**(B) 2m – l**

**Explanation:**

Let x be the lower class limit of a continuous frequency distribution

Let y be the upper class limit of a continuous frequency distribution

**Now, according to the question,**

Mid-point of a class = (x+y) /2

=m

x+y = 2m or, x+l=2m

It is given that, y = l = upper class limit

x=2m-l

Hence, the lower class limit of the class = 2m-l

**Hence, option (B) is the correct answer.**