If $f: R \rightarrow R$ is a function defined by $f(x)=[x] \cos \left(\frac{2 x-1}{2}\right) \pi$, where $[x]$ denotes the greatest integer function, then f is

(A) continuous for every real ×

(B) discontinuous only at $\mathrm{x}=0$

(C) discontinuous only at non-zero integral values of x

JEE Main · 2012 · 29 Apr 2012 · Offline · Continuity and Differentiability

f : R \to R

is a function defined by

f (x) = [x] cos (\frac{2 x - 1}{2}) π

, where

[x]

denotes the greatest integer function, then f is

Answer: (A)

Rewrite the cosine factor:

cos (\frac{2 x - 1}{2}) π = cos (π x - \frac{π}{2}) = sin (π x)

, so

f (x) = [x] sin (π x)

. At every integer the step function

[x]

jumps, but there the factor

sin (π x)

is exactly zero, so the product has no break; at non-integer points both factors are continuous. The function is therefore continuous for every real

x

JEE Main · 2012 · 29 Apr 2012 · Offline · Single correct option