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
Answers
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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