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1 Single Choice

A standard six-sided die is placed on a grid at the starting position 'S', as shown in the figure. On a standard die, the sum of the numbers on opposite faces is always 7. Initially, the face with '1' is on top, the face with '5' is pointing forward (in the direction of the

y

-axis), and the face with '3' is pointing to the right (in the direction of the

x

-axis). The die is then rolled one cell at a time along the path S → A → B → C. A roll consists of tipping the die 90° over one of its bottom edges. What is the number on the top face of the die when it reaches position 'C'?

A

1

B

2

C

4

D

6

View Solution

The problem requires tracking the orientation of a standard die as it is rolled along a specified path. A standard die has opposite faces summing to 7. **Step 1: Determine the initial state of the die at position S.** - Top face = 1, so Bottom face =

7 - 1 = 6

. - Front face = 5, so Back face =

7 - 5 = 2

. - Right face = 3, so Left face =

7 - 3 = 4

. Initial State (S): - Top: 1 - Bottom: 6 - Front: 5 - Back: 2 - Right: 3 - Left: 4 **Step 2: Analyze the first roll from S to A (roll to the right).** When the die is rolled to the right, it pivots on its bottom-right edge. The faces change as follows: - The current Top face (1) becomes the new Right face. - The current Right face (3) becomes the new Bottom face. - The current Bottom face (6) becomes the new Left face. - The current Left face (4) becomes the new Top face. - The Front and Back faces remain unchanged. State at A: - Top: 4 - Bottom: 3 - Front: 5 - Back: 2 - Right: 1 - Left: 6 **Step 3: Analyze the second roll from A to B (roll forward).** When the die is rolled forward, it pivots on its bottom-front edge. The faces change as follows: - The current Top face (4) becomes the new Front face. - The current Front face (5) becomes the new Bottom face. - The current Bottom face (3) becomes the new Back face. - The current Back face (2) becomes the new Top face. - The Right and Left faces remain unchanged. State at B: - Top: 2 - Bottom: 5 - Front: 4 - Back: 3 - Right: 1 - Left: 6 **Step 4: Analyze the third roll from B to C (roll to the right).** This is another roll to the right. The transformation is the same as in Step 2: - The current Top face (2) becomes the new Right face. - The current Right face (1) becomes the new Bottom face. - The current Bottom face (5) becomes the new Left face. - The current Left face (6) becomes the new Top face. - The Front and Back faces remain unchanged. Final State at C: - Top: 6 - Bottom: 1 - Front: 4 - Back: 3 - Right: 2 - Left: 5 After the sequence of rolls, the number on the top face of the die at position C is 6. Answer: \boxed{6}

2 Single Choice

Based on the official itinerary, the Prime Minister's flight ___________ from the international airport at 08:00 hours tomorrow morning, marking the start of his four-nation tour.

A

will be departing

B

departs

C

is departing

D

has departed

View Solution

1. **Analysis of the Sentence:** The sentence describes the departure of the Prime Minister's flight. The key phrases are "Based on the official itinerary" and the specific time "at 08:00 hours tomorrow morning". These phrases indicate a pre-arranged, scheduled event. 2. **Tense Rules:** One of the specific uses of the **Simple Present Tense** is to refer to future events that are part of a fixed schedule, timetable, or official program. For example, "The train arrives at 7 PM." 3. **Evaluation of Options:** * **1. will be departing**: This is the Future Continuous tense. It is used to describe an action that will be in progress at a specific time in the future. While grammatically possible, it is not the most precise choice for a scheduled event on an itinerary. * **2. departs**: This is the Simple Present Tense. Given the context of an "official itinerary," this is the most appropriate and standard tense to use for a scheduled future event. It conveys a sense of certainty and official planning. * **3. is departing**: This is the Present Continuous Tense. It is often used for future personal arrangements or plans (e.g., "I am meeting my friend tomorrow."). While it can be used for future events, the Simple Present is stylistically preferred for formal, fixed timetables. * **4. has departed**: This is the Present Perfect Tense, which indicates an action that has already happened. This is incorrect as the event is scheduled for "tomorrow morning." 4. **Conclusion:** For events that are part of a fixed schedule or official timetable, the Simple Present Tense is the most appropriate choice. Therefore, "departs" is the correct verb form to complete the sentence. Answer: \boxed{departs}

3 Single Choice

Although he has been working quite hard ______, he arrived ______ for the crucial project meeting this morning.

A

late, lately

B

lately, late

C

lately, later

D

later, lately

View Solution

This question tests the understanding of adverbs that have similar forms but distinct meanings. The two adverbs in question are `late` and `lately`. 1. **Lately**: This adverb means 'recently' or 'in the near past'. It describes a period of time leading up to the present. 2. **Late**: This adverb means 'after the expected, scheduled, or usual time'. It refers to a specific point in time relative to a deadline or schedule. 3. **Later**: This adverb is the comparative form of 'late' and means 'at a time in the future' or 'subsequently'. Let's analyze the sentence blanks: **Step 1: Analyze the first blank.** The sentence fragment is "...has been working quite hard ______". The use of the present perfect continuous tense ("has been working") indicates an action that started in the past and has continued up to the present. The adverb that fits this context is `lately`, meaning 'recently'. **Step 2: Analyze the second blank.** The sentence fragment is "...he arrived ______ for the crucial project meeting". This part of the sentence describes his arrival relative to the scheduled time of the meeting. The appropriate adverb to indicate he was not on time is `late`. **Step 3: Determine the correct sequence.** Based on the analysis of both blanks, the correct sequence of adverbs to fill the blanks is `lately` followed by `late`. **Step 4: Form the completed sentence.** The completed sentence reads: "Although he has been working quite hard **lately**, he arrived **late** for the crucial project meeting this morning." **Step 5: Evaluate other options.** * Option `late, lately` reverses the correct order. * Option `lately, later` is incorrect because "arrived later" would imply arriving after some other unspecified event, not necessarily being tardy for the meeting itself. * Option `later, lately` is incorrect in both meaning and order. Answer: \boxed{lately, late}

4 Multiple Select

Let

V

be a vector space over a field of scalars

F

. Let

\mathbf{u}, \mathbf{v}

be vectors in

V

and

c

be a scalar in

F

. Based on the ten fundamental axioms of a vector space, which of the following statements are provably true for any such vector space

V

?

A

The zero vector

\mathbf{0}

in

V

is unique.

B

For any

\mathbf{u} \in V

, its additive inverse

-\mathbf{u}

is unique.

C

For any scalar

c \in F

, the equality

c\mathbf{0} = \mathbf{0}

holds.

D

For any two vectors

\mathbf{u}, \mathbf{v} \in V

, if

\mathbf{u} + \mathbf{v} = \mathbf{0}

, then

\mathbf{u} = \mathbf{0}

and

\mathbf{v} = \mathbf{0}

.

View Solution

This question tests the understanding of properties that can be derived directly from the ten axioms of a vector space. **Analysis of Option 1: The zero vector

\mathbf{0}

in

V

is unique.** This statement is TRUE. The uniqueness of the zero vector is a fundamental theorem derived from the axioms. **Proof:** Assume there are two zero vectors,

\mathbf{0}_1

and

\mathbf{0}_2

, in

V

. By the definition of a zero vector (Axiom 4): 1. For any vector

\mathbf{u}

,

\mathbf{u} + \mathbf{0}_1 = \mathbf{u}

. Since

\mathbf{0}_2

is a vector, we have

\mathbf{0}_2 + \mathbf{0}_1 = \mathbf{0}_2

. 2. For any vector

\mathbf{u}

,

\mathbf{u} + \mathbf{0}_2 = \mathbf{u}

. Since

\mathbf{0}_1

is a vector, we have

\mathbf{0}_1 + \mathbf{0}_2 = \mathbf{0}_1

. By the commutativity of addition (Axiom 2),

\mathbf{0}_2 + \mathbf{0}_1 = \mathbf{0}_1 + \mathbf{0}_2

. Therefore,

\mathbf{0}_2 = \mathbf{0}_1

. The zero vector is unique. **Analysis of Option 2: For any

\mathbf{u} \in V

, its additive inverse

-\mathbf{u}

is unique.** This statement is TRUE. This is another fundamental theorem. **Proof:** Let

\mathbf{u}

be any vector in

V

. Assume it has two additive inverses,

\mathbf{v}_1

and

\mathbf{v}_2

. By the definition of an additive inverse (Axiom 5):

\mathbf{u} + \mathbf{v}_1 = \mathbf{0}

and

\mathbf{u} + \mathbf{v}_2 = \mathbf{0}

. Consider

\mathbf{v}_1

. We can write:

\begin{aligned}\mathbf{v}_1 & = \mathbf{v}_1 + \mathbf{0} & & \text{(Axiom 4: Additive Identity)} \\ & = \mathbf{v}_1 + (\mathbf{u} + \mathbf{v}_2) & & \text{(Since } \mathbf{u} + \mathbf{v}_2 = \mathbf{0}) \\ & = (\mathbf{v}_1 + \mathbf{u}) + \mathbf{v}_2 & & \text{(Axiom 3: Associativity)} \\ & = (\mathbf{u} + \mathbf{v}_1) + \mathbf{v}_2 & & \text{(Axiom 2: Commutativity)} \\ & = \mathbf{0} + \mathbf{v}_2 & & \text{(Since } \mathbf{u} + \mathbf{v}_1 = \mathbf{0}) \\ & = \mathbf{v}_2 & & \text{(Axiom 4: Additive Identity)}\end{aligned}

Thus,

\mathbf{v}_1 = \mathbf{v}_2

. The additive inverse is unique. **Analysis of Option 3: For any scalar

c \in F

, the equality

c\mathbf{0} = \mathbf{0}

holds.** This statement is TRUE. This property can be derived from the axioms. **Proof:**

\begin{aligned}c\mathbf{0} & = c(\mathbf{0} + \mathbf{0}) & & \text{(Axiom 4)} \\ c\mathbf{0} & = c\mathbf{0} + c\mathbf{0} & & \text{(Axiom 7: Distributivity)}\end{aligned}

Now, add the additive inverse of

c\mathbf{0}

, which is

-(c\mathbf{0})

, to both sides:

\begin{aligned}c\mathbf{0} + (-(c\mathbf{0})) & = (c\mathbf{0} + c\mathbf{0}) + (-(c\mathbf{0})) \\ \mathbf{0} & = c\mathbf{0} + (c\mathbf{0} + (-(c\mathbf{0}))) & & \text{(Axiom 3: Associativity)} \\ \mathbf{0} & = c\mathbf{0} + \mathbf{0} & & \text{(Axiom 5: Additive Inverse)} \\ \mathbf{0} & = c\mathbf{0} & & \text{(Axiom 4: Additive Identity)}\end{aligned}

This proves the statement. **Analysis of Option 4: For any two vectors

\mathbf{u}, \mathbf{v} \in V

, if

\mathbf{u} + \mathbf{v} = \mathbf{0}

, then

\mathbf{u} = \mathbf{0}

and

\mathbf{v} = \mathbf{0}

.** This statement is FALSE. The condition

\mathbf{u} + \mathbf{v} = \mathbf{0}

simply means that

\mathbf{v}

is the additive inverse of

\mathbf{u}

(i.e.,

\mathbf{v} = -\mathbf{u}

). This does not require either vector to be the zero vector. **Counterexample:** Consider the vector space

V = \mathbb{R}^2

over the field

\mathbb{R}

. Let

\mathbf{u} = (1, -2)

and

\mathbf{v} = (-1, 2)

. Here,

\mathbf{u} \neq \mathbf{0}

and

\mathbf{v} \neq \mathbf{0}

. However,

\mathbf{u} + \mathbf{v} = (1, -2) + (-1, 2) = (1-1, -2+2) = (0, 0) = \mathbf{0}

. The statement is therefore not true for all vectors in any vector space. **Conclusion:** The first three statements are provably true from the vector space axioms, while the fourth is false. Thus, the correct options are the first three. Answer:

\boxed{\text{Options 1, 2, and 3 are correct.}}

5 Numerical

Let

V = \mathbb{R}^2

be a vector space over the field of real numbers. Consider the subset

W

of

V

defined by:

W = \left\{ \begin{pmatrix} x \\ y \end{pmatrix} \in \mathbb{R}^2 \;\bigg|\; x - 2y = 0 \text{ and } 4x^2 - ky^2 = 0 \right\}

For what integer value of

k

is the set

W

a non-trivial subspace of

\mathbb{R}^2

?

Correct Answer: 16

View Solution

A subset

W

of a vector space

V

is a non-trivial subspace if it is a subspace and is not the set containing only the zero vector, i.e.,

W \neq \{ \mathbf{0} \}

. For

W

to be a subspace, it must contain the zero vector, be closed under vector addition, and be closed under scalar multiplication. **Step 1: Analyze the conditions defining the set

W

.** Any vector

\begin{pmatrix} x \\ y \end{pmatrix} \in W

must satisfy both equations: 1.

x - 2y = 0

2.

4x^2 - ky^2 = 0

From the first equation, we can express

x

in terms of

y

:

x = 2y

. We can substitute this relationship into the second equation to understand the constraint imposed by the parameter

k

. **Step 2: Substitute the linear constraint into the quadratic constraint.** Substituting

x = 2y

into the second equation:

\begin{aligned} 4(2y)^2 - ky^2 & = 0 \\ 4(4y^2) - ky^2 & = 0 \\ 16y^2 - ky^2 & = 0 \\ (16 - k)y^2 & = 0 \end{aligned}

This resulting equation,

(16 - k)y^2 = 0

, must hold for every vector in

W

. **Step 3: Analyze the cases based on the value of

k

.** We have two distinct possibilities for the expression

(16-k)

. **Case 1:**

16 - k \neq 0

(i.e.,

k \neq 16

) In this case, for the equation

(16 - k)y^2 = 0

to be true, we must have

y^2 = 0

, which implies

y=0

. Since

x = 2y

, it follows that

x=0

. Therefore, the only vector that satisfies both original conditions is the zero vector

\begin{pmatrix} 0 \\ 0 \end{pmatrix}

. The set is

W = \left\{ \begin{pmatrix} 0 \\ 0 \end{pmatrix} \right\}

, which is the trivial subspace. **Case 2:**

16 - k = 0

(i.e.,

k = 16

) In this case, the equation becomes

(0)y^2 = 0

, which simplifies to

0=0

. This equation is true for any value of

y

. This means that if

k=16

, the second condition

4x^2 - ky^2 = 0

becomes redundant for any vector that already satisfies the first condition

x - 2y = 0

. The set

W

is then defined solely by the linear equation:

W = \left\{ \begin{pmatrix} x \\ y \end{pmatrix} \in \mathbb{R}^2 \;\bigg|\; x - 2y = 0 \right\}

This is the equation of a line passing through the origin in

\mathbb{R}^2

. A line through the origin is a standard example of a non-trivial subspace of

\mathbb{R}^2

. It contains infinitely many vectors (e.g.,

\begin{pmatrix} 2 \\ 1 \end{pmatrix}

) and is closed under both vector addition and scalar multiplication. **Step 4: Conclusion.** The set

W

forms a non-trivial subspace of

\mathbb{R}^2

only when the second condition does not force the set to contain only the zero vector. This occurs when

k=16

. **Answer:** The required integer value of

k

is

\boxed{16}

.

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