IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

muskaanrelhan

Q-(2)

If is a bijection, then exists such that

for all

the result is a direct consequence of the property mentioned above

pankaj7math

(i)sqrt(x-[x]) is a periodic function and its graph will will get repeated after every one unit.On the interval [0,1) ,[x] being 0,the graph will be that of sqrt(x).

Repeat this graph on every integral interval and thus you have the graph of sqrt(x-[x]).

Plot the graph of sqrt(x-[x]) on the line y=x treating the line y=x like x-axis and you have the required graph.I would have drawn but i dont know how to upload the picture.

(ii) all values of x between 0 and pi including 0 and pi(recall definition of principal value range of cosine function

(iii)just wait for a while

reddevil_2009

u r wrong muskaanrelhan............................

cos^-1(cosx)=x in only principal domain ...................what about rest of graph???????????

muskaanrelhan

my answer was only fr

reddevil_2009

so u should have mentioned that.........................................

&&& by the way I have solved it...........so users dont answer 2nd part

Decoder

third part is challenging..brilliant ques.

look i use a result ...[x] = [x/2] +[(x+1)/2]

now use this result for 2x here..

when u get a suitable expresion ..convert everything to fractional parts ..and solve by graph ...( as only one tranformation {x +1} is required)
i have given u he concept..try ..well i will u detailed soln. later..

pankaj7math

reddevil you are mistaken about the (ii).plot the graph of LHS and RHS and you will find that 0<=x<=pi is the only solution.The value of LHS changes when x crosses pi.It is 2pi-x when pi<x<2pi and then it is x-2pi for 2pi<x<3pi and so on which obviously isn't equal to x.Refer to any standard textbook for the graph.

clear your fundamentals before you do such questions.If you have any difficulty then talk to me

There appears to be some misconception regarding cos^(-1)(cosx) in general and Muskan is right .The answer is

reddevil_2009

ok my mistake.............................................Now will someone draw graph of 2nd ..........plzzzzzzz

hsbhatt

Let x = I+f.

Two cases arise. Either f<1/2 or f>=1/2

Case I: f<1/2, then [2x] = 2I

Then the equation becomes $\frac{1}{I} + \frac{1}{2I} = f+\frac{1}{3} \Rightarrow f = \frac{3}{2I} - \frac{1}{3}$

f=0 yields no solutions

We must enforce the inequality 0<f<1/2 and from this we get I = 2,3 and 4

This gives for I = 2, f = 5/12 and hence x = 2+5/12 = 29/12

Similarly the other two solutions are 19/6 and 97/24

Case 2: f>=1/2, then [2x] = 2I+1

Then the equation is

The only solutions is I = 1 which gives f=1 is impossible

So the only solutions are 29/12, 19/6 and 97/24

tushar.vaidya

abt graph 1 they are periodic parabolas

determinedforiit

i used a graph making software and got this graph for Q ii) which u can see by downloading from

http://rapidshare.com/files/147707877/cosx.emf

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