Past Questions — 2010

Definition: complemented distributive lattice = Boolean algebra; self-complemented adds Boolean property.

y=ABCD+ABC'=ABC(D+D')=ABC*1=ABC.

Let's analyze the given regular expression and the options: 1. **Given Expression: 0*(10)*** - This expression generates strings that start with any number of 0s, followed by any number of repetitions of the sequence '10'. - Examples of strings in this set: ε (empty string), 0, 00, 10, 010, 0010, 1010, 01010. - Key Constraint: A '1' can never be followed by another '1', and a '1' must always be followed by a '0'. Also, the string cannot end with a '1'. 2. **Evaluating Option (a): (1*0)*1*** - This expression can generate strings that end with '1' (due to the trailing 1*). For example, the string '1' is accepted by (1*0)*1* but is NOT accepted by 0*(10)*. - It can also generate '110', which is not accepted by 0*(10)*. - Therefore, (a) is incorrect. 3. **Evaluating Option (b): 0 + (0 + 10)*** - The expression (0 + 10)* generates any combination of '0' and '10'. This is equivalent to saying 'no two 1s are adjacent and the string cannot end in 1'. - This is actually a broader set than 0*(10)*. For example, the string '0100' is accepted by (0 + 10)* because it is (0)(10)(0). However, '0100' is NOT accepted by 0*(10)* because in 0*(10)*, once the (10)* part begins, you cannot have a standalone '0' after a '10' unless it is part of another '10'. - Wait, let's re-verify: 0*(10)* means all 0s must come before all 10s. '0100' cannot be formed. But (0 + 10)* allows 0s anywhere. Thus, (b) is incorrect. 4. **Evaluating Option (c): (0 + 1)*10(0 + 1)*** - This expression represents all strings over {0, 1} that contain the substring '10' at least once. - This is a much larger set. For example, '1110' is accepted here but not by 0*(10)*. - Therefore, (c) is incorrect. Since none of the provided options match the language defined by 0*(10)*, the correct answer is 'None of the above'.

To determine which statement is true, we must check if the given sets satisfy the four group axioms under the specified operation: Closure, Associativity, Identity, and Inverse. 1. **Analysis of Option (a):** The set of all rational negative numbers under multiplication. - **Closure:** If we multiply two negative rational numbers (e.g., -1 * -1), the result is a positive rational number (1). Since the result is not a negative number, the set is not closed under multiplication. Therefore, it cannot be a group. 2. **Analysis of Option (b):** The set of all non-singular matrices (matrices with a non-zero determinant) under multiplication. - **Closure:** The product of two non-singular matrices $A$ and $B$ is also non-singular because $\det(AB) = \det(A)\det(B)$. Since $\det(A) \neq 0$ and $\det(B) \neq 0$, their product is non-zero. - **Associativity:** Matrix multiplication is inherently associative. - **Identity:** The identity matrix $I$ is non-singular ($\det(I) = 1$) and acts as the identity element. - **Inverse:** By definition, a non-singular matrix has a multiplicative inverse $A^{-1}$ such that $AA^{-1} = I$, and $A^{-1}$ is also non-singular. Since all four axioms are satisfied, this is a group (known as the General Linear Group $GL(n, F)$). 3. **Analysis of Option (c):** The set of all matrices under multiplication. - **Inverse:** For a set to be a group, every element must have an inverse. Singular matrices (matrices with determinant 0) do not have multiplicative inverses. Therefore, the set of all matrices cannot form a group under multiplication. 4. **Analysis of Option (d):** Since (c) is false, (d) must also be false. Conclusion: Only option (b) is mathematically correct.

In filter design, a 'prototype filter' (usually a low-pass prototype) is a normalized filter where the component values are chosen such that the characteristic impedance (Z0) is normalized to a specific value, typically 1 Ohm or a standard resistive value (like 50 or 600 Ohms). The purpose of the prototype is to define the mathematical transfer function and attenuation characteristics. For the filter to effectively pass signals in the passband without introducing phase distortion or reflection, the image impedance (Z0) must be purely resistive. If Z0 were inductive or capacitive in the passband, it would cause impedance mismatching and frequency-dependent phase shifts, which contradicts the goal of a normalized prototype filter. Therefore, Z0 is designed to be purely resistive in the passband to ensure maximum power transfer and a flat response.

1. Definition of a Complete Graph: A complete graph with n vertices, denoted as K_n, is a simple undirected graph in which every pair of distinct vertices is connected by a unique edge. 2. Definition of Chromatic Number: The chromatic number of a graph, denoted by χ(G), is the smallest number of colors needed to color the vertices of the graph such that no two adjacent vertices share the same color. 3. Analysis of K_n: In a complete graph K_n, every vertex is adjacent to every other vertex. Therefore, if we have n vertices, each vertex must be assigned a unique color because it is connected to all other (n-1) vertices. If any two vertices were to share the same color, it would violate the coloring rule since an edge exists between every possible pair. 4. Conclusion: Since each of the n vertices requires a distinct color, the chromatic number χ(K_n) = n. Thus, a complete graph with n vertices is n-chromatic. Evaluating options: - a. 2-chromatic: Only true for K_2 (or bipartite graphs). - b. n/2 chromatic: Incorrect; does not satisfy the adjacency requirement for all vertices. - c. n-1 chromatic: Incorrect; one vertex would be left without a unique color, causing a conflict with its neighbors. - d. n-chromatic: Correct.

In mathematics and digital electronics, when performing a subtraction operation (regardless of whether it is done using direct subtraction or 2's complement addition), the terms are defined as follows: 1. Minuend: The number from which another number is subtracted. 2. Subtrahend: The number that is being subtracted from the minuend. 3. Difference: The result of the subtraction. In the context of 2's complement subtraction, the operation 'A - B' is performed by adding the 2's complement of B to A. In this scenario, 'A' is the minuend and 'B' is the subtrahend. Therefore, the number which is subtracted from another number is known as the subtrahend. Analysis of other options: - Option (a) Carry: This refers to the bit that 'carries over' to the next higher column during addition. - Option (c) Minuend: This is the number that is being subtracted FROM, not the number being subtracted.

A flip-flop is a fundamental building block of digital electronics used to store one bit of data. It is called a 'bistable multivibrator' because it has two stable states: logic '0' (low) and logic '1' (high). It will remain in one of these states indefinitely until an external trigger (like a clock pulse or input change) forces it to switch to the other state. Analysis of options: - (a) Stable: This is too general; while it is stable in its states, the term 'stable' does not describe the circuit's behavior of switching between two states. - (b) Bistable: Correct. 'Bi' means two, and 'stable' refers to the states. A flip-flop is a bistable device. - (c) Tri stable: This is not a standard term in digital electronics. Multivibrators are typically categorized as astable (no stable states), monostable (one stable state), or bistable (two stable states). - (d) Not stable: Incorrect, as flip-flops are specifically designed to hold a state reliably.

To determine if a relation R on set A is reflexive, we must check if for every element x in A, (x, x) is in R. Given: 1. R is symmetric: If (x, y) ∈ R, then (y, x) ∈ R. 2. R is transitive: If (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∈ R. Common Misconception: Many assume that since (x, y) ∈ R implies (y, x) ∈ R (symmetry), and since (x, y) and (y, x) are in R, then (x, x) must be in R (transitivity). This logic only holds IF there exists some y such that (x, y) ∈ R. Counter-example: Let A = {1, 2} and R = {(1, 1)}. - R is symmetric: (1, 1) is the only pair, and its reverse is itself. - R is transitive: (1, 1) and (1, 1) implies (1, 1). - However, R is NOT reflexive because (2, 2) ∉ R. Another extreme case: Let R be the empty set (∅) on a non-empty set A. The empty relation is vacuously symmetric and transitive, but it is not reflexive. Analysis of Options: - Option (a): Incorrect. R is not necessarily reflexive, so it is not necessarily an equivalence relation. - Option (b): Incorrect. A partial order requires antisymmetry and reflexivity; R is symmetric and not necessarily reflexive. - Option (c): Incorrect. While R *might* not be reflexive, the statement 'R is not reflexive' as a general rule is false because R *could* be reflexive (e.g., if R is the identity relation). The correct logical conclusion is that R is 'not necessarily' reflexive. Since none of the definitive statements in a, b, or c are universally true for all symmetric and transitive relations, the correct choice is 'None of these'.

The number system used in daily human life is the Decimal Number System. It is a base-10 system, meaning it uses ten distinct digits (0, 1, 2, 3, 4, 5, 6, 7, 8, and 9) to represent any value. Analysis of other options: - a. Octal: A base-8 system (digits 0-7) used primarily in computing to simplify binary representations. - b. Binary: A base-2 system (digits 0 and 1) which is the fundamental language of digital electronic circuits and computers. - c. Hexadecimal: A base-16 system (digits 0-9 and letters A-F) used in computing for memory addressing and color codes.

Step-by-step reasoning: 1. **Definition of Context-Sensitive Grammar (CSG):** A grammar is context-sensitive if all its production rules are of the form αAβ → αγβ, where A is a non-terminal and γ is a non-empty string. A key property of CSGs is that they are 'non-contracting,' meaning the length of the string on the right side of a production is always greater than or equal to the length of the string on the left side (|α| ≤ |β|). 2. **Automata Correspondence:** Context-sensitive languages (CSLs) are exactly the languages accepted by Linear Bounded Automata (LBA). An LBA is a non-deterministic Turing machine whose tape is limited to the portion occupied by the input string. 3. **Decidability (Recursiveness):** A language is 'Recursive' if there exists a Turing machine that halts on all inputs (it accepts strings in the language and rejects strings not in the language). 4. **The Membership Problem for CSLs:** For any context-sensitive language L and a string w, we can determine if w ∈ L by exploring all possible derivations. Because the grammar is non-contracting, there are only a finite number of possible strings of length ≤ |w| that can be derived. Therefore, an algorithm (or an LBA) can check all possibilities and guaranteed to halt with a 'Yes' or 'No' answer. 5. **Conclusion:** Since the membership problem for any context-sensitive language is decidable, every context-sensitive language is a Recursive language. Comparison with options: - (a) Always Recursive: Correct, as proven above. - (b) Recursively enumerable but non-recursive: Incorrect. This describes Type-0 languages (Unrestricted Grammars) that are not Type-1. - (c) May be recursive: Incorrect. It is not just a possibility; it is a mathematical certainty for all CSGs. - (d) None of the above: Incorrect.

Bandwidth in the context of optical fibers refers to the range of frequencies (or the data rate) that the fiber can transmit without significant signal degradation. Optical fibers use light waves, which operate at extremely high frequencies (typically in the terahertz range). Because of this massive available spectrum, the achievable bandwidth (the capacity to carry data) is exceptionally high. While modern high-end fibers can reach Terabits per second (Tbps), the 'typical' bandwidth capability discussed in fundamental computer networking and telecommunications textbooks is characterized as being in the order of Gigahertz (GHz). Comparing the options: - Option (b) KHz (Kilohertz) and Option (c) Hz (Hertz) are far too low; these are typical of very basic analog signals or low-frequency radio, not high-speed data transmission. - Option (a) GHz (Gigahertz) correctly identifies the high-capacity nature of optical fiber communication, which allows for the transmission of billions of bits per second.

This question refers to one of the most famous theorems in graph theory: the Four Color Theorem. 1. Definition: A 'proper coloring' of a graph is an assignment of colors to the vertices such that no two adjacent vertices share the same color. 2. Planar Graph: A planar graph is a graph that can be embedded in the plane, meaning it can be drawn in such a way that its edges intersect only at their endpoints. 3. The Four Color Theorem: Proven by Kenneth Appel and Wolfgang Haken in 1976, the theorem states that any planar graph can be colored with no more than four colors such that no two adjacent vertices have the same color. 4. Why other options are incorrect: - Option (a) 2: Only bipartite graphs can be colored with 2 colors. Many planar graphs (like a triangle/K3) require more than 2. - Option (b) 3: Some planar graphs (like the complete graph K4) require 4 colors and cannot be colored with 3. - Option (d) 5: While it is true that any planar graph can be colored with 5 colors (the Five Color Theorem), the question asks for the minimum number required to color *every* planar graph. Since 4 is sufficient for all, 4 is the definitive answer.

In graph theory, an Eulerian graph (or Euler graph) is a graph that contains an Eulerian circuit—a closed trail that visits every edge of the graph exactly once and returns to the starting vertex. According to Euler's Theorem (first proved by Leonhard Euler in 1736), a connected graph G has an Eulerian circuit if and only if every vertex in the graph has an even degree. Reasoning for options: - (a) Same degree: This describes a 'regular graph'. While a regular graph with even degrees is Eulerian, not all regular graphs are Eulerian (e.g., a 3-regular graph is not), and not all Eulerian graphs are regular. - (b) Even degree: This is the necessary and sufficient condition for a connected graph to be Eulerian. If a vertex has an odd degree, any path entering and leaving that vertex will eventually leave it 'stranded' or require it to be a start/end point, making a closed circuit impossible. - (c) Odd degree: If a connected graph has exactly two vertices of odd degree, it has an Eulerian path (but not a circuit). If all vertices are of odd degree, it cannot be an Euler graph. - (d) Different degree: This is not a requirement for any specific standard graph type related to Eulerian circuits.

In the Theory of Computation, a Context-Free Language (CFL) is defined as 'inherently ambiguous' if every possible Context-Free Grammar (CFG) that generates that language is ambiguous. 1. **Definition of Ambiguity in CFG**: A grammar is ambiguous if there exists at least one string in the language it generates that has more than one leftmost derivation or more than one parse tree. 2. **Definition of Inherently Ambiguous CFL**: While some ambiguous grammars can be rewritten into unambiguous grammars for the same language, an 'inherently ambiguous' language is one for which no such unambiguous grammar exists. Evaluating the options: - **Option (a)**: While it is true that at least one ambiguous CFG exists, this is a property of all ambiguous languages, not the defining characteristic of 'inherently' ambiguous ones. - **Option (b)**: This contradicts the definition of inherent ambiguity. If an unambiguous CFG existed, the language would not be inherently ambiguous. - **Option (c)**: This is incorrect because by definition, a CFL must be generated by at least one CFG. - **Option (d)**: This is the correct definition. For an inherently ambiguous CFL, it is impossible to construct any CFG that is unambiguous.

To understand why a Turing Machine (TM) is more powerful than a Finite State Machine (FSM), we must look at their memory capabilities: 1. **Finite State Machine (FSM):** An FSM has a strictly finite amount of memory, represented by its states. It cannot 'count' or 'remember' an arbitrary number of symbols; it can only distinguish between a fixed number of conditions. For example, an FSM cannot recognize the language L = {a^n b^n | n >= 0} because it cannot remember exactly how many 'a's it has seen if 'n' exceeds the number of its states. 2. **Turing Machine (TM):** A TM possesses an infinite external tape that it can read from and write to. This tape serves as an unlimited memory source. Because the TM can move the head back and forth and modify the tape, it can store and retrieve arbitrarily long sequences of symbols, allowing it to recognize complex languages (Recursively Enumerable languages) that an FSM (Regular languages) cannot. **Analysis of Options:** - **Option (a):** Incorrect. Tape movement in a TM is bidirectional (Left and Right), whereas FSMs typically process input in one direction. Confining movement to one direction would actually make it less powerful. - **Option (b):** Incorrect. A Turing Machine *does* have a finite set of states (the control unit). The 'infinite' part is the tape, not the state set. - **Option (c):** Correct. The infinite tape allows the TM to store and recall an unlimited amount of information, which is the fundamental reason for its superior computational power over the FSM.

In graph theory, the 'nullity' (also known as the cyclomatic number or first Betti number) of a graph refers to the number of independent cycles in the graph. 1. For a connected graph (k=1) with n vertices and e edges, the number of edges in a spanning tree is (n-1). Any edge not in the spanning tree creates exactly one fundamental cycle. Therefore, the nullity is e - (n - 1) = e - n + 1. 2. For a graph with k connected components, each component i has n_i vertices and e_i edges. The nullity of each component is (e_i - n_i + 1). 3. The total nullity of the graph is the sum of the nullities of its components: Nullity = Σ (e_i - n_i + 1) Nullity = Σ e_i - Σ n_i + Σ 1 Nullity = e - n + k Comparing this to the options: - Option (a) n: Incorrect; this is the number of vertices. - Option (b) e–n+k: Correct; this is the standard formula for the cyclomatic number/nullity. - Option (c) n–1: Incorrect; this is the number of edges in a spanning tree for a connected graph. - Option (d) n+k: Incorrect; this has no standard meaning in this context.

In the Theory of Computation, a 'Transition Graph' (TG) is a representation of a Finite Automaton that allows for non-determinism and transitions labeled with strings (including the empty string ε). According to Kleene's Theorem and the fundamental properties of regular languages: 1. **Equivalence to NDFSM**: A Transition Graph is essentially a Non-deterministic Finite State Machine (NDFSM) with extended capabilities (like ε-transitions). Any TG can be directly represented as or converted into an NDFSM. 2. **Equivalence to DFSM**: Every NDFSM (and thus every TG) can be converted into an equivalent Deterministic Finite State Machine (DFSM) using the subset construction algorithm (powerset construction). Therefore, any language accepted by a TG is a regular language and can be accepted by a DFSM. 3. **Equivalence to Regular Expression**: According to Kleene's Theorem, a language is regular if and only if it can be described by a regular expression. Since TGs recognize regular languages, every TG has an equivalent regular expression (which can be derived using methods like State Elimination or Arden's Theorem). Since a Transition Graph is equivalent to an NDFSM, a DFSM, and a Regular Expression, the correct option is 'All of the above'.

The question describes a scenario where a file is updated (corrected) immediately after a transaction is input. This is the definition of 'On-line updating' (also known as Real-time processing). 1. **On-line updating (Correct):** In this method, the database or file is updated instantly as soon as the transaction occurs. This ensures that the system always reflects the most current data. Examples include ATM withdrawals or airline ticket bookings. 2. **Off-line updating (Incorrect):** Also known as 'Batch processing', this method involves collecting transactions over a period of time and processing them all at once later (e.g., end-of-day processing). The file is not corrected immediately. 3. **Batching (Incorrect):** This is the process of grouping transactions together to be processed as a single unit (batch), which is the opposite of immediate updating. 4. **Sorting (Incorrect):** Sorting is the process of arranging data in a specific order (alphabetical or numerical) and is not a method of updating files based on transaction input.

To find the regular expression that denotes all possible strings over the alphabet Σ = {a, b}, we are looking for the definition of the Kleene closure (Σ*). 1. **Option (a) a* b*:** This denotes strings where any number of 'a's are followed by any number of 'b's. It cannot generate strings where 'a' follows 'b' (e.g., 'ba' is not possible). Therefore, it does not represent all possible strings. 2. **Option (b) (a | b):** This denotes a language containing only two strings: 'a' or 'b'. It does not allow for strings of length 0 (empty string) or length greater than 1. 3. **Option (c) (ab)+:** This denotes strings consisting of one or more repetitions of the sequence 'ab' (e.g., 'ab', 'abab', 'ababab'). It cannot generate strings like 'aa', 'bb', or 'a'. 4. **Option (d) (a | b)*:** The pipe symbol '|' (or '+') denotes the union (OR), meaning 'a' or 'b'. The asterisk '*' (Kleene star) denotes zero or more repetitions of the preceding element. Thus, (a | b)* generates any combination of 'a's and 'b's of any length, including the empty string (ε). This is the definition of Σ* for Σ = {a, b}.

A complete graph, denoted as K_n, is a simple undirected graph in which every pair of distinct vertices is connected by a unique edge. Step-by-step derivation: 1. Let 'n' be the number of vertices in the graph. 2. Each vertex must be connected to every other vertex. Since there are 'n' vertices, each vertex is connected to (n - 1) other vertices. 3. If we multiply the number of vertices (n) by the number of edges connected to each vertex (n - 1), we get n(n - 1). 4. However, in an undirected graph, the edge connecting vertex A to vertex B is the same as the edge connecting vertex B to vertex A. Therefore, every edge has been counted twice. 5. To find the actual number of unique edges, we divide by 2: Total Edges = n(n - 1) / 2. Alternatively, this can be viewed as a combination problem: we are choosing 2 distinct vertices out of n to form an edge. The number of ways to do this is 'n choose 2', which is calculated as: C(n, 2) = n! / [2!(n - 2)!] = [n * (n - 1)] / 2. Analysis of options: - Option (a) n: This is the number of vertices, not edges. - Option (b) n/2: This does not represent a standard graph property for complete graphs. - Option (c) n^2 - 1: This is incorrect and grows much faster than the actual number of edges. - Option (d) n(n-1)/2: This matches the derived formula.

1. Definition of a Spanning Tree: A spanning tree of a connected, undirected graph G is a subgraph that is a tree (connected and acyclic) and includes all the vertices (nodes) of G. 2. Condition for Existence: By definition, every connected graph contains at least one spanning tree. If a graph is not connected, it has zero spanning trees, but the question explicitly states the graph is 'connected'. 3. Minimum Case: The minimum number of spanning trees occurs when the graph itself is already a tree. In a tree with n nodes, there are exactly n-1 edges and no cycles. Since removing any edge would disconnect the graph and adding any edge would create a cycle, the only spanning tree of a tree is the tree itself. 4. Analysis of Options: - (a) zero: Incorrect, because the graph is connected. - (b) 1: Correct, as a tree is a connected graph with exactly one spanning tree. - (c) n-1: Incorrect, this is the number of edges in a spanning tree, not the number of spanning trees. - (d) n/2: Incorrect, this has no theoretical basis for the minimum number of spanning trees.

Step-by-step reasoning: 1. **Analyze the expression**: The expression given is L1 ∩ L2 L3. In formal language theory, the juxtaposition of two languages (L2 L3) denotes the concatenation operation. Therefore, the expression is L1 ∩ (L2 ∘ L3). 2. **Closure under Concatenation**: Regular languages are closed under concatenation. Since L2 and L3 are regular languages, their concatenation (L2 ∘ L3) is also a regular language. Let's call this resulting language L_reg. 3. **Closure under Intersection with Regular Languages**: A fundamental theorem in the Theory of Computation states that the intersection of a Context-Free Language (CFL) and a Regular Language is always a Context-Free Language. 4. **Application**: Here, L1 is a CFL and L_reg (L2 ∘ L3) is a regular language. Therefore, L1 ∩ L_reg must be a Context-Free Language. Why other options are incorrect: - (b) Regular language: The intersection of a CFL and a regular language is not necessarily regular; it is only guaranteed to be context-free. - (c) Non context-free language: This contradicts the closure property mentioned above. - (d) None of the above: Since option (a) is mathematically proven, this is incorrect.

1. Definition of Chromatic Number: The chromatic number of a graph, denoted by χ(G), is the minimum number of colors required to color the vertices of the graph such that no two adjacent vertices share the same color. 2. Analysis of the Graph: The question describes a graph consisting of 'isolated n vertices'. An isolated vertex is a vertex with a degree of 0, meaning it is not connected to any other vertex by an edge. 3. Coloring Process: Since there are no edges in the graph, there are no pairs of adjacent vertices. Therefore, the condition that 'no two adjacent vertices share the same color' is vacuously satisfied regardless of the colors used. To minimize the number of colors, we can assign the same single color to all n vertices. 4. Conclusion: Because only one color is needed to satisfy the coloring condition, the graph is 1-chromatic. 5. Evaluation of Options: - (a) 1-chromatic: Correct, as no edges exist to force a second color. - (b) 2-chromatic: Incorrect, as this would imply the graph requires at least two colors (e.g., a bipartite graph with at least one edge). - (c) 3-chromatic: Incorrect, as this would imply the graph requires at least three colors (e.g., a graph containing an odd cycle). - (d) n-chromatic: Incorrect, as this would describe a complete graph (Kn) where every vertex is connected to every other vertex.

Step-by-step analysis of the language L = {aⁿb²ⁿ | n ≥ 1}: 1. **Definition of the Language**: The language consists of strings where a sequence of 'a's is followed by a sequence of 'b's, and the number of 'b's is exactly twice the number of 'a's (e.g., ab², aa b⁴, aaa b⁶). 2. **Checking for Regularity**: A regular language can be recognized by a Finite Automaton (FA). An FA has finite memory and cannot count an arbitrary number of 'a's to ensure that exactly twice as many 'b's follow. Since the value of 'n' is unbounded, this language requires memory (a stack), so it is NOT a regular language. (Option 'a' is incorrect). 3. **Checking for Context-Free Property**: A language is Context-Free (CFL) if it can be generated by a Context-Free Grammar (CFG) or recognized by a Pushdown Automaton (PDA). - **CFG Construction**: We can define a grammar G with the production rule: S → aSbb | ab². This grammar generates exactly two 'b's for every one 'a'. - **PDA Logic**: A PDA can push one symbol onto the stack for every 'a' read. Then, for every 'b' read, it can either pop a symbol or use a counter mechanism. Specifically, for every 'a', push two markers (or one marker and pop it once for every two 'b's). Since a PDA can handle this linear dependency, the language is Context-Free. 4. **Conclusion**: Because the language can be described by a CFG and recognized by a PDA, it is a Context-Free Language. It is not a 'Non context free language' because it does not require the power of a Turing Machine or a Linear Bounded Automaton (unlike languages like {aⁿbⁿcⁿ}). (Option 'c' is incorrect).