Question

Let [x] be the greatest integer less than or equal to x. Then at which of the following point(s) the function f(x) = x cos (π(x + [x])) is discontinuous?

(a) x = 2

(b) x = 0

(c) x = 1

(d) x = -1

 

 

Answer 1

Given f(x) = x cos (π(x + [x]))

At x = 2

limx→ 2-x cos (π(x + [x])) = 2 cos (π+2π)

= 2 cos 3π

= -2

limx→ 2+x cos (π(x + [x])) = 2 cos (2π+2π)

= 2 cos 4π

= 2

LHL ≠ RHL

So f(x) is discontinuous at x = 2.

 

At x = 0

limx→ 0-x cos (π(x + [x])) = 0 cos (-π+0)

= 0

limx→ 0+x cos (π(x + [x])) = 0

LHL = RHL

So f(x) is continuous at x = 0.

 

At x = 1

limx→ 1-x cos (π(x + [x])) = cos (π)

= -1

limx→ 1+x cos (π(x + [x])) = cos 2π

= 1

LHL ≠ RHL

So f(x) is discontinuous at x = 1.

 

At x = -1

limx→ -1-x cos (π(x + [x])) = -cos (-3π)

= 1

limx→ -1+x cos (π(x + [x])) = -cos 2π

= -1

LHL ≠ RHL

So f(x) is discontinuous at x = 1.

 

 

Function is discontinuous at x=2 , x=1 , x=-1