this is an even function and this is an odd function, try to see if we add them we get f of x, so I wrote f of x as a sum of even part and odd part. So that’s what this remark is all about. So whenever you are asked to write a function as a sum of even and odd function, you have to do it like this, okay.
Moving ahead we have next example that is this, we need to check if it is an even or odd function. First step is clear, you have to start with f of minus x which is this. Now writing e par minus x as 1 by e par x and taking LCM we have this. Now we add and subtract x in the numerator. Why? I will tell you. Try to see the first term will be broken into two terms, this. Advantage is 1 minus e par x cancels. Now why I did that, because I want to prove that f of minus x looks like fx or minus fx. So in numerator I do not need e par x as in f of x as there is no e par x in the numerator. So I wanted to cancel it, and hence I added and subtracted x. So moving like this equal to f of x so this f must be an even function.
At last we have this, I want you to prove this question at home, I give you two hints, I am your friend. Number one is this, that integer comes out of gint under addition and number 2 is gint of minus x is minus 1 minus gint of x for k not belonging to integer. In the example try to see x plus pi divide by pi. It will be x by pi plus 1, so 1 will come out, using Hint number 1. Then now try to see, you will have to start with f of minus x. So obviously in the denominator inside gint you will get minus x by pi. So this minus will be tackled using Hint number 2. Further you can solve it, I have full confidence on you.
So best of luck, take care, god bless.
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JEE MAIN & ADVANCE 12th PCM-Maths – Inverse Trigonometric Function Demo Videos
Continuing with some miscellaneous examples, next we have this. In the last module I told you we can also prove this without using formula of sin inverse, let’s prove it.
Say I call this angle A and this angle B.
Now A plus B equal to theta. We want to prove that this theta is equal to sin inverse of 77 by 85. Apply sin on both sides, so sin A cos B plus cos A sin B equal to sin theta. Now try to see, first of all I would like to express everything in terms of sin because angles are in terms of sin inverse. Now look, A is sin inverse 8 by 17, and B is sin inverse 3 by 5. So sin A will be 8 by 17 and sin B will be 3 by 5. So keeping the values here we get number one, A plus 17 then root of minus 9 by 25 plus 3 by 5 root of 1 minus 64 by 289 equal to sin theta. Simplying these values we get 77 by 85 on left hand side. Now finally we need to check because we directly applied sin so we need to check for extra solutions if it is correct or not.
We called A and B this, now try to see as 1 by root 2 is approximately 0.7, and both of these terms 8 by 17 and 3 by 5 are less than 0.7, so these angles A and B both must be less than pi by 4. So they belong to 0 to pi by 4. So theta which is A plus B lies in 0 to pi by 2. So this thing sin theta equal to 77 by 85 will give me theta equal to sin inverse 77 by 85.
So that’s the way to prove this thing without using formula of sin inverse.
Okay, next we have this. We want to prove this result for x positive, okay, let’s try. Number one, let sin inverse x be theta. As x is positive so this theta must lie in 0 to pi by 2. Now applying sin, x is equal to sin theta, always start with right hand side. So put x is equal to sin theta here in this expression. To get sin inverse 1 minus 2 sin square sin theta. I hope you remember this is just cos 2 theta, so it gives me cos inverse cos 2 theta. As theta is 0 to pi by 2, this 2 theta will lie in 0 to pi that is principle domain of cos and hence this gives me 2 theta, theta was sin inverse x. So it is equal to 2 sin inverse which we wanted to prove.
So with that I conclude this module. See you soon with more examples next module.
God bless take care.
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JEE MAIN & ADVANCE 12th PCM Maths -Definite Integration Demo Videos
Hello, students, let us take some examples on properties. So the first example says integration from minus 1 to 1 bracketed of x bracket of 1 plus sin pi x plus 1 dx.
Now over here in the example, students, we are going to use this property that is property number 3, that is I am going to break this integral into two parts. So let us break this integral. That is first integral is from minus 1 is equal to 0 f(x)dx plus integral from 0 to 1 f(x) dx. Now, students, note over here that in the first integral limits are given from minus 1 to 0. So 1 x is between minus 1 to 0. I am going to calculate sin pi x, so sin pi x will be from minus 1 to 0 because pi x, when x is between minus 1 to 0, pi x will be in third and fourth quadrant. And in third and fourth quadrant, sin pi x is from minus 1 to 0. And hence when I add 1 both the sides it will become 1 plus sin x will be from 0 to 1, substituting this using the bracketed value I add 1 plus sin pi x will be equal to zero. Because 1 plus sin pi x was from 0 to 1, so bracketed value was becoming equal to zero. Now substituting this value in the integral you can see over there the value becomes zero of the bracketed value. And now for the second integral limits are given from 0 to 1, so 1x between 0 to 1, sin pi x, since pi x is from 0 to pi, sin pi x will become 0 to 1. And now when I take, add 1 both the the side I will get 1 to 2, the range will become from 1 to 2. So bracketed of this value will become equal to 1. Now I am going to substitute this value 1 over there in the integral so it will become 1 over there in the integral. Now simplifying it further I will obtain I will be equal to integral minus 1 equal to 0. Now since the inner bracket was become equal to 0 so the integral of bracketed 1, bracket 1 is simple 1, so it becomes 1 dx. And for the second integral, it’s inner bracket has become equal to 1 but x is still over there so it becomes bracket of x plus 1.
Now, students, when I simplify it further, let us check how do we do it. For the first integral the value is very easy to simplify, that is integrating 1, we get x limits minus 1 to 0. For second integral, students, I hope you remember this that bracketed of x plus integer will be equal to bracketed x plus integer. Over here in the second integral bracket x plus 1 is there, 1 is the integer. So when I simplify it becomes 0 to 1 bracketed plus 1 dx. In the integral of bracketed limits are given from 0 to 1. Now since limit is from 0 to 1, bracketed x will become equal to 0. Hence the second integral becomes 0. And now when I simplify it further I get the answer as 2.
So, students, I hope you understood the question. Let’s move further.
Now the second integral, the second question which is given to us is I have to evaluate the value of this integral. You can see it is given in two parts. In both the parts limits are different. In first integral, limit is from minus 4 to minus 5, in second integral limit is from minus 2 to 1. Looking at this I can say we will have to use the substitution which we have studied, x is equal to b minus a into t plus a, that is this property which we studied. I convert both the limits from 0 to 1. So to do that let’s start with the first integral I have taken it as I1. So minus 4 to minus 5 sin of x square minus 3. Now substituting the limit, substituting x equal to b minus a into t plus a, over here b minus a that is upper limit minus lower limit plus a is the lower limit that is minus 4. So simplifying it x becomes minus t minus 4. And now when I substitute it in the given integral I get integral 0 to 1 using the property, students, x is replaced by minus t minus 4, taking the square of it, simplifying it further. When I simplify it further I expand the bracket, right, and then I am simplifying it and I1 becomes 0 to 1, sin of t square plus 8t plus 13. I have taken this integral aside. And I am now going to solve integral second that is I2, which is given as minus 2 to 1 integration of this. Again substituting x equal to b minus a into t plus a. b and a are the upper and the lower limits. So when I substitute the upper and lower limits over here x equals to t minus 2. Substituting this substitution in the integral I get 0 to 1 sin of substitution. And now simplifying it further I get I2 equal to 0 to 1 sin of the integral. And now simplifying it further 0 to 1 it is just a simplification I have done and obtained this.
You can see over here, friends, I1 and I2 are both of opposite signs. So when I add I1 and I2 it becomes equal to 0. So, students, the value of this I1 plus I2 becomes equal to 0.
I hope you understood this, thank you.
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JEE MAIN & ADVANCE 12th PCM Chemistry -Solution and Colligative Properties Module-1 Demo Videos
Hello, dear friends, we are studying the chapter, Solution and Colligative Properties. So in this module I will take you through the next colligative property which is elevation in boiling point, so let us see that.
So when I say elevation in boiling point, the first thing we need to know is what is boiling point. It is the temperature at which the vapour pressure of the liquid becomes equal to the external pressure. So that is what is boiling point.
Next, on addition of a non volatile solute the vapour pressure of the solvent will always decrease. That is what we learnt from the colligative properties definition itself earlier and which is in fact giving us the first one which is relative lowering of vapour pressure, right.
So, next thing, therefore to boil the solution, the required temperature will always be higher so it will take more amount of temperature or more amount of energy to boil the same solution. So let me show you this more graphically, instead of reading it out more like a point wise, let me show you the same thing graphically here. So we know that at boiling point vapour pressure of the solution will be equal to external pressure so taking that into consideration let us start with the graph here.
So, let me draw the graph between vapour pressure versus temperature, right. So now closely observe the vapour pressure line, I have drawn a line here which is the P external that is 1 Atm. Now for volatile solvent at equilibrium we know that solvent and solvent which is at liquid and which is at vapour, there is always an equilibrium established and we can use Clausius-Clapeyron Equation which is lnP is equal to delta H vap by RT. So therefore lnP versus T you can see how it is going to be. It is more like a logarithmic graph versus 1 by T. So, if at all I draw that between vapour pressure and T, the graph will look something like this. So this is that solvent graph, got it. So using that equation which we just mentioned.
Now observe it very carefully, if at all I draw this particular line and it hits at that point X axis and that point Tb0 is the boiling point of the solvent. And as I said earlier the first point, so vapour pressure of the liquid has just become equal to external pressure and that’s why it is called the boiling point.
Now, the next point is now on adding the non volatile solute I told you that the vapour pressure will decrease. Now if the vapour pressure will decrease we learnt this earlier the curve was supposed to shift onto this side, so now the curve will look more like this for solution, got it. Now observe again carefully, if at all I draw this line it will touch at Tb. Now this is a temperature which is higher, right. So to reach that 1 Atm which is the external pressure I am requiring more temperature here. So the temperature will be higher, so that is what I said. Now there is a certain difference which is established and that difference, delta Tb is what is the elevation in the boiling point, and therefore this point becomes the boiling point of solution. So to reach that 1 Atm, so we need Tb. So what I can conclude here that Delta Tb is equal to Tb minus Tb0 and that is what we call it as the elevation in boiling point.
So we will see how we can write it in the mathematical way. So elevation in the boiling point is directly related to molality, so I can write delta Tb is directly proportional to molality and molality will always have the unit of mol per kg. So therefore I can write delta Tb is equal to some constant Kb into m. And this constant Kb is called Ebullioscopic constant or Molal Elevation constant. Please remember very, very important constant using problem solving, got it.
So now, m depends upon number of mols of solute particles, that is more the number of solute higher will be the elevation in the boiling point, right, because it is directly related. So therefore, if m is equal to 1 I can say delta Tb will be equal to Kb. And units of Kb will be Kb is equal to delta Tb by m, so therefore unit will be Kelvin kg per mol or degree centigrade kg per mol. So units are also very important, students, so please remember that carefully, it is Kelvin kg per mol.
And Kb for water usually it is always given in the problems but still it is better to remember, it is 0.52 Kelvin kg per mol.
Now, let me take you ahead in determining Kb and this can be derived from enthalpy and boiling point. Now in this equation delta Tb is equal to Kb into m. We know that m is responsible for the property of solute, while Kb will be responsible for the property of the pure solvent. So that means Kb is all related with the solvent properties. So therefore, Kb is given as R Tb square M by delta H vap. What is this Tb, M and delta H vap, well these are all values for the solvent only. So what are those, let me write down that.
So R is the gas constant, we always take it as 8.314 joules per mol per Kelvin.
Tb is the boiling point temperature in Kelvin.
M is the molar mass in kg per mol.
Again please remember we have to write that in kg per mol and this is for solvent, not for solution.
And then delta H vaporization which is the enthalpy of vaporization in joules per mol or cal per mol as per whatever be the values of Kb.
So this is how we have to calculate Kb.
Kb can also be represented in another form which is R Tb square by 1000 into Lv. Now what is this Lv, we know that R is gas constant, Tb is boiling point, M is molar mass, but Lv is the latent heat of vaporization which is usually represented in joules per gram. So either the question might give you delta H vaporization or Lv. Using any of these we have to calculate Kb and accordingly we have to go for further applications.
As we go for the next modules we will see the applications of elevation in boiling point.
Thank you
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JEE MAIN & ADVANCE 12th PCM Chemistry -General Organic Chemistry Demo Videos
Hello, students, in the previous module we talked about the flipping of the chair conformation and I told you that, there will be several more conformation those comes in between these two extreme chair conformation. So, let’s actually study the conformational Analysis of Cyclohexane.
So, what are the several more conformation that we get during the process of flipping.
So, this is nothing but a chair conformation and in the process of flipping the rotation of the carbon-carbon sigma bond, first convert that into one of the momentary arrangement like this called as half chair conformation. The further rotation triggers the flipping of the bonds and converts this half chair into this kind of conformation called a twist boat. Where this twist boat on further rotation gets converted to a boat conformation and this boat further gets converted into a new conformation which is again a twist boat just like a mirror image of c. Whereas this twist boat gets converted to a half chair conformation and ultimately this half chair conformation gets converted to the other extreme end of a chair conformation.
Well, if we study energy changes taking place during the flipping of the chair conformation the graphical representation that we get is what you simply call as Conformational Analysis of Cyclohexane. So, let’s learn it actually the chair conformation. So, this is the graphical representation. As we have studied earlier the chair conformation of the cyclohexane is the most stable hence least energetic. Well, in the process when the flipping starts this (a) form that is chair conformation gets converted to the next one which is nothing but half chair. Where this half chair and chair conformations are separated by the energy barrier of 10 kilo calorie. This half chair in a flipping further gets converted to, that is the (c) one which is nothing but a twist boat and this twist boat is higher in energy as compared to chair conformation by 5.5 kilo calorie. Well, further flipping pushes this twist boat into the new conformation that is the boat conformation. Well this boat conformation is again important for us with the examination point of view. But one information that we need to know that boat conformation is separated by the energy barrier in comparison to a chair conformation by 6.9 kilo calorie. Why such a large difference in the energy between the chair and the boat. Let’s look at this, in this boat conformation if you observe these 8 hydrogen atoms which are highlighted. These 8 hydrogen atoms come exactly behind each other if you observe them in the Newman projection, which means due to the eclipsing of these bonds this boat conformation exhibit a torsional strain and not only that if you observe these two H’s now, which is point towards each other they start creating a steric strain. And since they look like a pole it is sometime also called as a Flag pole interaction. Due to presence of the steric strain and the torsional strain this boat conformation is having 6.9 kilo calorie higher than the chair conformation. Well, this boat as we have seen earlier then gets converted again to a twist boat and this twist boat in a flipping process further gets converted to a half chair and ultimately we get the extreme end, that is nothing but a chair conformation.
So, this analysis gives us clarification about what are the various conformations that we get in a process of a flipping of a chair conformation. That’s it from this module.
Thank you so much.
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JEE MAIN & ADVANCE 12th PCM Chemistry -Solution and Colligative Properties Module-2 Demo Videos
Hello, dear students, we are studying the chapter, Solution and Colligative Properties, so in this module let me take you through the understanding of vapour pressure versus the temperature, so let us go with that.
Now we will see the distribution of molecular speeds in liquid, as I said that it has similarities to that of gases, so let’s see that first. Well, liquid also has the random motion like gases as well as their distribution is similar to that of velocity distribution of gases. So we have learnt that in earlier modules also. So, and we learnt that this is because of the Maxwell-Boltzmann’s Distribution which is nothing but a plot between a number of molecules versus the kinetic energy. So if we plot that let us see what I can see here, number of molecules versus kinetic energy. Now at a certain temperature T1, this is how the graph varies, this was a similar graph for gases also. Similarly, if at all I increase the temperature, let me take it to T2 temperature, then the graph will vary in this way, that is it will slowly peak will start coming down and it will start becoming more uniform. So in that case now what I could suggest here is let us take a point or kinetic energy after which let’s say liquid particles are starting to escape, right, so this is the kinetic energy, what you can call is the minimum kinetic energy required for molecules to escape from the liquid state. So that is what it is, so it is the minimum kinetic energy. Now if this minimum kinetic energy if at all being drawn towards that curve they will start touching at two different points, then they will generate an area under the curve. So let us see what exactly it represents, so at temperature T1, area under the curve represents the number of particles which are escaping out to the vapour phase. So that yellow coloured is the area which shows that at temperature T1 how many number of particles have escaped. Similarly at temperature T2 you could see this blue coloured shaded part as well as the yellow coloured part, both of them combined are the total number of particles escaping out to the vapour state. So it is very clear that in vapour phase molecules are more at higher temperature, right, as we learnt this earlier also, as the temperature is increased vapour pressure will also increase, right. So what does that mean, you can see evidently here that at temperature T1 the number of molecules escaping are very less but at temperature T2, you could see number of particles escaping are more. So therefore what I could conclude here is low temperature is equivalent to say that it is less likely to vaporize that means it is low vapour pressure. While high temperature it is equivalent to say more likely to vaporize therefore it is at high vapour pressure. So you can now relate how vapour pressure and temperature are co-related to each other, right. So let us try to derive that, now we know that vapour pressure will only be defined when liquid-vapour equilibrium is existing so therefore let me consider that liquid vapour equilibrium is there where K1 is the equilibrium constant and P1 is the vapour pressure. Now therefore K1 is the equilibrium constant at temperature T1 and P1 is the vapour pressure of liquid at temperature at T2.
Now what I can say according to Vant Hoff’s equation we can always relate that as log of K2 by K1 is equal to delta H vaporization by 2.303R into 1 by T1 minus 1 by T2. This is Vant Hoff’s equation where Delta H vaporization is the enthalpy of vaporization, right.
Now observe it carefully, can we write K1 is equal to P1, because yes, liquid does not have any vapour content so therefore vapour phase as P1 so equilibrium constant K1 will be equal to P1. Similarly, if at temperature T2 I can write K2 is equal to P2. So can I change that log of K2 by K1 as P2 by P1. So in that case this Vant Hoff’s equation will be converted to Clausius-Clapeyron’s Equation. So, students, that is how we derive at Clausius-Calpeyron’s equation which is the relation between the vapour pressure of the liquid versus the temperature for a volatile solvent.
So as we go into the next module we will see its applications and more importantly how it will derive towards the Rour’s Law.
Thank you.
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