OPTION : (D)
The given query states the following conditions:
$$\boxed{\begin{align}\text{Sex} &= F \land \\ x&= M \land \\ \text{Marks} &\leq m \end{align}} \to (1)$$
Let the relation be $\text{Student} (\text{Name}, \text{Sex}, \text{Marks})$ $$\begin{array} {c c c} \hline \text{Name} & \text{Sex} & \text{Marks} \\\hline S1 & F & 30 \\\hline S2 & F & 10 \\\hline S3 & M & 20 \end{array}$$ $\text{Student} (\text{Name}, \text{Sex}, \text{Marks})$ Relation is renamed as $\text{Student}(n, x, m).$
Taking the cross product of the relations
$$\begin{array} {c c c} \hline \text{No.} & \text{Name} & \text{Sex} & \text{Marks} & n & x & m \\\hline 1 & S1 & F & 30 & S1 & F & 30 \\\hline 2 & S1 & F & 30 & S2 & F & 10 \\\hline 3 & S1 & F & 30 & S3 & M & 20 \\\hline 4 & S2 & F & 10 & S1 & F & 30 \\\hline 5 & S2 & F & 10 & S2 & F & 10 \\\hline {\color{DarkBlue}{6}} & {\color{DarkBlue}{S2}} & {\color{DarkBlue}{F}} & {\color{DarkBlue}{10}} & {\color{DarkBlue}{S3}} & {\color{DarkBlue}{M}} & {\color{DarkBlue}{20}} \\\hline 7 & S3 & M & 20 & S1 & F & 30 \\\hline 8 & S3 & M & 20 & S2 & M & 10 \\\hline 9 & S3 & M & 20 & S3 & M & 30 \end{array}$$
Selecting the tuple (row$\# 6$ from the above table), which satisfies the condition $(1)$ and PROJECTING $\Pi_{name} \implies \boxed{S2}$
$\Pi_{name} ( \sigma_{sex=F} (\text{Student}) ) = \boxed{\begin{matrix} S1 \\ S2 \end{matrix}}$
Hence, the query:
$\Pi_{name} \begin{bmatrix} \sigma_{sex=F} (student) \end{bmatrix} - \Pi_{name} \begin{bmatrix} student \bowtie \sigma_{x,x,m} (student) \\ sex = F \wedge \\ x= M \wedge \\ marks \leq m \end{bmatrix}$
$\boxed{\begin{matrix} S1 \\ S2 \end{matrix}} – \boxed{S2} = \boxed{S1}$
Let us take another relation data of $\text{Student}(\text{Name}, \text{Sex}, \text{Marks})$ $$\begin{array} {c c c} \hline \text{Name} & \text{Sex} & \text{Marks} \\\hline S1 & M & 100 & \text{> highest marks of M student} \\\hline S2 & F & 50 & \text{> highest marks of F student} \\\hline S3 & M & 40 \\\hline S4 & F & 30 \end{array}$$
Taking the cross product $$\begin{array} {c c c} \hline \text{No.} & \text{Name} & \text{Sex} & \text{Marks} & n & x & m \\\hline 1 & S1 & M & 100 & S1 & M & 100 \\\hline 2 & S1 & M & 100 & S2 & F & 50 \\\hline 3 & S1 & M & 100 & S3 & M & 40 \\\hline 4 & S1 & M & 100 & S4 & F & 30 \\\hline {\color{DarkBlue}{5}} & {\color{DarkBlue}{S2}} & {\color{DarkBlue}{F}} & {\color{DarkBlue}{50}} & {\color{DarkBlue}{S1}} & {\color{DarkBlue}{M}} & {\color{DarkBlue}{100}} \\\hline 6 & S2 & F & 50 & S2 & F & 50 \\\hline 7 & S2 & F & 50 & S3 & F & 40 \\\hline 8 & S2 & F & 50 & S4 & F & 30 \\\hline 9 & S3 & M & 100 & S1 & M & 100 \\\hline 10 & S3 & M & 100 & S2 & F & 50 \\\hline 11 & S3 & M & 100 & S3 & M & 40 \\\hline 12 & S3 & M & 100 & S4 & F & 30 \\\hline {\color{DarkBlue}{13}} & {\color{DarkBlue}{S4}} & {\color{DarkBlue}{ F}} & {\color{DarkBlue}{30}} & {\color{DarkBlue}{S1}} & {\color{DarkBlue}{M}} & {\color{DarkBlue}{100}} \\\hline 14 & S4 & F & 30 & S2 & F & 50 & \\\hline {\color{DarkBlue}{15}} & {\color{DarkBlue}{S4}} & {\color{DarkBlue}{ F}} & {\color{DarkBlue}{30}} & {\color{DarkBlue}{S3}} & {\color{DarkBlue}{M}} & {\color{DarkBlue}{40}} \\\hline 16 & S4 & F & 30 & S4 & F & 30 \\\hline \end{array}$$
Consider the row numbers $5, 13, 15$ from the above table,
$\boxed{\begin{matrix} S2 \\ S4 \end{matrix}} \implies$ Female students who scored less than equal to some Male students.
$\Pi_{name} [ \sigma_{sex=F} (\text{Student}) ] = \boxed{\begin{matrix} S2 \\ S4 \end{matrix}}$
Hence, the result of the query will be:
$\boxed{\begin{matrix} S2 \\ S4 \end{matrix}} -\boxed{ \begin{matrix} S2 \\ S4 \end{matrix}} = \{\}$
From the above relational data of table Student(Name, Sex, Marks)
(D) is the correct option
In short,
$\{ \geq \text{All boys} \} = \mid \text{universal}\mid - \mid < \text{some M} \mid$
$\{ > \text{All boys} \} = \mid \text{universal} \mid - \mid \leq \text{some M} \mid$
$\{ \geq \text{some boys} \} = \mid \text{universal} \mid - \mid < \text{all M} \mid$