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 $=\underset{x\to n^{-}}{\lim }\left[x\right]\cos \left(\frac{2x-1}{2}\right)\pi$
$=\left(n-1\right)\cos \left(\frac{2n-1}{2}\right)\pi =0$
( $\because$ [x] is the greatest integer function)
R.H.L $=\underset{x\to n^{+}}{\lim }\left[x\right]\cos \left(\frac{2x-1}{2}\right)\pi$
$=n\cos \left(\frac{2n-1}{2}\right)\pi =0$
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.