 Hello, good afternoon. Welcome to the J main problem solving session. So those who have joined in the session I would request you to type in your names in the chat box so that I know who all are attending this session Moreover just a brief up on what are these topics that I'm going to talk about today So we are going to talk about trigonometry with special focus on properties of triangles We are also going to talk about topics related to permutation combination and a bit of coordinate geometry So many of you had requested me to hold a session on these three topics so this topic is exclusively on these Problem solving for these topics only So those who have joined in the session, please type in your names in the chat box. Hello. Good afternoon, Vaishnavi All right, so let's get started with the very first problem of the day The number 24 factorial is divisible by Which of the following options? Rotten, Sondarya, Shruti Those who have written computer science exams yesterday hope everything went well Good, good to know that. Again, let's try to solve all the questions that you see today well within three minutes The lesser the better. Yeah, so I'll wait for response to come at least from three of you before I start solving the question Anyone? Okay for this question, I'll just give you a hint We all know that coefficient of any prime number P in n factorial is given by the formula GIF of n by P, GIF of n by P square GIF of n by P cube and so on till we start getting zeros correct Now look at the numbers that we have over here all these numbers if you see They're all made up of the prime factors 2 and 3 correct So try to see or try to find out What are the what is the exponent of the prime number 2 in 24 factorial and what is the exponent of the prime number 3 in? 24 factorial Okay, and let's say to see which of the following options is the best fit option for 24 factorial to be divisible by so if I talk about e2 e2 of 24 factorial that would be 24 by 2 24 by 2 square 24 by 2 cube 24 by 2 to the power 4 and beyond that there's no point going because I'll start getting zeros after that Right, so that will be around 12 6 3 plus 1 Right, that's going to be 22. So from 24 factorial we can extract The prime factor to 22 times Okay, let's talk about the exponent of the prime number 3 in 24 factorial so that will be 24 by 3 24 by 3 square which is 9 24 by 3 cube that would start becoming 0 correct So that will be 8 plus 2 only which is 10 So I can write this as this into 3 to the power of 10 correct, okay Now if you see we can make a 6 maximum of 6 factor can be made maximum 10 times so option 1 cannot be correct Okay Now let's try to make 24 Okay, to make a 24 I would need a 8 and a 3 correct, so we know that 24 is made up of 2 cube and 1 3 correct now Given that I have 2 to the power 22. I can extract a maximum of 2 to the power 3 7 number of times And of course 3 can also be extracted 7 number of times and we would be left with 1 2 and of course 3 3's right, but anyhow, this is going to be 24 to the power of 7 into 54 you can call it that means I can always extract 24 to the power 6 from this because if this number can divide if this number can divide 24 factorial so can 24 to the power 6 can okay, so option B is definitely going to be the right option So I am going to stop here because just a single option correct question. Is that fine? Hope you all have recalled this formula, which is useful Yes, option B is correct Kushal. Absolutely Let's move on to the next question. Now simple one find the value of 35 c8 Right plus summation of 42 minus rc7 from r equal to 1 to 7 and summation of 47 minus hc40 minus h From s equal to 1 to 5. I think this is an easy one because you have done this in the very Initial part of your permutation combination chapter Let me remind you once again. We are going to focus mainly on permutation combination Solution and properties of triangles Including a bit of trigonometric equation as well as a bit of coordinate geometry with special focus on circles So how many of you are writing the GMN exam on April 8th morning session? Anybody who's taking it on April 8th morning session. Oh Great. Okay. So just two of you. Yes guys are waiting the answer for this So we have been given 35 c8 and This is going to be 41 c7 40 c7 39 c7 38 c7 37 c7 36 c7 to 35 c7 Okay, and here we have we can actually write this term as 47 minus sc7 So that's going to be 46 c7 45 c7 44 c7 43 c7 till 42 c7 Okay Now the formula that I'm going to use over here is the Pascal's identity this says that Ncr minus 1 plus ncr is n plus 1 cr right So if you club the terms 35 c8, which 35 c7 Okay, so these two terms are going to generate 36 c8 Correct. So I'm going to get 36 c8 So 36 c8 and 36 c7 is going to give me 37 c8 Right, then this will give you 38 c8 39 c8 40 c8 41 c8 42 c8 43 c8 44 c8 45 c8 46 c8 47 c8 so answer is 47 c8 which is going to be your option number D. Is that fine? So, please do not forget this very famous and useful identity on binomial coefficients ncr minus 1 plus ncr is n plus 1 cr Great, so first one to answer this correctly or Spurik Well done. Let's move on to the next question number three 18 guests have to be seated half on each side of a long table for particular guests desire to sit on one particular side and Three others on the other side the number of ways in which the sitting arrangement can be made is which of the following Okay, anybody else So far this site has responded All right, so let's discuss this So first of all, let's say the four particular guests want to sit on that side where you have round chairs Okay, so let's say four guests want to sit on this side and Three particular guests want to sit on that side where you have square chairs Now you are not choosing them. They have already mentioned that there are some particular guests Okay, and one particular side. So you are not choosing them because they have already been decided correct yes or no now let's say these four guests have sat in these four chairs Right, so obviously we have to fill the side. Let's say I call this as side a and side b So I have to fill the side a with five more guests correct and out of 18 Okay, let me make let me make three guests also sit on this side So let's say this chair is gone this chair is gone and this chair is gone now out of 18 4 plus 3 that is 7 have already been seated that means 11 of them remain and Let me start by filling The side a of the table. So I need five more people to make it a set of 9 So I can choose it in 11 C 5 right and Whatever people remain they can be sent to the side B So if you want you can write 6 C 6, but this is immaterial because that's going to be water Okay But on each side we can now arrange the nine guests who have who are seated So side a the nine guests can be arranged in nine factorial ways side B also nine guests can be arranged in nine factorial ways and Therefore your answer is going to be option number B Okay, let me tell you this is your Adi Sharma question. It's a very simple one. Let's take another simple one The number of arrangements that can be made with the letters of the word Mathematics in which all the vowels come together Okay, remember the vowels are a e a I in this So a e a I So M M T T H C S so these are the vowels. These are the consonants So how many arrangements can be made with the letters of the word mathematics in which all the vowels come together very easy question I think you should be all able to answer this within 30 seconds Okay for C Oh B, okay Anybody else sigh perfect kushal. Okay, mostly people are going with B So this we are going to apply a string theory. We're going to tie this up So we have one two three four five six seven eight alphabets But out of that, I think two M and two T's are repeated And not only that you can make an inter arrangement of the vowels that can be done in four factorial divided by two factorial because two is repeated So that's going to be eight factorial nine four factorial by two factorial two factorial two factorial guys Let me tell you these are all very simple questions. I think they all come in the beginning itself later on you realize Questions are becoming slightly more difficult So you have to save your time and energy in these questions Again a simple one Question number five the letters of the word random are written in all possible orders and these words are written out as in a dictionary Find the rank of the word random Again should not take you more than 30 seconds Okay Alright, so let's slide down the number these letters in alphabetical order, which will be a d m and oh, okay A d m and or that's the alphabetical order in which you can write this the number of words beginning with a that would be five factorial But D also is five factorial with m also will be five factorial with n also five factorial with o also five factorial right So it's almost five times five factorial. That's going to be 600 Okay, now let us begin with our the first word will be Okay, our a random also start with our a so we'll have our a d and For that that will be three factorial words then our a M That also will be three factorial words then our a n Random also starts with our a n. So I have to fill the next one in order. So our a and d Okay, random also starts with our a and d right So I have to fill the next in order So our a Nd is gone. So next word will be m o. That's one and then our a and d o m That's another one. So let's add them up. So that's going to be six plus six twelve twelve plus two fourteen So the rank of the word random is going to be six one four that's option a very simple simple question and Most of you have done these type of questions before as well Next let's talk about this question question number six the sum of all five digit numbers that can be formed using 1234 Five when repetition of digit is not allowed. So we have to find the sum of all digits Some of all five digit numbers Okay anybody else All right, so let's see a five rated numbers. So how many of these numbers will end with a five? We all know four factorial such numbers will end with a five How many of them will end with a four again for factorial? How many of them will end with a three again for factorial and with the two is four factorial and with the one will be four factorial correct? So if you add all the units digit That's going to be one plus two plus three plus four plus five into four factorial right So that's going to be 15 times 24 15 times 24 is going to be 360 This is just the sum of the numbers or some of the digits in the units place right In a similar way can I say some of the digits in the tens place will also be 360? But since they are occurring on the tens place, they will have a place value of 10 correct, so you'll have 360 as the sum of the numbers in the units place Same would be true for the sum of the numbers in the tens place But I will be multiplying it with a 10 because of its place value of 10 So in a similar way we'll have 10 square That's the sum of the numbers in the hundreds place then 10 cube Right the unit 10 hundred thousand and this will be 360 into 10 to the power 4 Right right, so this is going to be my sum Okay, so this is going to be three six zero followed by four zeros and Again, we'll have three six zero followed by three zeros Again, three six zero followed by two zeros and finally three six zero. So that is going to be zero Six, I'm sorry. I think yeah, I made a mistake here. So there'll be a three six Double zero over here and three six zero. So zero six nine nine nine nine three That's going to be option number D in this case so yes Option D is correct and the first one to get this correct was Saundariya. Well done Let's move on to the next question The number of triangles whose vertices are on the vertices of an octagon But none of whose sides happen to come from the sides of the octagon Okay, this is a question number seven for you. So Please mark your response against this Yes, exactly. So they are made out of the diagonals Okay Okay, so I have changed the answer Kushal says see What about others? Okay, so first of all if you don't have any restrictions how many triangles can be formed So number of triangles without any restriction that would obviously be eight C three, right? Right. So you have to choose three vertices from eight of the vertices Right, that's eight C three Now let's subtract those cases where I find alpha I form such a triangle which has one side common Okay, which has one side common With the octagon. So first of all I can choose one side itself in eight C one way, right? Okay, so let's say I choose this side. Let's say I choose this side Okay, now exactly one side when I say one side means exactly one side Okay, now for the rest of this vertex I mean two of the vertices are already fixed for the remaining vertex. I cannot choose seven and four Correct. So I have to choose one from eight one two or three that can be done in four C one way Right. So I can say a total of 32 triangles would be such that it has exactly one side common with the octagon Sounds good Now what are the total number of triangles which has got exactly two sites? Let me not use the word exactly over here. It's just two sites Okay Now in that case you have to choose two continuous locations Right, so you can choose one two three. I can write it down one two three two three four three four five four five six five six seven six seven eight Seven eight one correct. So let's count it one two three four five six seven and I can also have eight one two Right, that's eight. So altogether eight such triangles are possible So now what I have to do is I have to do one Minus two minus three to get exactly those many number of triangles which has got no sites common with the octagon So that's going to be eight C three eight C three is going to be eight into seven into six by six Which is going to be 56. So 56 minus 40 is going to be answered and that's going to be 16 So option number D is correct next again a simple question question number eight the number of odd proper devices of Five zero four zero all right. So first. Let us factor is five zero four zero Okay, so five zero four zero can be factorized as two to the power four Three to the part two into five into seven now if you want all the factors to be odd, you should not select any factor to right So it's like selection of None or more objects from Three five and seven which are present in these following quantities. So think as if you have Two threes present you have one five present and you have one seven present and you have to select None or more from them So how many ways can I select a three I can select either one or a two or none So there are three ways to select a three Okay, that is nothing but two plus one In a similar way the number of ways to select a five will be one plus one a number of ways to select a seven will be one plus one Right, you understand why an additional one is over here because you may not even select it. So that's one more case right But let's if you don't select any one of all you are actually selecting one and that's actually Not a proper divisor. So I have to subtract the one So the answer is going to be three into two into two minus one. That's going to be eleven and That's going to be option C. That's correct. Is that fine next There are five balls again back to back the ninth question on permutation combination There are five balls of different colors and five boxes of colors same as those of the balls the number of ways in which The balls each one each in a box could be placed as that exactly one ball goes to a box of its own color Okay, okay, most of you are going with 9b. So let's say we have Boxes of these colors Okay, and even balls of the same color right violet indigo blue green and yellow now the question says the number of ways in which The balls one in each box could be placed says that exactly one ball goes to a box of its own color That means only one ball goes to the box of its own color Rest all the balls are in different color boxes. That means no other ball can go to the Box of its own color right? So it's very obvious that let's say I choose a ball which I want to send in the right color Let's say V goes into V That means the rest four should be deranged correct, so we have to derange these four balls and We know the formula for derangement. That is equal to n factorial 1 minus 1 by 1 factorial Plus 1 by 2 factorial minus 1 by 3 factorial and so on till minus 1 to the power of n 1 by n factorial, correct, right? So these four balls can be deranged in This way, which is okay So first two terms will actually get cancelled so 1 by 2 factorial minus 1 by 3 factorial plus 1 by 4 factorial that's going to be 12 minus 4 plus 1 that's going to be 9 ways correct for each of these nine ways Right each for each of these balls there are nine ways to derange the others So your answer is going to be five times nine. That's going to be 45 Option C is correct. I is nobody got this correct Surprisingly, is it clear now? So your answer is just five d4 d4 is nine. That's going to be 45 next 10th question The number of ways of number ways to give 16 different things to three persons ABC so that B gets one more than a and C gets two more than B is This should be easy Absolutely, so let's say a gets x items B gets x plus one items and C gets x plus three items together. They should be 16 items So it's very obvious that three x plus four should be 16 So three x is 12 so x is four Right, so a gets four items five items and seven items correct So the number of ways to distribute is 16 C4 Into 12 C5 right if you want you can do 7 C7 as well Right, that's just going to be 16 factorial by 4 factorial 5 factorial 7 factorial That's going to be option number a It's a less than 10 second problem guys Let's now take up some problems on Trignometry find the values of x satisfying this trigonometric equation Okay, again, the best way would be to just put these values Pi by 3 pi by 6 pi by 2 pi by 4 etc and see which of them satisfies Okay, but however if I have to do this in a rigorous way I would first try to convert cos of 2x as 2 cos square x minus 1, okay and Let me bring these terms out to the left-hand side so this will become negative of root 2 plus 1 by root 2 Times root 2 cos x minus 1 okay equal to 0 right Now this itself can be factorized as root 2 cos x plus 1 times a root 2 cos x minus 1 You can take root 2 cos x minus 1 as common correct and you'll end up getting root 2 cos x plus 1 minus 1 minus So this gives you root 2 cos x minus 1 times 2 cos x minus 1 equal to 0 So your cos x can be 1 by root 2 or your cos x can be half But in the question it is said that cos x is not half So this is rejected So we have to write the general solution for cos x equal to half, which is x equal to 2n pi plus minus pi by 4 and belonging to integer An option number D becomes the right option in this case Let's move on to the next Question number 12 if the value of this determinant is 0 then b is which of the following Okay So let's expand this first. So when you expand it you get cos a plus b times Cos a cos b minus sin a sin b that is again cos a plus b Then you get sin a plus b times Uh Sin a cos b plus sin b cos a which is again this And finally cos 2b times You get a sin square a plus cos square a which is going to be 1. So this is 0 So 1 plus cos of 2b is 0 Right, that means cos of 2b is negative 1 And in that case b is nothing but A odd multiple of pi Right, so b is going to be 2n plus 1 pi by 2 and belonging to integer Again, this question is just a 15 20 second question not more than that Let's move on to the next one question number 13 The smallest positive value of x and y satisfying x minus y is equal to pi by 4 And cot x plus cot y is equal to 2 You can look at the options and then you can see where you are getting that x minus y is equal to pi by 4 That would be a better way to solve this Yeah, it's a super simple question Option b only satisfies it, but let's say if you're to solve this rigorously. How will you solve this? Now assuming none of these is not an option. Okay, then only you can take this chance so basically if you know x is equal to uh y plus pi by 4 you can place it over here. So that becomes cot of Y plus pi by 4 plus cot y is equal to 2 Now we all know the formula for cot a plus b That's reciprocal of tan a plus b So we all know that it's going to be Tan a plus tan b by 1 minus tan a tan b So this is going to be 1 minus tan y by So let me just write as this is possible So that will become our 2t minus t square plus 1 is equal to 2t square plus 2t 2t 2t will go off. So you'll have 3t square as 1 that means T is going to be plus minus 1 by root 3 So the smallest positive value will come when you have tan theta is plus 1 by root 3 So theta has to be 5 by 6 or in fact you can say y has to be Y has to be 5 by 6 So that's only given by option b and y is 5 by 6. That means x has to be 5 5 by 12 So option number b is correct Let's move on to the next one question number 14 The number of values of x satisfying the equation mod of sine x is equal to sine x plus 3 in the interval 0 to 2 pi Absolutely, there would be no solution because Even if you take sine x as positive you'll get 0 equal to 3 if you take sine s as negative We'll get minus 2 sine x equal to 3 which cannot be possible because it's giving you sine x is equal to negative 3 by 2 zero solutions next question number 15 The number of values of theta in the interval minus pi by 2 to pi by 2 such that theta is not n pi by 5 For n is equal to 0 plus minus 1 plus minus 2 tan theta is tan 5 theta As well as sine 2 theta is cos 4 theta Yes any response Okay kushal Anybody else these type of any response Okay, so let's discuss this First of all if you look at this equation This as good as saying tan of theta into tan of 5 theta is equal to 1 That's tan theta or 1 minus tan theta tan 5 theta is going to be zero Which clearly implies Tan of 6 theta is undefined correct That means 6 theta has to be a multiple odd multiple of 5 by 2 this has to be an odd multiple of 5 by 2 right So if I just start with plus minus pi by 2 Okay So 6 theta is plus minus pi by 2 implies theta is plus minus pi by 12 I can also afford to take uh plus minus 3 pi by 2 that will give me plus minus 3 pi by 12 And I can also afford to take plus minus 5 pi by 12 because I have to maintain the interval of theta to be minus 5 by 2 to pi by 2 So these are the possible angles in this interval for which tan of 6 theta will become undefined Let me write here as an odd multiple of 5 by 2 Okay, plus minus or you can say 2 n plus plus 5 by 2, okay Let's talk about the second one So sin of 2 theta this I can write it as 1 minus 2 sin square 2 theta Okay So it's 2 sin square 2 theta Plus sin 2 theta minus 1 equal to zero Okay I think this can be factorized as 2 sin 2 theta minus 1 times Sin 2 theta plus 1 Okay equal to zero Which gives you two possibilities one is sin 2 theta is half and other is sin 2 theta is negative 1, right? Now sin 2 theta is half means 2 theta can be 5 by 6 Right, okay It can be 5 pi by 6 correct Minus pi means it can be Minus pi by 2 and it can be 3 pi by 2 Okay, but again 3 pi by 2 will exceed so I'll have to drop it Correct So theta can be pi by 12 5 pi by 12 And minus pi by 4 So I think these are the three values which are common to Or just two values are coming Minus pi by 4 is common No minus pi by 4 is not common So kushal, how are you getting three values? Yeah, so ideally there should be only two answers Do let me know if I'm missing out anything Pi by 12 is common So this is common Oh, I'm sorry. Sorry. Yeah So minus 3 pi by 12 is minus pi by 4. So yes, all the three values are common. That's why option number d is correct Okay Well, this was also an easy one, but should not take you more than three minutes Let's start with some problems on properties of triangles Consider a triangle a b c let a b c denote the length of the sides opposite to vertices a b and c respectively Suppose a is 6 b is 10 and the area of the triangle is 15 root c square units a c b is obtuse And r denotes the radius of the in-circle then r square is So this triangle is not drawn to a scale. Uh, remember the c is an obtuse angle Okay So what do I need over here? I need the in-circle radius. So which is known to be delta by s correct Hope everybody knows the properties of triangle basic formulas correct Now, how do I know s? I know a I know b But I don't know c So unless until I know c, I cannot know this right Now first of all, I'll use the Information given to me that the area of the triangle is 15 root c square units. That's half 10 into let's say I call this angle as c for the time being so half 10 into 6 Sign c is going to be 15 root 3 correct That means sign of c is going to be a root 3 by 2 So c can be 60 degrees or c can be 120 degrees But I will reject this because it has been categorically mentioned that c has to be an obtuse angle. So c has to be 120 degrees correct once I know my c Let's say I want to know my small c I can use the cosine rule cos of c is equal to 10 square plus 6 square minus c square by 2 into 10 into 6 And cos of 120 is negative half So that's going to be negative 60 Is going to be 136 minus c square correct So c square is going to be uh, if I'm not wrong 196 196 means c is going to be 14 So this is 14 s is going to be 15 So r is going to be 15 the root 3 by 15. That's going to be root 3 So r square is going to be 3 and hence option number c is correct Surprisingly only one of you could solve this Let's talk about the next question Question number seven Yes, can you see this we know Okay, so Maria says option c Okay, so let's discuss this. So first of all Let's take this condition into our account So this is 3 sin a cos b is equal to cos a sin b right And this is already given to us one fourth So I can say sin b cos a is given to you as one fourth Now if I just have to write the formula of sin a plus b That's nothing but sin a Cos b plus cos a sin b That would be going that would be Going to be one by four plus three by four. That's going to be one in other words a plus b is going to be 90 degrees Which implies c also has to be 90 degrees Which clearly means it's a right angle triangle at c right Again an easy one. Let's move on to the 18th question In a triangle a b c a d and b are the medians drawn to the angular points a and b respectively Angle d a b is twice of angle a b e is equal to 36 degrees each ad is six units Then the circum radius of the triangle is Okay Anybody else Guys, can you all see the screen? Okay, so let's discuss this So it's obvious that this angle has to be uh, 54 180 minus 54, which is going to be 126 degrees, correct Let's say this point is your centroid that is g We all know that a g is two third of ad So let's apply sin law Or sin rule On triangle a g b So can I say a g by sin 18 degree will be equal to a b by sin 126 degrees In other words, I can say a b is equal to sin 126 degrees By sin 18 degree into a g a g is two third ad and ad is six Correct. We have been given ad value as six over here, correct So this becomes four times sin 126. Now, what is the value of sin 126 degrees? Sin 126 degrees is nothing but Cost of 36 degrees, right? We all know the value as root five plus one by four correct and sin 18 degree This is root five minus one by four So it'll be four times root five plus one by root five minus one Correct. This is the value of a b is that fine Now if I know a b that means I know small c And we know that circum circle radius is c by sin of c correct Which is nothing but If you drop the factor of two from here, that's nothing but this And c is your a b So it's going to be two times root five plus one By root five minus one cosec of c cosec of c So let's rationalize the denominator over here. So that will become two times root five plus one the whole square by four By four means two cosec c That's five plus one going to be six six plus two root five by two cosec c That's going to be three plus root five cosec c That's going to be option number b is correct So let me see who all I've got the right answer No, nobody has got the right answer guys, what's going on Properties of triangle no wonder people who are facing difficulty in this because you don't know your basic formulas And what is the approach? So please do practice this in the coming week, especially people who are writing on 8th of april next question That's question number 19 for the day if the in radius of a circle inscribed in an isosceles triangle whose one angle is 2 pi by 3 is root 3 then the area of the triangle is Always make a diagram first So it has to be ab equal to ac And let's say without loss of generality. I consider this to be 120 degrees Yes, anyone Okay, so it's very obvious that this angle is going to be 30 degree and this angle is also going to be 30 degree Okay, let's call these sites as uh, let's say A Okay, so the area of this triangle Is going to be Half into a into a into sign of 120 degree. That's going to be root 3 by 4 a square now Let's say the other side is x Okay for the timing Can you apply sign rule over it? So if you apply sign rule, I'll get x By sign 120 degree is equal to a by sign 30 degree right That means 2x by root 3 is equal to 2a Which means x can be written as root 3a right Now what is in say in circle radius in circle radius is given as delta by s So delta you already have root 3 by 4 a square S is nothing but the semi perimeter semi perimeter will be a plus a plus root 3a by 2 So by 2 means 2 will go up Correct and this r value is given to us as root of 3 So root of 3 is going to be root 3 by 2a 1 plus root 3 So root 3 root 3 goes off. So a is going to be 2 plus 2 root 3 and x is going to be 2 root 3 plus 6 But I don't need that actually because I just have to find out the value of area So area is going to be root 3 by 4 a square that is going to be 1 plus root 3 square and 4 will go off So that's root 3 plus This is going to be 4 plus 2 root 3 Hope everything is fine so far x is root 3a that's fine. So a plus a plus root 3a by 2 delta by s is root of 3 So I'm getting the answer for this as 6 plus 4 root 3 6 plus 4 root 3 is not in the options Let me just quickly cross check a plus a plus root 3 a by 2 2 goes up, okay, that's 2a Okay, I got the mistake. I got the mistake. This has to be 2 over it a small mistake Yeah So let me just Yeah, complete this again. So a is going to be 4 plus 2 root 3 a is going to be 4 plus 2 root 3 So this will be root 3 times 2 plus root 3 square That's going to be root 3 times 7 plus This is going to be 2 into 4 root 3 So it's going to be 12 plus 7 root 3. That's going to be option number c Is that fine? Guys, nobody is answering that means you are very underconfident about this chapter So needs much practice in this Let's take another one Length of two sides of a triangle are given by the roots of this equation The angle between the sides is pi by 3. Find the perimeter of the triangle Okay, Purvik, what about others? All right, so let's discuss this So let's say we have this triangle abc Okay, so let's say I have been given these two sides b and c And the included angle here is 60 degree So from this I can say b and c Are the roots of this equation. So b plus c is 2 root 3 And bc is going to be 2 Correct If I use the cosine formula cos a which is b square plus c square minus a square by 2 bc I can actually write it as b plus c the whole square minus 2 bc minus a square by 2 bc Okay, cos a is going to be half b plus c is going to be 2 root 3 2 root c square is going to be 12 minus 4 minus a square By again a factor of 4 Right So 8 minus a square is going to be 2 so a square is going to be 6. That means a is root 6 Which clearly means the parameter of this triangle, which is a plus b plus c is going to be root 6 times root 6 plus b plus c b plus c is already known to us as 2 root 3 So yes option number b is the right option in this case The first one to get this correct was Purvik Next question number 21 If c1 and c2 are two circles touching each other and the coordinate axis is And if c1 is a smaller if c1 is smaller than c2 And its radius is two units Find the radius of c2 Yeah, that's correct. So I've drawn because touching the coordinate axis is Okay, so This is the diagram as of now Uh for for that Purvik you need to solve a lot of problems today. What I'll do is I will Send you some dpp's on the properties of triangles Okay, and request all of you to complete those dpp's After that you'll feel confident about the topic. Okay, so size says 21 c is the right answer. What about others? So guys, it's very obvious. Let's say I call this radius as r2 and I call this as r1 Okay, if you look at o2 c2 distance Correct o2 c2 distance is made up of o2 c1 c1 and then c1 to c2 Correct So o c2 is made up of o c1 plus c1 c2 Right, correct and o c2 is nothing but root 2 r2 because this is going to be 45 degrees And this is already r2 So this distance o2 c2 is going to be root 2 times r2 o c1 is again going to be root 2 times r1 Correct and this is going to be r1 plus r2 c1 c2 is going to be r1 plus r2 Now what is given to you is r1 is two units So I can put it over here So this will become a root 2 plus 1 into 2 Okay, plus r2 So this implies root 2 minus 1 r2 is 2 times root 2 plus 1 So r2 is going to be 2 times root 2 plus 1 by root 2 minus 1 If I rationalize this it becomes root 2 plus 1 square. That's going to be nothing but 3 plus 2 root 2 Correct, so the answer is going to be 6 plus 4 root 2 Option number a is going to be the right option in this case Is that fine now the same question can also be framed as There is a circle And from a point on the director circle Okay, two tangents are drawn and there is circle which touches the circle and these two tangents Correct, what is the radius of the smaller circle? They can ask you this question in the same way remember director circle Basically is the locus of the point somewhere you can draw Two perpendicular tangents so it becomes the same situation So irrespective of whether they state coordinate accesses or a point on the director circle from where you're drawing two tangents The situation doesn't change All right, let's move on to the next one Question 22 Find the image of this circle in the mirror 4x plus 7y plus 13 equal to 0 Okay, let's discuss In this case, what will happen the radius is not going to change in the image. Just the center is going to change Correct So first of all, I would like to know the image of the center which is minus 8 comma 12 About the line 4x plus 7y Plus 13 equal to 0 Okay So we know the formula. Let's say the image is x 1 comma y 1 So and let's say this point is a comma b. So I know the formula Or let me say p comma q and let's say this line is ax plus by plus c equal to 0 So x 1 minus p by a is equal to y 1 minus q by b is equal to negative 2 ax 1 by 1 C by a square plus b square correct So x 1 Plus 8 by a a is going to be 4 y 1 minus 12 by b b is going to be 7 is minus 2 ax 1 ax 1 is going to be minus 32 sorry uh ap Not ap and aq Yeah, so ap is going to be minus 32 bq is going to be 84 plus 13 by a square plus b square that's going to be 16 plus 49 So let's calculate this value over here So minus 32 plus 84 would be 52 52 plus 13 is going to be 65 65 by 65 that's going to be negative 2 only So x plus 8 is going to be negative 8 So x is going to be negative 16 Or x 1 is going to be negative 16 And y 1 is going to be 12 minus 14. That's going to be negative 2 So your center of the image circle would be at minus 16 comma minus 2 correct So these two cannot be your answer okay So minus 16 minus 2 means only option a will work fine over here Because even c cannot be your answer So absolutely correct Purvik option number a is the right option in this case Is that fine? So you can actually distinguish from the centers only so no need to go into the exact equation next question number 23 if 2 comma 4 is a point interior to this circle and the circle does not cut the Accesses at any point then lambda belongs to which interval okay Most of you have given the answer as b. Let's discuss this out See first of all If you see the center center of this circle is at 3 comma 5 Right And if it doesn't touch the coordinate accesses means radius has to be less than 3 Right and what is radius? Radius is going to be under root of 3 square plus 5 square minus lambda. So this has to be less than 3 Correct that means 5 square minus lambda has to be less than 0 that means lambda has to be greater than 25 correct Now secondly, it's given that 2.4 point is interior to this circle That means s1 should be less than 0 s1 is when you put the point in place of x and y So 2 square plus 4 square minus 6 into 2 minus 10 into 4 plus lambda should be less than 0 That means 20 minus 12 minus 40 plus lambda should be less than 0. That means lambda should be less than 32 So combining these two conditions, I can say that your lambda lies between 25 to 32 That means option number a is going to be correct Okay, so I think the first one to get this correct was Sondarya Absolutely, okay Let's move on to the next one That's question number 24 Can be a straight line passing through origin and l2 with a straight line x plus y equal to 1 If the intercept made by the circle on l1 and l2 are equal then which of the following equations can represent l1 Okay Anybody else Okay, so let's figure out the center and the radius of this circle So this circle center is at Uh half comma negative three by two correct And radius will be one fourth plus nine fourth mine. There is ten four. That is five by two under root correct now l2 is given to us And l1 let's say Is a line whose equation is y equal to mx right Now on any circle What is the length of the intercept? Let's say this is the radius Okay, this is the distance of the line Of center from the line So it will be two under root of r square minus d square correct so let's say from this circle The sorry from this line the distance of the center is going to be five by two minus If you put the center over here, it will be half minus three by two minus one by under root of two whole square Into two is that fine? So this is going to be two times Five by two and if I'm not mistaken this is just going to be uh two by That is root two square root two square is going to be two So this is going to be two into one by root two. That's going to be root two itself right so The intercept made by the circle on this line is root two And it's given in the question that The other line y equal to mx will also make the intercept of the same length So let's find out the length of the intercept made by y equal to mx per c. Sorry y equal to mx onto this circle So let me just erase everything and start with the basics. So y equal to mx So this is going to be a Half comma negative three by two so the distance will be Minus three by two minus m by two mod Under root of one plus m squared correct This will be the distance So root two will be two times under root five by two minus This I can write it as m plus three by two under root m square plus one square Okay So let's solve for m in this case So this will become half is equal to five by two minus m plus three whole square by four m square plus one Okay, so that's going to be negative of Two so plus two Is equal to m plus three whole square Okay By four m square plus one So this is going to be eight m square plus eight and this is going to be m square plus six m plus nine That's seven m square minus six m Minus one equal to zero and this is factorizable This is factorizable as seven m plus Okay So your m value can be one or your m value can be negative one by seven So it could either be line y equal to x or it could be light y is equal to one negative one seventh of x So I think option b is the only option that satisfies both these conditions So option b is correct and the first one to give this answer is Purvik So question number 24 No, sorry before that Kushal gives the answer 24 option b is the right option Let's look into question number 25 Okay Okay, yeah again, it's a very simple question guys you just have to Appreciate the fact that if you draw a tangent at a point Okay, let's say I draw a tangent at negative three comma zero Correct And is another tangent at four comma one So this is minus three comma zero. This is four comma one Okay, so we know that the normals to these lines will actually pass through the center Right, this will be a center So what are the normal to this line four x minus three y equal to negative 12 So four x minus three y is equal or plus 12 equal to zero the tangent the normal to this will be of the form three x Plus four y plus k equal to zero Now I can find the k by putting this point So this point if I put I'll get minus nine plus k equal to zero. That means k is going to be nine So the normal is going to be three x plus four y plus nine equal to zero in a similar way if I have to find out The normal to three x plus four y equal to 16 So it'll become four x minus three y plus some lambda equal to zero put the point four comma one So lambda is going to be negative 13 So the other one is four x minus three y negative 13 equal to zero So let's see where do these uh Lines meet By the way, if you look at those points itself the center is minus one comma Sorry one comma minus three One comma minus three does it satisfy this? Let's check. Let us satisfy both the equation. Let's check So one comma three satisfies this Uh one comma minus three. Oh, it also satisfies this. So basically let's not go any further No, in fact the second one also gives you the same answer. Okay, then in that case, uh, I have to also find out the radius So negative one comma sorry one comma negative three distance from To this point is going to be Two square plus three square under root Okay, that's going to be uh root 13 Right So So c is going to be So r is going to be under root of g square plus f square minus c so root of 13 Is going to be root of g square. That's one square plus nine minus c So it's very obvious that c is going to be No, everything is fine Minus c is going to be this. Yes. C is going to be 13 Is equal to 10 minus c Why am I getting the answer as? Oh the point is one comma Uh, this point is one comma three One comma minus three. Yeah So in that case my radius is slightly incorrect. So this will become a root of 25 which is going to be five. Yeah Yeah, in that case my c will come out to be c will come out to be 25 is equal to 10 minus c is going to be negative 15 So in that case option number a is going to be the right option Okay Well, great. So let's move on to the next one The angle between the pair of tangents drawn from a point p to this circle is to alpha Then find the locus of the point p Okay, just a second No, I think it's visible Yeah, can you see now everybody? So this circle has a center as minus two comma three And what is the radius radius is going to be under root of 4 plus 9 minus c So that's going to be under root of 13 minus 13 cos square minus 9 sine square That's going to be two sine alpha Okay Now if I want to find out the locus of h comma k Can I say this is root of s1? So sine of alpha would be in fact tan of alpha would be In fact, you can take sine of alpha also. You don't have to go into root of s1. You can take sine of alpha as r divided by distance of h comma k from 2 comma 3 you can write it as h plus 2 the whole square And k minus 3 the whole square right So let's substitute so your r is equal to sine of alpha to two sine of alpha So If I replace here with two sine of alpha Sine alpha sine alpha you can just cancel it off Right, so it's as good as saying h plus 2 the whole square plus k minus 3 the whole square is equal to 4 Right generalizing it you get x square plus 4x plus 4 plus y square minus 6 y plus 9 Okay, you can just cancel off this 4 So x square plus y square plus 4x minus 6 y plus 9 equal to 0 That's going to be your option number d D option is going to be the correct answer Okay, great. So first one to get this correct is sine here We don't have much time. So let's move on to the next one. That's question number 27 The angle between the tangents from alpha comma beta to this circle I'm sure this is going to be very easy question Done Yeah, that's a very simple question So from alpha beta Okay, you're drawing two tangents What's the angle between the tangents? So connect the center This is going to be your r r is nothing but your a Right, this is root s1 So tan of theta is going to be A by root s1 So theta is going to be tan inverse of a by root s1 So the angle between the two tangents will be two theta So the answer is going to be twice of tan inverse a by root s1 That's option number b. That's correct. Absolutely. That was a sitter. So let's move on to the next one That's your question number 28 for the day So from a point a one comma one on the circle Two chords a b and a c each of length two units are drawn find the equation of the chord bc Okay All right, so the center is going to be one comma one. Sorry two comma two, correct And you're drawing chords each of length two units Radius is going to be root of two correct So let's say I have These two chords a b and ac They are of length Two two units each Yes now If you see the diameter Okay, that would actually be containing your b and c because The radius is going to be two root two So that's going to meet the Pythagoras theorem two square plus two square is going to be two root two square Isn't it So from this diagram you can make out that actually your bc is the Diametrically opposite ends of this particular circle right, okay And not only that they the center and this basically are perpendicular So the line from the origin The line from the origin is containing the point one comma one as and the center both That means this line Bc is perpendicular to the given line y equal to x so this has to have a slope of negative one So line having a slope of negative one and passing through two comma two the equation will be This which is nothing but y minus two is negative x plus two that means x plus y is equal to four Correct And another quick way is to see which of the given line satisfies two comma two so only this This doesn't satisfy two comma two does it No, this doesn't satisfy two comma two This also doesn't so only be satisfied so common sense also says b is going to be the right answer So I just show you do the longer way in case You need it but in such cases you can always first check Which of them contains the center the moment you realize bc is your diameter is that fine All right, so let's move on to the next one. That's question number 29 for the day Two concentric circles of which the smallest is x square plus y square is equal to four have a difference in radii as d The line y equal to x plus one cuts the circles in real points Then d lies in the interval. Is it done guys? Okay anybody else So I think this line will always cut this in real points because if you substitute your y as x plus one You get something like 2x square plus 2x minus c equal to zero, correct And if you find the discriminant b square minus four ac that will be four Plus four into two into three That's always be positive, right? So it'll always have real roots Okay Now let's say the other circle is x square plus y square is equal to two plus d square Okay difference in radius is two. So two plus d square So in this case if you try to substitute x with x plus one y with x plus one you end up getting this So that's going to give you two x square plus two x Plus one minus two plus d the whole square equal to zero Now the discriminant of this will be b square minus four ac minus four ac This should be greater than zero That means one plus twice of two plus d square should be greater than zero Yeah So I can write this as d plus two the whole square. I think I did not take this one into account. Okay, this one will also come So this expression will slightly change. Yeah So b square minus four ac will have one minus two plus d square Okay, this should be greater than zero If I greater than equal to zero let's take Okay, this so one minus twice of one minus two plus d square should be greater than zero Right So two times two plus d square minus one should be greater than zero or divide by two And write it as root two square so that you can complete the square here itself So I can write this as two plus d minus one by root two And two plus d plus one by root two should be greater than zero Correct So d must be greater than the greatest and lesser than the least So d must be less than minus two minus one by root two and d must be greater than One by root two minus two So that is only given by option number a in this case. So option number a becomes the right option Okay, I think only poor we could answer this Okay, so without much waste of time, let's move on to the last question of the day If the lines this and this are diameters of the circle of radius 49 pi square units then find the equation This is super simple because you have been given the two diameters Obviously they will intersect at the center And radius is given as seven So just the last question before your j main exam for maths Simple one just complete it In fact, just see which of the points satisfy both the equations minus one comma one. Does it satisfy minus one comma one? No, it doesn't this is One comma minus one One comma minus one one comma minus one. Yes, this satisfies So it will have At least minus two x plus two y term and radius is seven. So radius is seven only when you have one plus one Minus of 47. So that will be minus of minus 47 under root So option number c becomes the right option. Okay all right, guys, so This might be your last session of maths before your j main exam But we'll definitely come back after j main with j advance questions Right and those questions would be of much much more difficulty level than the questions that we are doing right now A couple of things before your j main exam Please revise your chemistry very well because chemistry is going to be the game changer Okay Maths of course you need to do Couple of mock papers just time yourself out. Don't try to overdo it Because maths is quite dicey in competitive exams. So Even though you may put a lot of hard work. It may not fetch you know, um The same amount of marks which chemistry or physics may fetch So just revise your formulas. Of course today you saw that you guys were not comfortable with properties of triangles So just solve around 30 40 odd questions Well from any standard book, in fact, I will send you some dpp's also on that and that that should be fine You just have to get a hand of the formulas That means you should know your small r in terms of capital r r1 r2 r3 know the uh area formulas Right, what are the relationship between the x center radius and the in-center all those things should be known to you the basics And do not make silly mistakes. That's the worst thing that you can do If you are getting an answer correct and by mistake you mark the wrong answer That is the worst thing that can happen to you. So be very very careful Read the entire question papers can the entire question paper before you start attempting it I think this time you would have known which button is Is uh supposed to be best so that you see the entire paper. So please do look at all the questions and start with the confidence boosting questions the easier ones Okay, and you guys you should always start with your favorite subject Right So you may have more sessions on physics and chemistry, but for maths. This is the last session before your j e main exam We'll meet after the j e main with uh sessions on j e advance Meanwhile over and out from my side Thank you and all the best