Question

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

• A. continuous for every real x
• B. discontinuous only at x = 0
• C. discontinuous only at non-zero integral values of x
• D. continuous only at x = 0

Answer 1

Answer: (A)

Let $f\left(x\right) = \left[x\right] \cos\left(\frac{2x-1}{2}\right) $
Doubtful points are x = n, n $\in$ I
L.H.L $= \displaystyle\lim_{x \to n^-} \left[x\right]\cos\left(\frac{2x-1}{2}\right)\pi $
$= \left(n-1\right)\cos\left(\frac{2n-1}{2}\right)\pi= 0 $
( $\because$ [x] is the greatest integer function)
R.H.L $ = \displaystyle\lim_{x \to n^{+} } \left[x\right]\cos\left(\frac{2x-1}{2}\right) \pi $
$= n \cos\left(\frac{2n-1}{2}\right) \pi= 0 $
Now, value of the function at x = n is f(n) = 0
Since, L.H.L = R.H.L. = f(n)
$\therefore \, f(x) = [x] \cos \left( \frac{2x -1}{2} \right)$ is continuous for every real x.