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 $\text{Student} (\text{Name}, \text{Sex}, \text{Marks})$
Name

Sex

Marks

S1

F

30

S2

F

10

S3

M

20

$\text{Student} (\text{Name}, \text{Sex}, \text{Marks})$ Relation is renamed as $\text{Student}(n, x, m).$
Taking cross product of the relations
No

Name

Sex

Marks

n

x

m

1

S1

F

30

S1

F

30

2

S1

F

30

S2

F

10

3

S1

F

30

S3

M

20

4

S2

F

10

S1

F

30

5

S2

F

10

S2

F

10

6

S2

F

10

S3

M

20

7

S3

M

20

S1

F

30

8

S3

M

20

S2

F

10

9

S3

M

20

S3

M

20

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 $\text{Student}(\text{Name}, \text{Sex}, \text{Marks})$
Name

Sex

Marks


S1

M

100

> highest marks of M student

S2

F

50

> highest marks of F student

S3

M

40


S4

F

30


Taking the cross product
NO

Name

Sex

Marks

x

x

M

1

S1

M

100

S1

M

100

2

S1

M

100

S2

F

50

3

S1

M

100

S3

M

40

4

S1

M

100

S4

F

30

5

S2

F

50

S1

M

100

6

S2

F

50

S2

F

50

7

S2

F

50

S3

M

40

8

S2

F

50

S4

F

30

9

S3

M

100

S1

M

100

10

S3

M

100

S2

F

50

11

S3

M

100

S3

M

40

12

S3

M

100

S4

F

30

13

S4

F

30

S1

M

100

14

S4

F

30

S2

F

50

15

S4

F

30

S3

M

40

16

S4

F

30

S4

F

30

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,
$\{ \geq \text{All boys} \} = \mid \text{universal}\mid  \mid < \text{some M} \mid$
$\{ > \text{All boys} \} = \mid \text{universal} \mid  \mid \leq \text{some M} \mid$
$\{ \geq \text{some boys} \} = \mid \text{universal} \mid  \mid < \text{all M} \mid$