Some Conceptual questions on functions

1. Solve the equations,  and  where [.] denotes greatest integer function.

Sol:  we know that [ f(x) ] = I (Integer)

and [x + I] = [x] + I

[sin x + [sin x]] = Integer

and [sin x] = Integer

Therefore,

y = 1/3 [ sin x + [sin x + [sin x]]] 

  = 1/3 [sin x] + 1/3 [sin x + [sin x]]

  = 1/3 [sin x] + 1/3[sin x] + 1/3[sin x]

 y = [sin x]  .................(i)


And, [y + [y]] = 2cos x

[y] + [y] = 2cosx

[y] = cos x

or, [[sin x]] = cos x ( since from (i) y = [sin x])

or, [sin x] = cos x

You can compare the graph of [sin x] and cos x i.e. the below graph, it is clear that in a period of 2pi, [sin x] is never equal to cos x

Therefore, [sin x] = cos x has no solution.

Therefore, given eqautions have no solution.






2. If [x] = [x/2] +[(x + 1)/2] , where [.] denotes the greatest integer function and n be a positive integer then show,

[(n + 1)/2] + [(n + 2)/4] + [(n + 4)/8 + [(n + 8)/16] + ...... = n


Sol:

Since [x] = [x/2] +[(x + 1)/2] 

therefore, [n] = [n/2] +[(n + 1)/2]

or,  [(n + 1)/2] = [n] - [n/2]

similarly, put x = n/2 , n/4, n/8 ............

therefore, [(n + 2)/4] = [n/2] - [n/4]

[(n + 4)/8] = [n/4] - [n/8]

[(n + 8)/16] = [n/8] - [n/16]

.........................................
......................................
......................................

put these values,

Take L.H.S.

[(n + 1)/2] + [(n + 2)/4] + [(n + 4)/8 + [(n + 8)/16] + ..... 

= [n] -[n/2] + [n/2] - [n/4] + [n/4] - [n/8].......................

= [n]

= n ( since n is a positive INTEGER [n] = n )

= R.H.S.

Hence Proved





3. Find the integral solutions to the equation [x][y] = x + y. Show that all the non-integral solution lie on exactly two lines. Determine these lines.


Sol.

Let 

and 

Where  and  are integer part and  and  are fractional part.

and 

Therefore,

[x][y] = x + y

or, 

or, 

Left hand side equals to right hand side when

1. and  

, when  because fractional part cannot be negative

Therefore,  gives the integral solution

By hit and trial let put 

therefore, I + I - I.I = 0

or, 2I - I.I = 0

or, I(2 - I) = 0

or, I = 0 or I = 2

therefore integral solutions are x = 0 , y = 0 and x = 2, y = 2
these are the only integral solution (check) ( i.e. sum of two integers is equal to multiplication of same two integers)

2. For, non integral solution ,




since first three terms are integers and right hand side is 0 which is also a integer, therefore  must be an integer ( since  )

 can have only two integer values 0 and 1

when,  we get integral solution i.e. 1st case

therefore,  

or, 

or, 

By hit and trial method,

Put 



= 1
Therefore, [x] = 2, [y] = 3 or [x] = 3, [y] = 2 are the solutions

therefore, equation becomes x + y = [x][y] = 2*3 = 6

or, x + y = 6 (equation of line)

Put, 



= 1 

Therefore, [x] = 0, [y] = 1 or [x] = 1, [y] = 0 are the solutions

therefore, equation becomes x + y = [x][y] = 1*0 = 0

or, x + y = 0 (equation of line)

Therefore, non-integral solutions lies on two lines

x + y = 0 and x + y = 6