Some Conceptual questions on functions

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1. Solve the equations, y = \frac{1}{3}[\sin x + [\sin x + [\sin x]]] and [y + [y]] = 2\cos x 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 x = {I_1} + {f_1}

 and y = {I_2} + {f_2}

 Where {I_1} and {I_2} are integer part and {f_1} and {f_2} are fractional part.

 and 0 \le {f_1},{f_2} < 1

 Therefore,

 [x][y] = x + y

 or, {I_1}{I_2} = {I_1} + {f_1} + {I_2} + {f_2}

 or, {I_1} + {I_2} - {I_1}{I_2} + {f_1} + {f_2} = 0

 Left hand side equals to right hand side when

 1.{f_1} + {f_2} = 0 and {I_1} + {I_2} - {I_1}{I_2} = 0 

 {f_1} + {f_2} = 0, when {f_1},{f_2} = 0 because fractional part cannot be negative

 Therefore, {I_1} + {I_2} - {I_1}{I_2} = 0 gives the integral solution

 By hit and trial let put {I_1} = {I_2} = I

 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 ,
{f_1},{f_2} \ne 0

 {I_1} + {I_2} - {I_1}{I_2} + {f_1} + {f_2} = 0

 since first three terms are integers and right hand side is 0 which is also a integer, therefore {f_1} + {f_2} must be an integer ( since {f_1},{f_2} > 0 )

 {f_1} + {f_2} can have only two integer values 0 and 1

 when, {f_1} + {f_2} = 0 we get integral solution i.e. 1st case

 therefore,  {f_1} + {f_2} = 1

 or, {I_1} + {I_2} - {I_1}{I_2} + 1 = 0

 or, {I_1}{I_2} - {I_1} - {I_2} = 1

 By hit and trial method,

 Put {I_1} = 2,{I_2} = 3

 {I_1}{I_2} - {I_1} - {I_2} = 6 - 2 - 3

 = 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, {I_1} = 0,{I_2} = - 1

 {I_1}{I_2} - {I_1} - {I_2} = 0 - 0 + 1

 = 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





Comments

  1. Unknown28 May 2017 at 04:31

    Did u made these questions yourself??

    ReplyDelete
    Replies
    1. abhijatindia29 May 2017 at 08:09

      No. These are unsolved questions of a book on differential Calculus.

      Delete

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