Hello, good afternoon. Welcome to the YouTube live session. So those who are attending the session, I would request you to type in your names in the chat box so that I know who all are attending the session. So before we start, I would just like to set the agenda for today's session. It would be a mixed back problem solving from various topics, OK? And as we have been doing so far, I'll be waiting for a response to come from your side. At least three of you should respond before I start solving the problem. There may be some disturbance with respect to the sound because of some internet issue, but I'm sure you would be able to see the screen properly. That's not very convincing. Let's go ahead and get to them. Yeah. All right, so let's start with the first problem. f of x satisfies the requirement of Langranger's mean value theorem in the interval 0 to 2. And it's given that f0 is 0. And f dash x is less than or equal to half. For all x in the interval 0 to 2, then which of the following options are correct, OK? So Atme, Shohan, or it's mostly people are saying option B. So basically, this is the application of Langranger's mean value theorem. And let's say for a value of c lying between 0 to x, if I apply a Langranger's mean value theorem, then it'll be f dash c is equal to f of x minus f of 0 by x. Correct? Now since x is less than 2, in fact, c will also lie between 0 to 2, OK? So from this expression, I can say that f of x is equal to x times f dash c. So mod of f of x would be mod of x mod of f dash c, right? And it has been given that mod of f dash x for all x in the interval 0 to 2 is less than equal to half. So even this term would be less than equal to half, right? So I can say this entire expression will be less than equal to mod x into half. And the biggest value of x is going to be 2. So this expression will be further less than equal to 2 into half, which is equal to 1. So f dash x mod will be less than equal to 1. So option number b is the right option over here. And the first one to answer this correctly was Atmesh, OK? So now we'll move on to the next question. Hope there's no doubt regarding this. The value of f of 0 so that f of x given by this expression is continuous everywhere. And you may have seen this question in your preparation of board exams as well. It's a commonly asked question in continuity. Should I take more than one minute to solve this? Let's say this is question number 2. Please type the question number along with the response. Done? OK, so Amogh has given the response, OK? So if it has to be continuous everywhere, it has to be continuous at 0. So limit of this function as x tends to 0. So it should match with the value of the function at 0, OK? Now here I see a combination of three different types of limits. One is your limit of a to the power x minus 1 by x extending to 0. That's log of a to the base e. And you have limit of sine x by x extending to 0. That's going to be 1. And limit of log of 1 plus x by x that's going to be 1 again. So if I make these manipulations in this problem, I would get 4 to the power x minus 1 by x whole cube. In a similar way, I would divide this by x and I would divide this by x squared, OK? Now I need a 3 over here, so I'll multiply, divide with 3. I need a 4 over here, correct? So I'll multiply, divide with 4. Is that fine? OK, this 4 will go up, I guess. This 4 and this 3 will go up. Because you're dividing x by 4 to the denominator term, so this is as good as multiplying with 4 by x, OK? So this term here would become ln of 4 whole cube and is a factor of 12 coming up from here. And rest all these terms will become 1 and 1. That means your option number c will become the right option in this case. Is that fine? No question with respect to this? Let's move on to the next question. The next question says the number of ordered pairs alpha, comma, beta, where alpha and beta belong to the interval minus pi to pi. Satisfying these two equations, cos alpha minus beta is 1 and cos alpha plus beta is 1 by e, OK? How about others? So only Atmesh thinks it's slightly different. So if you look into this condition, first of all, cos of alpha minus beta equal to 1 implies only possibilities alpha minus beta has to be 0. That means alpha has to be equal to beta.  Now when you say cos of alpha is 1 by e, in what quadrant can your 2 alpha lie? Remember, your 2 alpha can actually lie between minus 2 pi to 2 pi, correct? If you're saying minus 2 pi to 2 pi, that means you have an option to take the first quadrant. Let's say I'm going from 0 to 2 pi. So these two are selected, correct? And if I'm going in the reverse direction also, I can select these two angles, right? That means there are four values of alpha that you can actually get, which implies there will be four values of 2 alpha, so there will be four values of alpha as well. Yeah, that's fine. It's in the first and the third quadrant. But your interval is from minus 2 pi to 2 pi. So altogether, you can actually take two complete terms. So you will get the same value for cos for different angles of alpha. And hence, the number of ordered pairs possible will be 4 in number. Is that fine? Now had alpha and beta been different quantities, then you would have got 8 as your answer. But in this case, alpha and beta are equal, so there's no point stopping the position. So whoever said d, I think the first person to answer this was Rohan. So d is the right answer in this case. Yeah, moving to the next question, which is from your mathematical reasoning, the statement negation of p or q or negation of p and q is logically equivalent to which of the following? All right, so I'm getting the response as b for most of you. Now this is as good as saying negation p and negation q. Or negation p and q. Now here I can write this as negation of p. We can take that as common. And negation q or q, correct? Now this is actually a tautology. This will always be giving you true as the response. So the response of this is going to be whatever is the response of negation p. So option b become the right option in this case. Absolutely correct. So most of you have answered this correctly. Shruti was the first person to answer this. Moving to the next question, if set a contains all those angles theta says that 2 cos square theta plus sine theta is less than equal to 2. And b contains all those values of theta, which is between 5 by 2 to 3, 5 by 2. Then we have to find the intersection of set a and b. This is your question number 5, let's say. OK, so I'm getting response from most of you as option c. All right, so let's check this out. So this I can, 2, 1 minus sine square theta plus sine theta is less than equal to 2, OK? So this can be written as 2 sine square theta minus sine theta is greater than equal to 0. That means we can write this as sine theta times 2 sine theta minus 1 greater than equal to 0. So if I make my wavy curve for this, my sine theta can either lie in this interval or this interval. Or you can even take cases. For example, you can take a case where sine theta is positive and 2 sine theta minus 1 is also positive, right? So in this case, your sine theta will be greater than equal to half, which implies your theta should belong to the interval of greater than equal to half if you quickly make the sine curve graph. So greater than equal to half would be covering this zone from this value, pi by 6 to 5 pi by 6. But we are only allowed to start from pi by 2. So for this case, your answer will be from pi by 2 to 5 pi by 6. Second case is where you can take your sine theta to be negative and 2 sine theta minus 1 also to be negative. That is, sine theta is negative. And sine theta is less than equal to half. So that interval would be simply the interval between pi 2, 3 pi by 2, pi 2, 3 pi by 2. Because we cannot exceed 3 pi by 2 over here, OK? So from the second instance, we get the interval of theta to be between pi and 3 pi by 2, OK? Because in this interval, my sine theta is negative, that is less than equal to 0, as well as less than equal to half. So this basically becomes a redundant condition, OK? So this is the dominating condition in this, just like this was the dominating condition over here. So in these two conditions, I get these two as my interval. So it can either be this or this. So option C is the one which discusses about these two limits of theta. So C option is correct. Let's move on to the sixth question now. The coefficient of x to the power 50 in 1 plus x to the power 41 times 1 minus x plus x square to the power of 40, all right? So Vaishnavi has given a response, OK? Go ahead and back up with the same answer. All right, so mostly, Janta is going with option A. Let's check.  So this becomes 1 plus x times 1 minus x plus x square whole to the power of 40. This is actually the formula of A plus B times A square minus AB plus B square, which we know that it's going to be 1 plus x cubed, OK? Now when you expand such a binomial term, you would realize that you get terms of this nature, OK? So all the powers that you would see over here, they would be multiples of 3 over here. So this will all contain multiples of 3 power, multiples of 3, OK? And here you have 1 plus x. That means you can either get a multiple of, you can get an expression with a multiple of 3, or a term which is of the nature or power of x can be, let me write it in this way. So your terms could be of the nature x to the power 3n or x to the power 3n plus 1, right? And none of them can be equal to 50. This will not give me a whole number value of n. Neither this will give me a whole number value of n. That means there is no such term, OK? That would contain x to the power 50. So your answer is going to be 0 in this case. Is that fine? Good. So the first one to answer this correctly was Vaishnavi. Next, just for a change, a simple log question. So log to the base 7 of under root of 7, under root of 7, under root of 7 is. All right, so mostly people are going for option number C. Let's quickly check this. This is like saying log to the base 7 of log to the base 7 of 7 to the power half plus 1 4th plus 1 8th, right? And this becomes actually 7 to the power of. If you add this, you'll get 4 by 8 plus 2 by 8 plus 1 by 8. That's equal to 7 by 8. So 7 to the power of 7 by 8, which gives you log of 7. And this will become 7 by 8. So I can write this as log of 7 to the base 7 minus log of 8 to the base 7, which is nothing but 1 minus. And this 8 can be done as 2 to the power 3. So it's 1 minus 3 log 2 to the base 7. Option number C becomes correct. A very easy question just to break the monotony. Next is evaluate this integral, where this represents the GIF function. And this is a normal bracket. This is the normal bracket. OK, so mostly people are going for option 8A. So I can write this as minus half to half GIF of x plus minus half to half ln of 1 plus x by 1 minus x. And we all know that this function is actually an odd function. So if this is an odd function, this becomes a 0. And GIF function changes its definition about 0. So from minus half to 0, it behave as negative 1. And from 0 to half, it will behave as 0. So this is just about integrating negative x from 0 to negative half. That's going to be 0 negative of negative of negative half. So that's going to be plus half. So that's going to be negative half as your answer. So absolutely option number A becomes the right option. So we'll move on to the next question now. This question says this is square OABC and is formed by the line pairs x, y equal to 0. And x, y plus 1 is equal to x plus y, where o is the origin. A circle with center C1 inside the square is drawn to touch the pair of lines x, y equal to 0. And another circle with center C2 and radius twice that of C1 is drawn to touch the circle o and the other line pair. So find the radius of the circle with center as C1. This is your question number 9. OK, Rohan. OK, you're getting 1 by 3, which is not in the options. I would just suggest recheck your calculation once again. No one else? Sure, sure, take your time. No hurry. OK, I need two more people to respond before I start solving this. Let's try to sketch the situation over here. Let's say I have these two circles. And this is enclosed within pair of lines. Now x, y equal to 0 is basically nothing but it's a pair of line which signifies your coordinate axis. So x, y equal to 0 signifies your x equal to 0 and y equal to 0. That's nothing but your y-axis and the x-axis, respectively. If you look at the other line, that is this line, this is like saying x, y minus x minus y plus 1 is equal to 0. This is x minus 1, y minus 1 equal to 0, which signifies x can be 1, y can be 1. So x equal to 1, y equal to 1. According to the given situation, it should be like this. So according to the statement mentioned in the question, one circle with center at c1 touches x, y equal to 0 pair. And other circle, the center has c2. And twice the radius touches the other circle. One thing is for sure, if they are touching like this, then the diagonal should be passing through the centers of these two circles. So this is, let's say, r. Then this is, let's say, 2r. And if I drop a perpendicular, this will also be 2r. And if I drop a perpendicular here, that is also going to be r. So let's say I call this point as o. Now, what is the diagonal length in this case? Diagonal length in this case is root of 2, correct? And this length is also equal to 2r. And this length is also equal to r. So this is actually 2 root 2r. And this is actually root 2r. So the total length from o to b, which is root 2, is actually root 2r plus 3r plus 2 root 2r. So this implies root 2 is equal to 3 root 2 plus 3r. So r is going to be root 2 divided by 3 plus 3 root 2. And that is given by option number c in this case. Now, let's talk about the next question. OK, so 10a. So it's given that f of x has a second-order derivative, f double dash x, which is equal to 6x minus 1. OK, so f dash x will be what? f dash x will be 3 times x minus 1 whole square plus, let's say, a constant c1, OK? Now, it is given that the graph has a tangent of y equal to 3x minus 5 through this point. That means f dash 2 is equal to 3, correct? That is the slope of the tangent at 2 is equal to 3. So this implies 3 times 2 minus 1 square plus c1 is equal to 3. That means c1 is equal to 0. So f dash x is going to be 3 times x minus 1 whole square only, OK? So f of x is going to be x minus 1 whole cube, plus, let's say, another constant. But it has been given that the curve is passing through 2 comma 1, right? So that means f of 2 is going to be 1, right? So 2 minus 1 cube plus c2 is going to be 1, which means 1 plus c2 is going to be 1, which means c2 is also equal to 0. So putting it back over here, this implies f of x is going to be x minus 1 whole cube. So option number A becomes the right option in this case again. So let's now talk about the 11th question. A natural number x is chosen at random from the first 100 natural numbers. Find the probability that this expression is less than 0. None of these, OK? Only Shruti thinks 7 by 25. All right, so let's discuss this. So if I go by the wavey curve sine scheme, so this is 20, 30, 40. So it will be plus, minus, plus, minus. So it will be negative when x is less than 20, or x lies between 30 and 40, correct? So when you say x is less than 20, means you can only choose natural numbers from 1 to 19. So x would belong from 1 to 19, OK? And this means x should belong from 31 to 39, OK? So I think there are 19 such numbers over here. And there are nine such numbers over here. So altogether there are 28 numbers which are going to meet this given inequality. And this becomes your number of favorable events. And sample space is just going to be 100 because you're choosing one number from 100. So your probability of this event is going to be 28 upon 100, which is 7 upon 25. So option C becomes the right option in this case. Is that fine? So the only person to give right answer in this case, no, many people gave. And the first person was Shruti. Shruti, Aditya, Ramcharan. Again, a simple question, but again, a small counting error can lead to wrong results in this case. Again, a simple question. Just like we had log question a little while ago, a simple one just to break the monotony. Yeah, so mostly people are going with option number B. That's actually a very simple one. This is going to be just r square minus r minus one whole square. So when you start putting the value of r as one and all, you start getting terms like this. So this will become a 2010 square minus 2009 square. And terms, alternately, will start getting canceled. So this will get canceled with the next term which is mod A to C, which is three square minus two square. And this will lead to cancellation of all the terms, okay, which is 2010 whole square. So option B is correct in this case. Next is question number 13, if Z is a non-real, IE by IE, let's get this one by IE. So this is an angle of a triangle for which of the following up, okay? All right, so first of all, Z minus one by IE should be completely real, right? Because sine inverse domain is minus one to one, so this should be completely real in nature, okay? And if I assume Z to be X plus IY, Z minus one by I will be like X minus one plus IY by IE. Okay? Which is nothing but X minus one by I plus Y and this should belong to real. If this has to be real, it implies this term should have existed. This term should have zero. That means X should be one. So it is clear that real part of Z should be one. Now, option A and B clearly states that. So let's further look into this. Now, next we know that sine inverse domain is between minus one to one. And by the way, this term is going to be just sine inverse of Y because your Z is actually one plus IY. This is your Z. So one and one gets canceled, IE and IE gets canceled, leaving you with just with Y, right? So Y should lie between actually minus one to one. Okay, screen, okay? Now, if I scrutinize the question a little bit more, if it has to be an angle of a triangle, that means my sine inverse Y should only be between zero and pi by two, correct? That means Y should have been between zero and one only. Okay, so I think this question, so it may be challenged and this option, ideally should have been from zero to one. But anyways, if I have to mark the best of these options, I would go with option number B. All right, so we have the next question over here, a limit question. Find the limit theta tending to zero of four theta tan theta minus two theta tan theta by one minus cross two theta. Amesh, I'm believing that it has to be an angle of a triangle and we have angles between zero to one at 80 degree only excluding zero and 180 degree. So sitting in between zero to pi by two only. Okay, so mostly people are going with option D. So basically this becomes a very easy question. So I can write it as eight theta square tan theta divided by one minus cross two theta is two sine square theta and this is also two sine square theta. Okay, so in this case, in the first term over here, okay, we can divide the numerator by theta square and divide the denominator also by theta square. Okay, whereas in this case, what will happen since even if you divide by theta square in the numerator and denominator, you would realize that still the denominator, numerator will have an extra tan of theta, which will make things go zero in this case, okay? While in this case, it will become four by two because tan theta by theta is going to be one. Sine square theta by theta square is again going to be one. So your answer is going to be two in this case, which is option number D is correct. Absolutely, so the first one to answer this correctly was Sondarya. Let's move on to the next question, which is the 15th question. The ends A and B of a rod of length root five are sliding along the curve YZ for two X square. If XA and XB be the coordinates of the ends, then at the moment when A is zero comma zero and B is one comma two, find DXB by DXA. Yeah, yeah, sure, sure, take your time. Okay, what about others? All right, so mostly people are going with option D. See, first of all, let's say this is the rod AB, whose length is given to us as root five. Let's say this coordinate is XA, YA, this coordinate is XB, YB. So I can say the distance between A and B is root of five. So XA minus XB whole square plus YA minus YB whole square and the root is root five, right? So XA minus XB square plus YA minus YB square is equal to five. Now, we know that they are lying on this parabola Y is equal to two X square. So I can say YA is two XA square and YB is two XB square, okay? So I can write this as XA minus XB square and this I can write it as two XA square minus two XB square square, okay? So that's given to us as five, okay? Now, let me differentiate with respect to XA, both the sides. So this will become two times XA minus XB into a derivative of this is one minus DXB by DXA. In a similar way, this can be done as two times, two XA square, two XB square into derivative of this term, which is going to be four XA, minus four XB DXB by DXA. And this is equal to zero, okay? Now, I want the value of DXB by DXA at the instant when XA is zero and XB is one. So what I'll do is wherever there is an XA, I will put zero and wherever there is an XB, I'll put a one. That becomes two, zero minus one, one minus, let me call this quantity as K for the timing. So let me call this quantity as K for the timing. So two times, again, it will be zero minus two, four into zero, which is again zero, minus four K. And this is equal to zero, right? So if you simplify this, it will give me two K minus two, minus 16 K is equal to zero. That means 18 K is equal to two. So K is going to be one by nine. That means DXB by DXA is going to be one by nine. So option number D becomes the right option. And the first one to answer this was Amok. So now let's move on to the next question. That's question number 16. So a couple of things I want to tell you guys. So if you think for one, one and a half minutes of trying, you are going nowhere, then please abandon the question. Do not invest any more time. Because if you have not been able to make any progress in one and a half minutes, it's like, the question may be beyond your scope or things may not be clicking to you at that moment of time. So don't unnecessarily waste time. I think max a question should take two and a half minutes or three minutes more, max to max to solve. If you're taking more than that time, that means you need to work out on your speech. Of course, without compromising on the accuracy. Okay, Aditya. So I've got three responses so far. Two of them are in favor of option B. Koshal says option A. Okay, so let's say this is our G of X function. So when you multiply, it becomes this term. Now what I'm doing is I'm just focusing on writing terms which are having even powers of X. I will not bother to write terms which are containing odd power of X because we know that for those cases, your answer is going to be zero because the limits of this integral are opposite in sign. So just evaluate this integral. So this is going to be P times two to the power, sorry, X to the power two N plus three by two N plus three. And this is going to be X to the power three by three. And this is going to be BNQ times X, okay? Or you can do two times this from zero to one, right? So this will give you two P by two N plus one, sorry, two N plus three. And you have two ANP by three and you have two BNQ. Now for all linear functions, PX plus Q, this integral has to be zero. That means it is independent of your PNQ. In other words, the coefficient of P and coefficient of Q should be zero each. So this should be zero and this should be zero, which clearly implies B has to be zero or BN has to be zero and AN has to be negative three by two N plus three. AN has to be negative three by two N plus three. So option number B says that, so B is the right option. Okay, so first one to answer this correctly was Aditya, great, well done. Let's go on to the next question. Find the, or match the conditions of column one with column two. Yeah, it's a fractional part. No, this entire thing is not a fractional part. This is a fractional part. Yeah, so let me tell you this bracket is your GIF, but this is the ordinary bracket, normal. Normal brackets. So basically it's GIF of X over here and this is going to be a fractional part of X. Okay, so mostly people are saying 17A. Okay, let's discuss. Sign of pi GIF of X is basically nothing but always a zero value, right? Because you're dealing with an integral value of pi. So this is always an integer, okay? So this has to be differentiable everywhere. So A will map with first option. Whereas if you see this function, we know that this function is going to behave as sine pi X when X is between zero and one, right? So sine pi X would be a graph which is like this from zero to one. And after that, it will be periodic with one. That means it'll keep repeating itself like this. So this would be your graph of sine of pi times fractional part of X. As you can see, this function will start having kinks or you can say cuffs. So these all points are like cuffs. And at cuffs, you know that the function is not going to be differentiable. So only at integer values. So this function will be not differentiable at one and minus one to state just two of the points. So B will map to option three. So this is given by option number A in this case. Next, the two vertices of a triangle are given to us four comma minus three and minus two comma five. If the ortho center of the triangle is at one comma two, find the third vertex. This is your 18th question. Okay, so 18th. So this is a very simple problem to solve because you can just plug in the points. For example, I know that this is going to be perpendicular, right? So k plus three by h minus four should be negative reciprocal of the slope of let's say this point BH. So it should be one by slope of BH. Slope of BH is going to be five minus two by minus two minus one, which is going to be three by minus three. That's going to be negative one. So that's going to be one, okay? So k plus three is going to be h minus four. So k plus three is equal to h minus four is only satisfied by option number B in this case, correct? Because 26 plus three is equal to 33 minus four, right? None of the other options satisfied. So option B is the right option. However, if you want to be sure that it is not none of these, you can again do the slope of h is negative reciprocal of the slope of BC but it will end up giving you the same answer. So it's an easy question. Let's not spend too much time on this. Next, normal AO, A1, A2 are drawn to the parabola y square is equal to 8x from the point h comma zero. If triangle OA1, A2 is equilateral, then the possible values of h. Okay, 19s. Anybody else? Okay, but that's not in the options for it. Possible values. That means one of the values can be equal to 8x from the point h comma zero. Okay, atmesh gets 19a. Okay, so let's discuss this. So OA is a normal, so is A1 and A2, correct? Okay, so basically these three points are actually called the co-normal points. So we have discussed them in our parabola chapter that these are called the co-normal points. And if you connect them, as per the question, this triangle is an equilateral triangle. This triangle is an equilateral triangle. So this angle has to be, correct? So what is the slope of OA1?  That means let's say this point is two t1 square comma four t1. And this point A2 is two t2 square comma four t2. So slope of OA1 is going to be four t1 by two t1 square. That's going to be two by t1. And this is given to us as tan of pi by six. So this slope is nothing but tan of the angle of inclination. That's going to be one by root three. Which means t1 could be safely written as two root three. In other words, what I'm trying to claim is that if I have a parametric equation of the normal, okay, which we all know is this, this is the equation of the normal at any parameter or any parametric point t, then one of the roots, then t1 is a root of this equation. t1 is a root of this equation, correct? Similarly, you can have this slope, let's say t2, then t2 is going to be negative of two by root three. So this is also going to be a root of this equation, okay? Anyways, so what am I claim is here that if let's say this normal is passing through h comma zero, so let me put y as zero and this will become our th, this will become two into two into t plus two into tq, which will actually give me, I'm writing down over here, th is equal to four t plus two tq, okay? Canceling the factor of th is going to be four plus two t square, right? Now, irrespective of whether you put t as t1 or t2, your answer will become four in plus two into two root three square. That's going to be four plus 12 into 224, that's going to be 28. So 28 is going to be your answer. That is nothing but option number C is correct. Now, we all know that the algebraic sum of the T's at the co-normal point is zero. So this is corresponding to your t equal to zero. So you can call this as t3 equal to zero point, okay? So that is automatically coming from here when you were cancelling the factor of t from both the sides, it was actually saying that t equal to zero could be a possible value of the root. All right, so let's move on to the next question, which is question number 20. Okay, Vaishnavi. So alpha, beta, gamma here are scalar quantities. A, B, C are unit vectors. Okay, Sai also says option B. So it's given that A dot B is zero and A dot C and B dot C is going to be cos of theta. And of course they are all unit vectors, so their moduli are equal to one each. So now I have been given that C is alpha A plus beta B plus gamma of A cross B. Okay, so let us take the dot product with A first. So C dot A is going to be alpha A dot A is going to be one, this is going to be zero and this is again going to be zero. That means alpha is going to be cos of theta. If you take dot product with B, it's going to be alpha cos theta. Sorry, it will be zero again because A dot B is zero. This will be beta and this again will be zero because A cross B dot A and A cross B dot B, they'll both will be zero. Okay, so even beta is going to be cos of theta. So it's very clear that option number B is not true. So you can just mark option B. Okay, but let's say we want to ascertain whether option A, C and D, none of them are incorrect. So let's check them out quickly. So if I now take C dot C, I'll get alpha A dot C plus beta B dot C, okay? Or you can do one thing, you can take the dot product of this vector with itself. So this dot product with itself. So this into this will give you alpha square, beta square and gamma square mod A cross B square. Rest, this into B will give you zero, this into this will give you zero, B into A would give you zero and this into this will give you zero. Similarly, this into this and this into this will give you zero again. So this is the only term that I would get. And mod C square is going to be one. So one is going to be alpha square plus beta square plus gamma square. And this is also going to be one. Now since alpha is equal to beta, two alpha square plus gamma square is going to be one. That means gamma square is one minus two alpha square, which is one minus two cos square theta. So that's going to be negative cos two theta. So this option is correct. So this cannot be my answer. This option is also correct. So this cannot be my answer. Okay. And from here I can write alpha square as one minus gamma square by two, okay? Which means beta square is going to be, so one minus gamma square is two cos square theta by two. Okay. And two cos square theta is going to be one plus cos two theta by two. So even option number D is saying the right thing. So this cannot be your answer. So only possibility is option number B. Okay, so had the question been a multiple option correct question and they would ask which of the options are true, then obviously A, C, D would have been the answer. Now let's talk about question number 21. Okay, Homo, where should I be? Easy question. So this becomes sine square theta plus 81 by 81 to the past sine square theta and that's equal to 30. So if you call this term as Y, it will become 81 by Y is equal to 30, which means Y square minus 30 Y. So this can be factorized as 27 and minus three. So Y could be three, Y could be 27. So 81 sine square theta if it is three, that means sine square theta has to be one fourth, correct? So sine theta has to be plus minus half and if sine, if 81 to the past sine square theta is equal to 27, that means sine square theta has to be three by fourth, which implies sine theta can be plus minus root three by two. So my possible values here could be, yeah, I think 120 can also be a correct answer. So of course 30 is a correct answer and root three by two could be obtained for 120 degree as well. So in this present case, both these options are correct. So option A and C both are correct. How is B, how is D correct? Yeah, D is also correct. D is also correct. So, okay, so they should have asked this question. Theta cannot be, I think that would be a right question to us. Theta cannot be then option number B is correct. And then in this case, in the present form A, C, and D will be correct. Next, find the sum of N terms of the following series. So here you can easily put some special values like N as two or something and you can get your answer, but I would like you to do it in a rigorous way. Hey. Let's go. Shit. Okay, all right. So basically what I'm going to do is I'm going to first multiply with one minus X and divide with one minus X. So if you do that, you get one minus X squared. This becomes one minus XQ and this term will actually become one minus X to the power four and so on. So it's going to form something like this N minus X plus X square plus X cube all the way till X to the power N, okay? Which again is a GP, okay? So this gives you the result like this. Let's take the LCM of one minus X inside. So it becomes one minus X square and one minus X minus X times one minus X to the power N. So option number C is the right option in this case. So this is a function from R to R defined as two X plus sine X, then F of X is, which of the following? One, one and onto, one, one, but not onto, onto, but not one, one, neither one, one, nor onto. Okay, Purvig says option B. Okay. Purvig wants to change his answer to A. What about others? Okay, it's very obvious that the derivative of this function is going to be two plus cos X. And we all know that cos X belongs to the interval minus one to one, right? So this will always be a positive term. That means your function is monotonically increasing. And if it is monotonically increasing function, it has to be a one, one function, okay? And if you take any value of X, let's say if X goes to infinity, even F of X goes to infinity. And if X goes to negative infinity, even F of X goes to negative infinity. And not only that, all the values of the function are going to be addressed by this function because it's a continuous function. Two X is a continuous function, sine X is a continuous function. So it has to be one, one and onto both. So option A is going to be correct. Next question is from statistics for this arithmetic progression, A, A plus D, A plus two D, till A plus two and D, the mean deviation from mean is. So we have to find out the mean deviation from mean. So Ramcharan has given the response, okay? Okay, anybody else? Okay, so mostly people are going with option number C. So we'll discuss this. First of all, we know that the mean deviation from the mean position is going to be one by N, summation of mod of Xi minus X bar. Okay, so the mean of the data itself would be the first data and the last data divided by two.  Undoubtedly. So the total number of data present over here is two N plus one. So I have to do summation of, okay? Each of the data, let's say I generalize the data with A plus R D and let's say from zero to two N minus A plus N D, okay? So basically we have to do one by two N plus ones, summation zero to two N, R equal to zero to two N, mod of R minus N times D. So this is nothing but one plus two plus three all the way till N, twice stop. Isn't it? Correct? So this is going to be D into two N plus one. This is going to be N into N plus one by two into two. So two and two gets canceled. So your answer will be N into N plus one D whole divided by two N plus one. So in that case, your option number B will become the right answer and not C. So none of you got it correct. Any question with respect to this? Did you get your mistake? Where you all went wrong? All right, so moving on to the next question. A dictionary is printed consisting of seven letter words. Only that can be made with the letters of the word cricket. If the words are printed in an alphabetical order, what is the number of words before the word cricket? So a very simple question that you would have done in your class 11 permutation combination.  Vashtami, the sum from N to zero would be N into N plus one by two only, right? Because zero is as good as starting from one itself. So you're going from one to N. Going from one to N means N into N plus one by two. So the first word will be CC. So let me call this question number as question number 25. So Aditya says, okay, option C. Okay, let's check. So first, all the words starting with CC will be before cricket, right? So that word will be five factorial. Then words starting with CE, okay? That will also be before cricket. Then again, will be five factorial. Then similarly, CI will be five factorial, okay? CK will also be five factorial. Then CR, when it comes to CR, so the first word will be CRC, okay? So CRC, it will be one, two, three, four. That's going to be four factorial, okay? Then CRE, that's again, will be four factorial. Then CRI, now cricket also start with CRI. So the next word will be CRIC. So cricket also starts with CRIC, correct? So CRIC. So the next word will be E. Then these two words can be filled in two factorial ways, okay? So it's E, KR, sorry, CRIC, KT. So next word will be CRIC after the next word will be K. So if K comes, the next word has to be ENT. So this is the word exactly after this. So let's count how many of them are there. 120 into four will be 480. This will be 16 into 16, 32 plus two. That is going to be 24 into two. That is 48, 48 plus two is going to be 50 again. So total words before cricket going to be 530, that's option number C is going to be correct. Absolutely, so first one to answer this correctly was Aditya. Okay, so next is a two and false question. So this is the direction for the two and false questions. All of you would have already gone through it. Statement one is true, two is true, two is the correct explanation. One is true, two is true, two is not the correct explanation. One is true, two is false. One is false, two is false, okay? So this is the first question. Consider a function g of x as f of x minus one and another function f of x plus f of one minus x is two. Statement one claims that g of x is symmetrical about half comma zero. Statement two says, if g of a minus x is negative g of a plus x, then g of x is symmetrical about a comma zero. So which of the options a, b, c and d are correct in this case? Let's say this is 26th question. Okay, 26th question, Amogha is giving the response. Okay, let's wait for one more person to respond. Okay, so let's discuss this. So if you look at the second statement, can I write this as f of x minus one plus f of one minus x minus one equal to zero, correct? This can be written in this way. Now this itself is g of x plus g of one minus x equal to zero, correct? That means g of x is negative of g of one minus x, okay? So it tries to say that whatever is the value of the function at x, it has the same value with a negative at one minus x. Let's say the value at zero is this. Then the value at one would be this. Correct? Now here I think one thing that they should have mentioned that f of x is a continuous function. Because if I do that, the value let's say at a point we can take one sixth is same as negative of the value at let's say five sixth. So this trend continues, you would realize that the graph will be in such a way that it will cut at half comma zero, okay? Because at half it will become g of half is equal to negative g of half. That means g of half should be equal to zero. I'm assuming in this case that my function is continuous function. So this function, if g of f of x is continuous will be symmetrical about half comma zero. So this is correct statement. And here if you replace your, now this means g of a minus x is negative of g of a plus x. See what does it mean? It means that, let me just make some space for myself. Let's say this is a, if you go x units left of a and x units right of a, they are negative of each other. So let's say I move x units to the left of a and I move a x units to the right of a then the graph would be negative of each other. That means even in this case you would realize that the graph will come out to be symmetrical about a comma zero. So this statement is also correct, okay? So this statement and this statement is also, both are correct. And two is helping me to, or two is explaining my statement number one. So option number a become the right option in this case. Is that fine? Okay, so one more thing you should remember that if g of a minus x is equal to g of a plus x, right? Then the graph will be symmetrical about x equal to a line. Okay? And if g of a minus x is equal to negative g of a plus x, the graph will be symmetrical about a comma zero point. Provided it's a continuous function. Yeah, so are the given lines coplanar? So this line is passing through one comma minus three comma two and this line is passing through two comma one comma minus three. Okay, so if these three lines are coplanar, we know that x one minus x two, let's say I call these as points x one, x two y, x one, y one, z one, x two, y two, z two. So, and let's say I call this as a one, b one, c one, a two, b two, c two. So this determinant should be zero if they are coplanar. So I can write this as a minus one comma minus four comma five, two one minus three one minus three two. So let's find out the value of this determinant. This will be negative one, two minus nine, which is going to be negative seven. By the way, if you look at statement two, it doesn't talk about all these things, right? So I'm just doing it just to make a confirmation whether these two lines are coplanar or not. Okay, so four plus three and five minus six. Minus one. So that's going to be minus plus seven and this is going to be 28 minus 35. Yes, they're coming out to be zero. But this is not the intended way that in which we are actually supposed to solve this question. If they're coplanar and you know that these are not proportional, these terms are not proportional, that means these two lines should actually meet, correct? Or they should actually intersect, right? So if they're not proportional, two of the conditions can happen. Either they are skew lines, right? Then only they will not meet, but skew lines are actually non-coplanar, okay? And if they're coplanar and they are parallel, then also they will not meet. But we know they are not parallel because two one and minus three is not proportional to minus one, sorry, one minus three and two, okay? So that means if these two lines are coplanar, they should actually meet.  Let's say they meet at x one, y one and z one, okay? That means x one, y one, z one should be satisfying both these lines, right? So x one is going to be two lambda plus one, y one is going to be lambda minus three, z one is going to be minus three lambda plus two. Now if you subtract these two equations, let's say I do two x one minus y one, or you can do one thing, you can just take according to the requirement of the question, you can call this as x one point itself and you can call all these points as y one itself, okay? x one, y one are some parameters. So here x will be two x one plus one, y will be, let me clear off everything else which is not required. So x one will be two x one plus one, sorry, x will be two x one plus one, y will be x one minus three, okay? And z will be negative three x one minus two. Similarly for the other one, it will be x equal to y one plus two, y will be equal to minus three y one plus one, and z will be equal to two y one minus three, okay? Now why I took this situation is because these two should be equal, these two should be equal and these two should be equal. That is two x one plus one should be equal to y one plus two, x one minus three should be minus three y one plus one, and minus three x one minus two will be two y one minus three, okay? So this will result into your first equation. So this will be your first equation. This would result into your second equation and this would result into your third equation. Now here your x one and y one value should be existing. That means you should be able to find the values for x and y one and that's why they should be consistent. So if you are able to find x one y one values, that means these two lines are intersecting and since they are not parallel, they have to be co-planar, okay? So statement one and statement two both are correct and statement two is still explaining statement one. So again in this case, option number A will be correct. Okay, so ignore this, this is just a repetition. Ignore this part, just do the question itself. Okay, so Amogh, see to answer your question, how is statement two explaining it? If you go back to the previous question, x one and y one are like your parameters over here. So if these values of x one and y one are found out, that means there exists some value of x one and y one. Okay, that means these two intersect. The moment any two lines intersect, right? We can always pass a plane through them and call it co-planar, right Amogh? So this statement two is actually trying to say that these two lines actually intersect and hence they are co-planar. Is that fine Amogh? If you have any other question, follow up question. Yeah, we realize that using the co-planar question which was not the intention of the question, the question was actually meant to be solved by using the fact that these two lines intersect, that's it. What will make these two lines intersect? That is statement number two is saying that. So they will intersect only when there is a solution for x one and y one. It should not happen that the answer from these two is not matching with the third one. If there is an answer, that means they intersect and hence they are co-planar. So we started by using the determinant methods, but that was not the intention of the question. So even statement number two could be used to prove that the lines were co-planar. Determinant method is another method to prove that the lines were co-planar. Yeah, so this question is easy. So let's take this up first. So the volume of the parallel pipette formed by this is maximum, okay? Okay, so this statement two is definitely correct. And the volume of the parallel pipette is given as modulus of the determinant formed by one a zero in this case and a one one and zero one a. Okay, so if you simplify this, it becomes a minus one and minus of a q. So db by da should be zero. That means one minus three a square should be zero. So a should be equal to plus minus one by root three. So double derivative of this is going to be minus six a. Which by the way, when I say determinant, so when a is equal to plus one by root three, this is going to be negative, right? When a is going to be minus one by root three, then double derivative is going to be positive. So this statement one is wrong. Statement two is correct. So d will be your right answer. That doesn't matter, Amok. Statement one is false, that's it. Whether it's explaining or not, it's immaterial because one is false. Oh, you were asking about the previous one. Okay, so this is clear. Oh, sorry, I meant I thought that it was for this question. So again, a question from differential equations. Okay, Aditya, Sondarya, all right. So if I talk about the first equation, this is actually a linear differential equation, right? So here the integrating factor is going to be e to the power integral of 2x dx. That's going to be e to the power x squared. So your solution is going to be y into e to the power x squared. So integral of 2 to the power, 2 into e to the power x squared, x squared. That's going to give you y into e to the power x squared is equal to 2x plus c. So statement one is correct, that's for sure. Now, statement two says that if y is equal to this, then dx, that means dx by dy have to show that it is e to the power minus x squared minus 2xy. So let's take this expression, differentiate both sides with respect to y. So it will be e to the power x squared plus y into 2x e to the power x squared dx by dy is equal to zero, okay? Unfortunately, I don't see this coming. I don't see this equivalent to this. So statement number two is going to be wrong. Statement number one is right, so option C is going to be correct. All right, so last question that is the 30th question for the day and we are done with the session then. Okay, so the distance of origin, this distance is given to us as, and from here, there's a point Q which is at a distance of one by P from this. And I have to find out the locus of Q, okay? So first of all, small P is going to be, if I'm not wrong, one by under root of cos square theta by a square plus sine square theta by b square. That's going to be a b under root of b square cos square theta plus a square sine square theta. This is going to be your P. Now, in order to find out the coordinates of Q, I would use the distance form of the equation of a line. I'm sure all of you are aware of the distance form that is x minus x one by cos theta equal to y minus y one by sine theta is equal to r. Where r is the distance of the unknown point x one y one from the known point x one y one. So Q is my unknown point, so let it be x. This point is a cos theta, okay? Now, the slope of the line, the slope of this tangent is nothing but b cos theta by minus a sine theta, right? So I can do one thing. I can write this minus a sine theta, okay? Or you can say, we can find, this is your tan of theta. This is your tan of, or you can say tan of phi. So cos of phi is going to be minus a sine theta by under root of a square sine square theta plus b square cos square theta, okay? So this expression I'm going to write over here. Minus a sine theta by under root of a square sine square theta plus b square cos square theta. This is equal to y minus b sine theta by sine phi. Now, let me call this as phi because I've started calling it phi over here. Theta has already been used. That's why I'm calling it as phi. So sine of phi will be b cos theta by again under root of a square sine square theta plus b square cos square theta. And this expression is one by p. This expression is one by p, right? Now let me rewrite this so that we can understand this. So we have x minus a cos theta divided by minus a sine theta by under root of a square sine square theta plus b square cos square theta. And this is equal to y minus b sine theta divided by b cos theta by under root of a square sine square theta plus b square cos square theta. And this term is one by p. One by p means under root of a square sine square theta plus b square cos square theta by a b. So from here I can say that x minus a cos theta by minus a sine theta is going to be one by a b. That means x is going to be a and a gets canceled. Or let it be at the place. So x is going to be a cos theta minus a sine theta by a b.  Okay, now I have to eliminate theta from here. I have to eliminate theta from these two expressions. Okay, so I can see one thing that if I do x by a square and y by b square, I get something cos square theta plus sine square theta by a square b square minus two.   This is going to be just b square.  Just a minute. I think I had already taken a comment down, right? This is going to be your a, this is going to be your b and b. So x by a is going to be a cos theta. Okay, so I should not have this thing over here. And this is a square b square, that's correct. Okay, and let me change here itself. So a, a and I'll bring a a over here. This also b, b and let me bring a b over here. Yeah, so I won't even have this b square term over here. And again, I'll have a square b square and this will be two a by b. So when adding these two will get canceled, these two will add up to give me one and these two will add up to give me one by a square b square. Okay, so my final locus will be this expression, which I believe is your option number a. So a becomes the right option in this case. Okay, this is slightly lengthy question, but the process was simple. You had to use a distance form and of course you should be knowing the equation of a tangent in parametric form. Okay, so guys, thank you so much for coming online. Okay, so over and out from my side and all the best for the exam, computer science exam, which is left off. We'll meet again with the CET session on Tuesday, next Tuesday.


QuickBooks Desktop 2023. Adjusting entries and reversing entries introduction. Let's do it within two-inch QuickBooks Desktop 2023. Support Accounting Instruction by clicking the link below, giving you a free month membership to all of the content on our website, broken out by category, further broken out by course. Each course then organized in a logical, reasonable fashion, making it much more easy to find what you need than can be done on a YouTube page. We also include added resources such as Excel practice problems, PDF files, and more like QuickBooks backup files when applicable. So once again, click the link below for a free month membership to our website and all the content on it. Here we are in QuickBooks Desktop. Get great guitars practice file. We started up in a prior presentation. Going through the setup process we do every time. Maximize the home page to the gray area in the view drop-down. We've got the hide icon bar, open windows lists checked off, open windows open on the left. Reports drop-down, company and financial. Let's open up the profit and loss, the P and L. We're going to go for the first two months on the range change. 010123 to 022823 and then customize it. Fonting and numbering, changing to 14. OK. Yes and OK. Reports drop-down. Again, company and financial. This time the balance sheet. Customize it. We're going to change it from 010123 to 022823. Fonts and numbers changing the font. Bringing it up to 14. OK. Yes and OK. That's the setup process that we do every time. Now let's talk about adjusting and reversing entries. Adjusting entries being entered as of the end of the period. Whether that be the end of the month or the end of the year, reversing entries, reverse some of the adjusting entries and are typically entered the day after the end of the period, which would be the end of the month or the end of the year. Let's first think about why adjusting and reversing entries are used when you might use them for your own bookkeeping and when you might use them if you are a bookkeeper, for example, doing multiple clients. How can you set up a system that would then work well? Adjusting and reversing entries are typically accrual concepts that are designed to get our financial statements reported as accurately as possible as of the cutoff date, that being the end of the month or the end of the year. You might think that if you're smaller business and if you're on a cash-based system that you don't need adjusting or reversing entries and that's not necessarily the case, although you might have different or less adjusting and reversing entry requirements than, say, a publicly traded company, for example. For example, down here we know that property, plant and equipment, depreciable assets is one area where we have to deviate from a cash basis method to an accrual basis method because if nothing else, the tax code basically requires us to do that. So the first way to think about the accrual process is to think about large companies that are publicly traded, for example, and have to deal with regulations on how they report stuff such as generally accepted accounting principles as opposed to small companies that have to deal with, if nothing else, the tax code if you're in the United States.  So most like classic adjusting entries have an income statement account, at least one income statement account, and at least one balance sheet account because they deal with timing to get the income and expenses reported in the proper period. Now, small companies might be on a cash-based method and you don't have to deal with generally accepted accounting principles. You don't have that same kind of requirement. And in some cases, you might be allowed to be on a cash-based method as opposed to an accrual method. In that case, the main purpose you're putting the financial statements together, at least from a regulatory standpoint, might be for taxes. You've got to fill out the taxes. If you're sole proprietorship, that'll just be a Schedule C. But you might have other requirements if you're some kind of corporation, S corporation or a partnership or LLC. So you're making your financial statements in order to deal with the tax code. Even the tax code, if you're going to try to put your financial statements on an essence of tax basis or one that you can use to, in essence, do your taxes as easy as possible, will require some adjustments, for example, as we saw with the depreciation. You also, as a small business, might have some requirements for reporting purposes to use an accrual basis, possibly if, for example, you have inventory, possibly from the tax code, and if you want to loan the bank, then might ask you to present your financial statements in whatever way the bank wants it in and possibly require you to have an audit or something like that. And in that case, you've got to conform to whatever the bank wants. And then, of course, on the internal side of things, just from a bookkeeping standpoint, the accrual method is typically thought to be more accurate than the cash method. That's why you have to deviate from the property, plant, and equipment, because if you report your expenses down here, let's run this income statement, for example, by month and say I'm going to hit the month dropdown. If I reported all the purchases of my equipment or my property, plant, and equipment in January, because I paid cash for like $100,000, then I couldn't really compare January to February. It's not a very fair comparison, because in reality, even if I paid cash for it, if I purchased a building, I'm going to use that for 30 years into the future. So for a comparative purpose, it doesn't really make sense to expense it. It doesn't really help in some ways, depending on the types of budgeting that we're going to do going forward. So the accrual concept was first come up with in order to have better reporting, so to have better internal reporting than you might want to be more on an accrual basis method. Now, the other thing to think about is, well, if I'm going to do these period end adjustments, when do I need to do them?   Maybe you're going to say, I'm just going to wait till the end of the year and do the adjusting entries at that time, because that's when I'm going to report the financial statements, and that's when I need them for the taxes. So possibly you're saying, okay, I know there's some timing issues at the end of January and the end of February, but most of them wash out because they're just timing issues. They're reported in January instead of February, but I'm not going to worry about the cutoff too much until I get to the year-end cutoff, because that's when I'm going to report my taxes. That's when I need to really get things honed down to report the taxes. So for small companies, you might be more on a year-end kind of basis. So you might be using a cash-based method. You might be working with a CPA who at the end of the year can do the adjusting entries at the end of the year. And that's the next thing to think about who's going to do the adjusting entries. So I think it's often useful to think about these processes as two separate processes. You might be doing both adjusting entries and data input, but it's useful to think of them as separated. And you might have someone else doing the adjusting entries, possibly like a CPA firm or your accountant, for example. So the general idea is what we want to do is on the home page here, these represent the day-to-day transactions that we've been putting in place, these forms that build the financial statements. What we want to do on the accounting side is make the data input as easy as possible. When I run payroll, for example, I don't want to have to worry about the cutoff, the fact that the payroll is going to run over some days before the cutoff at the end of the month or a year and some days after the cutoff at the end of the month or a year. What I want to do is run payroll as easy as possible. It's complicated enough. I don't want to make it more complicated. And if there's any adjustment then that needs to be made, then I'd like to make that adjustment using a period-end adjustment. So that's the idea that we're going to use. On the bookkeeping side, I want to make the data input as easy as possible. I would like to make the data input so easy that I can hire someone to just do the data input without having to worry about any other stuff by setting up the forms and all that kind of stuff. That ease of data input will result in some cases where my timing is not perfect, such as the payroll, for example, and that means I'm going to have timing differences where I'm reporting the expenses not in the proper period. And I'm just going to adjust for that at the end of the month or year. Now, we're going to adjust it as of the end of February, and you might ask why didn't we do it at the end of January? We could have, but I'm just going to do it as of February because I want to give the example of what you would do. Many small businesses might not do this at all until the end of the year because that's when they're going to basically do their taxes. Larger businesses might do it monthly or quarterly because that's when they have to report their financial statements. So that's the general idea. So what I want to do as of the cutoff, which for us is going to be the end of February, is try to get our financial statements as best as possible, as correct as possible on whatever basis that we're going to be on. We will imagine an accrual basis, but the same concepts will be there for a cashed base system. And then after I get the financial statements correct as of the cutoff date, I want to reverse anything that's going to mess up my accounting department which are just related to timing differences because the accounting department is doing fine with what they're doing. It's just they need a little tweak at the end. That's how we designed the system to be. It's not an error. We're just doing the little tweaks that we're supposed to do at the end of the period to make things correct. So let's just look at some of these accounts and see which ones might need adjustment. Cash is usually not something that we need an adjustment for. So that when it is what it is, we've reconciled it to the bank rec. So that shouldn't be something that needs an adjustment. With the accounts receivable, we saw that we're going to need an adjustment which is a little bit unusual. This is a little bit uncharacteristic of a normal adjusting entry because we had some deposits that we put in here that resulted in negative receivables which really should be liabilities. And when we put those in place, we said, hey, look, what I'm going to do is make it easy on the bookkeeper by using the accounts receivable to track the advanced deposits instead of trying to record those as unearned revenue. And then we're just going to fix it at the end of the period which might be the end of the month which we will do or at the end of the year if there's any timing differences breaking out the accounts receivable, increasing it, and recording the liability. Now notice if you're sole proprietor reporting the Schedule C and you're just doing this for taxes, you don't even really need to do that particular adjusting entry because you're only going to be reporting the income statement on the tax return. But if you're reporting the full balance sheet, then you would generally want to have the liability broken out. And then the inventory, you might have to adjust that to the inventory count. We also could have accounts receivable adjustment because we've got the wrong timing, for example. And that might happen if, for example, I'm on a job cost system and I basically try to get my staff's time and for a month and then I bill out the month after. So for example, if they were doing the work last month and then I invoiced and the invoice doesn't go out, in our case, until March, but the work was actually done in February, even though the invoice was created in March, I should technically record the income when we earned it and the work was actually done in February. So that's another thing and that does have an impact on the profit and loss. That's a traditional adjusting entry. Inventory, we might have to do the physical count, for example, for the inventory to make any adjustments, see if there's any shrinkage or anything like that that we need to account for. Prepaid insurance. The prepaid insurance is a classic adjusting entry process because you'll recall when we put the prepaid insurance on the books as an asset, we did the accrual thing similar to what we did with the furniture and fixture. Instead of expensing it, we put it on the books as an asset and then monthly, which is what we're going to do or at the end of the second month, yearly, you can make the adjustment to record the expense.  And then we've got the depreciation. This is a clear example of one that even a small business has to deal with and again, you might only deal with it yearly. What you're going to do with it is you're going to say, if I purchased large stuff, I've got to put it on the books as an asset. If I purchased a building, I can't just expense it even if I paid cash for it, even if I'm on a cash-based system, they're going to make me do an accrual thing. I've got to put it on the books as an asset and then most small businesses will work with their tax software or tax preparer to give them the adjustment and then you'll have to do adjusting entries for the accumulated depreciation. And then we've got the accounts payable and we've got the liabilities is another one, the, let's say the loan that we had. Where is it? The loans here. This one, we might have to break out the short-term and long-term portion of the loan and so we might set that up so we can break that out as well. Now, so those are just some examples of the items that we'll go through and make the adjustments for. Now, as we do this, if you're an accountant, what you want to do is say, what am I going to do on my side as the bookkeeper, whether you're doing your own books or whether you are a bookkeeper and what am I going to let the CPA firm do at the end of the period? Ideally, you would like a situation to say, hey, look, I'm going to make things as easy as possible because I'm doing the accounting side of things and maybe you have the CPA firm do the adjustments at the end of the period. And so you might want to, when we're going through these adjusting entries, for example, think about how you can do something as easily as possible just using bank feeds and making the data input automatic basically as easy as possible and then at the end of the year, possibly for small businesses, you give the information necessary such as loan documents, such as the purchases of large purchases and whatnot to your CPA firm or possibly you do it yourself at the end of the year and then just do the adjustments that are necessary periodically basically at the end of the year. That's how you kind of want to think of it from a bookkeeping standpoint. Logistically, you're thinking, who's going to do what, right? The accounting department data input or when I'm doing the data input, I want to make that as easy as possible. When I put my adjusting entry hat on or when I work with my CPA, whoever my group of team is that I'm working with to do this stuff, I want them to do the adjusting entry process and I want them to do the initial statements to do what I need to do, which means at least report the taxes, report possibly in a bookkeeping system that makes sense for my decision-making process and any other reporting that is necessary possibly for a loan or any other kind of requirements and then reverse it or fix it, make it go back to the easiest process possible for my bookkeeping, right? Don't mess me up with the adjusting entries. Do the adjusting entries up my bookkeeping. So we'll talk about that on those concepts. As we go through these line items one by one, you can look at some of these balance sheet accounts and just think about which ones might need adjustments and why. Those are the ones you want to focus in on payrolls, another one. Those are the ones you want to focus in on when you set up your bookkeeping system to think about who's going to do what, how can I make this logistically as easy as possible? What are the needs of the company? What are the taxes need? Do they have other reporting needs? Are we on a cash basis or a cruel basis? And then you could set up your system as easy as possible. Notice that if you're a bookkeeper too it makes it easier to pick up clients that are in a specialized area. So if you're in a system where you're saying I specialize in smaller businesses that are sole proprietorships and I work with a designated CPA or a tax preparer who knows how I do stuff and possibly I work with a designated payroll provider so I have a whole system set up for that type of business and then you might specialize in that type of business. Otherwise when you're widening things out obviously you're going to have to recreate the wheel with every business so you've got to think about where am I specializing in can I just scale what I'm doing or do I want to have a wider net with a whole bunch of different clients and do different stuff and then network with the people that can fill the gaps, CPA firms and tax firms and payroll providers if you're doing payroll outside of QuickBooks. Okay so we'll go through each of these one by one.


Everybody forced here ripping 22 23 impeccable basketball hobby Case number 28 Good luck everybody those results on your screen can be found anytime at latent sports cards comm results Quickly because the bottom of the website if you're on the home page and the search bars are great tool to search for Results from the past as well if you need to find them. Good luck y'all. Let's get started Got the whole case flawless and eminence next Be pumped for that No, every time we've done one of those it's it's it's get it gets kind of crazy I mean it's eminence and flawless, but like we've hit some actual monsters out of those Hopefully we can continue that Fallen. What's up Jeremiah? Happy Sunday evening. Anyone got the Sunday scaries? No three-day weekend this weekend. I got the Sunday scaries and I'll have the Monday scaries a Tuesday scaries the Wednesday scaries Then then Thursday Normal and Friday scaries Saturday cool, that's fine, and then Sunday scaries again Five out of seven scaries, dude Haunted house over here God's a Zion Williamson stainless stars numbered 14 of 99 a little bit of peel on the protective cover Get Sunday scaries at five on Friday. Yeah Her dad that's how it goes All right, let's get into some monsters. Good luck y'all fun day, but it's also a little scary, right? Sorry, it's Halliburton 16 of 75 Pacers Or Sam was a Halligod nice hardened 8 of 10 Sixers beard 8 of 10 James Harden Sixers Chris Middleton now for the Bucks numbered 18 of 99 and it hits the first hit goes the Clippers impeccable stars auto Elton brand at 25 of 25 Good stuff combo spot at Elton brand Yeah, fixies Magic on the impeccable rookie signatures Caleb Houston 84 of 99 Bucks rookie auto Marjahn boat champs six of 25. That's a sweet Marjahn. Oh Here watch that Jeremiah Those Braves man those Braves Boat Champ Chris stops for Zingas illustrious ink wizards to 99 We've got a jet home grand stainless stars rookie auto Shits to 49 we're cash and shits. That's what we're doing. Whoo cash and shits Heck yeah, okay, see That is Marvin. Hey congrats Marvin. Yeah, man so Jeremiah I drove over like the drive from here to Tampa is probably like 90 minutes with no traffic, I would say a little less well from here might be like an hour and a half pushing two hours But anyway, I took my mom my brother and my wife to Raise game last year We drove over there Jake I think that's how this one ends, but we drove over this last year Got a Keegan Murray stainless stars rookie very nice 21 to 49 Kings. Yep. I like that too with the purple Great card Kings drove over there It was rainy traffic, so no wasn't rainy The traffic was horrible literally it was bumper to bumper. I think the entire way to Tampa took us three hours to get to the top The park didn't get in game starts like immediately. You don't have any time to do anything except sit down and watch baseball Red Sox get no hit through nine innings, but don't give up any runs go to the go to the 10th, obviously Bobby Dalbeck triples to get in The first run of the game and the top of the 10th and then he comes in on the sack fly They're up to nothing Red Sox proceed to blow it in the bottom of the 10th after getting no hit for nine innings Getting our hopes up and destroying them. I watched a Kevin Kiermeyer walk-off Homer Do off Hansel Robles? Yeah, I've Hansel Robles last year. I was like one of his last appearances as a red So that game yeah 75 Jordan pool Warriors Kevin a Kevin Kier the only one in his career walk-off Homer Yeah, that was that was dude rookie auto Terry Easton of 49 Rockets Yeah, the drive to St. Pete is totally is totally horrible Kevin Porter, Jr. To 7 it's a 99 for the Rockets. That was That's exactly how the Red Sox season went all last year too, so Kennedy Chandler rookie auto to 25 for the Grizzlies That's why that's why nobody goes to raise games. It's because of the location. It's always been the location of the truck Elvin Hayes immortal ink to 25 Wizards, I know hawk you said you had the nugs, right? You got a Christian Brown rookie auto 9 to 25 That's good stuff right there Christian Brown silver to 25 rookie auto Dude serious Vaughn seriously when we stay in Boston and go to Fenway It's like the nicest thing ever just if it's nice you can walk a few miles if you're a little farther away Transportation everywhere. Oh my god. Oh, that's sick Elegance Jalen Brunson patch auto nicks game worn That is so fire Travis see with the Knicks that is a Mavericks patch, but that's a fantastic Knicks card What a hit we'll get that in a mag Such a such a great pickup for the Knicks and we've got Kings elegance rookie Jersey auto redemption Keegan Murray Kings had the stainless stars base parallel and now a Keegan Murray rookie Jersey auto redemption Wow, wow, wow Kings We the Kings Chop chop not great. Chop not great. I'm sorry Jake If Jake's lucky, they'll just move to Orlando Jake would literally probably quit working here just we go to every single home game He probably would He's shaking his head because he's thinking about it right now Here we go, we got the bar card as well Himmy Butler to 49 for the heat Keldon Johnson Spurs number to 75 Yeah, we were watching the Grand Slam live John Marant Grizzly is numbered 299 Cavaliers watercolor auto carousel vert to 99 Lakers elegance two of ten RPA max Chrissy very nice card Lakers Grizzly's Jake Laravia elegance RPA does 25 back-to-back RPA is a goal in a silver Impeccable rookie signatures hoax they take Griffin now. Oh, it's a really good really good box here stainless stars rookie auto Jalen Doran orange seven of 25 Jalen Doran that the mag and a monogenobly eight of ten indelible ink Spurs Man, oh man. That was a really good last box there like consistently Obi-Wan Jenobley Eight of ten indelible ink Spurs what a legends what a hit there's Kevin Durant shory ounce of fine silver four of 20 for 20 Kevin Durant Go to the Nets or you'll see with the Nets Give me a moment. We'll do a quick recap more mixer spots coming in nice nice nice with a nice Keegan 21 of 49 stainless stars purple autos of carousel avert Chris that's porzingus AJ Griffin Caleb Houston Elton brand Alvin Hayes Kennedy Chandler Christian Brown Jake Laravia Max Christie two of ten elegance and a monogenobly as well Kevin Durant for 22 ounce of fine silver Nets Redemption Kings elegance Ricky Jersey auto Keegan Murray Shad Holmgren of 49 rookie stainless stars auto Awesome chat to 25 Jalen Durant orange stainless stars Ricky auto pistons Alligance patch auto next Jalen Brunson to 25 game-worn patch auto and rookie auto bow champ to 25 or the bucks. Thanks again, everybody We're moving on


Hey, this is Brett and this 2020 Ram 2500 Crew Cab Short Box Big Horn Level 1 Night Edition is stock number 13693Z. I'm here at Summit Automotive in Fond du Lac, Wisconsin, your new and used heavy duty truck and Ram headquarters. This 2020 Ram 2500 has the 6.7 liter Cummins diesel engine. It's paired up with the 6 speed, 68 RFE automatic transmission. This truck has been fully safety and inspected by our service shop for the state of Wisconsin inspection process. It has a fresh oil and filter change all the fluid tab and checked and topped off and it is 100% ready to go. I'm going to go all the way around inside underneath, start it up and take a look under the hood to give you the most accurate representation of the truck possible. Diamond black pro coat is the color and I shoot all my videos in 4k so if you have HD capabilities on your computer, tablet, smartphone or television I recommend turn them on now because it is your best way to check out the quality condition options and cleanliness of the truck before seeing it in person. The only thing to note on the driver's side here is that it's got one little scratch that's been touched up on there and that's really the only major imperfection I saw on the whole truck but the rest of the box is in really nice shape. Cab and doors are in really nice condition as well, didn't see any dents or dings on there, front fender is in nice shape and if you like the video you can subscribe to my YouTube channel which is youtube.com slash summit auto. Click the bell notifications you'll get updates on the videos I do each and every day here at Summit Automotive as well as access to one of the largest catalogs of vehicle walk arounds on YouTube. Comes with the 20 inch painted alloy wheels and it has Firestone Transforce AT tires these are LT28560R20s and they have right around half the tread left on them. Nome and underbody is an exceptionally clean condition this is a one owner clean title history out of New Mexico. Night package gives you the blacked out headlight bezels, the painted front and rear bumpers, no major dents or dings on there and it does have the front bumper parking sensors. You also get the gloss black grille part of that night edition and the hood's in really nice shape I didn't see any major dents or dings on there. Passenger's side front fender looks really good as well, you do get the blacked out Cummins and Ram logos on there as well. Passenger's side front wheel, no scuffs or scrapes, also the blacked out 2500 logo and as you go down the passenger side here no major dents or dings on there take note of how clean that body is and reflective that paint is I take these HD videos so if you are far away or even if you're close by just cannot make the trip down but you're still interested in purchasing the vehicle you can see the truck here the truck and have confidence in the vehicle that you're looking at before you even get here so when you do get here there's absolutely no surprises and you can make a smart and informed buying decision from wherever you're at if this video helps you make that buying decision let your salesman know that you saw the video then it was helpful and that brought sent you get the wheel-to-wheel black painted side steps back wheels in excellent condition and the back tires have just as much tread as the front tires frame and underbody looks like it's brand new has all the remaining factory exhaust so it hasn't been altered in any way and lower rockers and cab corners look great as well what you'd expect on a truck from New Mexico once again no dents or dings on that box and as we come around to the back of the vehicle rear bumpers in great shape does have the backup parking sensors full towing package which includes the receiver hitch four pin and seven pin wiring and the tailgate is absolutely perfect looks really good comes with a spraying bed liner fifth wheel and gooseneck prep package seven pin and four pin wiring in the bed and you get the LED bed lights back here you know this one also has the cargo cam what's nice is they got a button back here to turn those lights on and off that sprained bed liner goes on to the tailgate which is really helpful as well and for full disclosure this back wheel just as nice as the rest down the rest is side of the truck once again doors all look really good has the heated fold-up tow mirrors built-in directional signals they fold up like so they also fold in like that inside the big horn level one package gives you the black and dark charcoal cloth bucket seats in the front no rips or tears on those seats they're in really nice shape this is a pretty nicely optioned out big horn level one power driver seat has the heated seats in front as well auto headlamps factory floor mats power pedals power windows locks and mirrors cloth bolsters wood grain trim these mirrors do power fold in I always like showing both sides so that you know that both sides are working properly back seats are just as clean as the front seats no rips or tears back here it does have the latch child safety system for any child car seats you may have fixed glass rear window you also get the speakers in the ceiling part of the alpine premium sound system you get a load floor back here which is nice you got muddy boots or tools put them on there you're not going to get your seats dirty or if you got pets put both sides down they got a nice flat place away and they're not going to get your seats dirty comes with a winter front as well and then on the other side is your factory subwoofer you also get in-floor storage with removable bins for easy cleaning you can put ice and drinks in there so you have too many coolers wherever you go or if you're not in Wisconsin just extra space side curtain airbags child safety locks on the back doors and the inside and bottoms of the doors all look really good didn't want to show you the vin sticker as long as I'm standing right here and then the tire and loading information sticker as well we'll hop inside check out the miles radio and everything that this truck has to offer on the interior you can see that this one has 23,030 miles you do get a digital speedometer and that instrument cluster is very nice and clean comes with the heated leather wrapped steering wheel cruise controls and gear selector on the right bluetooth and information center controls on the left and it also has the audio controls on the back of the steering wheel your six speed automatic shifter 68 RFE automatic transmission comes with the 8.4 for C radio this one does have the factory navigation system there's highway 41 so you can see that is working nicely AM FM and serious XM radio capabilities and you get your heated seats heated steering wheel buttons as well as the cargo cam and backup cam and you can zoom in on that backup cam over your receiver hitch to get hooked up to your boat camper trailer the first time every time you'll sub all your different apps on here including projection manager you can project your cell phone to a screen via Android Auto or Apple CarPlay which is a demo on Android Auto on an 8.4 for C radio if you want to check that out that's in the upper right hand part of screen down here is your push button four-wheel drive your heated seat heating steering wheel buttons your climate controls factory brake controller also as your factory exhaust brake tow-haul mode and can turn those front and rear parking sensors on and off two USBs two USB C's and an aux jack and then cell phone holder coin holder as well as cup holders keyless entry with your remote start on there that's in nice shape as well dual glow boxes and the passenger side format and seat are in excellent conditional rips or tears on there and the headliner is in very nice condition I don't think this truck's ever been smoked in smells very clean inside you get map lights up there assistant SOS buttons auto dimming mirror and your home link buttons for your garage door security systems and lighting systems we'll start it up and take a look under the hood starts right up no check engine lights or anything like that I would personally like to thank you for checking out the video today and hopefully from this HD video you've been able to verify the quality condition options and of course cleanliness of this truck all the way around inside and out under the hood we have the 6.7 liter Cummins diesel engine engine base very clean runs very smooth once again this truck has been fully safety to and inspected by our service shop as a fresh oil and filter change all the fluids have been checked and topped off and this truck is 100% we're ready to go there is the emissions sticker and I would highly recommend this truck for make quality and condition standpoint and to see more pictures of this truck or one of our other 550 new news cars trucks scvs minivans wranglers half tons three quarter tons one tons you name it we got to go to that website right there summitauto.com full pictures and descriptions of every single vehicle from two locations all at summitauto.com and if you'd like to check out more HD videos you can go to youtube.com slash summit auto click the bell notifications you'll get updates on the videos I do each and every day here at summit automotive in fact any second you receive link to subscribe my youtube channel the upper left link to more ram 2500 truck videos like this one the upright link to this vehicle on our website lower left then a link to one of our latest youtube videos and lower right click those checks out we're super excited to help you with this ultra clean 2020 ram 2500 crew cab short box big horn level one night edition and diamond black pistol pro coat thanks again for checking out the video remember to like subscribe and share on the youtube channel i really appreciate it thanks again


Education and learning. In Australia there are many different education opportunities for both children and adults. Education will probably be one of your first priorities so it is important that you get to know the opportunities and options available to you. For example you might be able to access part-time study or evening classes if you have work or family commitments during the day. In Australia education will be different depending on where you live and it will probably be different from what you are used to. It might take some time for you to understand the system. Learning English. Learning English will often be your first step towards settling in Australia. English language skills are very important. They can help you find a job and settle in the community. In Australia English classes are free and they are a great way to meet new people. There are lots of people at all levels so you don't need to feel embarrassed if you don't speak English well. Peter McMillan is an English language teacher. My advice would be to make the most of your English classes because life can be much more difficult if you don't speak the language in Australia. Learning English helps you to find a better job. It helps to make your life easier every day. There are many students who go to work part-time or even full-time and they come to study in the evenings just because you have a job it doesn't mean that you have to stop studying. You should enroll in English classes as soon as possible. Ask your case worker to help you find English language classes for yourselves and your children. Agok and Adit have both had to learn English when they got to Australia. It's important to study English in Australia so that it can get familiar with other people. Maybe if you are looking for a job you will not get it if you don't know English so English is really important for those things. Talk to each other about your lessons and try to help each other. Going to school. In Australia all children are expected to attend school. Schools offer safe and secure environments. Before starting school most 4 to 5 year olds go to preschool. Preschools help children develop physically, emotionally and socially in the year before school. Generally primary school is for children between the ages of 6 and 12 years. High school is for children aged between about 13 and 18. Children will be expected to attend classes from the morning until mid-afternoon. You should make sure that they eat a healthy breakfast and pack a healthy lunch and you should notify the school if they're sick. You will also have to pay some school fees and you might have to pay for some school uniforms, books and excursions. Children are placed in classes according to their age not by how much schooling they have had. Some of your children will be behind in the Australian school system others might be ahead. You need to be patient with your children and with the system. Your children might need to take extra efforts outside school time in order to catch up. In Australian schools students learn how to socialize with people their own age. They learn how to work in a team, complete projects and recognize the strengths of different people. Schools also encourage independent thinking, sharing opinions and open discussions. High schools might also teach your children about sex education to inform them about some potential dangers such as unwanted pregnancies and sexually transmitted diseases. Diane Fuller is a primary school teacher. We always talk about values, about respect, about responsibility, about valuing friendship. In school children need to be well behaved. They are not allowed to hurt other children. This is known as bullying. Bullying is something we don't accept. It's very important if someone is being bullied that they come and talk to someone that can help them. In Australia parents are often involved in their children's school. If you have children at school consider joining the Parent-Teacher Association and becoming part of the school community. This can help you meet other parents and understand the school system. Mr. Puteho knows how important it is to get involved. As a parent I always attend all the meetings that I'm invited to and participate and then at least you know get to understand and know what is happening around the school and what is expected of me as a parent and what is expected of my child at school. Your child might need your help with homework. Ask your children's teacher about ways to help. If you feel you can't give your children as much help as you would like, ask if there are any homework clubs or study groups available to help your children. Further education. There are three main options for further education in Australia. These are going to a university, studying at a technical and further education institute called TAFE or completing an apprenticeship to learn a trade. It is important to keep your mind open about these different options and different career paths. Some people are more suited to becoming a doctor or lawyer. Others are more suited to trades such as becoming a builder or mechanic. Different options require different types of training. Think about the most suitable options for you and your family. Johnson, Aminata and Igut are all completing further education. We are coming from Africa to Australia you might you know find the you know the way of education is completely different. To get into university is very hard because you know there are a number of people that are competing. You might have to do foundation course, you might have to do English language, diploma and if you really want to get into uni then you will but not all of us are meant to get into uni. I thought it would be a stepping stone for me to start with TAFE instead of just coming and straight off going to uni. At TAFE things are done more practical whereby at the university they have to do things theoretically. Mechanics, plumbers they make more money than the office people. What I've learned from TAFE helped me a lot. I went through the application forms and I saw one for nursing which I tried to do in my country before. So I said okay the best they can do is to just apply and see what goes up. So I did it and I got accepted. Apprentices you study and you get money at the same time and do things practically and then in the end you will get certificate. When I went to TAFE I thought I was too big to even go to TAFE again but when I went to TAFE on the first day we started school I met people that are even older than my mom in Africa I tell you. Ask your settlement service provider to help you find out more about studying at university, TAFE or about completing an apprenticeship. Did you know? Did you know that you may need to study to get the job that you want? There are no age limits on people going to TAFE or university and there are lots of courses available.