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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 whostudy and those who believe that they are studying but are actually not. Every time Istart taking a fresh batch, it hardly takes one or two classes to identify both types ofstudents. And it has nothing to do with intelligence or background. It has to do withattitude and fighting spirit. Every time I throw a problem in the class or post it in theforums, the students who are studying would be trying hard to solve the problem. Theywould arrive at an answer, whether right or wrong. On the other hand, the studentsare who are just striking a pose of studying would be waiting for me to give theanswers and the solutions. They would be acting as if they are thinking hard but inreality would not be thinking at all. On the internet, I frequently see students whowaste time in idle chit chat when they should be busy solving problems. They feelsatisfied that they have done something for their CAT by looking at problem or talkingabout CAT to others but in essence they have done nothing and wasted one more day

Students like these, as I te ll my serious students, are going to fill up the CAT form to give you your 99 pe rcentile. One thingI have 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 astrict rule for students to submit one book review per week. And sometimes we see the book review either copied from the netor written after hearing about the content from their friends . We even have strict rule for submission of quant or DIassignments. And many students turn in their assignments half-attempted. The serious students, on the other hand, finish theassignments before time and would frequently come to me with Sir aur material mil sakta hai kya?There is a world ofdifference between the two attitudes. Where I constantly hear from the whiners Sir aap solve kar dijiye, I also constantlyhear 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 whoscore at the top. They are the ones who are fighting the hardest, whether they are weak or strong in the subject does notmatter. And for students on the internet who want to do idle chit chat in the forums or engage in socializing, you would findTotalGadha.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 havebeen tinkering with our camera and trying to make some good video lessons for our CBT Club students. We have uploaded onesuch 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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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 thearticle on function is incredible.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 Gadha

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Re: Functionsby 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


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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 .

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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 rea lly inspirable.........n also basic knowledge abt function isvery 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 relationbetween 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 s implify 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,cozit helped a lot in physics also.sir what ever u hav explained ovr here is abt ana lyzin a function,which m vry gud at,but it has nvr helpd me in sketchin dgraphs....i was nvr able to correlate both analyzin and sketchin.plz tell me how i can do tha t vry eas ily,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 is R


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Doubt In Functionsby A B C - Saturday, 4 July 2009, 01:43 PM

HI TG...

I have a doubt he re..

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 c ontradict ion?

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 a lso very inspirational.In my opinion any seriousCAT aspirant who reads your article would become more dedicated towards his/her CAT preparation.Your explanation offunctions was a lso very useful.I just want 1 advice from you.I'm not natural with quant.Even though i've been practising allalong,i'm not sure if i can think as analytically as most of the o ther aspirants who are strong in quant .so far i've tried to coverall 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,settheory,progressions ,probability,logarithms,surds and indices.although i'm practising problems from P&C,Mensuration but i getstuck when the sum is slightly twisted.this is worrying me.Regardsmanjushaa

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Re: Doubt In Functionsby manjushaa srinivasan - Saturday, 4 July 2009, 11:05 PM

I also have a problem visualising 3 D figures especially cubes w hich is hampering my speed and accuracy in mensuration to agreat extent.What exactly should i do about this?

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

Nice Article...Tg

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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 manne

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

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Re: Doubt In Functionsby 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 withnumber system,geometry,P&L,percentages,functions,averages,alligations,progressions,log,algebra.With the new pattern do whave cover the entire syllabus?how is it possible to solve DI directly from computer screen especially the ones which involvetables where we need to calculate and give rankings etc?Since skimming and scanning of RCs,skillful selection of sitterquestions 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 buttoday i have rea lised that my previous efforts were not true, now I promise you that i will seriously prepare for this exam,andwill perform well,thank you so much for such a wonderful article...

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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 loopho les regarding one o f the mostimportant 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/

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Re: Functionsby 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???????????

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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)=38x=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 poss ible

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=+-2So 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!!

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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 o f 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 a re reading this can actually do something still cozthere is still time but even after this if they cannot do anything then trust me they are helping the winners more n more lol ROF!!!

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

Hi TG,

Simply superb article...I think I have fallen in love w ith functions...can you please add more problems on functions? or may beone set of problems? or may be 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 o f yours, BUT I am

very disappointed w ith this article of yours, and hence writing to you for the first time. I guess, you started this whole exerciseof totalgadha.com to ridicule the jargon taught by famous coaching classes in their curriculum to show their superiority amongthe students community, TG i belive you started the exercise to change the way education system works at this level. But itseems 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.


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

Keep up the good work

Regards

Surinder

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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 da t i can just make the conceptsmore 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 de rivative f'(x) is pos itive, 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 de termine if the function is increas ing or decreasing.

for your example f(x) = x 3-4x^2+1f'(x) = 3x2 - 8x

f''(x) = 6x-8


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saurabh

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Re: Functionsby 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 EYEBUT MY VIEW & I M SERIOUSLY SERIOUS NOW. JZ TEL ME THAT CAN SUM1 OR HAS HAS1 CLEARED CAT WITH PREPARATION INSUCH 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 NEVERLATE". Work honestly and yu will 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: Functionsby 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 w ill 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=and then the function decreases as x increases, same is true for negative values of x, in the negative side minimum value of y -1/2(at x= -1) and then y increases as x decreases . so range o f the function should be -1/2 x = {1+-sqrt(1-4y^2)}/2*y.Hence range will be, -1/2


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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 coe ft ofx^2 can't be zero hence with this approach we may not get y=0 as a feasible. It has to checked separately.

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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...

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Re: Doubt In Functionsby 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

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Re: Doubt In Functionsby 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


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Ankur

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Re: Doubt In binomialsby amit kheterpa l - 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

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Re: Doubt In binomialsby Ankur Bansa l - 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!!!!!

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

Nice one,

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Re: Functionsby 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 booon algebra...can u plz suggest 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 ; 0x1/4-4x + 2 ; 1/4x1/24x - 2 ; 1/2x3/4


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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 te ll 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 o f d 1st quest of f(f(f(x)))) where domain of x was 0


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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 SharmaShow parent | Reply

Re: Functionsby Kamal Joshi - Friday, 14 October 2011, 09:07 PM

Dear TG sir,

The links such as these return a fata l error through out the s ite. Kindly reso lve the issue. can't afford to miss the Gyan from yo


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

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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 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 wit

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: Functionsby 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


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at 0, 1/4, 1/2, 3/4 & 1.

y-axis||| /\ /\ /\ /\| / \ / \ / \ / \| / \ / \ / \ / \

__|/______\/______\/______\/______\______ x axis0 1/4 1/2 3/4 1

answer = 4 . (1/2 . 1/4 . 1) = 4. 1/8 = 1/2 remains the same though...

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Re: Functionsby 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 p lease inform/provide a link to..Thanks in advance...

The KING

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Re: Functionsby Kamal Lohia - Tuesday, 1 November 2011, 04:11 PM

Ravi

Thanks for the correction. It'll be updated in the article soon.

Kamal Lohia

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Re: Functionsby 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.

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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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