Question

If f: R→R is a function defined by f(x) = `[x - 1]cos((2x - 1)/2)π`, where [ ] denotes the greatest integer function, then f is ______.

Options

• discontinuous only at x = 1

• discontinuous at all integral values of x except at x = 1

• continuous only at x = 1

• continuous for every real x

Answer 1

If f: R→R is a function defined by f(x) = `[x - 1]cos((2x - 1)/2)π`, where [ ] denotes the greatest integer function, then f is continuous for every real x.

Explanation:

Given: f(x) = `[x - 1]cos((2x - 1)/2)π` [x – 1] is discontinuous at all points where x – 1 is an integer, `cos((2x - 1)/2)π` is always continuous.

So, check whether f(x) is discontinuous at x = n, n ∈ I

L.H.L = `lim_(x→"n"^-)f(x)`

= `lim_(x→"n"^-)[x - 1]cos((2x - 1)/2)π`

= `lim_("h"→0)["n" - "h" - 1]cos((2"n" - 2"h" - 1)/2)π`

= `("n" - 1)cos(2"n" - 1)π/2` = 0  ...`((∵ "cos x is equal to zero when"),("x is odd integral multiple of"  π/2))` 

R.H.L = `lim_(x→"n"^+)"f"(x)`

= `lim_(x→"n"^+)[x - 1]cos((2x - 1)/2)π`

= `lim_("h"→0)["n" + "h" - 1]cos((2"n" + 2"h" - 1)/2)π`

= `("n" - 1)cos(2"n" - 1)π/2`

= 0

f(n) = `["n" - 1]cos(2"n" - 1)π/2`

= 0

So, f(n^–) = f(n⁺) = f(n) ∀n ∈ I,

Hence, f(x) is continuous for every real x.