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Functions by Total Gadha - Thursday, 2 July 2009, 10:33 AM In a CAT classroom or on the internet, there are two kinds of students- those who study and those who believe that they are studying but are actually not. Every time I start taking a fresh batch, it hardly takes one or two classes to identify both types of students. And it has nothing to do with intelligence or background. It has to do with attitude and fighting spirit. Every time I throw a problem in the class or post it in the forums, the students who are studying would be trying hard to solve the problem. They would arrive at an answer, whether right or wrong. On the other hand, the students are who are just ‘striking a pose of studying’ would be waiting for me to give the answers and the solutions. They would be acting as if they are thinking hard but in reality would not be thinking at all. On the internet, I frequently see students who waste time in idle chit chat when they should be busy solving problems. They feel satisfied that they have done ‘something’ for their CAT by looking at problem or talking about CAT to others but in essence they have done nothing and wasted one more day. “Students like these,” as I tell my serious students, “are going to fill up the CAT form to give you your 99 percentile.” One thing I have learnt from my past experiences is that you can never teach attitude to students. Either they have it or they don’t. Whereas winners focus on completing their tasks, whiners have excuses and ways of work avoidance. In TathaGat, we have a strict rule for students to submit one book review per week. And sometimes we see the book review either copied from the net or written after hearing about the content from their friends. We even have strict rule for submission of quant or DI assignments. And many students turn in their assignments half-attempted. The serious students, on the other hand, finish their assignments before time and would frequently come to me with “Sir aur material mil sakta hai kya?” There is a world of difference between the two attitudes. Where I constantly hear from the whiners “Sir aap solve kar dijiye”, I also constantly hear from the winners “Sir abhi answer mat batayiye, solve karne dijiye.” The only sane advice for the instructors in this field, who sometimes feel as frustrated as I do, is- spend 80% of your time with your winners. And winners are not the ones who score at the top. They are the ones who are fighting the hardest, whether they are weak or strong in the subject does not matter. And for students on the internet who want to do idle chit chat in the forums or engage in socializing, you would find TotalGadha.com too boring a place. We are serious about studies here. For our students on TG.com, we would like to share one of our endeavors in CAT CBT Club- Video lessons. Dagny and I have been tinkering with our camera and trying to make some good video lessons for our CBT Club students. We have uploaded one such video- Divisors of a number- for our TG junta. Do have a look and give us your feedback. Unfortunately, to watch the rest of the lessons that we are making, you will have to become a member of our exclusive club.
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Reply Re: Functions by Rohit Ghosh - Thursday, 2 July 2009, 01:31 PM Nice article TG, as usual supreme quality. Show parent | Reply Re: Functions by abhik chakraborty - Thursday, 2 July 2009, 04:12 PM tg sir, i am puzzled of what category i am till now. but one thing i know for sure what i should try to be or at least try to do. and the article on function is incredible. thank you. Show parent | Reply Re: Functions by Total Gadha - Thursday, 2 July 2009, 04:38 PM
Hi Abhik, I really do not know about that but I think you (and everyone else) should read thishttp://totalgadha.com/tgtown/amitkumar/2009/05/16/are-u-really-preparing/ Total Gadha
Show parent | Reply Re: Functions by Veena Binya - Thursday, 2 July 2009, 04:35 PM Hi TG.. The range of the function 1/1+root(x) has been given as [0,1] Should it not be (0,1] ? Because it can only tend to 0, but can never become 0.... Regards Veena Show parent | Reply
Re: Functions by Nikhil Rajagopalan - Thursday, 2 July 2009, 06:00 PM nice one Show parent | Reply Re: Functions by CATching CAT - Thursday, 2 July 2009, 06:24 PM Lesson is as usual excellent. But the sentence that stole the show is totalgadha.com/mod/forum/discuss.php?d=5440
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“Students like these,” as I tell my serious students, “are going to fill up the CAT form to give you your 99 percentile.” A wake-up call . Show parent | Reply Re: Functions by SHEKHAR SHARMA - Thursday, 2 July 2009, 07:56 PM hi tg sir...... really this article abt to know urself that in which catagory i am, is really inspirable.........n also basic knowledge abt function is very good. thank you sir Show parent | Reply Re: Functions by William Wallace - Friday, 3 July 2009, 02:07 PM Awesome article as usual between two sequence. for example in question f(1)=3600 and f(1) + f(2) + ....f(n)=n^2f(n) u have calculated f(n+1) to simplify but in question f(x)=9^x/(9^x + 3) u have calculated f(1-x). I guess this decision comes with practise but still i want to ask you any tips on how to decide which value to calculate. Please suggest Show parent | Reply Re: Functions by ayush roy - Friday, 3 July 2009, 03:56 PM sir it is a vry nice article.... sir m an engineering student,have passed out this year only,a vry serious follower of this site,n a big fan of ur articles..... wen i was preparin for IIT JEE,the most serious prb which i was facin was to sketch the graphs, i always wanted to do that,coz it helped a lot in physics also. sir what ever u hav explained ovr here is abt analyzin a function,which m vry gud at,but it has nvr helpd me in sketchin d graphs....i was nvr able to correlate both analyzin and sketchin. plz tell me how i can do that vry easily,without puttin values into a function. thank you Show parent | Reply Re: Functions by shiva chepuri - Friday, 3 July 2009, 06:58 PM it is a wonderful article !!! please discuss types of function if possible sir Sqrt(4-x2 ) range should be (-infinity , 2] because sqrt(x) outcome is R but i dont know when to calculate f(x+1) and when to f(1-x) or f(x-1) so as to get a relation
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Show parent | Reply Doubt In Functions by A B C - Saturday, 4 July 2009, 01:43 PM HI TG... I have a doubt here.. f(x) = odd and g(x) = even then f(g(x)) = odd.. now consider f(x) = x^3 and g(x) = x^2 than f(g(x)) = x^6..which is an even function and hence the statement... Am i missing some concept here ..? Show parent | Reply Re: Functions by Total Gadha - Saturday, 4 July 2009, 06:18 PM
contradicts
Hi Shiva, Give me one example when it is more than 2. Total Gadha
Show parent | Reply Re: Doubt In Functions by Total Gadha - Saturday, 4 July 2009, 06:20 PM
Hi Shailendra, I don't see the statement you have mentioned. The case you are mentioning, I have specified that it is an even function. Where is the contradiction? Total Gadha
Show parent | Reply Re: Doubt In Functions by manjushaa srinivasan - Saturday, 4 July 2009, 11:02 PM Hi TG i found your article very thought-provoking and clear cut.Himanshu's blog was also very inspirational.In my opinion any serious CAT aspirant who reads your article would become more dedicated towards his/her CAT preparation.Your explanation of functions was also very useful.I just want 1 advice from you.I'm not natural with quant.Even though i've been practising all along,i'm not sure if i can think as analytically as most of the other aspirants who are strong in quant .so far i've tried to cover all chapters but i would like to know if i'd be able to crack the quant section if i'm well versed with number system,geometry,averages,percentages,alligations,profit and loss,interest,functions,set theory,progressions,probability,logarithms,surds and indices.although i'm practising problems from P&C,Mensuration but i get stuck when the sum is slightly twisted.this is worrying me. Regards manjushaa Show parent | Reply Re: Doubt In Functions by manjushaa srinivasan - Saturday, 4 July 2009, 11:05 PM I also have a problem visualising 3 D figures especially cubes which is hampering my speed and accuracy in mensuration to a great extent.What exactly should i do about this? Show parent | Reply Re: Functions by Mohit Sharma - Sunday, 5 July 2009, 01:40 AM Nice Article...Tg Show parent | Reply
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Re: Doubt In Functions by Total Gadha - Sunday, 5 July 2009, 04:31 AM
Hi Manjusha, The trick to get good at quant is to try and solve a lot of problems on your own. Try to solve them in any which manner that you can- by putting values, by generalizing, by longer methods or by using theorems etc. But solve as much as you can. You will get good at this as you go along. Total Gadha
Show parent | Reply Re: Doubt In Functions by Veena Binya - Sunday, 5 July 2009, 09:46 AM Hi TG.. The range of the function 1/1+root(x) has been given as [0,1] Should it not be (0,1] ? Because it can only tend to 0, but can never become 0.... Regards Veena Show parent | Reply Re: Doubt In Functions by manjushaa srinivasan - Sunday, 5 July 2009, 02:34 PM hi TG Thanks for the reply. previously when cat was CBT,it was feasible to leave few tough topics like P&C,inequalities,TSD etc if we were strong with number system,geometry,P&L,percentages,functions,averages,alligations,progressions,log,algebra.With the new pattern do we have cover the entire syllabus?how is it possible to solve DI directly from computer screen especially the ones which involve tables where we need to calculate and give rankings etc?Since skimming and scanning of RCs,skillful selection of sitter questions are redundant now,what should be our new strategy? Show parent | Reply Re: Doubt In Functions by amit jain - Monday, 6 July 2009, 10:45 PM can some one explain the question, area between the coordinate axis and fff(x). I am nt getting it. Show parent | Reply Re: Functions by Prakash jain - Wednesday, 8 July 2009, 07:48 AM sir you have realy openend my eyes, i belong to the 1st category, i have been pretending to prepare for CAT since 2 year but today i have realised that my previous efforts were not true, now I promise you that i will seriously prepare for this exam,and will perform well,thank you so much for such a wonderful article... Show parent | Reply Re: Functions by Dipanjan Biswas - Wednesday, 8 July 2009, 12:18 PM At first I want to give you thanks for writing such a helluva article..It helped me to find my loopholes regarding one of the most important sections of quant. I would like to draw your attention on a few typo errors.. (1)How to draw a graph of F(x): Point number three..Latter one should be for maximum.. (2)f(x)=x^3(x+2) the second derivative of this function is positive.. Sir, I find it hard to draw any rectangular hyperbola graph..Can you elucidate it further.. TG Rockzzz\m/ Show parent | Reply Re: Functions by dishant dhokia - Thursday, 9 July 2009, 11:58 AM
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For how many values of x is f(f(f(x))) = 19?
In this sum
f(f(f(x))) = 19
The next step is f(f(x)) = 38………….
I am not able to deduce this………..
Please someone elaborate??????????? Show parent | Reply
Re: Functions by Dipanjan Biswas - Friday, 10 July 2009, 08:21 AM Hi Dishant, Look at the function.. See f(x) = x/2 if x is even = x+1 if x is odd In other way x=2f(x) if f(x) is either even or odd x=f(x)-1 if f(x) is even As f(x)=19=odd so x=38 likewise f(x)=38 x=76 or 37 likewise f(x)=76 x=152 or 75 for f(x)=37 x=74 so {x}=>{74,75,152} Hope it helps Cheers!! Show parent | Reply Got a doubt by Varun Agrawal - Friday, 10 July 2009, 03:01 PM Please discuss types of function if possible sir Sqrt(4-x2 ) range should be [-2 , 2] because sqrt(x) of a number gives 2 values. Positive and negative. Can you please elucidate how it is [0,2] Thanks
Show parent | Reply Re: Got a doubt by Dipanjan Biswas - Saturday, 11 July 2009, 01:38 PM Varun please do go back to the question.. It does not want the range of x rather the range of f(x). Max of f(x)=2 when x=0 Min of f(x)=0 when x=+-2 So f(x) [0,2] Please don't be in haste before raising your doubts. TG family takes utmost care before publishing these lessons..
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Hope you have accepted my comment on a positive note!! All the best!! Show parent | Reply Re: Functions by shiva chepuri - Saturday, 11 July 2009, 08:00 PM yes sir got it but it should be [-2,2] if i am wrong plz correct Show parent | Reply Re: Functions by Varun Goenka - Sunday, 12 July 2009, 10:22 PM the func. is basically y=sqrt(4-x^2) i.e. y^2=4-x^2 this is equation of a circle...x^2 + y^2 = 4...with origin as centre and 2 as radius... but the given f(x) is just the upper half of the circle.. y=sqrt(4 - x^2)...is for upper half of the circle... y= -sqrt(4 - x^2)...is for lower half of the circle... our question is related to upper half of the circle...so range is [0,2] and domain is [-2,2].. I think this should help... Regards, Varun Goenka Show parent | Reply
Re: Functions by Siddharth sahoo - Monday, 13 July 2009, 03:36 PM @sahil, As discussed by you, x^2 * x^3 will be x^5. X^m * x^n = x(m+n).x5 is an odd function. Show parent | Reply Re: Doubt In Functions by soumen paul - Tuesday, 14 July 2009, 06:59 PM f(x)=x^3 is not a odd function if x=2,4,6........ g(x)=x^2 is not a even function if x=3,5,7........ f(x)=2x+1 is a odd function for 0,1,2..... g(x)=2x+2 is a even function for 0,1,2...... so here f[g(x)]=4x+5 is a odd function Show parent | Reply Re: Functions by Ashish Kumar - Thursday, 16 July 2009, 12:23 AM lol PATAKKA article....but what i loved more is those who are whiners and are reading this can actually do something still coz there is still time but even after this if they cannot do anything then trust me they are helping the winners more n more lol ROFL !!! Show parent | Reply
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Re: Functions by A S - Thursday, 16 July 2009, 04:19 AM Hi TG,
Simply superb article...I think I have fallen in love with functions...can you please add more problems on functions? or may be one set of problems? or may be some book....anything will do.... Show parent | Reply
Re: Functions by A S - Thursday, 16 July 2009, 04:24 AM
hi dishant, take it this way...can f(x) ever be odd? when x is odd, f(x)=x+1...so it will always give you even f(x).. when x is even, f(x)=x/2; this is only chance when you can get odd f(x)..
here f(m) [m=f(f(x)) ] is odd.....so m must be even... that means m/2=38 so, m=76 i.e. f(f(x))=76..
hope it helps Show parent | Reply Re: Functions by piyush dixit - Friday, 17 July 2009, 09:39 PM HI TG, I am a frequent reader of your post, and following this site almost since its inception. I am an adherent fan of yours, BUT I am very disappointed with this article of yours, and hence writing to you for the first time. I guess, you started this whole exercise of totalgadha.com to ridicule the jargon taught by famous coaching classes in their curriculum to show their superiority among the students community, TG i belive you started the exercise to change the way education system works at this level. But it seems you are getting infected by the same BUG. I am sorry, if i have misunderstood you. Hats off for all your other posts, which I have been through. PIYUSH Show parent | Reply Re: Functions by Total Gadha - Sunday, 19 July 2009, 10:49 PM
Hi Piyush, I am infected by what I do, down to my last nail. ignore everything else. Total Gadha
Show parent | Reply Re: Functions by Surinder Pal Singh - Saturday, 25 July 2009, 09:43 AM
My opinion should hardly matter to you. Take what is good and
Hi TG Sir, Excellent article. But the link above is infected with a Trojan Horse.
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Kindly verify its integrity as I was just saved by my firewall.
Keep up the good work Regards Surinder
Show parent | Reply Re: Functions by gaurav kaushal - Sunday, 26 July 2009, 11:35 PM hi tg sir , thx for this gr8 article..... now dat i v read this whole article ...where can i get some questions related to this article...so dat i can just make the concepts more clear thanks a lot..sir Show parent | Reply Re: Functions by vivek ghiria - Monday, 27 July 2009, 10:09 AM Hi TG I have few doubts in Functions. If the first derivative f'(x) is positive, then the function f(x) is increasing. If the first derivative f'(x) is negative, then the function f(x) is decreasing. For f(x) = x^3-4x^2+1 f'(x) = 3x^2-8x ....(1) Now how to know whether expression (1) is +ve or -ve to determine increasing or decreasing charactersitic? Also I recall there is some way to determine whether f(x) is continuous or not from first derivative of f(x). Please explain that rule also. Thanks Vivek
Show parent | Reply Re: Doubt In Functions by dilip rathore - Tuesday, 4 August 2009, 04:12 PM g(x) is calculated for some x and some value is obtained. later "this value" is put into f(x). so f(g(x)) is not actually x^6. e.g. for x= 2, g(x) = g(2) = 4 so f(g(x)) = f(4) = 4^3. now (4)^3 and (-4)^3 are not same right :P Show parent | Reply Re: Functions by dilip rathore - Tuesday, 4 August 2009, 04:34 PM hi vivek, i think we check f''(x) to determine if the function is increasing or decreasing. for your example f(x) = x^3-4x^2+1 f'(x) = 3x2 - 8x f''(x) = 6x-8
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for x > 4/3 , function is increasing and for less than 4/3 , its decreasing (slope). pls let me know if i am wrong Show parent | Reply Re: Functions by himanshu gupta - Saturday, 8 August 2009, 07:28 PM thank u sir 4 such a good article.sir is thr any oter approach to solve th last 2 ques?? Show parent | Reply Re: Functions by abhishek tripathi - Sunday, 9 August 2009, 04:14 AM well sir i am a new user over here i found dis site preety useful but regarding first question to find out the range can u exactly tell me how do we aarive at 38 when (fx)=19 since it is odd and according 2 function definaion in da question it should be x+1=20 me will b very glad if u tell me how 2 solve dese type of questions. Show parent | Reply Re: Functions by Jiggs Sanyal - Monday, 17 August 2009, 10:03 AM Hello TG Sir, I am a hardcore fan of your articles and of the TG website. It is the first website that I open everyday. But for the last few weeks, I have been sort of disappointed coming here, because you have completely stopped writing articles for people like me who are not a part of the CAT CBT club. In one of the discussion forums, I had seen you mention that TG is for those who have the zeal to make their dreams come true. I have the requisite zeal but then TG now seems to have become more and more commercialised. I am not against commercialisation but you should keep in mind those who have the zeal but not the bulky pockets to pay. Sorry If I hurt you in any way. you are the ideal motivator to me.
Show parent | Reply Re: Functions by Jai Prakash - Saturday, 22 August 2009, 07:39 PM Thanks TG. I've been a guy mentioned by you. but now i promise that i'll work hard and..... whatever directed by you and necessary for excelling. It's my 1st post. TG had been rocking but this time it made rhe rivers of truth flow from eyes. I never posted on TG as i have been getting the answers of my questions from orher's posts. I've made the sentence of TG Sir " YOU ARE NEVER LATE" the ultimate source of inspiration. Thanks a ton for showing me the mittor. Hope the other guys too will correct themselves. Sir due to time constraints i'm not posting some querries. Will do it later on.........
"ITS BETTER TO FAIL IN DOING SOMETHING THAN TO EXCEL IN DOING NOTHING." Show parent | Reply Re: Functions by Vivek .. - Saturday, 5 September 2009, 01:30 PM hi,, i m a regular visitor of totalgadha.com & i really appreciate ur work.. all i want to know is,,if sum1 starts prep now for cat,,does he has any chance of clearing it. i m clear wid 50 % of the concepts pof all the 3 subs. but prob is i m never able to get marks or %ile upto my satistfaction or wat is reqd.,,so i m confused of whther wil i b able to do it or nt? in the last 3 mocks of tathagat i scored 24,25 & 15 marks & i m very ashamed of this score. It wud be helpful if u give me ur sincere advice of shud i stop following the dream or nt? thank you. Show parent | Reply Re: Functions by saurabh chauhan - Saturday, 5 September 2009, 11:27 PM Hi Jay... nice quote... "ITS BETTER TO FAIL IN DOING SOMETHING THAN TO EXCEL IN DOING NOTHING." i'll make this my driving factor rom today...and many thanks to TG..the article.."Are u really prepaing?" inspired me a lot...nw i'll really pepare and get through......my prep. starts now.................3 2 1......go....
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saurabh Show parent | Reply Re: Functions by Vivek .. - Sunday, 6 September 2009, 09:19 AM INFACT U CAN SAY THAT I WAS A 2ND TYPE OF STUDENT UPTIL NOW BUT YOUR ARTICLE HAS REALLY OPENED NOT ONLY MY EYES BUT MY VIEW & I M SERIOUSLY SERIOUS NOW. JZ TEL ME THAT CAN SUM1 OR HAS HAS1 CLEARED CAT WITH PREPARATION IN SUCH A SHORT SPAN.?? ANXIOUSLY WAITIN FOR YOUR REPLY. Show parent | Reply Re: Functions by Jai Prakash - Monday, 7 September 2009, 04:51 PM Dear vivek saini it seems (perhaps) you haven't rushed through the various posts of TG. What TG sir say to such question is "YOU ARE NEVER LATE". Work honestly and yu will make it.... Best wishes... Jai Praqas Show parent | Reply Re: Functions by hbk Not joining IIML - Monday, 14 September 2009, 05:36 PM Forget function guys, this guy(TG) can inject attitude in U. A really serious one. Hats off. ****************I'm gonna get back soon************************ Show parent | Reply Re: Functions by subhadip mahalanabish - Wednesday, 16 September 2009, 01:45 PM Dear TG sir, consider the function y=x/(x^2+1). We have to find the range of this function. If we directly put x=0, we will get y=0, then if we put x=1 then y=1/2, for x=3, y = 3/10, for y=4, x=4/17 and so on, so the maximum value of the function is 1/2 at x=1 and then the function decreases as x increases, same is true for negative values of x, in the negative side minimum value of y is -1/2(at x= -1) and then y increases as x decreases. so range of the function should be -1/2 <= y <= +1/2 including y = 0. But if we calulate in this way, y = x/x^2 + 1 => x^y -x + y= 0 => x = {1+-sqrt(1-4y^2)}/2*y. Hence range will be, -1/2<=y<=+1/2 excluding y = 0. Here we are seeing that y cant be equal to zero. But previously we already saw that y can be 0 for x= 0. How can we remove the confusion of which result is to be used? regards, Subhadip. Show parent | Reply Re: Functions by subhadip mahalanabish - Wednesday, 16 September 2009, 03:45 PM hi all, On giving a second look I found myself that the function is discontinuous at x=0.Probably this is the reason why y=0 is not to be considered in the range of the the function. regards, Subhadip. Show parent | Reply Re: Functions by hbk Not joining IIML - Saturday, 19 September 2009, 06:18 PM @subhadip
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In your approach we are considering the eqn (x^2)y -x+y as a quadratic in x having coft of y. By defn of quadratic eqn coeft of x^2 can't be zero hence with this approach we may not get y=0 as a feasible. It has to checked separately. Show parent | Reply thanx..... by raghav sharma - Sunday, 27 September 2009, 03:34 PM sir... your adomitable aproach towards conspt has really given a lot of helf to me solve my probs.... thanx a ton... Show parent | Reply Re: Doubt In Functions by Kishan Kumar - Thursday, 12 November 2009, 12:09 AM f(x) = odd and g(x) = even then "f(g(x)) = odd.." you went wrong at this part it is not odd its even Show parent | Reply Re: Doubt In Functions by Subhash Medhi - Saturday, 22 May 2010, 05:45 PM Dear TG sir, I have a doubt in the solution to the question, Let f(x) be a function with domain 0<=x<=1 defined as 2x if 0<=x<=1/2 f(x) = { -2x + 2 if 1/2<=x<=1 Find the area of the curve y = f(f(f(x))) and the coordinate axis. in the second line, 0 <= f(f(f(x))) <= 1 - - > 1/2 <= f(f(x)) <= 1 - - > 1/4 <= f(x) <= 1/2 - - - >1/8<=x<=1/4 The doubt is that if 1/2 <= f(f(x)) <= 1 then shouldn't it imply that 1/2 <= - 2f(x) + 2 <= 1 which would imply 1/2 <= f(x) <= 3/4 ? Since f(f(x)) lies in the interval [1/2,1], shouldn't we be using the second part of the definition of f(x) ?.Kindly clarify my doubt. I am having very similar doubts in the subsequent lines of the solution.Kindly help me out. Regards, Subhash Show parent | Reply Re: Doubt In Functions by shail mishra - Thursday, 10 June 2010, 07:44 PM Hello TG I also couldn't understand the above mentioned question. Kindly explain it. Regards SM Show parent | Reply Re: Doubt In binomials by Ankur Bansal - Friday, 2 July 2010, 08:17 AM Hello TG Sir and hi to ALL Can anyone help me with binomial theorems and help me in finding out the coefficients. For ex. Find the coefficient of x in (1+2x+3x^2)^15(1+x)^5 ? And i am really interested in knowing the concept for these type of problems??? And TG Sir, if you can then pls come up with one good article on Binomial Theorems. Thanks in advance.
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Show parent | Reply Re: Doubt In binomials by amit kheterpal - Friday, 2 July 2010, 11:52 AM Find the coefficient of x in (1+2x+3x^2)^15(1+x)^5 ? 2x+3x^2=a
(1+a)^15=1+5c1*a+15c2*a^2+...........15c15*a15
(1+x)^5=1+15c1*x+5c2*x^2+.......5c5*x^5 now,(1+15c1*a+15c2*a^2+...........15c15*a15)*(1+5c1*x+5c2*x^2+.......5c5*x^5) putting the value of a so coffecient of x will be 5c1+(15c1)*2 =35
Show parent | Reply Re: Doubt In binomials by Ankur Bansal - Friday, 2 July 2010, 12:57 PM Hi Amit,
Thanks for reply And for the same problem if we have to find the coefficient of x^4 then how we will proceed?
Thanks in advance!!!!! Show parent | Reply Re: Functions by G D - Tuesday, 6 July 2010, 05:27 PM Nice one, Show parent | Reply
Re: Functions by Tasmai Merchant - Monday, 26 July 2010, 05:14 PM Hey Guys .....I wanna kno if buying books from Tg like number system, or geometry wud be of any help....nd i can't find any book on algebra...can u plz suggest me some options of buying one.. Show parent | Reply Re: Doubt In Functions by Ritesh Raman - Thursday, 5 August 2010, 03:25 AM TG Sir, Thanks a lot for your out of the world kind of materials. I am just giving some lines on find the area of f(F(f(x)))...ques. OOOPS late to reply but may be useful for guys who look from now on. First we will find f(f(x)) It should come as 4x ; 0≤x≤1/4 -4x + 2 ; 1/4≤x≤1/2 4x - 2 ; 1/2≤x≤3/4
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-4x + 4 ; 3/4≤x≤1 Then find f(f(f(x))).This shuld be 8x ; 0≤x≤1/8 -8x + 2 ; 1/8≤x≤1/4 8x - 2 ; 1/4≤x≤3/8 -8x + 4 ; 3/8≤x≤1/2 8x - 4 ; 1/2≤x≤5/8 -8x + 6 ; 5/8≤x≤3/4 8x - 6 ; 3/4≤x≤7/8 -8x + 8 ; 7/8≤x≤1 Now the min value of each funct(straight line ) is 0 and max is 1. You can verify it by sub. values This gives us the graph that TG sir provided. Hence area = 8 * (1/2*1/8*1) = 1/2 sq unit. I hope there is no mistake. Is it correct Sir??? Show parent | Reply Re: Functions by Deepak Kumar - Thursday, 5 August 2010, 12:26 PM Hi TG Sir, Its always feel nice to read your articals. For the last problem for calculating f(94), I approached this way. As f(x)+ f(x-1)=x^2...........(1) on replacing x with x+1; f(x+1)+f(x)=(x+1)^2........(2) Now subtracting (1) from (2) f(x+1)-f(x-1)=2x+1..........(3) On putting values of x as 93,89........19 will give LHS as f(94)-f(18). f(18) can be easily calculated from (1) when f(19) is given. This approach will reduce RHS to an easy airthmatic progression. thanks a lot for your help sir. Show parent | Reply
Re: Functions by himanshu jha - Thursday, 5 August 2010, 06:40 PM hi tg sir, i know this is not the place to contact you..but i couldnt find any other way...try to reach u thrgh ur admin mail but dat too in vain...i ve just got one query i want to join cbt club..i want to pay thrgh demand draft but in it "payable at" option is given..so what to fill in dat new delhi r something else....if any other person knows it then please let me know or any other way to contact tg sir... thanxs in advance... Show parent | Reply Re: Functions by wishy wishy - Friday, 5 August 2011, 05:28 PM Please correct me if I am wrong: In the last question,f(x) + f(x-1) = x^2 then f(1) = 1, f(2) = 3 , f(3) = 6, f(4) = 10 i.e, f(n) = n(n+1)/2 => f(19) = 190 and hence f(94)= 4565 Show parent | Reply Re: Functions by Total Gadha - Friday, 5 August 2011, 10:02 PM hi himanshu. it should be New Delhi Show parent | Reply
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Re: Functions by Gregory Matchado - Thursday, 18 August 2011, 05:39 PM Hi TG.. Thank you so much for all the articles.. !!! And let me also tell you that you're videos on divisors, remainders and co primes are just tooooooo brilliant... Thanks a lot.. Show parent | Reply Re: Functions by abhinay dutta - Thursday, 8 September 2011, 06:50 PM hi, i have calculated the last question in post and arrived at 4181. guys please chk , and correct me if i m wrong.... Show parent | Reply
Re: Functions by nitin sharma - Friday, 16 September 2011, 11:58 AM Hi TG sir i didnot get the explanation of d 1st quest of f(f(f(x)))) where domain of x was 0<x<1..som1 plz help me..i men how did u draw the graph.. Show parent | Reply Re: Functions by IIM gadha - Friday, 16 September 2011, 12:26 PM
@TG sir,
The graph drawn for f(f(f(x))) is incorrect. Till 0<x<1/2, the graph is of 2x but from 1/2<x<1 the graph should be of -2X + 2 i.e the one with negative slope not positive as shown in your graph. Though, the area remains unaffected but the graph is wrong. Please correct me or yourself. Show parent | Reply Re: Functions by Kamal Lohia - Friday, 16 September 2011, 04:47 PM
Vibha Function given is for f(x) but the area is asked for f(f(f(x))). Please check before posting. Kamal Lohia
Show parent | Reply Re: Functions by IIM gadha - Friday, 16 September 2011, 04:57 PM
@Kamal sir Apologies! Show parent | Reply Re: Functions by ketav sharma - Saturday, 17 September 2011, 06:49 PM hanji dutta sahib its 4181 Show parent | Reply
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Re: Functions by Rahul Sharma - Tuesday, 11 October 2011, 08:13 PM HI TG, the below mentioned link that u shared is net getting opened, could you please help me with it.
http://totalgadha.com/tgtown/amitkumar/2009/05/16/are-u-really-preparing/
Rahul Sharma
Show parent | Reply
Re: Functions by Kamal Joshi - Friday, 14 October 2011, 09:07 PM Dear TG sir, The links such as these return a fatal error through out the site. Kindly resolve the issue. can't afford to miss the Gyan from you. http://totalgadha.com/tgtown/amitkumar/2009/05/16/are-u-really-preparing/ Thanks to you, Kamal sir, Dagny ma'm and rest of the awesome team! Show parent | Reply Re: Functions by vasu r - Tuesday, 18 October 2011, 01:03 PM how to sketch t graph in t question of f(f(fx)))? y divide into 8 parts?? cud u pl explain ? Show parent | Reply Re: Functions by Kamal Lohia - Tuesday, 18 October 2011, 01:32 PM
Hi Vasu Just don't look at the solution and only try to see/analyse the question and thge function given. See if f(x) is some linear function what about the degree of f(f(x))???? It'll also remain a linear function as you are just replacing 'x' in f(x) by f(x). So degree of f(f(x)) is not changing. Similarly f(f(f(x))) is also a linear function. But as f(x) was a different function in deferent periods, so is the case with f(f(f(x))) also. Now it will be better that you try yourself to findout period and then graph of new function. Kamal Lohia
Show parent | Reply Re: Functions by Ravi Kumar - Tuesday, 1 November 2011, 02:43 PM Re: Functions by IIM gadha - Friday, 16 September 2011, 12:26 PM
@TG sir,
The graph drawn for f(f(f(x))) is incorrect. Till 0<x<1/2, the graph is of 2x but from 1/2<x<1 the graph should be of -2X + 2 i.e the one with negative slope not positive as shown in your graph. Though, the area remains unaffected but the graph is wrong. Please correct me or yourself.
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a 0 14 12 34&1 t , /, /, / . yai -xs | | | / \ / \ / \ / \ | / \ / \ / \ / \ |/ \ / \ / \ / \ _|___\___\___\___\___ xai _/___/___/___/______ xs 0 14 / 12 / 34 / 1
ase =4.(/ .14.1 =4 18 nwr 12 / ) . /
=12rmistesm tog.. / ean h ae huh.
Show parent | Reply
Re: Functions by King Casanov - Tuesday, 1 November 2011, 02:29 PM @ TG Sir and Kamal Saar...would this suffice the need of CAT'11 Function part of QA or anything more req.? If so please inform/ provide a link to.. Thanks in advance...
The KING Show parent | Reply Re: Functions by Kamal Lohia - Tuesday, 1 November 2011, 04:11 PM
Ravi Thanks for the correction. It'll be updated in the article soon. Kamal Lohia
Show parent | Reply Re: Functions by ayush lakhotia - Monday, 7 November 2011, 06:29 PM Kamal Sir, f(a+b)=F(a)+F(b) and f(5)=12. Find F(12) Please help me in solving this. Show parent | Reply Re: Functions by Kamal Lohia - Tuesday, 8 November 2011, 02:42 PM
Hi Ayush Given is: f(a + b) = f(a) + f(b) For a = b, we have: f(2a) = 2f(a) Similarly if we put b = 2a in given equation, we get: f(3a) = 3f(a) using second equation also. So in general, f(na) = nf(a). Now given is: f(5) = 12. So f(12) = f((12/5)×5) = (12/5)×f(5) = (12/5)×12 = 144/5.
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Kamal Lohia
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MBA CAT GMAT You are not logged in. (Login)
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Functions by Total Gadha - Thursday, 2 July 2009, 10:33 AM In a CAT classroom or on the internet, there are two kinds of students- those who study and those who believe that they are studying but are actually not. Every time I start taking a fresh batch, it hardly takes one or two classes to identify both types of students. And it has nothing to do with intelligence or background. It has to do with attitude and fighting spirit. Every time I throw a problem in the class or post it in the forums, the students who are studying would be trying hard to solve the problem. They would arrive at an answer, whether right or wrong. On the other hand, the students are who are just ‘striking a pose of studying’ would be waiting for me to give the answers and the solutions. They would be acting as if they are thinking hard but in reality would not be thinking at all. On the internet, I frequently see students who waste time in idle chit chat when they should be busy solving problems. They feel satisfied that they have done ‘something’ for their CAT by looking at problem or talking about CAT to others but in essence they have done nothing and wasted one more day. “Students like these,” as I tell my serious students, “are going to fill up the CAT form to give you your 99 percentile.” One thing I have learnt from my past experiences is that you can never teach attitude to students. Either they have it or they don’t. Whereas winners focus on completing their tasks, whiners have excuses and ways of work avoidance. In TathaGat, we have a strict rule for students to submit one book review per week. And sometimes we see the book review either copied from the net or written after hearing about the content from their friends. We even have strict rule for submission of quant or DI assignments. And many students turn in their assignments half-attempted. The serious students, on the other hand, finish their assignments before time and would frequently come to me with “Sir aur material mil sakta hai kya?” There is a world of difference between the two attitudes. Where I constantly hear from the whiners “Sir aap solve kar dijiye”, I also constantly hear from the winners “Sir abhi answer mat batayiye, solve karne dijiye.” The only sane advice for the instructors in this field, who sometimes feel as frustrated as I do, is- spend 80% of your time with your winners. And winners are not the ones who score at the top. They are the ones who are fighting the hardest, whether they are weak or strong in the subject does not matter. And for students on the internet who want to do idle chit chat in the forums or engage in socializing, you would find TotalGadha.com too boring a place. We are serious about studies here. For our students on TG.com, we would like to share one of our endeavors in CAT CBT Club- Video lessons. Dagny and I have been tinkering with our camera and trying to make some good video lessons for our CBT Club students. We have uploaded one such video- Divisors of a number- for our TG junta. Do have a look and give us your feedback. Unfortunately, to watch the rest of the lessons that we are making, you will have to become a member of our exclusive club.
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Reply Re: Functions by Rohit Ghosh - Thursday, 2 July 2009, 01:31 PM Nice article TG, as usual supreme quality. Show parent | Reply Re: Functions by abhik chakraborty - Thursday, 2 July 2009, 04:12 PM tg sir, i am puzzled of what category i am till now. but one thing i know for sure what i should try to be or at least try to do. and the article on function is incredible. thank you. Show parent | Reply Re: Functions by Total Gadha - Thursday, 2 July 2009, 04:38 PM
Hi Abhik, I really do not know about that but I think you (and everyone else) should read thishttp://totalgadha.com/tgtown/amitkumar/2009/05/16/are-u-really-preparing/ Total Gadha
Show parent | Reply Re: Functions by Veena Binya - Thursday, 2 July 2009, 04:35 PM Hi TG.. The range of the function 1/1+root(x) has been given as [0,1] Should it not be (0,1] ? Because it can only tend to 0, but can never become 0.... Regards Veena Show parent | Reply
Re: Functions by Nikhil Rajagopalan - Thursday, 2 July 2009, 06:00 PM nice one Show parent | Reply Re: Functions by CATching CAT - Thursday, 2 July 2009, 06:24 PM Lesson is as usual excellent. But the sentence that stole the show is totalgadha.com/mod/forum/discuss.php?d=5440
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“Students like these,” as I tell my serious students, “are going to fill up the CAT form to give you your 99 percentile.” A wake-up call . Show parent | Reply Re: Functions by SHEKHAR SHARMA - Thursday, 2 July 2009, 07:56 PM hi tg sir...... really this article abt to know urself that in which catagory i am, is really inspirable.........n also basic knowledge abt function is very good. thank you sir Show parent | Reply Re: Functions by William Wallace - Friday, 3 July 2009, 02:07 PM Awesome article as usual between two sequence. for example in question f(1)=3600 and f(1) + f(2) + ....f(n)=n^2f(n) u have calculated f(n+1) to simplify but in question f(x)=9^x/(9^x + 3) u have calculated f(1-x). I guess this decision comes with practise but still i want to ask you any tips on how to decide which value to calculate. Please suggest Show parent | Reply Re: Functions by ayush roy - Friday, 3 July 2009, 03:56 PM sir it is a vry nice article.... sir m an engineering student,have passed out this year only,a vry serious follower of this site,n a big fan of ur articles..... wen i was preparin for IIT JEE,the most serious prb which i was facin was to sketch the graphs, i always wanted to do that,coz it helped a lot in physics also. sir what ever u hav explained ovr here is abt analyzin a function,which m vry gud at,but it has nvr helpd me in sketchin d graphs....i was nvr able to correlate both analyzin and sketchin. plz tell me how i can do that vry easily,without puttin values into a function. thank you Show parent | Reply Re: Functions by shiva chepuri - Friday, 3 July 2009, 06:58 PM it is a wonderful article !!! please discuss types of function if possible sir Sqrt(4-x2 ) range should be (-infinity , 2] because sqrt(x) outcome is R but i dont know when to calculate f(x+1) and when to f(1-x) or f(x-1) so as to get a relation
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Show parent | Reply Doubt In Functions by A B C - Saturday, 4 July 2009, 01:43 PM HI TG... I have a doubt here.. f(x) = odd and g(x) = even then f(g(x)) = odd.. now consider f(x) = x^3 and g(x) = x^2 than f(g(x)) = x^6..which is an even function and hence the statement... Am i missing some concept here ..? Show parent | Reply Re: Functions by Total Gadha - Saturday, 4 July 2009, 06:18 PM
contradicts
Hi Shiva, Give me one example when it is more than 2. Total Gadha
Show parent | Reply Re: Doubt In Functions by Total Gadha - Saturday, 4 July 2009, 06:20 PM
Hi Shailendra, I don't see the statement you have mentioned. The case you are mentioning, I have specified that it is an even function. Where is the contradiction? Total Gadha
Show parent | Reply Re: Doubt In Functions by manjushaa srinivasan - Saturday, 4 July 2009, 11:02 PM Hi TG i found your article very thought-provoking and clear cut.Himanshu's blog was also very inspirational.In my opinion any serious CAT aspirant who reads your article would become more dedicated towards his/her CAT preparation.Your explanation of functions was also very useful.I just want 1 advice from you.I'm not natural with quant.Even though i've been practising all along,i'm not sure if i can think as analytically as most of the other aspirants who are strong in quant .so far i've tried to cover all chapters but i would like to know if i'd be able to crack the quant section if i'm well versed with number system,geometry,averages,percentages,alligations,profit and loss,interest,functions,set theory,progressions,probability,logarithms,surds and indices.although i'm practising problems from P&C,Mensuration but i get stuck when the sum is slightly twisted.this is worrying me. Regards manjushaa Show parent | Reply Re: Doubt In Functions by manjushaa srinivasan - Saturday, 4 July 2009, 11:05 PM I also have a problem visualising 3 D figures especially cubes which is hampering my speed and accuracy in mensuration to a great extent.What exactly should i do about this? Show parent | Reply Re: Functions by Mohit Sharma - Sunday, 5 July 2009, 01:40 AM Nice Article...Tg Show parent | Reply
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Re: Doubt In Functions by Total Gadha - Sunday, 5 July 2009, 04:31 AM
Hi Manjusha, The trick to get good at quant is to try and solve a lot of problems on your own. Try to solve them in any which manner that you can- by putting values, by generalizing, by longer methods or by using theorems etc. But solve as much as you can. You will get good at this as you go along. Total Gadha
Show parent | Reply Re: Doubt In Functions by Veena Binya - Sunday, 5 July 2009, 09:46 AM Hi TG.. The range of the function 1/1+root(x) has been given as [0,1] Should it not be (0,1] ? Because it can only tend to 0, but can never become 0.... Regards Veena Show parent | Reply Re: Doubt In Functions by manjushaa srinivasan - Sunday, 5 July 2009, 02:34 PM hi TG Thanks for the reply. previously when cat was CBT,it was feasible to leave few tough topics like P&C,inequalities,TSD etc if we were strong with number system,geometry,P&L,percentages,functions,averages,alligations,progressions,log,algebra.With the new pattern do we have cover the entire syllabus?how is it possible to solve DI directly from computer screen especially the ones which involve tables where we need to calculate and give rankings etc?Since skimming and scanning of RCs,skillful selection of sitter questions are redundant now,what should be our new strategy? Show parent | Reply Re: Doubt In Functions by amit jain - Monday, 6 July 2009, 10:45 PM can some one explain the question, area between the coordinate axis and fff(x). I am nt getting it. Show parent | Reply Re: Functions by Prakash jain - Wednesday, 8 July 2009, 07:48 AM sir you have realy openend my eyes, i belong to the 1st category, i have been pretending to prepare for CAT since 2 year but today i have realised that my previous efforts were not true, now I promise you that i will seriously prepare for this exam,and will perform well,thank you so much for such a wonderful article... Show parent | Reply Re: Functions by Dipanjan Biswas - Wednesday, 8 July 2009, 12:18 PM At first I want to give you thanks for writing such a helluva article..It helped me to find my loopholes regarding one of the most important sections of quant. I would like to draw your attention on a few typo errors.. (1)How to draw a graph of F(x): Point number three..Latter one should be for maximum.. (2)f(x)=x^3(x+2) the second derivative of this function is positive.. Sir, I find it hard to draw any rectangular hyperbola graph..Can you elucidate it further.. TG Rockzzz\m/ Show parent | Reply Re: Functions by dishant dhokia - Thursday, 9 July 2009, 11:58 AM
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For how many values of x is f(f(f(x))) = 19?
In this sum
f(f(f(x))) = 19
The next step is f(f(x)) = 38………….
I am not able to deduce this………..
Please someone elaborate??????????? Show parent | Reply
Re: Functions by Dipanjan Biswas - Friday, 10 July 2009, 08:21 AM Hi Dishant, Look at the function.. See f(x) = x/2 if x is even = x+1 if x is odd In other way x=2f(x) if f(x) is either even or odd x=f(x)-1 if f(x) is even As f(x)=19=odd so x=38 likewise f(x)=38 x=76 or 37 likewise f(x)=76 x=152 or 75 for f(x)=37 x=74 so {x}=>{74,75,152} Hope it helps Cheers!! Show parent | Reply Got a doubt by Varun Agrawal - Friday, 10 July 2009, 03:01 PM Please discuss types of function if possible sir Sqrt(4-x2 ) range should be [-2 , 2] because sqrt(x) of a number gives 2 values. Positive and negative. Can you please elucidate how it is [0,2] Thanks
Show parent | Reply Re: Got a doubt by Dipanjan Biswas - Saturday, 11 July 2009, 01:38 PM Varun please do go back to the question.. It does not want the range of x rather the range of f(x). Max of f(x)=2 when x=0 Min of f(x)=0 when x=+-2 So f(x) [0,2] Please don't be in haste before raising your doubts. TG family takes utmost care before publishing these lessons..
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Hope you have accepted my comment on a positive note!! All the best!! Show parent | Reply Re: Functions by shiva chepuri - Saturday, 11 July 2009, 08:00 PM yes sir got it but it should be [-2,2] if i am wrong plz correct Show parent | Reply Re: Functions by Varun Goenka - Sunday, 12 July 2009, 10:22 PM the func. is basically y=sqrt(4-x^2) i.e. y^2=4-x^2 this is equation of a circle...x^2 + y^2 = 4...with origin as centre and 2 as radius... but the given f(x) is just the upper half of the circle.. y=sqrt(4 - x^2)...is for upper half of the circle... y= -sqrt(4 - x^2)...is for lower half of the circle... our question is related to upper half of the circle...so range is [0,2] and domain is [-2,2].. I think this should help... Regards, Varun Goenka Show parent | Reply
Re: Functions by Siddharth sahoo - Monday, 13 July 2009, 03:36 PM @sahil, As discussed by you, x^2 * x^3 will be x^5. X^m * x^n = x(m+n).x5 is an odd function. Show parent | Reply Re: Doubt In Functions by soumen paul - Tuesday, 14 July 2009, 06:59 PM f(x)=x^3 is not a odd function if x=2,4,6........ g(x)=x^2 is not a even function if x=3,5,7........ f(x)=2x+1 is a odd function for 0,1,2..... g(x)=2x+2 is a even function for 0,1,2...... so here f[g(x)]=4x+5 is a odd function Show parent | Reply Re: Functions by Ashish Kumar - Thursday, 16 July 2009, 12:23 AM lol PATAKKA article....but what i loved more is those who are whiners and are reading this can actually do something still coz there is still time but even after this if they cannot do anything then trust me they are helping the winners more n more lol ROFL !!! Show parent | Reply
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Re: Functions by A S - Thursday, 16 July 2009, 04:19 AM Hi TG,
Simply superb article...I think I have fallen in love with functions...can you please add more problems on functions? or may be one set of problems? or may be some book....anything will do.... Show parent | Reply
Re: Functions by A S - Thursday, 16 July 2009, 04:24 AM
hi dishant, take it this way...can f(x) ever be odd? when x is odd, f(x)=x+1...so it will always give you even f(x).. when x is even, f(x)=x/2; this is only chance when you can get odd f(x)..
here f(m) [m=f(f(x)) ] is odd.....so m must be even... that means m/2=38 so, m=76 i.e. f(f(x))=76..
hope it helps Show parent | Reply Re: Functions by piyush dixit - Friday, 17 July 2009, 09:39 PM HI TG, I am a frequent reader of your post, and following this site almost since its inception. I am an adherent fan of yours, BUT I am very disappointed with this article of yours, and hence writing to you for the first time. I guess, you started this whole exercise of totalgadha.com to ridicule the jargon taught by famous coaching classes in their curriculum to show their superiority among the students community, TG i belive you started the exercise to change the way education system works at this level. But it seems you are getting infected by the same BUG. I am sorry, if i have misunderstood you. Hats off for all your other posts, which I have been through. PIYUSH Show parent | Reply Re: Functions by Total Gadha - Sunday, 19 July 2009, 10:49 PM
Hi Piyush, I am infected by what I do, down to my last nail. ignore everything else. Total Gadha
Show parent | Reply Re: Functions by Surinder Pal Singh - Saturday, 25 July 2009, 09:43 AM
My opinion should hardly matter to you. Take what is good and
Hi TG Sir, Excellent article. But the link above is infected with a Trojan Horse.
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Kindly verify its integrity as I was just saved by my firewall.
Keep up the good work Regards Surinder
Show parent | Reply Re: Functions by gaurav kaushal - Sunday, 26 July 2009, 11:35 PM hi tg sir , thx for this gr8 article..... now dat i v read this whole article ...where can i get some questions related to this article...so dat i can just make the concepts more clear thanks a lot..sir Show parent | Reply Re: Functions by vivek ghiria - Monday, 27 July 2009, 10:09 AM Hi TG I have few doubts in Functions. If the first derivative f'(x) is positive, then the function f(x) is increasing. If the first derivative f'(x) is negative, then the function f(x) is decreasing. For f(x) = x^3-4x^2+1 f'(x) = 3x^2-8x ....(1) Now how to know whether expression (1) is +ve or -ve to determine increasing or decreasing charactersitic? Also I recall there is some way to determine whether f(x) is continuous or not from first derivative of f(x). Please explain that rule also. Thanks Vivek
Show parent | Reply Re: Doubt In Functions by dilip rathore - Tuesday, 4 August 2009, 04:12 PM g(x) is calculated for some x and some value is obtained. later "this value" is put into f(x). so f(g(x)) is not actually x^6. e.g. for x= 2, g(x) = g(2) = 4 so f(g(x)) = f(4) = 4^3. now (4)^3 and (-4)^3 are not same right :P Show parent | Reply Re: Functions by dilip rathore - Tuesday, 4 August 2009, 04:34 PM hi vivek, i think we check f''(x) to determine if the function is increasing or decreasing. for your example f(x) = x^3-4x^2+1 f'(x) = 3x2 - 8x f''(x) = 6x-8
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for x > 4/3 , function is increasing and for less than 4/3 , its decreasing (slope). pls let me know if i am wrong Show parent | Reply Re: Functions by himanshu gupta - Saturday, 8 August 2009, 07:28 PM thank u sir 4 such a good article.sir is thr any oter approach to solve th last 2 ques?? Show parent | Reply Re: Functions by abhishek tripathi - Sunday, 9 August 2009, 04:14 AM well sir i am a new user over here i found dis site preety useful but regarding first question to find out the range can u exactly tell me how do we aarive at 38 when (fx)=19 since it is odd and according 2 function definaion in da question it should be x+1=20 me will b very glad if u tell me how 2 solve dese type of questions. Show parent | Reply Re: Functions by Jiggs Sanyal - Monday, 17 August 2009, 10:03 AM Hello TG Sir, I am a hardcore fan of your articles and of the TG website. It is the first website that I open everyday. But for the last few weeks, I have been sort of disappointed coming here, because you have completely stopped writing articles for people like me who are not a part of the CAT CBT club. In one of the discussion forums, I had seen you mention that TG is for those who have the zeal to make their dreams come true. I have the requisite zeal but then TG now seems to have become more and more commercialised. I am not against commercialisation but you should keep in mind those who have the zeal but not the bulky pockets to pay. Sorry If I hurt you in any way. you are the ideal motivator to me.
Show parent | Reply Re: Functions by Jai Prakash - Saturday, 22 August 2009, 07:39 PM Thanks TG. I've been a guy mentioned by you. but now i promise that i'll work hard and..... whatever directed by you and necessary for excelling. It's my 1st post. TG had been rocking but this time it made rhe rivers of truth flow from eyes. I never posted on TG as i have been getting the answers of my questions from orher's posts. I've made the sentence of TG Sir " YOU ARE NEVER LATE" the ultimate source of inspiration. Thanks a ton for showing me the mittor. Hope the other guys too will correct themselves. Sir due to time constraints i'm not posting some querries. Will do it later on.........
"ITS BETTER TO FAIL IN DOING SOMETHING THAN TO EXCEL IN DOING NOTHING." Show parent | Reply Re: Functions by Vivek .. - Saturday, 5 September 2009, 01:30 PM hi,, i m a regular visitor of totalgadha.com & i really appreciate ur work.. all i want to know is,,if sum1 starts prep now for cat,,does he has any chance of clearing it. i m clear wid 50 % of the concepts pof all the 3 subs. but prob is i m never able to get marks or %ile upto my satistfaction or wat is reqd.,,so i m confused of whther wil i b able to do it or nt? in the last 3 mocks of tathagat i scored 24,25 & 15 marks & i m very ashamed of this score. It wud be helpful if u give me ur sincere advice of shud i stop following the dream or nt? thank you. Show parent | Reply Re: Functions by saurabh chauhan - Saturday, 5 September 2009, 11:27 PM Hi Jay... nice quote... "ITS BETTER TO FAIL IN DOING SOMETHING THAN TO EXCEL IN DOING NOTHING." i'll make this my driving factor rom today...and many thanks to TG..the article.."Are u really prepaing?" inspired me a lot...nw i'll really pepare and get through......my prep. starts now.................3 2 1......go....
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saurabh Show parent | Reply Re: Functions by Vivek .. - Sunday, 6 September 2009, 09:19 AM INFACT U CAN SAY THAT I WAS A 2ND TYPE OF STUDENT UPTIL NOW BUT YOUR ARTICLE HAS REALLY OPENED NOT ONLY MY EYES BUT MY VIEW & I M SERIOUSLY SERIOUS NOW. JZ TEL ME THAT CAN SUM1 OR HAS HAS1 CLEARED CAT WITH PREPARATION IN SUCH A SHORT SPAN.?? ANXIOUSLY WAITIN FOR YOUR REPLY. Show parent | Reply Re: Functions by Jai Prakash - Monday, 7 September 2009, 04:51 PM Dear vivek saini it seems (perhaps) you haven't rushed through the various posts of TG. What TG sir say to such question is "YOU ARE NEVER LATE". Work honestly and yu will make it.... Best wishes... Jai Praqas Show parent | Reply Re: Functions by hbk Not joining IIML - Monday, 14 September 2009, 05:36 PM Forget function guys, this guy(TG) can inject attitude in U. A really serious one. Hats off. ****************I'm gonna get back soon************************ Show parent | Reply Re: Functions by subhadip mahalanabish - Wednesday, 16 September 2009, 01:45 PM Dear TG sir, consider the function y=x/(x^2+1). We have to find the range of this function. If we directly put x=0, we will get y=0, then if we put x=1 then y=1/2, for x=3, y = 3/10, for y=4, x=4/17 and so on, so the maximum value of the function is 1/2 at x=1 and then the function decreases as x increases, same is true for negative values of x, in the negative side minimum value of y is -1/2(at x= -1) and then y increases as x decreases. so range of the function should be -1/2 <= y <= +1/2 including y = 0. But if we calulate in this way, y = x/x^2 + 1 => x^y -x + y= 0 => x = {1+-sqrt(1-4y^2)}/2*y. Hence range will be, -1/2<=y<=+1/2 excluding y = 0. Here we are seeing that y cant be equal to zero. But previously we already saw that y can be 0 for x= 0. How can we remove the confusion of which result is to be used? regards, Subhadip. Show parent | Reply Re: Functions by subhadip mahalanabish - Wednesday, 16 September 2009, 03:45 PM hi all, On giving a second look I found myself that the function is discontinuous at x=0.Probably this is the reason why y=0 is not to be considered in the range of the the function. regards, Subhadip. Show parent | Reply Re: Functions by hbk Not joining IIML - Saturday, 19 September 2009, 06:18 PM @subhadip
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In your approach we are considering the eqn (x^2)y -x+y as a quadratic in x having coft of y. By defn of quadratic eqn coeft of x^2 can't be zero hence with this approach we may not get y=0 as a feasible. It has to checked separately. Show parent | Reply thanx..... by raghav sharma - Sunday, 27 September 2009, 03:34 PM sir... your adomitable aproach towards conspt has really given a lot of helf to me solve my probs.... thanx a ton... Show parent | Reply Re: Doubt In Functions by Kishan Kumar - Thursday, 12 November 2009, 12:09 AM f(x) = odd and g(x) = even then "f(g(x)) = odd.." you went wrong at this part it is not odd its even Show parent | Reply Re: Doubt In Functions by Subhash Medhi - Saturday, 22 May 2010, 05:45 PM Dear TG sir, I have a doubt in the solution to the question, Let f(x) be a function with domain 0<=x<=1 defined as 2x if 0<=x<=1/2 f(x) = { -2x + 2 if 1/2<=x<=1 Find the area of the curve y = f(f(f(x))) and the coordinate axis. in the second line, 0 <= f(f(f(x))) <= 1 - - > 1/2 <= f(f(x)) <= 1 - - > 1/4 <= f(x) <= 1/2 - - - >1/8<=x<=1/4 The doubt is that if 1/2 <= f(f(x)) <= 1 then shouldn't it imply that 1/2 <= - 2f(x) + 2 <= 1 which would imply 1/2 <= f(x) <= 3/4 ? Since f(f(x)) lies in the interval [1/2,1], shouldn't we be using the second part of the definition of f(x) ?.Kindly clarify my doubt. I am having very similar doubts in the subsequent lines of the solution.Kindly help me out. Regards, Subhash Show parent | Reply Re: Doubt In Functions by shail mishra - Thursday, 10 June 2010, 07:44 PM Hello TG I also couldn't understand the above mentioned question. Kindly explain it. Regards SM Show parent | Reply Re: Doubt In binomials by Ankur Bansal - Friday, 2 July 2010, 08:17 AM Hello TG Sir and hi to ALL Can anyone help me with binomial theorems and help me in finding out the coefficients. For ex. Find the coefficient of x in (1+2x+3x^2)^15(1+x)^5 ? And i am really interested in knowing the concept for these type of problems??? And TG Sir, if you can then pls come up with one good article on Binomial Theorems. Thanks in advance.
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Show parent | Reply Re: Doubt In binomials by amit kheterpal - Friday, 2 July 2010, 11:52 AM Find the coefficient of x in (1+2x+3x^2)^15(1+x)^5 ? 2x+3x^2=a
(1+a)^15=1+5c1*a+15c2*a^2+...........15c15*a15
(1+x)^5=1+15c1*x+5c2*x^2+.......5c5*x^5 now,(1+15c1*a+15c2*a^2+...........15c15*a15)*(1+5c1*x+5c2*x^2+.......5c5*x^5) putting the value of a so coffecient of x will be 5c1+(15c1)*2 =35
Show parent | Reply Re: Doubt In binomials by Ankur Bansal - Friday, 2 July 2010, 12:57 PM Hi Amit,
Thanks for reply And for the same problem if we have to find the coefficient of x^4 then how we will proceed?
Thanks in advance!!!!! Show parent | Reply Re: Functions by G D - Tuesday, 6 July 2010, 05:27 PM Nice one, Show parent | Reply
Re: Functions by Tasmai Merchant - Monday, 26 July 2010, 05:14 PM Hey Guys .....I wanna kno if buying books from Tg like number system, or geometry wud be of any help....nd i can't find any book on algebra...can u plz suggest me some options of buying one.. Show parent | Reply Re: Doubt In Functions by Ritesh Raman - Thursday, 5 August 2010, 03:25 AM TG Sir, Thanks a lot for your out of the world kind of materials. I am just giving some lines on find the area of f(F(f(x)))...ques. OOOPS late to reply but may be useful for guys who look from now on. First we will find f(f(x)) It should come as 4x ; 0≤x≤1/4 -4x + 2 ; 1/4≤x≤1/2 4x - 2 ; 1/2≤x≤3/4
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-4x + 4 ; 3/4≤x≤1 Then find f(f(f(x))).This shuld be 8x ; 0≤x≤1/8 -8x + 2 ; 1/8≤x≤1/4 8x - 2 ; 1/4≤x≤3/8 -8x + 4 ; 3/8≤x≤1/2 8x - 4 ; 1/2≤x≤5/8 -8x + 6 ; 5/8≤x≤3/4 8x - 6 ; 3/4≤x≤7/8 -8x + 8 ; 7/8≤x≤1 Now the min value of each funct(straight line ) is 0 and max is 1. You can verify it by sub. values This gives us the graph that TG sir provided. Hence area = 8 * (1/2*1/8*1) = 1/2 sq unit. I hope there is no mistake. Is it correct Sir??? Show parent | Reply Re: Functions by Deepak Kumar - Thursday, 5 August 2010, 12:26 PM Hi TG Sir, Its always feel nice to read your articals. For the last problem for calculating f(94), I approached this way. As f(x)+ f(x-1)=x^2...........(1) on replacing x with x+1; f(x+1)+f(x)=(x+1)^2........(2) Now subtracting (1) from (2) f(x+1)-f(x-1)=2x+1..........(3) On putting values of x as 93,89........19 will give LHS as f(94)-f(18). f(18) can be easily calculated from (1) when f(19) is given. This approach will reduce RHS to an easy airthmatic progression. thanks a lot for your help sir. Show parent | Reply
Re: Functions by himanshu jha - Thursday, 5 August 2010, 06:40 PM hi tg sir, i know this is not the place to contact you..but i couldnt find any other way...try to reach u thrgh ur admin mail but dat too in vain...i ve just got one query i want to join cbt club..i want to pay thrgh demand draft but in it "payable at" option is given..so what to fill in dat new delhi r something else....if any other person knows it then please let me know or any other way to contact tg sir... thanxs in advance... Show parent | Reply Re: Functions by wishy wishy - Friday, 5 August 2011, 05:28 PM Please correct me if I am wrong: In the last question,f(x) + f(x-1) = x^2 then f(1) = 1, f(2) = 3 , f(3) = 6, f(4) = 10 i.e, f(n) = n(n+1)/2 => f(19) = 190 and hence f(94)= 4565 Show parent | Reply Re: Functions by Total Gadha - Friday, 5 August 2011, 10:02 PM hi himanshu. it should be New Delhi Show parent | Reply
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Re: Functions by Gregory Matchado - Thursday, 18 August 2011, 05:39 PM Hi TG.. Thank you so much for all the articles.. !!! And let me also tell you that you're videos on divisors, remainders and co primes are just tooooooo brilliant... Thanks a lot.. Show parent | Reply Re: Functions by abhinay dutta - Thursday, 8 September 2011, 06:50 PM hi, i have calculated the last question in post and arrived at 4181. guys please chk , and correct me if i m wrong.... Show parent | Reply
Re: Functions by nitin sharma - Friday, 16 September 2011, 11:58 AM Hi TG sir i didnot get the explanation of d 1st quest of f(f(f(x)))) where domain of x was 0<x<1..som1 plz help me..i men how did u draw the graph.. Show parent | Reply Re: Functions by IIM gadha - Friday, 16 September 2011, 12:26 PM
@TG sir,
The graph drawn for f(f(f(x))) is incorrect. Till 0<x<1/2, the graph is of 2x but from 1/2<x<1 the graph should be of -2X + 2 i.e the one with negative slope not positive as shown in your graph. Though, the area remains unaffected but the graph is wrong. Please correct me or yourself. Show parent | Reply Re: Functions by Kamal Lohia - Friday, 16 September 2011, 04:47 PM
Vibha Function given is for f(x) but the area is asked for f(f(f(x))). Please check before posting. Kamal Lohia
Show parent | Reply Re: Functions by IIM gadha - Friday, 16 September 2011, 04:57 PM
@Kamal sir Apologies! Show parent | Reply Re: Functions by ketav sharma - Saturday, 17 September 2011, 06:49 PM hanji dutta sahib its 4181 Show parent | Reply
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Re: Functions by Rahul Sharma - Tuesday, 11 October 2011, 08:13 PM HI TG, the below mentioned link that u shared is net getting opened, could you please help me with it.
http://totalgadha.com/tgtown/amitkumar/2009/05/16/are-u-really-preparing/
Rahul Sharma
Show parent | Reply
Re: Functions by Kamal Joshi - Friday, 14 October 2011, 09:07 PM Dear TG sir, The links such as these return a fatal error through out the site. Kindly resolve the issue. can't afford to miss the Gyan from you. http://totalgadha.com/tgtown/amitkumar/2009/05/16/are-u-really-preparing/ Thanks to you, Kamal sir, Dagny ma'm and rest of the awesome team! Show parent | Reply Re: Functions by vasu r - Tuesday, 18 October 2011, 01:03 PM how to sketch t graph in t question of f(f(fx)))? y divide into 8 parts?? cud u pl explain ? Show parent | Reply Re: Functions by Kamal Lohia - Tuesday, 18 October 2011, 01:32 PM
Hi Vasu Just don't look at the solution and only try to see/analyse the question and thge function given. See if f(x) is some linear function what about the degree of f(f(x))???? It'll also remain a linear function as you are just replacing 'x' in f(x) by f(x). So degree of f(f(x)) is not changing. Similarly f(f(f(x))) is also a linear function. But as f(x) was a different function in deferent periods, so is the case with f(f(f(x))) also. Now it will be better that you try yourself to findout period and then graph of new function. Kamal Lohia
Show parent | Reply Re: Functions by Ravi Kumar - Tuesday, 1 November 2011, 02:43 PM Re: Functions by IIM gadha - Friday, 16 September 2011, 12:26 PM
@TG sir,
The graph drawn for f(f(f(x))) is incorrect. Till 0<x<1/2, the graph is of 2x but from 1/2<x<1 the graph should be of -2X + 2 i.e the one with negative slope not positive as shown in your graph. Though, the area remains unaffected but the graph is wrong. Please correct me or yourself.
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a 0 14 12 34&1 t , /, /, / . yai -xs | | | / \ / \ / \ / \ | / \ / \ / \ / \ |/ \ / \ / \ / \ _|___\___\___\___\___ xai _/___/___/___/______ xs 0 14 / 12 / 34 / 1
ase =4.(/ .14.1 =4 18 nwr 12 / ) . /
=12rmistesm tog.. / ean h ae huh.
Show parent | Reply
Re: Functions by King Casanov - Tuesday, 1 November 2011, 02:29 PM @ TG Sir and Kamal Saar...would this suffice the need of CAT'11 Function part of QA or anything more req.? If so please inform/ provide a link to.. Thanks in advance...
The KING Show parent | Reply Re: Functions by Kamal Lohia - Tuesday, 1 November 2011, 04:11 PM
Ravi Thanks for the correction. It'll be updated in the article soon. Kamal Lohia
Show parent | Reply Re: Functions by ayush lakhotia - Monday, 7 November 2011, 06:29 PM Kamal Sir, f(a+b)=F(a)+F(b) and f(5)=12. Find F(12) Please help me in solving this. Show parent | Reply Re: Functions by Kamal Lohia - Tuesday, 8 November 2011, 02:42 PM
Hi Ayush Given is: f(a + b) = f(a) + f(b) For a = b, we have: f(2a) = 2f(a) Similarly if we put b = 2a in given equation, we get: f(3a) = 3f(a) using second equation also. So in general, f(na) = nf(a). Now given is: f(5) = 12. So f(12) = f((12/5)×5) = (12/5)×f(5) = (12/5)×12 = 144/5.
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Kamal Lohia
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