Step 1: The function \( f(x) = 5x \) maps each real number to a unique value in \( \mathbb{R} \), ensuring it is a one-to-one mapping.

Step 2: Since the function \( f \) covers the entire real number set \( \mathbb{R} \), it is onto.

Step 3: \( f \) is not many-one because no two distinct values of \( x \) map to the same value of \( f(x) \). Hence, \( f(x) = 5x \) is onto and one-one. Quick Tip: To determine if a function is onto, check whether every element in the codomain has a preimage in the domain.

The order of a differential equation is determined by the highest derivative present in the equation. In this case, the highest derivative is \(\frac{d^3y}{dx^3}\), which indicates that the order of the equation is 3.
Quick Tip: The order of a differential equation is always the highest derivative, regardless of its power or coefficients.

The given integral is a standard form: \[ \int \frac{dx}{1+x^2} = \tan^{-1}(x). \]
Applying limits: \[ \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{dx}{1+x^2} = \tan^{-1}(\sqrt{3}) - \tan^{-1}\left(\frac{1}{\sqrt{3}}\right). \] \[ \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}, \quad \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}. \] \[ Result: \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}. \]
Quick Tip: Use standard trigonometric integrals and apply limits step by step.

The dot product of a unit vector with itself is equal to 1: \[ \hat{i} \cdot \hat{i} = 1, \quad \hat{j} \cdot \hat{j} = 1, \quad \hat{k} \cdot \hat{k} = 1. \]
Adding these values: \[ 1 + 1 + 1 = 3. \]
Quick Tip: The dot product of a unit vector with itself is always 1, and with others, it's 0.

The inverse of the product of two invertible matrices follows the rule: \[ % Option (AB)^{-1} = B^{-1}A^{-1}. \]
The order of inversion is reversed due to matrix multiplication properties.
Quick Tip: For invertible matrices, the product of inverses is taken in the reverse order of multiplication.

The number of elements in \( A \) is \( |A| = 3 \) and in \( B \) is \( |B| = 2 \).
The total number of relations from \( A \) to \( B \) is given by: \[ 2^{|A| \cdot |B|} = 2^{3 \cdot 2} = 2^6 = 64. \] Quick Tip: The total number of relations is \( 2^{m \cdot n} \), where \( m \) and \( n \) are the sizes of the sets \( A \) and \( B \), respectively.

When two coins are tossed, the sample space is: \[ S = \{HH, HT, TH, TT\}. \]
The event of getting both tails is \( E = \{TT\} \).
The probability is given by: \[ P(E) = \frac{Number of favorable outcomes}{Total number of outcomes} = \frac{1}{4}. \] Quick Tip: The probability of an event is calculated as the ratio of favorable outcomes to the total outcomes.

Two vectors are perpendicular if their dot product is zero: \[ \vec{v_1} \cdot \vec{v_2} = 0. \]
Substituting the given vectors: \[ % Option (3)(1) + (2)(-4) + (1)(\lambda) = 0. \] \[ 3 - 8 + \lambda = 0 \quad \Rightarrow \quad \lambda = 5. \] Quick Tip: For perpendicular vectors, their dot product is always zero.

The conditional probability is given by: \[ P(B/A) = \frac{P(A \cap B)}{P(A)}. \]
Substituting the given values: \[ P(B/A) = \frac{\frac{2}{13}}{\frac{3}{13}} = \frac{2}{3}. \] Quick Tip: Conditional probability is calculated as \( P(B/A) = \frac{P(A \cap B)}{P(A)} \).

Given \( y = \log_e(\tan x) \), differentiating with respect to \( x \): \[ \frac{dy}{dx} = \frac{1}{\tan x} \cdot \frac{d}{dx}(\tan x). \]
Since \( \frac{d}{dx}(\tan x) = \sec^2 x \), we have: \[ \frac{dy}{dx} = \frac{\sec^2 x}{\tan x}. \] Quick Tip: Apply the chain rule carefully when differentiating logarithmic functions.

The function \( f(x) = |x - 1| \) can be written as: \[ f(x) = \begin{cases} x - 1 & if x \geq 1,
1 - x & if x < 1. \end{cases} \]
To check continuity at \( x = 1 \), we evaluate:
1. \( f(1) = |1 - 1| = 0. \)
2. Left-hand limit (\( \lim_{x \to 1^-} f(x) \)): \[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (1 - x) = 0. \]
3. Right-hand limit (\( \lim_{x \to 1^+} f(x) \)): \[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x - 1) = 0. \]
Since \( f(1) = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) \), the function is continuous at \( x = 1 \). Quick Tip: For continuity, ensure \( f(a) = \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) \).

We know: \[ \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}, \quad \cot^{-1}(-\sqrt{3}) = \tan^{-1}(-\frac{1}{\sqrt{3}}). \]
Since \( \tan^{-1}(-x) = -\tan^{-1}(x) \): \[ \cot^{-1}(-\sqrt{3}) = -\tan^{-1}(\frac{1}{\sqrt{3}}) = -\frac{\pi}{6}. \]
Thus: \[ \tan^{-1}(\sqrt{3}) - \cot^{-1}(-\sqrt{3}) = \frac{\pi}{3} - \left(-\frac{\pi}{6}\right) = \frac{\pi}{3} + \frac{\pi}{6} = \frac{\pi}{2}. \] Quick Tip: Use trigonometric identities and inverse function properties for simplifications.

Given \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \), we have: \[ \vec{c} = -(\vec{a} + \vec{b}). \]
Now calculate the dot products: \[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = \vec{a} \cdot \vec{b} + \vec{b} \cdot (-\vec{a} - \vec{b}) + (-\vec{a} - \vec{b}) \cdot \vec{a}. \]
Simplify: \[ \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{a} - \vec{b} \cdot \vec{b} - \vec{a} \cdot \vec{a} - \vec{b} \cdot \vec{a} = -(\vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b}). \]
Since \( \vec{a}, \vec{b}, \vec{c} \) are unit vectors: \[ \vec{a} \cdot \vec{a} = \vec{b} \cdot \vec{b} = 1, \quad so \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -2. \] Quick Tip: Use unit vector properties and dot product identities for simplifications.

Using integration by parts:
Let \( u = \log x \) and \( dv = dx \). Then \( du = \frac{1}{x} dx \) and \( v = x \). \[ \int \log x \, dx = x \log x - \int x \cdot \frac{1}{x} dx = x \log x - \int dx. \] \[ \int \log x \, dx = x \log x - x + C. \] Quick Tip: For \( \int \log x \, dx \), apply integration by parts with \( u = \log x \) and \( dv = dx \).

We are given: \[ x = a(\theta + \sin \theta), \quad y = a(1 - \cos \theta). \]
Differentiating \( x \) and \( y \) with respect to \( \theta \): \[ \frac{dx}{d\theta} = a(1 + \cos \theta), \quad \frac{dy}{d\theta} = a\sin \theta. \]
Using the chain rule, \( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \): \[ \frac{dy}{dx} = \frac{a\sin \theta}{a(1 + \cos \theta)} = \frac{\sin \theta}{1 + \cos \theta}. \] Quick Tip: For parametric equations, use \( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \) and simplify carefully.

Let \( y = x^x \). Taking the natural logarithm on both sides: \[ \ln y = x \ln x. \]
Differentiating both sides with respect to \( x \): \[ \frac{1}{y} \frac{dy}{dx} = \ln x + 1. \]
Multiply by \( y \) to get \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = x^x (\ln x + 1). \] Quick Tip: For \( x^x \), take the natural logarithm to simplify differentiation.

Using the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \): \[ \int_{-\pi/2}^{\pi/2} \sin^2 x \, dx = \int_{-\pi/2}^{\pi/2} \frac{1 - \cos 2x}{2} \, dx. \]
Separate the terms: \[ \int_{-\pi/2}^{\pi/2} \sin^2 x \, dx = \frac{1}{2} \int_{-\pi/2}^{\pi/2} 1 \, dx - \frac{1}{2} \int_{-\pi/2}^{\pi/2} \cos 2x \, dx. \]
The first term evaluates to: \[ \frac{1}{2} \int_{-\pi/2}^{\pi/2} 1 \, dx = \frac{1}{2} \left[ x \right]_{-\pi/2}^{\pi/2} = \frac{1}{2} (\pi - (-\pi)) = \frac{\pi}{2}. \]
The second term evaluates to 0 because \( \cos 2x \) is an odd function. Thus: \[ \int_{-\pi/2}^{\pi/2} \sin^2 x \, dx = \frac{\pi}{2}. \] Quick Tip: Use trigonometric identities like \( \sin^2 x = \frac{1 - \cos 2x}{2} \) to simplify integrals.

The function \( f(x) = |x| \) can be written as: \[ f(x) = \begin{cases} x & if x \geq 0,
-x & if x < 0. \end{cases} \]
The left-hand derivative at \( x = 0 \): \[ f'(0^-) = \lim_{h \to 0^-} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^-} \frac{-h - 0}{h} = -1. \]
The right-hand derivative at \( x = 0 \): \[ f'(0^+) = \lim_{h \to 0^+} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h - 0}{h} = 1. \]
Since \( f'(0^-) \neq f'(0^+) \), \( f(x) \) is not differentiable at \( x = 0 \). Quick Tip: Check both left-hand and right-hand derivatives to determine differentiability at a point.

The dot product of two vectors is given by: \[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta, \]
where \( \theta \) is the angle between the vectors. The absolute value of \( \cos \theta \) is always less than or equal to 1, i.e., \( |\cos \theta| \leq 1 \). Therefore: \[ |\vec{a} \cdot \vec{b}| = |\vec{a}| |\vec{b}| |\cos \theta| \leq |\vec{a}| |\vec{b}|. \] Quick Tip: For proving inequalities involving vectors, use the properties of the dot product and trigonometric bounds.

Expand the determinant using row or column operations. First, subtract the first column from the second and third columns: \[ \begin{vmatrix} 1 + a & 1 & 1
1 & 1 + b & 1
1 & 1 & 1 + c \end{vmatrix} \rightarrow \begin{vmatrix} 1 + a & b & c
1 & b & c
1 & b & c \end{vmatrix}. \]
Using the properties of determinants and simplifying yields: \[ abc \left( 1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right). \] Quick Tip: For determinant proofs, use row and column operations to simplify the matrix.

Rewriting the equation: \[ \frac{dy}{dx} = \frac{1 + y^2}{\tan^{-1} y - x}. \]
Use substitution \( z = \tan^{-1} y - x \), then differentiate and solve. The solution is: \[ z = C, \quad where z = \tan^{-1} y - x. \]
Thus: \[ \tan^{-1} y - x = C. \] Quick Tip: Substitution is a powerful tool for solving first-order differential equations.

Take the natural logarithm of both sides: \[ \ln (\cos x)^y = \ln (\cos y)^x. \]
Differentiating both sides: \[ y \ln (\cos x) + \frac{y}{\cos x} \frac{dy}{dx} = x \ln (\cos y) + \frac{x}{\cos y}. \]
Simplify to find \( \frac{dy}{dx} \). Quick Tip: Logarithmic differentiation simplifies equations involving powers of variables.

Differentiating both sides with respect to \( x \): \[ \frac{dy}{dx} = \cos (a + y) - x \sin (a + y) \frac{dy}{dx}. \]
Rearranging terms: \[ \frac{dy}{dx} = \frac{\cos (a + y)}{1 + x \sin (a + y)}. \] Quick Tip: Implicit differentiation is essential when the dependent variable appears inside a trigonometric function.

For a die, the numbers that are odd are \( 1, 3, 5 \). The probability of rolling an odd number is: \[ P(odd) = \frac{3}{6} = \frac{1}{2}. \]
The probability of rolling an even number is also \( \frac{1}{2} \). Using the binomial probability formula, for exactly one odd number in three rolls: \[ P(one odd) = \binom{3}{1} \left( \frac{1}{2} \right)^1 \left( \frac{1}{2} \right)^2 = 3 \cdot \frac{1}{8} = \frac{3}{8}. \] Quick Tip: Use the binomial theorem for probabilities involving repeated independent trials.

Plot the constraints on a graph to form the feasible region. The vertices of the feasible region are determined by solving the intersection points: \[ 2x + y = 3 \quad and \quad x + 2y = 6. \]
Substitute the vertices into \( Z = x + 2y \): \[ Z(0, 3) = 0 + 2(3) = 6, \quad Z(1, 2) = 1 + 2(2) = 5, \quad Z(3, 0) = 3 + 2(0) = 3. \]
The minimum value of \( Z \) is \( 3 \). Quick Tip: Linear programming problems are solved by evaluating the objective function at the vertices of the feasible region.

Factorize the denominator: \[ x^3 - x^2 - x + 1 = (x - 1)(x^2 + 1). \]
Use partial fractions: \[ \frac{3x + 5}{x^3 - x^2 - x + 1} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1}. \]
Find \( A, B, \) and \( C \), then integrate: \[ \int \frac{A}{x - 1} \, dx + \int \frac{Bx + C}{x^2 + 1} \, dx. \] Quick Tip: Partial fractions simplify the integration of rational functions.

First, compute \( A^2 \): \[ A^2 = \begin{bmatrix} 3 & 1
-1 & 2 \end{bmatrix} \cdot \begin{bmatrix} 3 & 1
-1 & 2 \end{bmatrix}. \]
Then verify: \[ A^2 - 5A + 7I = 0. \]
For \( A^{-1} \), use: \[ A^{-1} = \frac{1}{det(A)} adj(A). \] Quick Tip: Matrix inverses can be found using the adjoint formula or Gaussian elimination.

For a cone, volume \( V = \frac{1}{3} \pi r^2 h \). Using the slant height \( l \): \[ r^2 + h^2 = l^2. \]
Substitute and differentiate to maximize \( V \). Solve for \( \theta \) where \( \tan \theta = \frac{r}{h} \). The result is: \[ \theta = \tan^{-1} (\sqrt{2}). \] Quick Tip: Optimization problems often involve substituting constraints into the objective function.

For two skew lines: \[ \vec{r}_1 = \vec{a}_1 + \lambda \vec{b}_1, \quad \vec{r}_2 = \vec{a}_2 + \mu \vec{b}_2, \]
the shortest distance \( d \) is given by: \[ d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}. \]
Here: \[ \vec{a}_1 = (-1, -1, -1), \quad \vec{b}_1 = (7, -6, 1), \quad \vec{a}_2 = (3, 5, 7), \quad \vec{b}_2 = (1, -2, 1). \]
Compute \( \vec{b}_1 \times \vec{b}_2 \), \( (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) \), and substitute in the formula. Quick Tip: Use the vector triple product and the cross product formula to calculate the shortest distance between skew lines.

For two lines: \[ \vec{b}_1 = (3, 5, 4), \quad \vec{b}_2 = (1, 1, 2). \]
The angle \( \theta \) is given by: \[ \cos \theta = \frac{\vec{b}_1 \cdot \vec{b}_2}{|\vec{b}_1||\vec{b}_2|}. \]
Substitute \( \vec{b}_1 \cdot \vec{b}_2 = 3 \cdot 1 + 5 \cdot 1 + 4 \cdot 2 = 16 \), \( |\vec{b}_1| = \sqrt{3^2 + 5^2 + 4^2} = \sqrt{50} \), \( |\vec{b}_2| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6} \). Then: \[ \cos \theta = \frac{16}{\sqrt{50} \cdot \sqrt{6}}. \] Quick Tip: The dot product helps calculate angles between vectors effectively.

Let the vertices of the triangle be \( A(x_1, y_1, z_1) \), \( B(x_2, y_2, z_2) \), \( C(x_3, y_3, z_3) \). The midpoints are: \[ M_1 = \left(\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}, \frac{z_2 + z_3}{2}\right), \quad M_2 = \left(\frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2}, \frac{z_1 + z_3}{2}\right), \] \[ M_3 = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right). \]
Equating midpoints: \[ (1, 5, -1) = \left(\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}, \frac{z_2 + z_3}{2}\right), \quad \dots \]
Solve for \( x_1, y_1, z_1, x_2, y_2, z_2, x_3, y_3, z_3 \). Quick Tip: Midpoint formulas are useful for reconstructing triangle vertices from given midpoints.

The given system of equations can be written in matrix form as: \[ \begin{bmatrix} 2 & -3 & 5
3 & 2 & -4
1 & 1 & -2 \end{bmatrix} \begin{bmatrix} x
y
z \end{bmatrix} = \begin{bmatrix} 11
-5
-3 \end{bmatrix}. \]
Let: \[ A = \begin{bmatrix} 2 & -3 & 5
3 & 2 & -4
1 & 1 & -2 \end{bmatrix}, \quad X = \begin{bmatrix} x
y
z \end{bmatrix}, \quad B = \begin{bmatrix} 11
-5
-3 \end{bmatrix}. \]
Then: \[ AX = B \quad \Rightarrow \quad X = A^{-1}B. \]
Find \( A^{-1} \) using the formula: \[ A^{-1} = \frac{1}{\det(A)} adj(A). \]
Compute \( \det(A) \), \( adj(A) \), and then \( A^{-1}B \) to solve for \( X \). Quick Tip: To solve systems of linear equations using the matrix method, calculate the inverse of the coefficient matrix if it is non-singular.

To find \( A^{-1} \), compute: \[ A^{-1} = \frac{1}{\det(A)} adj(A), \]
where \( \det(A) \) is the determinant of \( A \) and \( adj(A) \) is the adjugate matrix of \( A \).

Step 1: Compute \( \det(A) \): \[ \det(A) = 2\begin{vmatrix} 2 & -4
1 & -2 \end{vmatrix} - (-3)\begin{vmatrix} 3 & -4
1 & -2 \end{vmatrix} + 5\begin{vmatrix} 3 & 2
1 & 1 \end{vmatrix}. \]
Simplify to find \( \det(A) \).

Step 2: Compute \( adj(A) \) by finding cofactors of \( A \).

Step 3: Compute \( A^{-1} = \frac{1}{\det(A)} adj(A) \). Quick Tip: The inverse of a matrix exists only if its determinant is non-zero. Use cofactor expansion for determinant and adjugate calculations.

To solve \( \int_{0}^{\frac{\pi}{2}} \frac{x \sin x}{1 + \cos^2 x} \, dx \):

Let \( I = \int_{0}^{\frac{\pi}{2}} \frac{x \sin x}{1 + \cos^2 x} \, dx \).
Substitute \( \cos x = t \), so \( -\sin x \, dx = dt \).
Change the limits accordingly:
\[ When x = 0, \, t = 1; \quad When x = \frac{\pi}{2}, \, t = 0. \]
Rewrite the integral:
\[ I = \int_{1}^{0} \frac{-x}{1 + t^2} \, dt. \]
Substitute back and simplify to evaluate \( I = \frac{\pi^2}{4} \). Quick Tip: Use substitution techniques and symmetry properties to simplify trigonometric integrals.

To evaluate \( \int_{0}^{\frac{\pi}{2}} \frac{x \, dx}{a^2 \cos^2 x + b^2 \sin^2 x} \):

Use the substitution \( \cos^2 x = t \), so \( -\sin 2x \, dx = dt \).
Change the limits accordingly:
\[ When x = 0, \, t = 1; \quad When x = \frac{\pi}{2}, \, t = 0. \]
Rewrite the integral and solve using standard results for rational functions.

The final result simplifies to: \[ \int_{0}^{\frac{\pi}{2}} \frac{x \, dx}{a^2 \cos^2 x + b^2 \sin^2 x} = \frac{\pi}{2\sqrt{a^2 + b^2}}. \] Quick Tip: For integrals involving \( \cos^2 x \) and \( \sin^2 x \), try trigonometric substitutions or use standard integral formulas.