Question

let f(x)=max.{sint ;0<=t<=x} g(x)=min.{sint;0<=t<=x} and h(x)=[f(x)-g(x)]where [.] denotes the greatest integer function then the range of h(x) is

Answer 1

Given : f(x)=max.{sint ;0<=t<=x} g(x)=min.{sint;0<=t<=x} and h(x)=[f(x)-g(x)]  [.] denotes the greatest integer function  

To find :  range of h(x)

Solution:

f(x)=max.{sint ;0<=t<=x}

g(x)=min.{sint;0<=t<=x}

Let say x = 0

=> f(0) = Sin0 = 0      as  0 ≤ t ≤ 0(x)  

   g(0) = Sin0 = 0       as  0 ≤ t ≤ 0(x)

h(x) = [ 0 - 0 ]  = [ 0 ] = 0

x = π/2

=> f(π/2) = Sinπ/2 = 1     as  0 ≤ t ≤ π/2  (x)    , f(x) is maximum value

   g(π/2) = Sin0 =  0       as  0 ≤ t ≤ π/2 (x)     ,g(x) is minimum value

h(x) = [ 1 - 0 ]  = [ 1 ] = 1

x = π

=> f(π/2) = Sinπ/2 = 1     as  0 ≤ t ≤ π   (x)    , f(x) is maximum value

   g(π or 0) = Sin0 or Sinπ  =  0  as  0 ≤ t ≤  π(x)     ,g(x) is minimum value

h(x) = [ 1 - 0 ]  = [ 1 ] = 1

x = 3π/2

=> f(π/2) = Sinπ/2 = 1     as  0 ≤ t ≤ 3π/2  (x)

   g(3π/2) = Sin3π/2 =  -1       as  0 ≤ t ≤  3π/2 (x)

h(x) = [ 1 - (-1) ]  = [ 2 ] = 2

x = 2π

=> f(π/2) = Sinπ/2 = 1     as  0 ≤ t ≤ 2π   (x)    , f(x) is maximum value

   g(3π/2) = Sin3π/2 =  -1      as  0 ≤ t ≤  2π(x)     ,g(x) is minimum value

h(x) = [ 1 - (-1) ]  = [2 ] = 2

Range of h(x)  = { 0 , 1  , 2  }

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