Statement 1: Let m be the mid-value and x be the upper limit of a class in the continuous frequency distribution, then the lower limit of this class is 2m - x.

Statement 2: For a given class: $\dfrac{\text{lower limit + upper limit}}{2}$ = mid-value of the class

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Given, mid-value = m

Upper limit of a class = x

As we know,

Mid-value = $\dfrac{\text{lower class-limit + upper class-limit}}{2}$

So, statement 2 is true.

⇒ m = $\dfrac{\text{lower class-limit} + x}{2}$

⇒ 2m = lower class-limit + x

⇒ Lower class-limit = 2m - x

So, statement 1 is true.

∴ Both the statements are true.

Hence, option 1 is the correct option.