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Functionsby 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 andthose who believe that they are studying but are actually not. Every time I start taking a freshbatch, it hardly takes one or two classes to identify both types of students. And it has nothing todo with intelligence or background. It has to do with attitude and fighting spirit. Every time Ithrow a problem in the class or post it in the forums, the students who are studying would betrying hard to solve the problem. They would arrive at an answer, whether right or wrong. Onthe other hand, the students are who are just striking a pose of studying would be waitingfor me to give the answers and the solutions. They would be acting as if they are thinking hardbut in reality would not be thinking at all. On the internet, I frequently see students who wastetime in idle chit chat when they should be busy solving problems. They feel satisfied that theyhave 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 myserious students, are going to fill up the CAT form to give you your 99 percentile. One thing Ihave learnt from my past experiences is that you can never teach attitude to students. Either

they have it or they dont. 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 orwritten after hearing about the content from their friends. We even have strict rule for submission of quant or DI assignments. And manystudents turn in their assignments half-attempted. The serious students, on the other hand, finish their assignments before time and wouldfrequently come to me with Sir aur material mil sakta hai kya?There is a world of difference between the two attitudes. Where I constantlyhear 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 yourwinners.And winners are not the ones who score at the top. They are the ones who are fighting the hardest, whether they are weak orstrong 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 tinkeringwith our camera and trying to make some good video lessonsfor our CBT Clubstudents. We have uploaded one such video- Divisors of anumber-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: Functionsby Rohit Ghosh- Thursday, 2 July 2009, 01:31 PM

Nice article TG, as usual supreme quality.

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Re: Functionsby 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 isincredible.thank you.

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Re: Functionsby 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 this-

http://totalgadha.com/tgtown/amitkumar/2009/05/16/are-u-really-preparing/

Total GadhaShow parent| Reply
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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....

RegardsVeena

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Re: Functionsby Nikhil Rajagopalan- Thursday, 2 July 2009, 06:00 PM

nice one

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Re: Functionsby CATching CAT- Thursday, 2 July 2009, 06:24 PM

Lesson is as usual excellent.

But the sentence that stole the show is

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 .

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Re: Functionsby 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

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Re: Functionsby William Wallace- Friday, 3 July 2009, 02:07 PM

Awesome article as usual 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 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

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Re: Functionsby 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 physicsalso.

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 tocorrelate both analyzin and sketchin.plz tell me how i can do that vry easily,without puttin values into a function.

thank you
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Re: Functionsby 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 isR

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Doubt In Functionsby 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 contradicts

the statement...

Am i missing some concept here ..?

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Re: Functionsby Total Gadha- Saturday, 4 July 2009, 06:18 PM

Hi Shiva,

Give me one example when it is more than 2.

Total Gadha

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Re: Doubt In Functionsby 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

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Re: Doubt In Functionsby manjushaa srinivasan- Saturday, 4 July 2009, 11:02 PM

Hi TGi found your article very thought-provoking and clear cut.Himanshu's blog was also very inspirational.In my opinion any serious CAT aspirant who readsyour article would become more dedicated towards his/her CAT preparation.Your explanation of functions was also very useful.I just want 1 advice fromyou.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 arestrong 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 numbersystem,geometry,averages,percentages,alligations,profit and loss,interest,functions,set theory,progressions,probability,logarithms,surds andindices.although i'm practising problems from P&C,Mensuration but i get stuck when the sum is slightly twisted.this is worrying me.Regardsmanjushaa

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Re: Doubt In Functions
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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 exactlyshould i do about this?

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Re: Functionsby Mohit Sharma- Sunday, 5 July 2009, 01:40 AM

Nice Article...TgShow parent| Reply

Re: Doubt In Functionsby 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

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Re: Doubt In Functionsby 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....

RegardsVeena

Show parent| Reply

Re: Doubt In Functions

by manjushaa srinivasan- Sunday, 5 July 2009, 02:34 PM

hi TGThanks 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 numbersystem,geometry,P&L,percentages,functions,averages,alligations,progressions,log,algebra.With the new pattern do we have cover the entire syllabus?howis it possible to solve DI directly from computer screen especially the ones which involve tables where we need to calculate and give rankings etc?Sinceskimming and scanning of RCs,skillful selection of sitter questions are redundant now,what should be our new strategy?

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Re: Doubt In Functionsby 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.

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Re: Functionsby 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 myprevious 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 wonderfularticle...

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Re: Functionsby 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 ofquant.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..
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Sir,I find it hard to draw any rectangular hyperbola graph..Can you elucidate it further..

TG Rockzzz\m/

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Re: Functionsby dishant dhokia- Thursday, 9 July 2009, 11:58 AM

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???????????

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Re: Functionsby 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 oddIn other wayx=2f(x) if f(x) is either even or oddx=f(x)-1 if f(x) is evenAs f(x)=19=oddso x=38likewise f(x)=38

x=76 or 37likewise f(x)=76x=152 or 75for f(x)=37x=74so {x}=>{74,75,152}

Hope it helps

Cheers!!

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Got a doubtby 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

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Re: Got a doubtby 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=0Min of f(x)=0 when x=+-2
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So f(x) [0,2]Please don't be in haste before raising your doubts. TG family takes utmost care before publishing these lessons..

Hope you have accepted my comment on a positive note!!All the best!!

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Re: Functionsby 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

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Re: Functionsby 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

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Re: Functionsby 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.

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Re: Doubt In Functionsby 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

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Re: Functionsby 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 evenafter this if they cannot do anything then trust me they are helping the winners more n more lol ROFL !!!

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Re: Functionsby A S- Thursday, 16 July 2009, 04:19 AM

Hi TG,
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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 maybe some book....anything will do....

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Re: Functionsby 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

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Re: Functionsby 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 thisarticle 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 byfamous coaching classes in their curriculum to show their superiority among the students community, TG i belive you started the exercise to change theway 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

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Re: Functionsby Total Gadha- Sunday, 19 July 2009, 10:49 PM

Hi Piyush,

I am infected by what I do, down to my last nail. My opinion should hardly matter to you. Take what is good and

ignore everything else.

Total Gadha

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Re: Functionsby Surinder Pal Singh- Saturday, 25 July 2009, 09:43 AM

Hi TG Sir,

Excellent article.

But the link above is infected with a Trojan Horse.

Kindly verify its integrity as I was just saved by my firewall.

Keep up the good work

Regards

Surinder

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Re: Functionsby 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

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Re: Functionsby 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+1f'(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

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Re: Doubt In Functionsby 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

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Re: Functionsby 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+1f'(x) = 3x2 - 8x

f''(x) = 6x-8

for x > 4/3 , function is increasing and for less than 4/3 , its decreasing (slope).

pls let me know if i am wrong

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Re: Functionsby 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??

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Re: Functionsby 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 at38 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 desetype of questions.
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S ow parent Rep y

Re: Functionsby 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 beensort 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 ofthe 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 thenTG 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.

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Re: Functionsby 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 my1st 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 myquestions 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 themittor. 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."

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Re: Functionsby 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 probis 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 stopfollowing the dream or nt? thank you.

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Re: Functionsby 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 getthrough......my prep. starts now.................3 2 1......go....

saurabh

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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 MSERIOUSLY 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.

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Re: Functionsby 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 yuwill make it....

Best wishes...

Jai Praqas

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Re: Functionsby 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************************

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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 theny=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 thefunction should be -1/2 x = {1+-sqrt(1-4y^2)}/2*y.Hence range will be, -1/2


19/24



in the second line,0


20/24



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Re: Functionsby G D- Tuesday, 6 July 2010, 05:27 PM

Nice one,

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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 plzsuggest me some options of buying one..

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Re: Doubt In Functionsby 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 as4x ; 0!x!1/4-4x + 2 ; 1/4!x!1/24x - 2 ; 1/2!x!3/4-4x + 4 ; 3/4!x!1

Then find f(f(f(x))).This shuld be8x ; 0!x!1/8-8x + 2 ; 1/8!x!1/48x - 2 ; 1/4!x!3/8-8x + 4 ; 3/8!x!1/28x - 4 ; 1/2!x!5/8-8x + 6 ; 5/8!x!3/48x - 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. valuesThis 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???

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Re: Functionsby 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 asf(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.

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Re: Functionsby 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 onequery 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
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any other person knows it then please let me know or any other way to contact tg sir...

thanxs in advance...

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Re: Functionsby 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

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Re: Functionsby Total Gadha- Friday, 5 August 2011, 10:02 PM

hi himanshu.

it should be New Delhi

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Re: Functionsby 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..

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Re: Functionsby 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....

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Re: Functionsby 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


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by tgdel123 tg- Friday, 16 September 2011, 04:57 PM

@Kamal sir

Apologies!

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Re: Functionsby ketav sharma- Saturday, 17 September 2011, 06:49 PM

hanji dutta sahib its 4181

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Re: Functionsby 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.


Rahul Sharma

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Re: Functionsby 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.


Thanks to you, Kamal sir, Dagny ma'm and rest of the awesome team!Show parent| Reply

Re: Functionsby 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 ?

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Re: Functionsby TG Team- 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

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Re: Functions

by Ravi Kumar- Tuesday, 1 November 2011, 02:43 PM

Re: Functionsby IIM gadha- Friday, 16 September 2011, 12:26 PM
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@TG sir,The graph drawn for f(f(f(x))) is incorrect. Till 0


24/24


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.

Kamal Lohia

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Re: Functionsby ashutosh kumar- Thursday, 14 June 2012, 10:25 AM

The above Link is showing error..Could u plz post the correct link...

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Re: Functionsby ashutosh kumar- Thursday, 14 June 2012, 10:26 AM

The above Link is showing error..Could u plz post the correct link...

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Re: Doubt In Functionsby Parthasarathi Boyal- Tuesday, 9 October 2012, 12:10 AM

if f(f(f(X)))=19 how f (f (x))will be 38?

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Re: Functionsby Rachit Pande- Wednesday, 9 April 2014, 10:14 PM

Hi TG,Kindly help solve the last expression of alternate + and -. unable to solve it.Rachit

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