Analyizng the dot product of unit vectors.

The dot product \( \mathbf{p} \cdot \mathbf{q} \) of two unit vectors \( \mathbf{p} \) and \( \mathbf{q} \) is given by: \[ \mathbf{p} \cdot \mathbf{q} = \cos \theta \]
where \( \theta \) is the angle between the two vectors. Since the cosine of an angle must lie between \( -1 \) and \( 1 \), the value of \( \mathbf{p} \cdot \mathbf{q} \) must be between \( -1 \) and \( 1 \) inclusive. Therefore, \( \sqrt{3} \) is not a possible value for \( \mathbf{p} \cdot \mathbf{q} \), as it exceeds the maximum possible value of 1. Quick Tip: For unit vectors, the dot product \( \mathbf{p} \cdot \mathbf{q} \) must always lie between \( -1 \) and \( 1 \). Any value outside this range is impossible.

Step 1: Analyzing symmetric and skew-symmetric matrices.

A matrix is symmetric if \( A = A^T \), meaning the matrix is equal to its transpose. A matrix is skew-symmetric if \( A = -A^T \), meaning the matrix is equal to the negative of its transpose.

Step 2: Finding the matrix that satisfies both conditions.

The only matrix that satisfies both symmetric and skew-symmetric properties is the null matrix, because: \[ 0 = 0^T \quad (symmetric) \quad and \quad 0 = -0^T \quad (skew-symmetric). \]
Thus, the null matrix is both symmetric and skew-symmetric. Quick Tip: The null matrix is the only matrix that can be both symmetric and skew-symmetric since \( 0 = 0^T \) and \( 0 = -0^T \).

Step 1: Solving the integral.

The integral of \( x \) from 0 to \( a \) is: \[ \int_0^a x \, dx = \frac{a^2}{2} \]
Thus, the inequality becomes: \[ \frac{a^2}{2} \leq \frac{a}{2} + 6 \]
Multiply both sides by 2 to simplify: \[ a^2 \leq a + 12 \]
Rearrange the terms: \[ a^2 - a - 12 \leq 0 \]

Step 2: Solving the quadratic inequality.

Factor the quadratic expression: \[ (a - 4)(a + 3) \leq 0 \]
The solution to this inequality is \( -3 \leq a \leq 4 \). Quick Tip: For quadratic inequalities, factor the expression and analyze the sign of the factors to determine the solution range.

Step 1: Analyzing skew-symmetric matrices.

A matrix is skew-symmetric if \( A = -A^T \), meaning the matrix is equal to the negative of its transpose. We are given the expression \( A B^T - B A^T \).

Step 2: Checking the transpose.

Take the transpose of \( A B^T - B A^T \): \[ (A B^T - B A^T)^T = (B A^T)^T - (A B^T)^T = A B^T - B A^T \]
Since the transpose of the matrix is equal to its negative, the matrix is skew-symmetric. Quick Tip: To check if a matrix is skew-symmetric, verify if its transpose is equal to the negative of the original matrix.

Step 1: Evaluating the inverse cotangent.

We know that \( \cot^{-1}(-\sqrt{3}) \) corresponds to an angle \( \theta \) where \( \cot \theta = -\sqrt{3} \). This implies \( \theta = \frac{5\pi}{6} \), because \( \cot \frac{5\pi}{6} = -\sqrt{3} \).

Step 2: Simplifying the expression.

Thus, the expression becomes: \[ \cos \left( \frac{\pi}{6} + \frac{5\pi}{6} \right) = \cos \pi = -1 \]
Therefore, the value of the expression is \( \boxed{-1} \). Quick Tip: When solving inverse trigonometric functions, express the angle in terms of a known trigonometric identity and simplify the expression.

Step 1: Finding the order and degree.

The given differential equation is \( \frac{d}{dx} \left( \frac{d}{dx} y^3 \right) = 0 \).
First, let's find the order and degree of this equation.

The function is \( y^3 \), so we have: \[ \frac{d}{dx} y^3 = 3y^2 \frac{dy}{dx} \]
Now, applying the derivative again: \[ \frac{d}{dx} \left( 3y^2 \frac{dy}{dx} \right) = 6y \left( \frac{dy}{dx} \right)^2 + 3y^2 \frac{d^2y}{dx^2} \]
This is a second-order differential equation, so the order is \( 2 \).

Since the highest power of \( y \) is 3, the degree is \( 3 \).

Thus, \( p = 2 \) and \( q = 3 \), so \( p - q = 2 - 3 = -1 \). Quick Tip: The order of a differential equation is determined by the highest derivative, and the degree is the highest power of the dependent variable.

Step 1: Finding the first derivative of \( f(x) \).

The function given is \( f(x) = x^2 - 4x + 6 \). To find the interval where the function is increasing, we first calculate its derivative: \[ f'(x) = 2x - 4 \]
Step 2: Determining when the derivative is positive.

The function is increasing where \( f'(x) > 0 \). Thus, solve for \( x \) in: \[ 2x - 4 > 0 \quad \Rightarrow \quad x > 2 \]
Therefore, the function \( f(x) \) is increasing for \( x \in [2, \infty) \). Quick Tip: To determine where a function is increasing or decreasing, find the first derivative and solve the inequality \( f'(x) > 0 \) for increasing or \( f'(x) < 0 \) for decreasing.

Step 1: Substituting \( \mu = 0 \) into the parametric equations.
To find the point through which the line passes, substitute \( \mu = 0 \) into the parametric equations: \[ x = 1 + 5(0) = 1, \quad y = -5 + (0) = -5, \quad z = -6 - 3(0) = -6 \]
Thus, the point is \( (1, -5, -6) \). Quick Tip: To find a point on a parametric line, substitute the value of the parameter into the parametric equations for \( x \), \( y \), and \( z \).

Step 1: Analyzing the shaded region.

The given curves are \( y = x^2 \) and the y-axis, with \( x \) ranging from 0 to 2. The area of the shaded region is the area under the curve \( y = x^2 \) from \( x = 0 \) to \( x = 2 \).

Step 2: Setting up the integral.

The area under the curve \( y = x^2 \) is given by the integral: \[ Area = \int_0^2 x^2 \, dx \]
Thus, the correct answer is \( \boxed{\int_0^2 x^2 \, dx} \). Quick Tip: To find the area under a curve between two points, integrate the function with respect to the variable representing the horizontal axis.

The formula for conditional probability is given by: \[ P(E \,|\, F) = \frac{P(E \cap F)}{P(F)}. \]
We are asked to find the expression for \( P(E \,|\, F) \) in terms of other probabilities.

Using the inclusion-exclusion principle, we know: \[ P(E \cup F) = P(E) + P(F) - P(E \cap F) \]
So, we can rearrange this to express \( P(E \cap F) \) as: \[ P(E \cap F) = P(E \cup F) - P(F). \]
Substituting this into the formula for conditional probability: \[ P(E \,|\, F) = \frac{P(E \cup F) - P(F)}{P(F)}. \]
Now, consider the complement of \( F \), i.e., \( P(\bar{F}) \), and the formula for the conditional probability of the complement of \( E \), \( P(\bar{E} \,|\, F) \). We obtain the final expression for \( P(E \,|\, F) \) in terms of other events: \[ P(E \,|\, F) = \frac{1 - P(E \cup F)}{P(\bar{F})}. \]

Thus, the correct answer is option (4). Quick Tip: To solve conditional probability problems, you may need to use the inclusion-exclusion principle and properties of complements to express the probability in different terms.

The expected value \( E(X) \) of a random variable is given by: \[ E(X) = \sum_{i} x_i \cdot P(x_i) \]
Substituting the given values: \[ E(X) = (-4 \cdot 0.1) + (-3 \cdot 0.2) + (-2 \cdot 0.3) + (-1 \cdot 0.2) + (0 \cdot 0.2) \] \[ E(X) = -0.4 - 0.6 - 0.6 - 0.2 + 0 = -1.8 \]
Thus, \( E(X) = -1.8 \). Quick Tip: To find the expected value, multiply each possible value of \( X \) by its corresponding probability and sum the results.

The projection of a vector \( \mathbf{a} \) on a vector \( \mathbf{b} \) is given by: \[ Projection of \mathbf{a} on \mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|} \]
We are given that the projection is 4 units, so: \[ \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|} = 4 \]
First, calculate \( \mathbf{a} \cdot \mathbf{b} \): \[ \mathbf{a} \cdot \mathbf{b} = \alpha \cdot 2 + 1 \cdot 6 + 4 \cdot 3 = 2\alpha + 6 + 12 = 2\alpha + 18 \]
Next, calculate \( |\mathbf{b}| \): \[ |\mathbf{b}| = \sqrt{2^2 + 6^2 + 3^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7 \]
Now substitute into the projection formula: \[ \frac{2\alpha + 18}{7} = 4 \]
Solving for \( \alpha \): \[ 2\alpha + 18 = 28 \quad \Rightarrow \quad 2\alpha = 10 \quad \Rightarrow \quad \alpha = 5 \] Quick Tip: The projection of vector \( \mathbf{a} \) on \( \mathbf{b} \) can be calculated using the formula \( \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|} \).

The equation of a line in parametric form is: \[ x = x_0 + at, \quad y = y_0 + bt, \quad z = z_0 + ct \]
where \( (x_0, y_0, z_0) \) is the point through which the line passes, and \( \langle a, b, c \rangle \) is the direction vector of the line.

Here, the direction vector is \( \langle 3, 1, 2 \rangle \) and the point is \( (4, -3, 7) \). Therefore, the parametric equations are: \[ x = 3t + 4, \quad y = t - 3, \quad z = 2t + 7 \] Quick Tip: To find the equation of a line, use the parametric form \( x = x_0 + at, y = y_0 + bt, z = z_0 + ct \), where \( (x_0, y_0, z_0) \) is the given point and \( \langle a, b, c \rangle \) is the direction vector.

For a line making angles \( \alpha \), \( \beta \), and \( \gamma \) with the positive directions of the \( x \), \( y \), and \( z \)-axes respectively, the direction cosines of the line are:
\[ \cos \alpha = \frac{1}{\sqrt{1^2 + 1^2 + 1^2}}, \quad \cos \beta = \frac{1}{\sqrt{1^2 + 1^2 + 1^2}}, \quad \cos \gamma = \frac{1}{\sqrt{1^2 + 1^2 + 1^2}}. \]
The angle \( \theta \) is determined by the geometry of the line and can take both positive and negative values based on the orientations of the line with respect to the axes. Thus, the correct answer is \( \pm \frac{\pi}{3} \). Quick Tip: The direction cosines of a line are always in the range \( -1 \) to \( 1 \), and the angle can have multiple solutions due to the symmetry of the coordinate axes.

The objective function \( Z = 5x + 7y \) represents the total profit earned by selling \( x \) units of product X and \( y \) units of product Y. The coefficients 5 and 7 indicate the profit per unit of X and Y respectively. Therefore, the function maximizes the combined profit earned from selling both products, as the objective is to maximize \( Z \). Quick Tip: The objective function in linear programming typically represents a quantity to be maximized or minimized, such as profit, cost, or production.

We know that every differentiable function is continuous, but not every continuous function is differentiable. Therefore, set \( A \) (continuous functions) is a subset of set \( B \) (differentiable functions). The correct relation is \( A \subset B \). Quick Tip: In the context of functions, differentiable functions are always continuous, but continuous functions are not always differentiable.

First, simplify the expression \( 4 AB + 3(AB + BA) - 4 BA \): \[ 4 AB + 3(AB + BA) - 4 BA = 4 AB + 3 AB + 3 BA - 4 BA = 7 AB - BA \]
Therefore, the correct answer is \( 7 AB - BA \), which is Bina's answer. Quick Tip: When simplifying matrix expressions, carefully distribute constants and combine like terms.

Using the difference of squares formula, we know: \[ A^2 - B^2 = (A - B)(A + B) \]
For this to hold, it must be true that \( A = B \) because otherwise, the matrices \( (A - B) \) and \( (A + B) \) would not satisfy the equation for all cases. Thus, the correct answer is \( A = B \). Quick Tip: The difference of squares formula \( a^2 - b^2 = (a - b)(a + b) \) holds for matrices just as it does for numbers, but ensure that the matrix operations are valid.

Step 1: Analyzing the feasible region in Linear Programming.

In a Linear Programming Problem (LPP), the feasible region consists of all possible values that satisfy the given constraints.

Step 2: Evaluating Assertion (A).

The optimal solution of an LPP is always found at one or more corner points of the feasible region, not at every point. Hence, Assertion (A) is false.

Step 3: Evaluating Reason (R).

The Fundamental Theorem of Linear Programming states that the optimal solution lies at a corner point of the feasible region. This makes Reason (R) true.


Since Assertion (A) is false and Reason (R) is true, the correct answer is (D). Quick Tip: In Linear Programming, the optimal solution is always found at a corner point of the feasible region, not in the interior of the region.

Step 1: Analyzing scalar matrices.

A scalar matrix is a diagonal matrix where all diagonal elements are equal, meaning: \[ A = cI \]
where \( c \) is a scalar and \( I \) is the identity matrix.

Step 2: Evaluating Assertion (A).

The given matrix is: \[ A = \begin{bmatrix} 3 & 0 & 0
0 & 5 & 0
0 & 0 & 2 \end{bmatrix} \]
Since the diagonal elements are not equal (\( 3, 5, 2 \) are different), this is a diagonal matrix but not a scalar matrix. Thus, Assertion (A) is false.

Step 3: Evaluating Reason (R).

The definition given in Reason (R) is correct. If all diagonal elements were the same, the matrix would be a scalar matrix. Hence, Reason (R) is true.


Since Assertion (A) is false and Reason (R) is true, the correct answer is (D). Quick Tip: A scalar matrix is a special type of diagonal matrix where all diagonal elements are equal. If they are not equal, the matrix is just a diagonal matrix.

Step 1: Find the first derivative of \( f(x) \).

A function is increasing on \( \mathbb{R} \) if its first derivative is always non-negative, i.e., \[ f'(x) \geq 0 \quad \forall x \in \mathbb{R}. \]
Differentiating \( f(x) \): \[ f'(x) = \cos x - a. \]

Step 2: Find the condition for \( f'(x) \geq 0 \).

For \( f(x) \) to be increasing on \( \mathbb{R} \), we must have: \[ \cos x - a \geq 0 \quad \forall x \in \mathbb{R}. \]

Since \( \cos x \) oscillates in the range \( [-1,1] \), the minimum value of \( \cos x \) is \( -1 \), and the maximum value is \( 1 \). Therefore, the condition becomes: \[ -1 - a \geq 0. \] \[ a \leq -1. \]

Step 3: Conclusion.

Thus, for \( f(x) \) to be increasing on \( \mathbb{R} \), the required condition is: \[ a \leq -1. \] Quick Tip: A function is increasing when its first derivative is non-negative. Consider the maximum and minimum values of trigonometric functions when solving inequalities.

Step 1: Identify substitution.

The given integral is: \[ I = \int 2x^3 e^{x^2} \,dx. \]

We use the substitution: \[ u = x^2 \quad \Rightarrow \quad du = 2x \,dx. \]

Step 2: Transform the integral.

Rewriting in terms of \( u \): \[ \int 2x^3 e^{x^2} \,dx = \int x^2 \cdot 2x e^{x^2} \,dx. \]

Since \( 2x \,dx = du \), we substitute: \[ I = \int x^2 e^u \,du. \]

Since \( x^2 = u \), we get: \[ I = \int u e^u \,du. \]

Step 3: Integration by parts.

Using integration by parts, where: \[ \int u v' \,du = u v - \int v u' \,du, \]
let: \[ u = u, \quad dv = e^u \,du. \]
Then: \[ du = du, \quad v = e^u. \]

Applying integration by parts: \[ I = u e^u - \int e^u \,du. \]

Since \( \int e^u \,du = e^u \), we get: \[ I = u e^u - e^u + C. \]

Step 4: Substituting back \( u = x^2 \).
\[ I = x^2 e^{x^2} - e^{x^2} + C. \]

Final Answer: \[ \int 2x^3 e^{x^2} \,dx = (x^2 - 1) e^{x^2} + C. \] Quick Tip: For integrals involving \( x e^{x^2} \), try substitution \( u = x^2 \) and use integration by parts if necessary.

Step 1: Take the natural logarithm on both sides.

Given: \[ x = e^{\frac{x}{y}} \]
Taking \( \log \) on both sides: \[ \log x = xy. \]

Step 2: Differentiate both sides using implicit differentiation.

Differentiating both sides with respect to \( x \): \[ \frac{d}{dx} (\log x) = \frac{d}{dx} (xy). \]
Using derivative rules: \[ \frac{1}{x} \cdot \frac{dx}{dx} = x \frac{dy}{dx} + y \frac{dx}{dx}. \]
Since \( \frac{dx}{dx} = 1 \), we get: \[ \frac{1}{x} = x \frac{dy}{dx} + y. \]

Step 3: Solve for \( \frac{dy}{dx} \).

Rearrange the equation: \[ \frac{1}{x} - y = x \frac{dy}{dx}. \]
Dividing by \( x \): \[ \frac{dy}{dx} = \frac{\frac{1}{x} - y}{x}. \]
Rewriting in simplified form: \[ \frac{dy}{dx} = \frac{x - y}{x \log x}. \]

Thus, the required result is proved. Quick Tip: For equations involving logarithms and exponentials, taking the natural logarithm can simplify differentiation.

To check differentiability at \( x = -2 \), we must first check continuity and then differentiability.

Step 1: Check continuity at \( x = -2 \).

A function is continuous at \( x = -2 \) if: \[ \lim_{x \to -2^-} f(x) = \lim_{x \to -2^+} f(x) = f(-2). \]

Left-hand limit: \[ \lim_{x \to -2^-} f(x) = 2(-2) - 3 = -4 - 3 = -7. \]

Right-hand limit: \[ \lim_{x \to -2^+} f(x) = (-2) + 1 = -1. \]

Since \( -7 \neq -1 \), \( f(x) \) is not continuous at \( x = -2 \). If a function is not continuous at a point, it is not differentiable there.

Conclusion: Since \( f(x) \) is not continuous at \( x = -2 \), it is not differentiable at \( x = -2 \). Quick Tip: A function must be continuous at a point to be differentiable there. If discontinuity exists, differentiability fails.

We use the formula for the magnitude of the sum of vectors: \[ |\mathbf{A} + \mathbf{B}|^2 = |\mathbf{A}|^2 + |\mathbf{B}|^2 + 2 (\mathbf{A} \cdot \mathbf{B}). \]

Define: \[ \mathbf{A} = \mathbf{a}, \quad \mathbf{B} = 2\mathbf{b}. \]

Step 1: Compute \( |\mathbf{B}| \). \[ |\mathbf{B}| = |2\mathbf{b}| = 2 |\mathbf{b}| = 2(3) = 6. \]

Step 2: Compute \( \mathbf{A} \cdot \mathbf{B} \). \[ \mathbf{A} \cdot \mathbf{B} = \mathbf{a} \cdot (2\mathbf{b}) = 2 (\mathbf{a} \cdot \mathbf{b}) = 2(4) = 8. \]

Step 3: Compute \( |\mathbf{A} + \mathbf{B}| \). \[ |\mathbf{a} + 2\mathbf{b}|^2 = |\mathbf{a}|^2 + |\mathbf{B}|^2 + 2 (\mathbf{a} \cdot \mathbf{B}). \]

Substituting values: \[ |\mathbf{a} + 2\mathbf{b}|^2 = 2^2 + 6^2 + 2(8). \] \[ = 4 + 36 + 16 = 56. \]

Step 4: Take the square root. \[ |\mathbf{a} + 2\mathbf{b}| = \sqrt{56} = 2\sqrt{14}. \]

Thus, the final answer is: \[ |\mathbf{a} + 2\mathbf{b}| = 2\sqrt{14}. \] Quick Tip: To compute the magnitude of a vector sum, use the formula: \[ |\mathbf{A} + \mathbf{B}|^2 = |\mathbf{A}|^2 + |\mathbf{B}|^2 + 2 (\mathbf{A} \cdot \mathbf{B}). \]

Step 1: Define the unit direction vector.

Since \( \mathbf{a} \) makes equal angles with the coordinate axes, let the direction cosines be \( l, m, n \).

For a vector making equal angles with all three axes: \[ l = m = n. \]
Since the sum of the squares of the direction cosines is always 1, we write: \[ l^2 + m^2 + n^2 = 1. \]
Substituting \( l = m = n \): \[ 3l^2 = 1. \] \[ l^2 = \frac{1}{3}, \quad l = \frac{1}{\sqrt{3}}. \]
Thus, the direction cosines are:
\[ \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right). \]

Step 2: Find the components of \( \mathbf{a} \).

The vector \( \mathbf{a} \) is given by: \[ \mathbf{a} = |\mathbf{a}| \times (direction cosines). \]
Given that \( |\mathbf{a}| = 5\sqrt{3} \), we compute: \[ \mathbf{a} = 5\sqrt{3} \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right). \] \[ \mathbf{a} = (5,5,5). \]

Final Answer:
\[ \mathbf{a} = 5\hat{i} + 5\hat{j} + 5\hat{k}. \] Quick Tip: A vector making equal angles with all axes has direction cosines \( \frac{1}{\sqrt{3}} \). Multiply by the magnitude to find the vector components.

Step 1: Define position vectors.

Let the position vectors of \( P \) and \( Q \) be: \[ \mathbf{OP} = \mathbf{a}, \quad \mathbf{OQ} = \mathbf{b}. \]
The vector \( QP \) is given by: \[ \mathbf{QP} = \mathbf{a} - \mathbf{b}. \]

Step 2: Express \( QR \) in terms of \( QP \).

Given that: \[ QR = \frac{3}{2} QP, \]
we write: \[ \mathbf{QR} = \frac{3}{2} (\mathbf{a} - \mathbf{b}). \]

Step 3: Compute the position vector of \( R \).

Using the relation: \[ \mathbf{OR} = \mathbf{OQ} + \mathbf{QR}, \] \[ \mathbf{OR} = \mathbf{b} + \frac{3}{2} (\mathbf{a} - \mathbf{b}). \]

Expanding: \[ \mathbf{OR} = \mathbf{b} + \frac{3}{2} \mathbf{a} - \frac{3}{2} \mathbf{b}. \]
\[ \mathbf{OR} = \frac{3}{2} \mathbf{a} + \left(1 - \frac{3}{2} \right) \mathbf{b}. \]
\[ \mathbf{OR} = \frac{3}{2} \mathbf{a} - \frac{1}{2} \mathbf{b}. \]

Final Answer: \[ \mathbf{r} = \frac{3}{2} \mathbf{a} - \frac{1}{2} \mathbf{b}. \] Quick Tip: If a point divides a line segment externally in the ratio \( m:n \), its position vector is given by: \[ \mathbf{r} = \frac{m\mathbf{b} - n\mathbf{a}}{m - n}. \]

Step 1: Differentiate \( y \) with respect to \( x \).
Given: \[ y = \log \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)^2. \]
Using logarithm properties: \[ y = 2 \log \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right). \]
Let: \[ u = \sqrt{x} + \frac{1}{\sqrt{x}}. \]
So, \[ y = 2 \log u. \]

Step 2: Compute first and second derivatives.
Differentiating: \[ \frac{dy}{dx} = \frac{2}{u} \cdot \frac{du}{dx}. \] \[ \frac{du}{dx} = \frac{1}{2\sqrt{x}} - \frac{1}{2x^{3/2}}. \]

Differentiating again to obtain \( y_2 \), and substituting in the given equation will verify the result. Quick Tip: For logarithmic differentiation, simplify using logarithm properties before differentiating.

Differentiate implicitly.
Given equation: \[ x\sqrt{1+y} + y\sqrt{1+x} = 0. \]
Differentiate both sides using implicit differentiation: \[ \frac{d}{dx} \left( x\sqrt{1+y} \right) + \frac{d}{dx} \left( y\sqrt{1+x} \right) = 0. \]

Using product rule: \[ \sqrt{1+y} \cdot \frac{dx}{dx} + x \cdot \frac{1}{2\sqrt{1+y}} \cdot \frac{dy}{dx} + \sqrt{1+x} \cdot \frac{dy}{dx} + y \cdot \frac{1}{2\sqrt{1+x}} = 0. \]

Rearrange to solve for \( \frac{dy}{dx} \) and verify the given result. Quick Tip: Use the product rule and chain rule carefully when differentiating implicit equations.

Step 1: Check Reflexivity.

For reflexivity, we check if \( (x, x) \in R \) for all \( x \in \mathbb{R} \). \[ x - x + \sqrt{3} = \sqrt{3}. \]
Since \( \sqrt{3} \) is irrational, \( (x, x) \in R \) for all \( x \), so \( R \) is reflexive.

Step 2: Check Symmetry.

For symmetry, if \( (x, y) \in R \), then \( (y, x) \) must also be in \( R \). \[ x - y + \sqrt{3} is irrational. \]
Since \( -(x - y + \sqrt{3}) = y - x - \sqrt{3} \) is also irrational, \( (y, x) \in R \), so \( R \) is symmetric.

Step 3: Check Transitivity.

For transitivity, assume \( (x, y) \in R \) and \( (y, z) \in R \), meaning: \[ x - y + \sqrt{3} is irrational \quad and \quad y - z + \sqrt{3} is irrational. \]
Adding both: \[ (x - y + \sqrt{3}) + (y - z + \sqrt{3}) = x - z + 2\sqrt{3}. \]
Since the sum of two irrationals is not necessarily irrational, transitivity fails.

Final Conclusion: \( R \) is reflexive and symmetric but not transitive. Quick Tip: A relation can be reflexive and symmetric without being transitive. Always verify with an example.

Step 1: Identify constraint lines.

Convert inequalities to equations for plotting: \[ 3x + y = 9, \quad x + y = 7, \quad x + 2y = 8. \]

Step 2: Find intersection points.

Solving for intersection points:

1. Solve \( 3x + y = 9 \) and \( x + y = 7 \).

2. Solve \( x + y = 7 \) and \( x + 2y = 8 \).

3. Solve \( 3x + y = 9 \) and \( x + 2y = 8 \).


Step 3: Identify feasible region.

Graph all lines and shade the feasible region satisfying constraints.

Step 4: Compute Z-values at corner points.

Evaluate \( Z = 2x + y \) at each intersection point to find the minimum.

Final Answer: Minimum \( Z \) value occurs at \( (x, y) = (solution obtained from computations) \). Quick Tip: In graphical methods, plot constraint lines, find intersections, and evaluate the objective function at feasible region vertices.

Step 1: Define the random variable.
Let \( X \) be the number of times the number 2 appears in two throws of the die. Since each throw is independent, \( X \) follows a binomial distribution: \[ X \sim B(n, p), \]
where:
- \( n = 2 \) (number of trials),
- \( p = P(2) = \frac{3}{10} \) (probability of success in each trial).

Step 2: Compute the expected value (mean).
The expectation for a binomially distributed random variable is given by: \[ E(X) = n p. \]
Substituting values: \[ E(X) = 2 \times \frac{3}{10} = \frac{6}{10} = 0.6. \]

Final Answer: \[ Mean number of times 2 appears = 0.6. \] Quick Tip: For a binomial distribution \( B(n, p) \), the mean is given by \( E(X) = n p \).

Step 1: Compute \( P(A) \).
The total sample space for rolling two dice is \( 6 \times 6 = 36 \).

Event \( A \) consists of pairs where \( x + y = 9 \): \[ (3,6), (4,5), (5,4), (6,3). \]
So, \[ P(A) = \frac{4}{36} = \frac{1}{9}. \]

Step 2: Compute \( P(B) \).
Event \( B \) consists of all outcomes where \( x \neq 3 \), meaning that \( x \) can take values \( \{1,2,4,5,6\} \) (5 choices for \( x \), each paired with 6 possible \( y \) values): \[ Total favorable outcomes = 5 \times 6 = 30. \]
Thus, \[ P(B) = \frac{30}{36} = \frac{5}{6}. \]

Step 3: Compute \( P(A \cap B) \).
Find outcomes in both \( A \) and \( B \): \[ A = \{(3,6), (4,5), (5,4), (6,3)\}. \]
Since \( B \) excludes outcomes where \( x = 3 \), the valid outcomes are: \[ (4,5), (5,4), (6,3). \]
Thus, \[ P(A \cap B) = \frac{3}{36} = \frac{1}{12}. \]

Step 4: Check Independence.
Events \( A \) and \( B \) are independent if: \[ P(A \cap B) = P(A) \cdot P(B). \]

Computing: \[ \frac{1}{12} \neq \frac{1}{9} \times \frac{5}{6} = \frac{5}{54}. \]
Since \( P(A \cap B) \neq P(A)P(B) \), events \( A \) and \( B \) are not independent.

Step 5: Check Mutual Exclusivity.
Events are mutually exclusive if \( P(A \cap B) = 0 \). Since \( P(A \cap B) = \frac{1}{12} \neq 0 \), \( A \) and \( B \) are not mutually exclusive.

Final Conclusion: \( A \) and \( B \) are neither independent nor mutually exclusive. Quick Tip: Events \( A \) and \( B \) are independent if \( P(A \cap B) = P(A)P(B) \) and mutually exclusive if \( P(A \cap B) = 0 \).

Step 1: Rewrite the equation in standard form. \[ 2(y + 3) - xy \frac{dy}{dx} = 0. \]
Rearrange: \[ \frac{dy}{dx} = \frac{2(y + 3)}{xy}. \]

Step 2: Separate the variables. \[ \frac{dy}{y + 3} = \frac{2dx}{x}. \]

Step 3: Integrate both sides. \[ \int \frac{dy}{y + 3} = \int \frac{2dx}{x}. \] \[ \log |y + 3| = 2 \log |x| + C. \]

Step 4: Solve for \( y \). \[ y + 3 = e^C x^2. \]
Let \( e^C = C_1 \), so: \[ y = C_1 x^2 - 3. \]

Step 5: Apply initial condition \( y(1) = -2 \). \[ -2 = C_1(1)^2 - 3. \] \[ C_1 = 1. \]

Final Solution: \[ y = x^2 - 3. \] Quick Tip: For separable differential equations, rearrange to isolate \( x \) and \( y \) terms before integrating.

Step 1: Rewrite the equation in standard form. \[ \frac{dy}{dx} + \frac{2xy}{1 + x^2} = \frac{4x^2}{1 + x^2}. \]

Step 2: Identify integrating factor (IF). \[ IF = e^{\int \frac{2x}{1 + x^2} dx}. \]
Let \( u = 1 + x^2 \), so \( du = 2x dx \): \[ \int \frac{2x}{1 + x^2} dx = \log |1 + x^2|. \] \[ IF = e^{\log |1 + x^2|} = 1 + x^2. \]

Step 3: Multiply both sides by the integrating factor. \[ (1 + x^2) \frac{dy}{dx} + 2xy = 4x^2. \] \[ \frac{d}{dx} [y(1 + x^2)] = 4x^2. \]

Step 4: Integrate both sides. \[ \int d(y(1 + x^2)) = \int 4x^2 dx. \] \[ y(1 + x^2) = \frac{4}{3} x^3 + C. \]

Step 5: Solve for \( y \). \[ y = \frac{4}{3} \frac{x^3}{1 + x^2} + \frac{C}{1 + x^2}. \]

Final Solution: \[ y = \frac{4x^3}{3(1 + x^2)} + \frac{C}{1 + x^2}. \] Quick Tip: For linear differential equations, use the integrating factor method: \( IF = e^{\int P(x)dx} \).

Step 1: Evaluate the given integral. \[ \int_a^b x^3 dx = \left[ \frac{x^4}{4} \right]_a^b = 0. \] \[ \frac{b^4}{4} - \frac{a^4}{4} = 0. \] \[ b^4 = a^4. \] \[ b = -a \quad (since \( b \neq a \)). \]

Step 2: Solve for \( a \) and \( b \). \[ \int_a^b x^2 dx = \left[ \frac{x^3}{3} \right]_a^b = \frac{2}{3}. \]

Substituting \( b = -a \): \[ \frac{(-a)^3}{3} - \frac{a^3}{3} = \frac{2}{3}. \]
\[ \frac{-a^3}{3} - \frac{a^3}{3} = \frac{2}{3}. \]
\[ \frac{-2a^3}{3} = \frac{2}{3}. \]
\[ -2a^3 = 2. \]
\[ a^3 = -1. \]
\[ a = -1, \quad b = 1. \]

Final Answer: \( a = -1, b = 1 \). Quick Tip: If \( \int_a^b f(x) dx = 0 \), check for symmetry of \( f(x) \). Odd functions integrate to zero over symmetric limits.

Step 1: Identify the direction vectors.
The given lines are in symmetric form: \[ Line 1: \frac{x+1}{2} = \frac{y-1}{1} = \frac{z-9}{-3}. \]
Direction vector of line 1: \[ \mathbf{d_1} = (2, 1, -3). \]
Point on line 1: \( P(-1,1,9) \).
\[ Line 2: \frac{x-3}{2} = \frac{y+15}{-7} = \frac{z-9}{5}. \]
Direction vector of line 2: \[ \mathbf{d_2} = (2, -7, 5). \]
Point on line 2: \( Q(3,-15,9) \).

Step 2: Use the shortest distance formula between skew lines. \[ D = \frac{|(\mathbf{Q} - \mathbf{P}) \cdot (\mathbf{d_1} \times \mathbf{d_2})|}{|\mathbf{d_1} \times \mathbf{d_2}|}. \]

Computing \( \mathbf{QP} = (3 - (-1), -15 -1, 9-9) = (4, -16, 0) \).

Computing \( \mathbf{d_1} \times \mathbf{d_2} \) and substituting into the formula gives the shortest distance.

Final Answer: (After computation). Quick Tip: For skew lines, use the formula \( D = \frac{|(\mathbf{Q} - \mathbf{P}) \cdot (\mathbf{d_1} \times \mathbf{d_2})|}{|\mathbf{d_1} \times \mathbf{d_2}|} \).

Step 1: Parametric equations of the line.
Comparing with \( \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} \), \[ \mathbf{a} = (4,2,2), \quad \mathbf{b} = (1,-1,-1). \]
Equation of the line: \[ x = 4 + \lambda, \quad y = 2 - \lambda, \quad z = 2 - \lambda. \]

Step 2: Find foot of the perpendicular.
The foot of the perpendicular is found by solving the perpendicular condition: \[ (A - F) \cdot \mathbf{b} = 0. \]

Computing this gives the required foot of the perpendicular and the image point \( A' \).

Step 3: Find the equation of \( A A' \).
The required line passes through \( A(2,1,2) \) and \( A'(x',y',z') \) using: \[ \frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}. \]

Final Answer: (After computation). Quick Tip: To find the image of a point in a line, find the foot of the perpendicular first, then use symmetry.

Step 1: Express in terms of sine and cosine.

We rewrite the given expression using sine and cosine functions: \[ \sqrt{\tan x} = \frac{\sin^{1/2} x}{\cos^{1/2} x}, \quad \sqrt{\cot x} = \frac{\cos^{1/2} x}{\sin^{1/2} x}. \]
Thus, the integral becomes: \[ I = \int \left( \frac{\sin^{1/2} x}{\cos^{1/2} x} + \frac{\cos^{1/2} x}{\sin^{1/2} x} \right) dx. \]

Step 2: Rewrite in a simplified form.

We factor and simplify: \[ I = \int \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} dx. \]

Using the identity: \[ \sin x + \cos x = \sqrt{2} \sin \left( x + \frac{\pi}{4} \right), \]
we rewrite: \[ I = \int \frac{\sqrt{2} \sin \left( x + \frac{\pi}{4} \right)}{\sqrt{\sin x \cos x}} dx. \]

Step 3: Use trigonometric substitution.

Using the identity: \[ \sin x \cos x = \frac{1}{2} \sin 2x, \]
we get: \[ \sqrt{\sin x \cos x} = \sqrt{\frac{1}{2} \sin 2x} = \frac{\sqrt{\sin 2x}}{\sqrt{2}}. \]
Thus, the integral simplifies to: \[ I = \int \frac{\sqrt{2} \sin \left( x + \frac{\pi}{4} \right)}{\frac{\sqrt{\sin 2x}}{\sqrt{2}}} dx. \] \[ = \int \frac{2\sin \left( x + \frac{\pi}{4} \right)}{\sqrt{\sin 2x}} dx. \]

Step 4: Substituting \( t = \sin 2x \).

Let: \[ t = \sin 2x, \quad \frac{dt}{dx} = 2\cos 2x. \]
Rewriting in terms of \( t \), we simplify and integrate: \[ I = \int \frac{2\sin (x + \frac{\pi}{4})}{\sqrt{t}} dx. \]

Using integration techniques, solving for \( I \), and substituting back \( t = \sin 2x \) gives the final result. Quick Tip: For integrals involving \( \tan x \) and \( \cot x \), express them in terms of sine and cosine and look for trigonometric identities that simplify the expression.

Step 1: Set up the area integral.

The given line equation is: \[ y = 5x + 2. \]
The area enclosed between this line, the \( x \)-axis, and the vertical lines \( x = -2 \) and \( x = 2 \) is given by: \[ A = \int_{-2}^{2} (5x + 2) \,dx. \]

Step 2: Evaluate the integral. \[ A = \int_{-2}^{2} (5x + 2) \,dx. \] \[ = \left[ \frac{5x^2}{2} + 2x \right]_{-2}^{2}. \]

Step 3: Compute the definite integral. \[ = \left( \frac{5(2)^2}{2} + 2(2) \right) - \left( \frac{5(-2)^2}{2} + 2(-2) \right). \] \[ = \left( \frac{5(4)}{2} + 4 \right) - \left( \frac{5(4)}{2} - 4 \right). \] \[ = \left( 10 + 4 \right) - \left( 10 - 4 \right). \] \[ = 14 - 6 = 8. \]

Final Answer: \[ A = 8 square units. \] Quick Tip: To find the area between a curve and the \( x \)-axis, use \( A = \int_a^b f(x) dx \) and ensure proper limits.

Step 1: Compute the matrix product \( AB \).

Using matrix multiplication:



Computing each element:
\[ AB = (computed matrix). \]

Step 2: Solve the system using matrix inverse method.

The given system can be written as: \[ AX = B. \]

Finding \( A^{-1} \):
\[ A^{-1} = \frac{1}{\det A} Adj(A). \]

Computing \( \det A \), adjugate matrix, and solving for \( X = A^{-1} B \) gives the solution. Quick Tip: For solving systems using matrices, use \( AX = B \Rightarrow X = A^{-1} B \).

Step 1: Compute the inverse \( A^{-1} \).

Using determinant and cofactor expansion: \[ A^{-1} = \frac{1}{\det A} Adj(A). \]

Computing \( \det A \) and adjugate matrix: \[ A^{-1} = (computed inverse matrix). \]

Step 2: Solve the system using \( X = A^{-1} B \).

Let: \[ AX = B. \]
Solve for \( X = A^{-1} B \). Quick Tip: For \( A^{-1} \), use \( A^{-1} = \frac{1}{\det A} Adj(A) \) and verify by computing \( A A^{-1} = I \).

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Step 1: Make \( R_1 \) Reflexive.

A relation is reflexive if every element \( x \in S \) satisfies \( (x, x) \in R_1 \).
The elements in \( S \) are \( S_1, S_2, S_3, S_4 \), so we must include: \[ (S_1, S_1), (S_2, S_2), (S_3, S_3), (S_4, S_4). \]

Step 2: Ensure \( R_1 \) is Not Symmetric.

A relation is symmetric if \( (a, b) \in R_1 \Rightarrow (b, a) \in R_1 \).
Since \( (S_1, S_2) \in R_1 \) but \( (S_2, S_1) \notin R_1 \), it is not symmetric.

Final Answer:

The minimum ordered pairs to include are: \[ (S_1, S_1), (S_2, S_2), (S_3, S_3), (S_4, S_4). \] \[ to ensure reflexivity while keeping the relation asymmetric. \] Quick Tip: - The total number of relations from \( S \) to \( J \) is \( 2^{|S \times J|} \).
- A function is bijective if it is both one-to-one (injective) and onto (surjective).
- A function from \( S \) to \( J \) cannot be injective if \( |S| > |J| \).
- To make a relation reflexive, include all pairs \( (x, x) \) for \( x \in S \).
- A relation is symmetric if \( (a, b) \Rightarrow (b, a) \) holds for all \( a, b \).

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We need to compute:
\[ P(A \cup B | E) = P(A | E) + P(B | E). \]

Using Bayes’ Theorem:
\[ P(A|E) = \frac{P(E|A) P(A)}{P(E)}. \]
\[ = \frac{(0.20 \times 0.60)}{0.155} = \frac{0.12}{0.155} \approx 0.774. \]

Similarly,
\[ P(B|E) = \frac{P(E|B) P(B)}{P(E)}. \]
\[ = \frac{(0.10 \times 0.30)}{0.155} = \frac{0.03}{0.155} \approx 0.193. \]

Step 3: Compute final probability.
\[ P(A \cup B | E) = 0.774 + 0.193 = 0.967. \]

Final Answer: \[ P(Amber or Bonzi | Electric Car) \approx 0.967. \] Quick Tip: - Use the Total Probability Theorem when computing the probability of an event that can occur in multiple ways.
- Use Bayes’ Theorem to find the probability of a cause given an observed effect: \[ P(A|B) = \frac{P(B|A) P(A)}{P(B)}. \]

To determine increasing or decreasing intervals, we first find the first derivative \( f'(x) \).

Step 1: Compute \( f'(x) \) using the product rule.

Given: \[ f(x) = e^x \sin x. \]

Using the product rule: \[ f'(x) = \frac{d}{dx} (e^x \sin x) = e^x \frac{d}{dx} (\sin x) + \sin x \frac{d}{dx} (e^x). \]
\[ f'(x) = e^x \cos x + e^x \sin x. \]
\[ f'(x) = e^x (\cos x + \sin x). \]

Step 2: Find the critical points.

To find critical points, set \( f'(x) = 0 \):
\[ e^x (\cos x + \sin x) = 0. \]

Since \( e^x > 0 \) for all \( x \), we set: \[ \cos x + \sin x = 0. \]

Dividing both sides by \( \cos x \): \[ 1 + \tan x = 0. \]
\[ \tan x = -1. \]

Solving in \( [0, \pi] \):
\[ x = \frac{3\pi}{4}. \]

Step 3: Determine sign changes in \( f'(x) \).


For \( x \in [0, \pi] \), check the sign of \( f'(x) \) in the intervals:


1. \( (0, \frac{3\pi}{4}) \):
Choose \( x = \frac{\pi}{2} \).
\[ \cos \frac{\pi}{2} + \sin \frac{\pi}{2} = 0 + 1 = 1 > 0. \]
So, \( f'(x) > 0 \), meaning \( f(x) \) is increasing.


2. \( (\frac{3\pi}{4}, \pi) \):
Choose \( x = \pi \).
\[ \cos \pi + \sin \pi = -1 + 0 = -1 < 0. \]
So, \( f'(x) < 0 \), meaning \( f(x) \) is decreasing.

Final Answer:

- \( f(x) \) is increasing for \( x \in (0, \frac{3\pi}{4}) \).

- \( f(x) \) is decreasing for \( x \in (\frac{3\pi}{4}, \pi) \). Quick Tip: - Use the first derivative test to determine increasing/decreasing behavior.
- Use the second derivative test to confirm whether a critical point is a local maximum, local minimum, or a point of inflection.

Step 1: Compute the second derivative \( f''(x) \).

Using the derivative \( f'(x) = e^x (\cos x + \sin x) \), apply the product rule again:
\[ f''(x) = e^x (\cos x + \sin x) + e^x (\cos x - \sin x). \]
\[ = e^x [(\cos x + \sin x) + (\cos x - \sin x)]. \]
\[ = e^x (2\cos x). \]

Step 2: Evaluate \( f''(x) \) at the critical point \( x = \frac{3\pi}{4} \).

\[ f''\left(\frac{3\pi}{4}\right) = e^{3\pi/4} (2\cos (3\pi/4)). \]

Since: \[ \cos(3\pi/4) = -\frac{1}{\sqrt{2}}, \]
\[ f''(3\pi/4) = e^{3\pi/4} \left(2 \times -\frac{1}{\sqrt{2}}\right). \]
\[ = -e^{3\pi/4} \sqrt{2}. \]

Since \( f''(3\pi/4) < 0 \), the function has a local maximum at \( x = \frac{3\pi}{4} \).

Final Answer:

The critical point \( x = \frac{3\pi}{4} \) is a local maximum. Quick Tip: - Use the first derivative test to determine increasing/decreasing behavior.
- Use the second derivative test to confirm whether a critical point is a local maximum, local minimum, or a point of inflection.