For the function $f:(0,1)\to R,f\left(x\right)=\left[2^{n}x\right]+\left\{2^{m}x\right\},(n,m\in N)$ , the number of points of discontinuity of the function can be (where [.] {} represent G.I.F. and fractional part of x respectively)

 $f\left(x\right)=\left[2^{m}x\right]+\left[2^{n}x\right]$
    $\downarrow$                         $\downarrow$
Discont              discont
    $\frac{1}{2^n},\frac{2}{2^n},\frac{3}{2^n},\mathrm{....}\frac{2^n-1}{2^n}$                           $\frac{1}{2^m},\frac{2}{2^m},\frac{3}{2^m},\mathrm{.....}\frac{2^m-1}{2^m}$
$p=2^n-1$  points                                          $q=2^m-1$ points
If n=m then ln is cont  $n\ne m$
P will be common in q is (p<q) vice and verse and at common point cont so discontinuous at no. of points  $=2^m-1-2^n+1$
 $$\begin{gathered} 2^m-2^n=24(m=5,n=3) \\ =28(m=5,n=2) \\ =496(m=8,n=4) \end{gathered}$$
 Hence, (A), (B) and (D) are correct