CBSE Class 12 Mathematics Question Paper 2024 (Set 2- 65/5/2) Available - Download Solution PDF with Answer Key

Step 1: {Find \( \frac{d}{dx} e^{\sin^2 x} \)

Let \[ y = e^{\sin^2 x}. \]
Differentiating both sides with respect to \( x \): \[ \frac{dy}{dx} = e^{\sin^2 x} \cdot \frac{d}{dx} (\sin^2 x). \]

Using the chain rule: \[ \frac{d}{dx} (\sin^2 x) = 2 \sin x \cos x. \]

Thus, \[ \frac{dy}{dx} = e^{\sin^2 x} \cdot 2 \sin x \cos x. \]

Step 2: {Find \( \frac{dy}{d(\cos x)} \)

Using \( u = \cos x \), we have: \[ \frac{du}{dx} = -\sin x. \]

Applying the chain rule: \[ \frac{dy}{d(\cos x)} = \frac{\frac{dy}{dx}}{\frac{du}{dx}}. \]

Substituting values: \[ \frac{dy}{d(\cos x)} = \frac{2 \sin x \cos x e^{\sin^2 x}}{-\sin x}. \]

Simplifying: \[ \frac{dy}{d(\cos x)} = -2 \cos x e^{\sin^2 x}. \]

Step 3: {Verify the options

Thus, the correct answer is \( -2 \cos x e^{\sin^2 x} \), which matches option (C).
Quick Tip: For derivatives involving exponentials, apply the chain rule and product rule carefully.

The determinant of a scalar multiple of a matrix is given by: \[ |cA| = c^n |A| \]
where \( c \) is a scalar and \( n \) is the order of the matrix. Since \( A \) is a 2x2 matrix, \( n = 2 \).

Therefore: \[ |5A'| = 5^2 |A| = 25 \times (-2) = -50 \]
Thus, the value of \( |5A'| \) is \( -50 \). Quick Tip: For any square matrix \( A \), the determinant of its adjugate is given by \( |adj A| = |A|^{n-1} \).

Step 1: {Find the first derivative \( f'(x) \)
\[ f(x) = \frac{x}{2} + \frac{2}{x}. \]
Differentiating with respect to \( x \): \[ f'(x) = \frac{1}{2} - \frac{2}{x^2}. \]

Step 2: {Find critical points

Setting \( f'(x) = 0 \): \[ \frac{1}{2} - \frac{2}{x^2} = 0. \]
Solving for \( x \): \[ \frac{1}{2} = \frac{2}{x^2}. \] \[ x^2 = 4. \] \[ x = \pm 2. \]

Step 3: {Find the second derivative \( f''(x) \)
\[ f''(x) = \frac{4}{x^3}. \]

Step 4: {Evaluate \( f''(x) \) at critical points

At \( x = 2 \): \[ f''(2) = \frac{4}{2^3} = \frac{4}{8} = \frac{1}{2} > 0. \]
Since \( f''(2) > 0 \), \( x = 2 \) is a point of local minima.

Step 5: {Verify the options

Thus, the local minima occurs at \( x = 2 \), which matches option (A).
Quick Tip: To find local minima or maxima, use the first derivative test and confirm with the second derivative test.

Step 1: {Find the first derivative (Slope of the curve)
\[ y = 7x - x^3. \]
Differentiating with respect to \( x \): \[ \frac{dy}{dx} = 7 - 3x^2. \]
This represents the slope of the curve.

Step 2: {Find the second derivative (Rate of change of slope)
\[ \frac{d^2y}{dx^2} = -6x. \]

Step 3: {Rate of change of slope with respect to time

Using the chain rule: \[ \frac{d}{dt} \left( \frac{dy}{dx} \right) = \frac{d^2y}{dx^2} \cdot \frac{dx}{dt}. \]
Given \( \frac{dx}{dt} = 2 \), we substitute: \[ \frac{d}{dt} \left( \frac{dy}{dx} \right) = (-6x) \times (2). \]

Step 4: {Evaluate at \( x = 5 \)
\[ \frac{d}{dt} \left( \frac{dy}{dx} \right) = (-6 \times 5) \times 2. \] \[ = -60. \]

Step 5: {Verify the options

Thus, the rate at which the slope is changing is \( -60 \) units/sec, which matches option (A).
Quick Tip: To find the rate of change of slope, differentiate the first derivative and multiply by \( \frac{dx}{dt} \).

Step 1: {Understand matrix multiplication

If matrix \( P \) is of order \( m \times n \) and matrix \( Q \) is of order \( p \times q \), then for the product \( PQ \) to be defined: \[ Number of columns of P = Number of rows of Q. \]
Given that \( Q \) has order \( 3 \times 2 \), we must have: \[ Columns of P = 3. \]

Step 2: {Check the requirement for a diagonal matrix

The resulting product \( PQ \) is a diagonal matrix, which means the resulting order must be square (i.e., \( n \times n \)). Since the number of columns of \( P \) is 3, the resulting matrix must be of order \( 2 \times 2 \), so: \[ Rows of P = 2. \]
Thus, \( P \) must be of order: \[ 2 \times 3. \]

Final Answer: \( 2 \times 3 \) (Option C). Quick Tip: The product of two matrices \( A \) and \( B \) is defined only when the number of columns in \( A \) matches the number of rows in \( B \).

Step 1: {Check injectivity (One-one function)

A function is injective if: \[ f(x_1) = f(x_2) \Rightarrow x_1 = x_2. \]
We differentiate \( f(x) \) to check for monotonicity: \[ f'(x) = \frac{d}{dx} (x^2 - 4x + 5) = 2x - 4. \]
Setting \( f'(x) = 0 \), we solve: \[ 2x - 4 = 0 \Rightarrow x = 2. \]
Since \( f'(x) \) changes sign around \( x = 2 \) (negative to positive), \( f(x) \) is not strictly increasing or decreasing, implying it is not injective.

Step 2: {Check surjectivity (Onto function)

A function is surjective if for every \( y \in \mathbb{R} \), there exists an \( x \) such that \( f(x) = y \). We check the range by completing the square: \[ f(x) = (x - 2)^2 + 1. \]
Since \( (x - 2)^2 \geq 0 \), we have: \[ f(x) \geq 1, \quad \forall x \in \mathbb{R}. \]
Thus, \( f(x) \) never attains values less than 1, so it is not surjective.

Step 3: {Conclusion

Since \( f(x) \) is neither injective nor surjective, the correct answer is (D).
Quick Tip: A quadratic function \( ax^2 + bx + c \) is not injective if \( a \neq 0 \) and is not surjective if its range does not cover all \( \mathbb{R} \).

Step 1: {Differentiate both sides

Given equation: \[ \sin(xy) = 1. \]
Differentiating both sides with respect to \( x \), using the chain rule: \[ \cos(xy) \cdot \frac{d}{dx} (xy) = 0. \]

Step 2: {Apply the product rule

Since \( \frac{d}{dx} (xy) \) requires the product rule: \[ \cos(xy) \cdot \left( x \frac{dy}{dx} + y \right) = 0. \]

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

Since \( \cos(xy) \neq 0 \), dividing by \( \cos(xy) \) on both sides: \[ x \frac{dy}{dx} + y = 0. \]
Rearranging: \[ x \frac{dy}{dx} = -y. \]
Dividing by \( x \): \[ \frac{dy}{dx} = -\frac{y}{x}. \]

Final Answer: \( -\frac{y}{x} \) (Option D).
Quick Tip: Use the chain rule when differentiating trigonometric functions involving product terms.

Step 1: Understanding the given matrix equation
We are given a square matrix \( A \) and its inverse \( A^{-1} \), and we need to determine the value of \( \lambda \) using the property: \[ A \cdot A^{-1} = I \]
where \( I \) is the identity matrix.

Step 2: Representing the given matrices

The given matrix \( A \) is: \[ A = \begin{bmatrix} 7 & -3 & -3
-1 & 1 & 0
-1 & 0 & 1 \end{bmatrix} \]

The given inverse matrix \( A^{-1} \) is: \[ A^{-1} = \begin{bmatrix} 1 & 3 & 3
1 & \lambda & 3
1 & 3 & 4 \end{bmatrix} \]

Step 3: Matrix Multiplication Calculation

Using the property \( A \cdot A^{-1} = I \), we compute: \[ \begin{bmatrix} 7 & -3 & -3
-1 & 1 & 0
-1 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 3 & 3
1 & \lambda & 3
1 & 3 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0
0 & 1 & 0
0 & 0 & 1 \end{bmatrix} \]

Performing matrix multiplication:

First row computations: \[ (7 \times 1) + (-3 \times 1) + (-3 \times 1) = 7 - 3 - 3 = 1 \] \[ (7 \times 3) + (-3 \times \lambda) + (-3 \times 3) = 21 - 3\lambda - 9 = 12 - 3\lambda \] \[ (7 \times 3) + (-3 \times 3) + (-3 \times 4) = 21 - 9 - 12 = 0 \]

Second row computations: \[ (-1 \times 1) + (1 \times 1) + (0 \times 1) = -1 + 1 + 0 = 0 \] \[ (-1 \times 3) + (1 \times \lambda) + (0 \times 3) = -3 + \lambda + 0 = \lambda - 3 \] \[ (-1 \times 3) + (1 \times 3) + (0 \times 4) = -3 + 3 + 0 = 0 \]

Third row computations: \[ (-1 \times 1) + (0 \times 1) + (1 \times 1) = -1 + 0 + 1 = 0 \] \[ (-1 \times 3) + (0 \times \lambda) + (1 \times 3) = -3 + 0 + 3 = 0 \] \[ (-1 \times 3) + (0 \times 3) + (1 \times 4) = -3 + 0 + 4 = 1 \]

Step 4: Comparing the resultant matrix with the identity matrix

Equating corresponding elements, we get: \[ 12 - 3\lambda = 0 \] \[ \lambda - 3 = 1 \]

Step 5: Solving for \( \lambda \)

Solving the first equation: \[ 12 - 3\lambda = 0 \quad \Rightarrow \quad 3\lambda = 12 \quad \Rightarrow \quad \lambda = 4 \]

Solving the second equation: \[ \lambda - 3 = 1 \quad \Rightarrow \quad \lambda = 4 \]

Thus, the value of \( \lambda \) is: \[ \lambda = 4 \] Quick Tip: To verify the inverse of a matrix, multiply it by the given matrix and ensure that the product equals the identity matrix. Compare corresponding elements to solve for unknowns.

Step 1: {Construct the Given Matrix \( A \)

Since \( a_{ij} = \max(i,j) - \min(i,j) \), compute each element: \[ A = \begin{bmatrix} \max(1,1) - \min(1,1) & \max(1,2) - \min(1,2)
\max(2,1) - \min(2,1) & \max(2,2) - \min(2,2) \end{bmatrix} \] \[ A = \begin{bmatrix} 0 & 1
1 & 0 \end{bmatrix}. \]

Step 2: {Compute \( A^2 \)
\[ A^2 = A \times A = \begin{bmatrix} 0 & 1
1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1
1 & 0 \end{bmatrix}. \]

Perform matrix multiplication:

First row, first column: \( (0 \times 0) + (1 \times 1) = 1 \).

First row, second column: \( (0 \times 1) + (1 \times 0) = 0 \).

Second row, first column: \( (1 \times 0) + (0 \times 1) = 0 \).

Second row, second column: \( (1 \times 1) + (0 \times 0) = 1 \).


Thus, \[ A^2 = \begin{bmatrix} 1 & 0
0 & 1 \end{bmatrix}. \]

Step 3: {Verify the options

Thus, \( A^2 \) is \( \begin{bmatrix} 1 & 0
0 & 1 \end{bmatrix} \), which matches option (C).
Quick Tip: For matrix exponentiation, calculate \( A^2 \) by multiplying \( A \) with itself using matrix multiplication rules.

Step 1: {Relation between determinant and adjugate

For any square matrix \( A \) of order \( n \), the determinant of its adjugate matrix is given by: \[ | adj A | = | A |^{n-1}. \]
Since the given order of matrix \( A \) is \( 3 \), we substitute: \[ | adj A | = | A |^2. \]

Step 2: {Solve for \( | A | \)

We are given: \[ | adj A | = 8. \]
Thus: \[ | A |^2 = 8. \]
Taking square root on both sides: \[ | A | = \pm \sqrt{8} = \pm 2\sqrt{2}. \]

Step 3: {Property of determinant of transpose

We use the property: \[ | A^T | = | A |. \]
Thus, we conclude: \[ | A^T | = \pm 2\sqrt{2}. \]

Final Answer: \( 2\sqrt{2} \) (Option D).
Quick Tip: Use the property \( | adj A | = | A |^{n-1} \) for square matrices to relate determinants of matrices and their adjugates.

We are given the integral: \[ I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot \theta \csc^2 \theta \, d\theta \]
To solve this, we use the substitution method. Notice that the derivative of \(\csc \theta\) is \(-\cot \theta \csc \theta\). Therefore, we can choose the substitution: \[ u = \csc \theta \quad so that \quad du = -\cot \theta \csc \theta \, d\theta. \]
This simplifies our integral to: \[ I = \int_{\csc(\frac{\pi}{4})}^{\csc(\frac{\pi}{2})} -u \, du \]
The limits change because when \(\theta = \frac{\pi}{4}\), \(\csc \frac{\pi}{4} = \sqrt{2}\), and when \(\theta = \frac{\pi}{2}\), \(\csc \frac{\pi}{2} = 1\). Therefore, the integral becomes: \[ I = -\int_{\sqrt{2}}^{1} u \, du \]

Now, we integrate: \[ I = -\left[ \frac{u^2}{2} \right]_{\sqrt{2}}^{1} \]

Evaluating the limits: \[ I = -\left( \frac{1^2}{2} - \frac{(\sqrt{2})^2}{2} \right) \] \[ I = -\left( \frac{1}{2} - \frac{2}{2} \right) \] \[ I = -\left( \frac{1}{2} - 1 \right) \] \[ I = \frac{1}{2} \]

Thus, the value of the integral is \(\frac{1}{2}\). Quick Tip: Use trigonometric identities to simplify complex integrals before direct evaluation.

Step 1: {Identify the standard integral form

We compare with: \[ \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1} \left( \frac{x}{a} \right) + c. \]
Given: \[ \int \frac{dx}{\sqrt{9 - 4x^2}}, \]
we rewrite it as: \[ \int \frac{dx}{\sqrt{3^2 - (2x)^2}}. \]
Setting \( a = 3 \) and \( u = 2x \), we get: \[ du = 2 dx \Rightarrow dx = \frac{du}{2}. \]

Step 2: {Solve the integral
\[ I = \int \frac{dx}{\sqrt{9 - 4x^2}} = \int \frac{\frac{du}{2}}{\sqrt{9 - u^2}}. \] \[ = \frac{1}{2} \int \frac{du}{\sqrt{9 - u^2}}. \] \[ = \frac{1}{2} \sin^{-1} \left( \frac{u}{3} \right) + c. \]
Substituting \( u = 2x \): \[ = \frac{1}{2} \sin^{-1} \left( \frac{2x}{3} \right) + c. \]

Final Answer: \( \frac{1}{2} \sin^{-1} \left( \frac{2x}{3} \right) + c \) (Option B).
Quick Tip: Use trigonometric substitution to simplify integrals of the form \( \int \frac{dx}{\sqrt{a^2 - x^2}} \).

Step 1: {Identify the given equation and limits

The given curve is: \[ y^2 = 4x. \]
We need to find the area bounded by this curve from \( x = 0 \) to \( x = 1 \).

Step 2: {Area formula for a parabola

The area enclosed by the parabola and the x-axis is given by: \[ A = \int_{0}^{1} y \, dx. \]
From the equation \( y^2 = 4x \), solving for \( y \): \[ y = \pm 2\sqrt{x}. \]
Since we take the area in the first quadrant (positive \( y \)), we consider: \[ A = 2 \int_0^1 \sqrt{x} \, dx. \]

Step 3: {Evaluate the integral

Using the integration formula: \[ \int x^n \, dx = \frac{x^{n+1}}{n+1}, \]
we compute: \[ \int \sqrt{x} \, dx = \int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2}. \]

Step 4: {Calculate definite integral

Evaluating from 0 to 1: \[ \frac{2}{3} \Big[ x^{3/2} \Big]_0^1 = \frac{2}{3} \times (1 - 0) = \frac{2}{3}. \]

Step 5: {Multiply by 2
\[ A = 2 \times \frac{2}{3} = \frac{4}{3}. \]
Since the region includes both the upper and lower parts of the parabola (symmetry about the x-axis), the total area is: \[ A_{total} = 2 \times \frac{4}{3} = \frac{8}{3}. \]

Final Answer: \( \frac{8}{3} \) (Option B).
Quick Tip: For parabolic regions, use the symmetry and direct integration for accurate area calculations.

Step 1: {Rewrite the differential equation
\[ \frac{dy}{dx} = e^{x+y}. \]
This is a separable differential equation.

Step 2: {Rearrange terms
\[ e^{-y} \, dy = e^x \, dx. \]

Step 3: {Integrate both sides
\[ \int e^{-y} \, dy = \int e^x \, dx. \]
The left integral: \[ \int e^{-y} \, dy = -e^{-y}. \]
The right integral: \[ \int e^x \, dx = e^x. \]

Step 4: {Solve for the general solution
\[ - e^{-y} = e^x + C. \]
Multiplying by -1: \[ e^{-y} + e^x = C. \]

Final Answer: \( e^x + e^{-y} = c \) (Option A).
Quick Tip: For separable differential equations, rewrite in the form \( f(y) dy = g(x) dx \) and integrate both sides.

Step 1: {Extract direction ratios

The given equation of the line is: \[ \frac{x}{1} = \frac{y}{-1} = \frac{z}{0}. \]
This implies that the direction ratios of the line are: \[ (1, -1, 0). \]

Step 2: {Find the angle with the Y-axis

The angle \( \theta \) made with the Y-axis is given by: \[ \cos\theta = \frac{direction ratio of y}{magnitude of the direction vector}. \] \[ \cos\theta = \frac{-1}{\sqrt{1^2 + (-1)^2 + 0^2}}. \]

Step 3: {Compute the magnitude
\[ \sqrt{1 + 1 + 0} = \sqrt{2}. \] \[ \cos\theta = \frac{-1}{\sqrt{2}}. \]

Step 4: {Find \( \theta \)
\[ \theta = \cos^{-1} \left( \frac{-1}{\sqrt{2}} \right). \]
\begin{align*
\cos^{-1\left(-\frac{1{\sqrt{2\right) &= \cos^{-1\left(- \cos \frac{\pi{4\right)

We know that \cos(\pi - \theta) &= -\cos \theta

\text{Therefore, &= \cos^{-1\left(\cos(\pi - \frac{\pi{4)\right)

&= \pi - \frac{\pi{4

&= \frac{3\pi{4

\text{Hence, this is the answer.
\end{align* \[ \theta = \frac{3\pi{4}. \]

Step 5: {Verify the options

Thus, the correct angle is \( \frac{3\pi}{4} \), which matches option (B).
Quick Tip: For a line given by \( \frac{x}{a} = \frac{y}{b} = \frac{z}{c} \), the angles it makes with the coordinate axes are found using: \[ \cos\alpha = \frac{a}{\sqrt{a^2 + b^2 + c^2}}, \quad \cos\beta = \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \quad \cos\gamma = \frac{c}{\sqrt{a^2 + b^2 + c^2}}. \]

Step 1: {Find the direction ratios of the given line

The given line equation in vector form is: \[ \vec{r} = (2 + \lambda) \hat{i} + \lambda \hat{j} + (2\lambda -1) \hat{k}. \]
Comparing with the standard parametric form: \[ \vec{r} = \vec{a} + \lambda \vec{b}, \]
we identify:
- Point on the line: \( (2, 0, -1) \),
- Direction vector: \( (1, 1, 2) \).

Thus, the direction ratios of the given line are (1,1,2).

Step 2: {Equation of a line through a given point

The equation of a line passing through \( (x_0, y_0, z_0) \) and having direction ratios \( (a, b, c) \) is given by: \[ \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}. \]

Step 3: {Substituting the values

The required line passes through \( (1, -3, 2) \) and has direction ratios \( (1, 1, 2) \), so: \[ \frac{x - 1}{1} = \frac{y + 3}{1} = \frac{z - 2}{2}. \]

Step 4: {Verify the options

Thus, the correct Cartesian equation of the required line is: \[ \frac{x - 1}{1} = \frac{y + 3}{1} = \frac{z - 2}{2}, \]
which matches option (D).
Quick Tip: The equation of a line in Cartesian form is derived using the point-direction form: \[ \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}, \] where \( (x_0, y_0, z_0) \) is the point through which the line passes and \( (a, b, c) \) are the direction ratios.

Step 1: {Definition of Conditional Probability

The conditional probability of \( A \) given \( B \) is: \[ P(A|B) = \frac{P(A \cap B)}{P(B)}. \]
Similarly, the conditional probability of \( B \) given \( A \) is: \[ P(B|A) = \frac{P(A \cap B)}{P(A)}. \]

Step 2: {Given Condition and Its Implication

It is given that: \[ P(A|B) = P(B|A). \]
Substituting the formulas: \[ \frac{P(A \cap B)}{P(B)} = \frac{P(A \cap B)}{P(A)}. \]

Step 3: {Cross Multiplying
\[ P(A) P(A \cap B) = P(B) P(A \cap B). \]
Since \( P(A \cap B) \neq 0 \) (otherwise, the conditional probabilities would be undefined), we can divide by \( P(A \cap B) \): \[ P(A) = P(B). \]

Step 4: {Verify the options

Thus, the correct conclusion is \( P(A) = P(B) \), which matches option (D).
Quick Tip: If \( P(A|B) = P(B|A) \), then \( P(A) = P(B) \). This condition implies that both events are equally probable.

Step 1: {Find the position vector of \( R \)

Using the section formula, the position vector of \( R \) dividing \( PQ \) in the ratio \( 3:1 \) is: \[ \vec{r} = \frac{3\vec{q} + 1\vec{p}}{3+1} = \frac{3\vec{q} + \vec{p}}{4}. \]

Step 2: {Find the position vector of \( S \)

Since \( S \) is the midpoint of \( PR \), its position vector is given by the midpoint formula: \[ \vec{s} = \frac{\vec{p} + \vec{r}}{2}. \]
Substituting \( \vec{r} = \frac{3\vec{q} + \vec{p}}{4} \): \[ \vec{s} = \frac{\vec{p} + \frac{3\vec{q} + \vec{p}}{4}}{2}. \]

Step 3: {Simplify the expression
\[ \vec{s} = \frac{\frac{4\vec{p} + 3\vec{q} + \vec{p}}{4}}{2}. \] \[ = \frac{\frac{5\vec{p} + 3\vec{q}}{4}}{2}. \] \[ = \frac{5\vec{p} + 3\vec{q}}{8}. \]

Step 4: {Verify the options

Thus, the correct answer is \( \frac{5\vec{p} + 3\vec{q}}{8} \), which matches option (D).
Quick Tip: The section formula states that if a point divides a line in the ratio \( m:n \), its position vector is given by: \[ \vec{r} = \frac{m\vec{B} + n\vec{A}}{m+n}. \]

Step 1: {Check if vectors form a triangle

Three vectors represent the sides of a triangle if their sum is zero or sum of any two is equal to the third: \[ \vec{a} + \vec{b} = \vec{c}. \]
Computing: \[ (6\hat{i} + 2\hat{j} - 8\hat{k}) + (10\hat{i} - 2\hat{j} - 6\hat{k}) = 16\hat{i} - 10\hat{k}. \]
Since this is not equal to \( \vec{c} \), the vectors do not sum to zero, but they do form a valid triangle.

Step 2: {Check if it is a right-angled triangle

For a right-angled triangle, the Pythagorean theorem holds: \[ |\vec{a}|^2 + |\vec{b}|^2 = |\vec{c}|^2. \]
Calculating magnitudes: \[ |\vec{a}|^2 = 6^2 + 2^2 + (-8)^2 = 36 + 4 + 64 = 104. \] \[ |\vec{b}|^2 = 10^2 + (-2)^2 + (-6)^2 = 100 + 4 + 36 = 140. \] \[ |\vec{c}|^2 = 4^2 + (-4)^2 + 2^2 = 16 + 16 + 4 = 36. \] \[ |\vec{a}|^2 + |\vec{b}|^2 = 104 + 140 = 244. \]
Since this does not satisfy \( |\vec{c}|^2 \), the assertion that they form a right-angled triangle is true.

Step 3: {Check if Reason (R) is a correct explanation

Reason (R) states a general property of three vectors forming a triangle, but it does not specifically explain why these vectors form a right-angled triangle.

Step 4: {Verify the options

Thus, both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A), which matches option (B).
Quick Tip: Three vectors form a triangle if their sum is zero or if the sum of any two vectors equals the third.

Step 1: {Understanding the domain of \( y = \cos^{-1}(x) \)

The inverse cosine function is defined for: \[ x \in [-1,1]. \]
Thus, the domain of \( y = \cos^{-1}(x) \) is correctly stated in Assertion (A).

Step 2: {Understanding the range of \( y = \cos^{-1}(x) \)

The principal value branch of \( \cos^{-1}(x) \) is: \[ [0, \pi]. \]
The Reason (R) incorrectly states that \( \frac{\pi}{2} \) is excluded, which is false.

Step 3: {Verify the options

Thus, Assertion (A) is true, but Reason (R) is false, which matches option (C).
Quick Tip: The domain of \( \cos^{-1}(x) \) is \( [-1,1] \) and the principal value range is \( [0,\pi] \).

Step 1: {Solve for \( a \)

We compute each inverse function separately: \[ \sin^{-1} \left( \frac{\sqrt{2}}{2} \right) = \frac{\pi}{4}, \quad \cos^{-1} \left( \frac{-1}{2} \right) = \frac{2\pi}{3}. \]
Thus: \[ a = \frac{\pi}{4} + \frac{2\pi}{3}. \]

Step 2: {Find common denominator and simplify \( a \)
\[ a = \frac{3\pi}{12} + \frac{8\pi}{12} = \frac{11\pi}{12}. \]

Step 3: {Solve for \( b \)

We compute: \[ \tan^{-1} (\sqrt{3}) = \frac{\pi}{3}, \quad \cot^{-1} \left( \frac{-1}{\sqrt{3}} \right) = \frac{-2\pi}{3}. \]
Thus: \[ b = \frac{\pi}{3} - \left( \frac{-2\pi}{3} \right) = \frac{\pi}{3} + \frac{2\pi}{3}. \]

Step 4: {Find common denominator and simplify \( b \)
\[ b = \frac{4\pi}{12}. \]

Step 5: {Compute \( a + b \)
\[ a + b = \frac{11\pi}{12} - \frac{4\pi}{12} = \frac{7\pi}{12}. \]

Final Answer: \( a + b = \frac{7\pi}{12} \).
Quick Tip: For inverse trigonometric functions, always check standard angles and their corresponding values carefully.

Step 1: {Simplify the given integral

Using the property of logarithm and exponent: \[ e^{\log \sin x} = \sin x. \]
Thus, the given integral reduces to: \[ I = \int \cos^3 x \sin x \ dx. \]

Step 2: {Substituting \( t = \cos x \)

Let: \[ t = \cos x, \quad so that \quad dt = -\sin x \ dx. \]

Rewriting the integral: \[ I = \int t^3 (-dt). \] \[ = -\int t^3 dt. \]

Step 3: {Evaluate the integral
\[ I = -\frac{t^4}{4} + C. \]

Step 4: {Substituting back \( t = \cos x \)
\[ I = -\frac{\cos^4 x}{4} + C. \]

Final Answer: \( -\frac{\cos^4 x}{4} + C \).
Quick Tip: Use substitution to simplify complex trigonometric integrals involving exponentials.

Step 1: {Rewrite the denominator
\[ 5 + 4x - x^2 = -(x^2 - 4x - 5). \]

Factorizing: \[ -(x - 2)^2 + 3^2. \] \[ = 3^2 - (x - 2)^2. \]

Step 2: {Use standard integral formula
\[ \int \frac{1}{a^2 - (x - b)^2} \ dx = \frac{1}{2a} \log \left| \frac{a + (x - b)}{a - (x - b)} \right| + C. \]

Here, \( a = 3 \) and \( b = 2 \), so: \[ \int \frac{1}{3^2 - (x - 2)^2} \ dx = \frac{1}{6} \log \left| \frac{3 + (x - 2)}{3 - (x - 2)} \right| + C. \]

Step 3: {Simplify the expression
\[ = \frac{1}{6} \log \left| \frac{1 + x}{5 - x} \right| + C. \]

Final Answer: \( \frac{1}{6} \log \left| \frac{1 + x}{5 - x} \right| + C \).
Quick Tip: Use the standard integral formula for quadratic denominators of the form \( a^2 - (x - b)^2 \).

Step 1: {Define variables

Let \( V \) be the volume, \( h \) be the height, and \( r \) be the radius of the cone.

Since the problem states that the height of the cone is always one-third of the radius: \[ h = \frac{r}{3} \Rightarrow r = 3h. \]

We also know that the rate of change of volume is given: \[ \frac{dV}{dt} = 15 \quad cm^3/min. \]

Step 2: {Express volume in terms of height

The volume of a cone is given by: \[ V = \frac{1}{3} \pi r^2 h. \]

Substituting \( r = 3h \): \[ V = \frac{1}{3} \pi (3h)^2 h = \frac{1}{3} \pi (9h^2) h = 3\pi h^3. \]

Step 3: {Differentiate with respect to \( t \)
\[ \frac{d}{dt} (3\pi h^3) = 15. \]

Applying the chain rule: \[ 3\pi \cdot 3h^2 \frac{dh}{dt} = 15. \]

Step 4: {Solve for \( \frac{dh}{dt} \) at \( h = 4 \)
\[ 9\pi h^2 \frac{dh}{dt} = 15. \]

Substituting \( h = 4 \): \[ 9\pi (4)^2 \frac{dh}{dt} = 15. \]
\[ 144\pi \frac{dh}{dt} = 15. \]
\[ \frac{dh}{dt} = \frac{15}{144\pi} = \frac{5}{48\pi}. \]

Final Answer: \( \frac{dh}{dt} = \frac{5}{48\pi} \) cm/min.
Quick Tip: When solving related rates problems, always express all variables in terms of a single variable before differentiating.

Step 1: {Find the direction cosines

Since the line makes equal angles with the coordinate axes, let the direction cosines be \( \cos \alpha = \cos \beta = \cos \gamma = l \).

Using the property: \[ l^2 + l^2 + l^2 = 1. \]

Step 2: {Solve for \( l \)
\[ 3l^2 = 1 \Rightarrow l^2 = \frac{1}{3} \Rightarrow l = \frac{1}{\sqrt{3}}. \]

Step 3: {Find direction ratios

Direction cosines: \[ \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right). \]
Thus, the direction ratios are: \[ (1,1,1). \]

Step 4: {Write the vector equation

The vector equation of the line passing through \( (2,3,-5) \) with direction ratios \( (1,1,1) \) is: \[ \vec{r} = 2\hat{i} + 3\hat{j} - 5\hat{k} + \lambda (\hat{i} + \hat{j} + \hat{k}). \]

Final Answer: \[ \vec{r} = 2\hat{i} + 3\hat{j} - 5\hat{k} + \lambda (\hat{i} + \hat{j} + \hat{k}). \] Quick Tip: For a line making equal angles with coordinate axes, the direction cosines satisfy \( l^2 + m^2 + n^2 = 1 \).

Step 1: {Find \( \lim\limits_{x \to 0} f(x) \)
\[ \lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} x \sin \left( \frac{1}{x} \right). \]

Since \( \sin \left( \frac{1}{x} \right) \) oscillates between \( -1 \) and \( 1 \), we have: \[ - x \leq x \sin \left( \frac{1}{x} \right) \leq x. \]

Applying the Squeeze theorem: \[ \lim\limits_{x \to 0} x \sin \left( \frac{1}{x} \right) = 0. \]

Step 2: {Compare with \( f(0) \)
\[ f(0) = 0. \]

Step 3: {Check continuity condition

Since: \[ \lim\limits_{x \to 0} f(x) = f(0), \] \( f(x) \) is continuous at \( x = 0 \).

Final Answer: \( f(x) \) is continuous at \( x = 0 \).
Quick Tip: Use the Squeeze theorem when a function involves oscillatory terms like \( \sin \left( \frac{1}{x} \right) \).

Step 1: {Find Left-Hand Derivative (LHD)
\[ LHD = \lim\limits_{x \to 5^-} \frac{|x - 5| - 0}{x - 5}. \]

For \( x < 5 \), \( |x - 5| = -(x - 5) \), so: \[ LHD = \lim\limits_{x \to 5^-} \frac{-(x - 5)}{x - 5} = -1. \]

Step 2: {Find Right-Hand Derivative (RHD)
\[ RHD = \lim\limits_{x \to 5^+} \frac{|x - 5| - 0}{x - 5}. \]

For \( x > 5 \), \( |x - 5| = x - 5 \), so: \[ RHD = \lim\limits_{x \to 5^+} \frac{x - 5}{x - 5} = 1. \]

Step 3: {Compare LHD and RHD

Since: \[ LHD \neq RHD \quad (-1 \neq 1), \] \( f(x) \) is not differentiable at \( x = 5 \).

Final Answer: \( f(x) \) is not differentiable at \( x = 5 \).
Quick Tip: A function is not differentiable at points where its left-hand derivative and right-hand derivative are not equal.

Step 1: {Identify standard form

The given equation is a first-order linear differential equation of the form: \[ \frac{dy}{dx} + P(x)y = Q(x), \]
where \( P = -2x \) and \( Q = 3x^2 e^{x^2} \).

Step 2: {Find the integrating factor
\[ I.F = e^{\int -2x \ dx} = e^{-x^2}. \]

Step 3: {Multiply throughout by \( I.F \)
\[ y e^{-x^2} = \int e^{-x^2} \cdot 3x^2 e^{x^2} dx. \]

Since \( e^{-x^2} e^{x^2} = 1 \), the integral simplifies to: \[ y e^{-x^2} = \int 3x^2 dx. \]

Step 4: {Evaluate the integral
\[ \int 3x^2 dx = x^3 + C. \]

Thus, the general solution is: \[ y e^{-x^2} = x^3 + C. \]

Step 5: {Find the particular solution using \( y(0) = 5 \)
\[ 5 e^{0} = 0 + C \Rightarrow C = 5. \]

Final Answer: \[ y = (x^3 + 5)e^{x^2}. \] Quick Tip: For first-order linear differential equations, always compute the integrating factor and use it to solve.

Step 1: {Rewrite the equation
\[ \frac{dy}{dx} = -\frac{y}{x} - \left(\frac{y}{x}\right)^2. \]

Step 2: {Substituting \( v = \frac{y}{x} \)

Let \( y = vx \), so: \[ \frac{dy}{dx} = v + x \frac{dv}{dx}. \]

Substituting in the equation: \[ v + x \frac{dv}{dx} = -v - v^2. \]

Step 3: {Separate the variables
\[ \frac{1}{v^2 + 2v} dv = -\frac{1}{x} dx. \]

Rewriting: \[ \int \frac{1}{(v+1)^2 - 1} dv = \int -\frac{1}{x} dx. \]

Step 4: {Integrating both sides
\[ \frac{1}{2} \log \left| \frac{v}{v+2} \right| = \log \left| \frac{C}{x} \right|. \]

Final Answer: \[ \left| \frac{y}{y+2x} \right| = \frac{C^2}{x^2}, \quad or \quad x^2 y = k(y + 2x). \] Quick Tip: For homogeneous differential equations, use substitution \( v = \frac{y}{x} \) to simplify.

Step 1: {Condition for Differentiability

A function is differentiable at \( x = -1 \) if:
- It is **continuous** at \( x = -1 \).
- The **left-hand derivative (LHD) equals the right-hand derivative (RHD)** at \( x = -1 \).

Step 2: {Check Continuity at \( x = -1 \)
\[ \lim_{x \to -1^-} f(x) = \lim_{x \to -1^+} f(x). \] \[ \lim_{x \to -1^-} (bx^2 - 3) = \lim_{x \to -1^+} (ax + b). \]
Substituting \( x = -1 \): \[ b(-1)^2 - 3 = a(-1) + b. \] \[ b - 3 = -a + b. \]
Solving for \( a \): \[ b - b = -a + 3. \] \[ a = 3. \]

Step 3: {Check Differentiability at \( x = -1 \)

The derivative of \( f(x) \) from the left: \[ \lim_{x \to -1^-} \frac{(bx^2 - 3) - (b - 3)}{x + 1}. \]
Differentiating \( bx^2 - 3 \): \[ \lim_{x \to -1^-} b(2x) = b(2(-1)) = -2b. \]

The derivative of \( f(x) \) from the right: \[ \lim_{x \to -1^+} \frac{(ax + b) - (b - 3)}{x + 1}. \]
Differentiating \( ax + b \): \[ \lim_{x \to -1^+} a. \]

Setting LHD = RHD: \[ -2b = a. \]
Substituting \( a = 3 \): \[ -2b = 3. \] \[ b = -\frac{3}{2}. \]

Final Answer: \( a = 3, \quad b = -\frac{3}{2} \).
Quick Tip: For a function to be differentiable at a point, it must be continuous, and the left-hand derivative must equal the right-hand derivative.

Step 1: {Take logarithm on both sides
\[ \log \left( (\cos x)^y \right) = \log \left( (\cos y)^x \right). \]
Using logarithm property: \[ y \log \cos x = x \log \cos y. \]

Step 2: {Differentiate both sides

Applying implicit differentiation: \[ \frac{d}{dx} \left( y \log \cos x \right) = \frac{d}{dx} \left( x \log \cos y \right). \]
Using product rule: \[ y \cdot \frac{d}{dx} (\log \cos x) + \frac{dy}{dx} \cdot \log \cos x = x \cdot \frac{d}{dx} (\log \cos y) + \log \cos y \cdot \frac{dy}{dx}. \]

Step 3: {Solve for \( \frac{dy}{dx} \)
\[ y (-\tan x) + \frac{dy}{dx} \log \cos x = x (-\tan y) + \frac{dy}{dx} \log \cos y. \]
Rearranging: \[ \frac{dy}{dx} (\log \cos x + x \tan y) = \log \cos y + y \tan x. \]
\[ \frac{dy}{dx} = \frac{\log \cos y + y \tan x}{\log \cos x + x \tan y}. \]

Final Answer: \[ \frac{dy}{dx} = \frac{\log \cos y + y \tan x}{\log \cos x + x \tan y}. \] Quick Tip: When both variables appear as exponents, logarithm differentiation is a powerful tool.

Step 1: {Define substitutions

Let: \[ x = \sin A, \quad y = \sin B. \]
Then: \[ A = \sin^{-1} x, \quad B = \sin^{-1} y. \]

Step 2: {Rewrite the given equation
\[ \sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y). \]
Rewriting using trigonometry: \[ \cos A + \cos B = a(\sin A - \sin B). \]

Using the sum-to-product formulas: \[ 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2} = 2a \cos \frac{A+B}{2} \sin \frac{A-B}{2}. \]

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

Dividing both sides: \[ \cot \frac{A - B}{2} = a. \] \[ A - B = 2 \cot^{-1} a. \]

Differentiating both sides with respect to \( x \): \[ \frac{1}{\sqrt{1 - x^2}} - \frac{1}{\sqrt{1 - y^2}} \frac{dy}{dx} = 0. \]

Step 4: {Rearrange for \( \frac{dy}{dx} \)
\[ \frac{dy}{dx} = \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}}. \]

Final Answer: \[ \frac{dy}{dx} = \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}}. \] Quick Tip: For equations involving square roots, substitution using trigonometric identities often simplifies differentiation.

Step 1: {Consider a transformation

Let: \[ I = \int_0^{\pi} \frac{e^{\cos x}}{e^{\cos x} + e^{-\cos x}} dx. \]

Applying the property: \[ I = \int_0^{\pi} \frac{e^{\cos(\pi - x)}}{e^{\cos(\pi - x)} + e^{-\cos(\pi - x)}} dx. \]

Since \( \cos(\pi - x) = -\cos x \), we get: \[ I = \int_0^{\pi} \frac{e^{-\cos x}}{e^{-\cos x} + e^{\cos x}} dx. \]

Step 2: {Add both integrals

Adding both forms of \( I \): \[ 2I = \int_0^{\pi} dx. \]
\[ 2I = \pi. \]

Step 3: {Solve for \( I \)
\[ I = \frac{\pi}{2}. \]

Final Answer: \[ I = \frac{\pi}{2}. \] Quick Tip: For symmetric integrals, apply transformations such as \( f(a - x) = f(x) \).

Step 1: {Decompose the fraction

Using partial fraction decomposition: \[ \frac{2x+1}{(x+1)^2 (x-1)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x-1}. \]

Solving for coefficients: \[ A = -\frac{3}{4}, \quad B = -\frac{1}{2}, \quad C = \frac{3}{4}. \]

Step 2: {Integrate each term
\[ \int \frac{A}{x+1} dx = -\frac{3}{4} \log |x+1|. \] \[ \int \frac{B}{(x+1)^2} dx = -\frac{1}{2(x+1)}. \] \[ \int \frac{C}{x-1} dx = \frac{3}{4} \log |x-1|. \]

Step 3: {Combine the results
\[ I = -\frac{3}{4} \log |x+1| - \frac{1}{2(x+1)} + \frac{3}{4} \log |x-1| + C. \]

Rewriting: \[ I = \frac{3}{4} \log \left| \frac{x - 1}{x + 1} \right| - \frac{1}{2(x + 1)} + C. \]

Final Answer: \[ \frac{3}{4} \log \left| \frac{x - 1}{x + 1} \right| - \frac{1}{2(x + 1)} + C. \] Quick Tip: For rational function integrals, use partial fractions to split the terms.

Step 1: {Find \( \mathbf{a} \times \mathbf{b} \)

The cross product of \( \mathbf{a} \) and \( \mathbf{b} \) is given by: \[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & -1 & 1
3 & 0 & -1 \end{vmatrix} \]
Expanding along the first row: \[ \mathbf{a} \times \mathbf{b} = \hat{i} \begin{vmatrix} -1 & 1
0 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 1
3 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -1
3 & 0 \end{vmatrix} \] \[ = (-1)(-1) - (1)(0) - \hat{j} [(2)(-1) - (1)(3)] + \hat{k} [(2)(0) - (-1)(3)] \] \[ = (1) \hat{i} + (5) \hat{j} + (3) \hat{k} \]
Thus, \[ \mathbf{a} \times \mathbf{b} = \hat{i} + 5\hat{j} + 3\hat{k}. \]

Step 2: {Find \( \lambda \) using \( \mathbf{c} \cdot \mathbf{d} = 3 \)

Let \( \mathbf{d} = \lambda (\hat{i} + 5\hat{j} + 3\hat{k}) \). \[ \mathbf{c} \cdot \mathbf{d} = (2\hat{i} + \hat{j} - 2\hat{k}) \cdot (\lambda \hat{i} + 5\lambda \hat{j} + 3\lambda \hat{k}) \] \[ = 2\lambda + 5\lambda - 6\lambda = 3. \]
Solving for \( \lambda \): \[ \lambda = 3. \]
Step 3: {Find \( \mathbf{d} \)
\[ \mathbf{d} = 3(\hat{i} + 5\hat{j} + 3\hat{k}) = 3\hat{i} + 15\hat{j} + 9\hat{k}. \]

Final Answer: \( \mathbf{d} = 3\hat{i} + 15\hat{j} + 9\hat{k} \).

\begin{quicktipbox
Use the cross product to find a perpendicular vector, then solve for the scalar multiplier using the given dot product condition.
\end{quicktipbox Quick Tip: Use the cross product to find a perpendicular vector, then solve for the scalar multiplier using the given dot product condition.

Step 1: {Define Events

Let:

\( E_1 \): Two balls transferred from Bag I are Red.
\( E_2 \): Two balls transferred from Bag I are Black.
\( E_3 \): Two balls transferred from Bag I are Red and Black.
\( A \): The ball drawn from Bag II is Red.


Step 2: {Find Probabilities of Events
\[ P(E_1) = \frac{1}{7}, \quad P(E_2) = \frac{2}{7}, \quad P(E_3) = \frac{4}{7}. \] \[ P(A|E_1) = \frac{7}{9}, \quad P(A|E_2) = \frac{5}{9}, \quad P(A|E_3) = \frac{6}{9}. \]

Step 3: {Use Law of Total Probability
\[ P(A) = P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2) + P(E_3) \cdot P(A|E_3). \] \[ P(A) = \left( \frac{1}{7} \times \frac{7}{9} \right) + \left( \frac{2}{7} \times \frac{5}{9} \right) + \left( \frac{4}{7} \times \frac{6}{9} \right). \] \[ P(A) = \frac{7}{63} + \frac{10}{63} + \frac{24}{63} = \frac{41}{63}. \]

Final Answer: \( P(A) = \frac{41}{63} \).

\begin{quicktipbox
Use the law of total probability to account for all possible scenarios in probability problems involving multiple events.
\end{quicktipbox Quick Tip: Use the law of total probability to account for all possible scenarios in probability problems involving multiple events.

The standard form of the equation of the line is \[ \frac{x - 4}{-2} = \frac{y - 1}{6} = \frac{z - 1}{-3} \]

Let the foot of the perpendicular from the point A \((2, 3, -8)\) to the given line be B \((-2\lambda + 4, 6\lambda - 3, -3\lambda + 9)\).

The direction ratios of AB are: \[ -2\lambda + 2, \, 6\lambda - 3, \, -3\lambda + 9 \]

As AB is perpendicular to the given line: \[ -2(-2\lambda + 2) + 6(6\lambda - 3) - 3(-3\lambda + 9) = 0 \]
Simplifying the equation: \[ 4\lambda - 4 + 36\lambda - 18 + 9\lambda - 27 = 0 \] \[ 49\lambda - 49 = 0 \] \[ \Rightarrow \lambda = 1 \]

Therefore, the foot of the perpendicular is: \[ B(2, 6, -2) \]

The perpendicular distance is: \[ AB = 3\sqrt{5} \]

Final Answer: \[ Foot of perpendicular = (2,6,-2), \quad Perpendicular Distance = 3\sqrt{5}. \] Quick Tip: To find the foot of a perpendicular, use the parametric equation of the line and set the dot product to zero.

Step 1: {Find direction vectors

Direction vectors: \[ \vec{a}_1 = 2\hat{i} - \hat{j} + \hat{k}, \quad \vec{b}_1 = \hat{i} + \hat{j} + 3\hat{k}. \]
\[ \vec{a}_2 = \hat{i} + \hat{j} - 2\hat{k}, \quad \vec{b}_2 = 2\hat{j} - \hat{k}. \]

Step 2: {Find shortest distance formula
\[ \vec{a}_2 - \vec{a}_1 = (-\hat{i} + 2\hat{j} - 3\hat{k}). \]
\[ \vec{b}_1 \times \vec{b}_2 = (-7\hat{i} + \hat{j} + 2\hat{k}). \]

Step 3: {Compute shortest distance
\[ Shortest Distance = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}. \]
\[ = \frac{1}{\sqrt{6}}. \]

Final Answer: \[ \frac{1}{\sqrt{6}}. \] Quick Tip: For shortest distance between skew lines, use the formula involving cross products.

Step 1: {Find \( \det(A) \)
\[ |A| = 1(6) - 2(3) - 3(4) = -12 \neq 0. \]

Since \( \det(A) \neq 0 \), \( A^{-1} \) exists.

Step 2: {Find adjoint matrix and inverse
\[ adj(A) = \begin{bmatrix} 6 & -6 & -6
-3 & 3 & 3
4 & 0 & -4 \end{bmatrix}. \]
\[ A^{-1} = \frac{1}{-12} \begin{bmatrix} 6 & -6 & -6
-3 & 3 & 3
4 & 0 & -4 \end{bmatrix}. \]

Step 3: {Solve the system using matrix method
\[ AX = B, \quad X = A^{-1} B. \]
\[ X = A^{-1} \begin{bmatrix} 1
2
3 \end{bmatrix}. \]

Computing: \[ X = \begin{bmatrix} 2
\frac{1}{2}
\frac{2}{3} \end{bmatrix}. \]

Final Answer: \[ x = 2, \quad y = \frac{1}{2}, \quad z = \frac{2}{3}. \] Quick Tip: To solve a system using the inverse method, compute \( A^{-1} \) and multiply by the constant matrix.

Step 1: {Multiply the matrices
\[ \begin{bmatrix} 1 & 2 & -3
2 & 3 & 2
3 & -3 & -4 \end{bmatrix} \begin{bmatrix} -6 & 17 & 13
14 & 5 & -8
-15 & 9 & -1 \end{bmatrix} = \begin{bmatrix} 67 & 0 & 0
0 & 67 & 0
0 & 0 & 67 \end{bmatrix}. \]

Step 2: {Solve the system using inverse method
\[ A^{-1} = \frac{1}{67} \begin{bmatrix} -6 & 17 & 13
14 & 5 & -8
-15 & 9 & -1 \end{bmatrix}. \]
\[ X = A^{-1} B. \]

Computing: \[ X = \begin{bmatrix} 3
-2
1 \end{bmatrix}. \]

Final Answer: \[ x = 3, \quad y = -2, \quad z = 1. \] Quick Tip: If the matrix product results in a diagonal matrix, solving for unknowns becomes straightforward.

Step 1: {Rewrite the given equation in standard ellipse form
\[ \frac{x^2}{9} + \frac{y^2}{36} = 1. \]

This represents an ellipse centered at the origin with semi-major axis \(6\) along the \(y\)-axis and semi-minor axis \(3\) along the \(x\)-axis.

Step 2: {Use integration to find the bounded area

The area of one quadrant of the ellipse is given by: \[ \int_0^3 \sqrt{9 - x^2} \ dx. \]

Using the standard formula: \[ Area = 4 \times \frac{6}{3} \int_0^3 \sqrt{9 - x^2} \ dx. \]

Solving using trigonometric substitution: \[ = 8 \left[ \frac{x}{2} \sqrt{9 - x^2} + \frac{9}{2} \sin^{-1} \frac{x}{3} \right]_0^3. \]

Step 3: {Evaluate the definite integral
\[ = 8 \left[ \frac{3}{2} \times 0 + \frac{9}{2} \times \frac{\pi}{2} \right]. \]
\[ = 8 \times \frac{9\pi}{4} = 18\pi. \]

Final Answer: \[ Area = 18\pi. \] Quick Tip: For ellipses, use standard area formulas or integration with trigonometric substitution.

Step 1: {Graphical Representation

The given constraints are plotted on a coordinate plane, and the feasible region is identified as the shaded area.




Step 2: {Finding Corner Points

From the graphical solution, the feasible region is bounded by the points: \[ A(0,4), \quad B(0,6), \quad C(4,4), \quad D(6,0), \quad E(5,0) \]

Step 3: {Evaluating \( Z \) at the Corner Points
\[ Z(A) = 300(0) + 600(4) = 2400 \] \[ Z(B) = 300(0) + 600(6) = 3600 \] \[ Z(C) = 300(4) + 600(4) = 3600 \] \[ Z(D) = 300(6) + 600(0) = 1800 \] \[ Z(E) = 300(5) + 600(0) = 1500 \]

Step 4: {Finding the Maximum Value

From the table: \[ \begin{array}{|c|c|} \hline \textbf{Corner Points} & \textbf{Value of } Z = 300x + 600y
\hline A(0,4) & 2400
B(0,6) & 3600
C(4,4) & 3600
D(6,0) & 1800
E(5,0) & 1500
\hline \end{array} \]

Thus, the maximum value of \( Z \) is 3600 at all points on the line segment \( BC \).

Final Answer: The maximum value of \( Z = 3600 \) at all points on the line segment \( BC \). Quick Tip: To solve Linear Programming Problems graphically, identify the feasible region, determine the corner points, and evaluate the objective function at each point.

Step 1: {Define the probability values

From the problem statement, 70% of customers pay their first-month bill in time. Thus: \[ P(E_1) = \frac{7}{10} = 0.7. \]
Similarly, the probability of not paying in time is: \[ P(E_2) = \frac{3}{10} = 0.3. \]

Final Answer: \[ P(E_1) = 0.7, \quad P(E_2) = 0.3. \] Quick Tip: The sum of probabilities of mutually exclusive and exhaustive events always equals 1.

Step 1: {Define conditional probabilities

The problem states:
- If the customer pays the first-month bill in time, the probability of paying next month is 0.8.
- If the customer does not pay in time, the probability of paying next month is 0.4.

Thus, we write: \[ P(A | E_1) = 0.8, \quad P(A | E_2) = 0.4. \]

Final Answer: \[ P(A | E_1) = 0.8, \quad P(A | E_2) = 0.4. \] Quick Tip: Conditional probability \( P(A|B) \) represents the likelihood of event \( A \) occurring given that \( B \) has already occurred.

Step 1: {Use the Law of Total Probability

The total probability theorem states: \[ P(A) = P(E_1) P(A | E_1) + P(E_2) P(A | E_2). \]

Step 2: {Substituting given values
\[ P(A) = (0.7 \times 0.8) + (0.3 \times 0.4). \]

Step 3: {Compute the values
\[ P(A) = 0.56 + 0.12 = 0.68. \]
In fraction form: \[ P(A) = \frac{17}{25}. \]

Final Answer: \[ P(A) = \frac{17}{25} = 0.68. \] Quick Tip: The Law of Total Probability states that if \( B_1, B_2, ..., B_n \) form a partition of the sample space, then: \[ P(A) = \sum_{i=1}^{n} P(B_i) P(A | B_i). \]

Step 1: {Use Bayes’ Theorem

By Bayes’ theorem: \[ P(E_1 | A) = \frac{P(E_1) P(A | E_1)}{P(A)}. \]

Step 2: {Substituting given values
\[ P(E_1 | A) = \frac{(0.7 \times 0.8)}{0.68}. \]

Step 3: {Compute the values
\[ P(E_1 | A) = \frac{0.56}{0.68} = \frac{14}{17}. \]

Final Answer: \[ P(E_1 | A) = \frac{14}{17}. \] Quick Tip: Bayes' Theorem helps update probabilities based on new information: \[ P(B | A) = \frac{P(A | B) P(B)}{P(A)}. \]

Step 1: {Definition of symmetry

A relation \( R \) is symmetric if: \[ (l_1, l_2) \in R \Rightarrow (l_2, l_1) \in R. \]

Step 2: {Check symmetry for parallel lines

Since \( R \) is defined as: \[ R = \{(l_1, l_2) : l_1 \parallel l_2\}. \]
If \( l_1 \parallel l_2 \), then it follows that \( l_2 \parallel l_1 \), which implies: \[ (l_2, l_1) \in R. \]
Thus, \( R \) is a symmetric relation.

Final Answer: \( R \) is a symmetric relation. Quick Tip: A relation is symmetric if whenever \( (a, b) \) is in the relation, then \( (b, a) \) must also be in the relation.

Step 1: {Definition of transitivity

A relation \( R \) is transitive if: \[ (l_1, l_2) \in R and (l_2, l_3) \in R \Rightarrow (l_1, l_3) \in R. \]

Step 2: {Check transitivity for parallel lines

Since \( R \) is defined as: \[ R = \{(l_1, l_2) : l_1 \parallel l_2\}. \]
If \( l_1 \parallel l_2 \) and \( l_2 \parallel l_3 \), then by the transitive property of parallelism: \[ l_1 \parallel l_3. \]
This implies: \[ (l_1, l_3) \in R. \]
Thus, \( R \) is a transitive relation.

Final Answer: \( R \) is a transitive relation. Quick Tip: A relation is transitive if whenever \( (a, b) \) and \( (b, c) \) are in the relation, then \( (a, c) \) must also be in the relation.

Step 1: {Equation of given rail line

The given equation is: \[ y = 3x + 2. \]
This represents a straight line with slope \( 3 \).

Step 2: {Find parallel lines

A line is parallel to \( y = 3x + 2 \) if it has the same slope, meaning its equation is of the form: \[ y = 3x + c, \quad c \in \mathbb{R}. \]
Thus, the set of all rail lines related to the given line in \( R \) is: \[ \{ l : l is a line of type y = 3x + c, c \in \mathbb{R} \}. \]

Final Answer: The set is \( \{l : l is a line of type y = 3x + c, c \in \mathbb{R} \} \). Quick Tip: All lines parallel to a given line will have the same slope but different intercepts.

Step 1: {Definition of symmetry

A relation \( S \) is symmetric if: \[ (l_1, l_2) \in S \Rightarrow (l_2, l_1) \in S. \]
Since \( S \) is defined as: \[ S = \{(l_1, l_2) : l_1 \perp l_2\}. \]
If \( l_1 \perp l_2 \), then \( l_2 \perp l_1 \), so: \[ (l_2, l_1) \in S. \]
Thus, \( S \) is a symmetric relation.

Step 2: {Definition of transitivity

A relation \( S \) is transitive if: \[ (l_1, l_2) \in S and (l_2, l_3) \in S \Rightarrow (l_1, l_3) \in S. \]
If \( l_1 \perp l_2 \) and \( l_2 \perp l_3 \), then \( l_1 \) may not necessarily be perpendicular to \( l_3 \), meaning: \[ (l_1, l_3) \notin S. \]
Thus, \( S \) is not transitive.

Final Answer: \( S \) is symmetric but not transitive. Quick Tip: Perpendicularity is symmetric but not always transitive. For example, if \( A \) is perpendicular to \( B \) and \( B \) is perpendicular to \( C \), it does not necessarily mean \( A \) is perpendicular to \( C \).

We are given that \( xy = 24 \), which represents the area of the visiting card. The expression for the area is expanded as follows: \[ A(x) = (x+3)(y+2) \]
Expanding the product: \[ A(x) = x(y+2) + 3(y+2) = xy + 2x + 3y + 6 \]
Substituting \( xy = 24 \) into the equation, we get: \[ A(x) = 2x + 3y + 24 + 6 = 2x + 3y + 30 \]
Thus, the area of the visiting card is \( A(x) = 2x + 3y + 30 \). Quick Tip: When given area constraints, express one variable in terms of the other to form a single-variable function.

From the given derivative of the area function: \[ A'(x) = 2 - \frac{72}{x^2} \]
Setting \( A'(x) = 0 \) to find the critical point: \[ 2 - \frac{72}{x^2} = 0 \quad \Rightarrow \quad \frac{72}{x^2} = 2 \quad \Rightarrow \quad x^2 = 36 \quad \Rightarrow \quad x = 6 \]
Thus, \( x = 6 \) is a critical point.

Next, we check if it corresponds to a minimum by computing the second derivative: \[ A''(x) = \frac{144}{x^3} \]
Substituting \( x = 6 \) into \( A''(x) \): \[ A''(6) = \frac{144}{6^3} = \frac{144}{216} = \frac{2}{3} > 0 \]
Since \( A''(6) > 0 \), the area is minimum at \( x = 6 \) and \( y = 4 \).

The dimension of the card with minimum area is Length 9 cm = , Breadth 6 cm Quick Tip: To find the minimum area, differentiate the function and set the derivative to zero. Verify using the second derivative test.