CBSE Class 12 Mathematics Set 3- (65/1/3) Question Paper 2026 with Solution PDF

Concept:

Lagrange's Identity: For any two vectors \( \vec{a} \) and \( \vec{b} \), the relationship between their cross product and dot product is given by:
\[ |\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 \]
Alternatively, using the definitions:
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \quad and \quad \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]



Step 1: List the given magnitudes

We are provided with the following values: \[ |\vec{a}| = 8 \] \[ |\vec{b}| = 3 \] \[ |\vec{a} \times \vec{b}| = 12 \]


Step 2: Apply Lagrange's Identity to relate the products

Using the identity: \[ (12)^2 + (\vec{a} \cdot \vec{b})^2 = (8)^2 \times (3)^2 \] \[ 144 + (\vec{a} \cdot \vec{b})^2 = 64 \times 9 \]


Step 3: Calculate the square of the dot product
\[ 144 + (\vec{a} \cdot \vec{b})^2 = 576 \]
Subtract 144 from both sides: \[ (\vec{a} \cdot \vec{b})^2 = 576 - 144 \] \[ (\vec{a} \cdot \vec{b})^2 = 432 \]


Step 4: Find the magnitude of the dot product

To find \( |\vec{a} \cdot \vec{b}| \), take the square root of 432: \[ |\vec{a} \cdot \vec{b}| = \sqrt{432} \]
Factorize 432 to simplify the radical: \[ |\vec{a} \cdot \vec{b}| = \sqrt{144 \times 3} \] \[ |\vec{a} \cdot \vec{b}| = 12\sqrt{3} \] Quick Tip: Tip 1: Lagrange's Identity is the most efficient way to solve problems involving both dot and cross products.
Tip 2: Always look for perfect square factors like 144 when simplifying square roots in competitive exams.

Concept:

The equation of the line is \( \frac{x}{0} = \frac{y}{0} = \frac{z}{1} \).
This line passes through the origin \( (0, 0, 0) \) and has direction ratios \( (0, 0, 1) \), which identifies it as the \( z \)-axis.
The perpendicular distance from a point \( (x, y, z) \) to the \( z \)-axis is \( \sqrt{x^2 + y^2} \).



Step 1: Identify the target line

The denominators of the symmetric form are \( 0, 0, 1 \).
This indicates the line is parallel to the vector \( \hat{k} \).
Since it passes through \( (0, 0, 0) \), the line is exactly the \( z \)-axis.


Step 2: Determine the foot of the perpendicular

Let the given point be \( P(3, 4, 2) \).
The foot of the perpendicular \( M \) from \( P \) onto the \( z \)-axis will have coordinates \( (0, 0, z_P) \).
Therefore, \( M = (0, 0, 2) \).


Step 3: Calculate the distance between the point and the foot

Using the distance formula: \[ Length = \sqrt{(3 - 0)^2 + (4 - 0)^2 + (2 - 2)^2} \] \[ Length = \sqrt{3^2 + 4^2 + 0^2} \]


Step 4: Final computation
\[ Length = \sqrt{9 + 16} \] \[ Length = \sqrt{25} = 5 \] Quick Tip: Tip 1: Distance from \( z \)-axis is \( \sqrt{x^2 + y^2} \).
Tip 2: Distance from \( y \)-axis is \( \sqrt{x^2 + z^2} \).
Tip 3:Distance from \( x \)-axis is \( \sqrt{y^2 + z^2} \).

Concept:

Corner Point Theorem: The optimal (maximum or minimum) value of an objective function in an LPP occurs at the vertices (corner points) of the feasible region.



Step 1: Identify the corner points of the shaded region

From the provided graph, the vertices are:

\( (0, 0) \)
\( (0, 2) \)
\( (3, 4) \)
\( (7, 0) \)



Step 2: Evaluate \( Z = 5x + 7y \) at each corner point

At \( (0, 0) \): \( Z = 5(0) + 7(0) = 0 \)

At \( (0, 2) \): \( Z = 5(0) + 7(2) = 14 \)

At \( (3, 4) \): \( Z = 5(3) + 7(4) = 15 + 28 = 43 \)

At \( (7, 0) \): \( Z = 5(7) + 7(0) = 35 \)


Step 3: Identify the maximum and minimum values

From the calculated values: \[ Z_{max} = 43 \quad at (3, 4) \] \[ Z_{min} = 0 \quad at (0, 0) \]


Step 4: Calculate the difference
\[ Difference = Z_{max} - Z_{min} \] \[ Difference = 43 - 0 = 43 \] Quick Tip: Tip 1: Always check the origin first for minimization problems if it's part of the feasible region.
Tip 2: Linear objective functions change most rapidly in the direction of their coefficients.

Concept:



Linear Programming (LP) is a method to achieve the best outcome in a mathematical model whose requirements are represented by linear relationships.
A linear relationship or function is a polynomial of degree 1.



Step 1: Define the objective function form


In any standard LPP, the objective function \( Z \) is typically written as: \[ Z = ax + by \]


where \( a \) and \( b \) are constants and \( x, y \) are decision variables.



Step 2: Determine the degree of the expression


The degree of a term is the sum of the exponents of the variables in that term.


In \( ax \), the exponent of \( x \) is 1.


In \( by \), the exponent of \( y \) is 1.


Since the highest power of the variables is 1, the degree is 1. Quick Tip: Tip 1: The word 'Linear' in Linear Programming itself specifies that the degree of the function and constraints must be 1.
Tip 2: If the degree were higher, it would fall under Non-Linear Programming.

Concept:

The Range (Principal Value Branch) of \( f(x) = \sin^{-1}x \) is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).
This means for any defined \( x \), \( -\frac{\pi}{2} \leq \sin^{-1}x \leq \frac{\pi}{2} \).



Step 1: Isolate the inverse trigonometric term

Given the equation: \[ \sin^{-1}x + \pi = y \]
Subtract \( \pi \) from both sides to get: \[ \sin^{-1}x = y - \pi \]


Step 2: Apply the standard range constraint

We know the bounds for the arcsine function: \[ -\frac{\pi}{2} \leq \sin^{-1}x \leq \frac{\pi}{2} \]


Step 3: Substitute and solve the inequality for \( y \)

Replace \( \sin^{-1}x \) with \( y - \pi \): \[ -\frac{\pi}{2} \leq y - \pi \leq \frac{\pi}{2} \]


Step 4: Add \( \pi \) to all parts of the inequality
\[ \pi - \frac{\pi}{2} \leq y \leq \pi + \frac{\pi}{2} \] \[ \frac{\pi}{2} \leq y \leq \frac{3\pi}{2} \] Quick Tip: Tip 1: Always memorize the principal value branches of all six inverse trigonometric functions.
Tip 2: When an equation involves an inverse function, the bounds of that function dictate the bounds of the other side of the equation.

Concept:

A scalar matrix is a diagonal matrix in which all the diagonal elements are equal.
For a matrix \( A = [a_{ij}] \), it is a scalar matrix if \( a_{ij} = 0 \) for \( i \neq j \) and \( a_{ii} = k \) for some constant \( k \).
A matrix is symmetric if \( A = A^T \), i.e., \( a_{ij} = a_{ji} \) for all \( i, j \).



Step 1: Write the general form of a \( 3 \times 3 \) scalar matrix

Let \( A \) be a scalar matrix with diagonal elements \( k \): \[ A = \begin{bmatrix} k & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & k \end{bmatrix} \]


Step 2: Find the transpose of the matrix

The transpose \( A^T \) is obtained by interchanging rows and columns: \[ A^T = \begin{bmatrix} k & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & k \end{bmatrix} \]


Step 3: Compare the matrix with its transpose

Since \( A = A^T \), the matrix is symmetric.
Note that it is not necessarily an identity matrix (unless \( k=1 \)) or a null matrix (unless \( k=0 \)).
It is not skew-symmetric because diagonal elements of a skew-symmetric matrix must be zero. Quick Tip:Tip 1: Every diagonal matrix (and thus every scalar and identity matrix) is always a symmetric matrix.
Tip 2: The only scalar matrix that is also skew-symmetric is the Zero matrix.

Concept:



Properties of Transpose of a Matrix:
\( (A^T)^T = A \)
\( (kA)^T = k A^T \)
\( (A \pm B)^T = A^T \pm B^T \)
\( (AB)^T = B^T A^T \) (Reversal Law)



Step 1: Evaluate property (i)


By the addition property of transposes, \( (A + B)' = A' + B' \).


This is True.



Step 2: Evaluate property (ii)


By property, \( (A - B)' = A' - B' \).


The given expression \( B' - A' \) is only true if \( A' = B' \).


This is False in general.



Step 3: Evaluate property (iii)


By the reversal law of transposes, \( (AB)' = B'A' \).


The given expression \( A'B' \) is only true if the matrices commute under transpose.


This is False in general.



Step 4: Evaluate property (iv)


Using the constant property and the reversal law:

\[ (k AB)' = k (AB)' = k (B'A') \]
This matches the given property.


This is True. Quick Tip:Tip 1: Always remember the reversal law: the transpose of a product is the product of the transposes in reverse order.
Tip 2: Transpose is distributive over addition and subtraction but not over multiplication.

Concept:



The determinant of a diagonal matrix is the product of its diagonal elements.
Interchanging any two rows (or columns) of a determinant changes its sign.
If a scalar is multiplied to a row, the determinant value is multiplied by that scalar.



Step 1: Calculate the value of \( \Delta_1 \)

\( \Delta_1 \) is a diagonal matrix: \[ \Delta_1 = 1 \times 2 \times 3 = 6 \]



Step 2: Calculate the value of \( \Delta_2 \)


Expand \( \Delta_2 \) along the first row: \[ \Delta_2 = 0 - 2(1 \times 6 - 0) + 0 \] \[ \Delta_2 = -2(6) = -12 \]




Step 3: Relate the two values


We have \( \Delta_1 = 6 \) and \( \Delta_2 = -12 \).
To find the relation, divide \( \Delta_2 \) by \( \Delta_1 \): \[ \frac{\Delta_2}{\Delta_1} = \frac{-12}{6} = -2 \] \[ \Delta_2 = -2\Delta_1 \] Quick Tip: Tip 1: For diagonal or triangular matrices, don't waste time expanding; just multiply the diagonal.
Tip 2: In \( \Delta_2 \), observe it's almost \( \Delta_1 \) with \( R_1, R_2 \) swapped and one element doubled.

Concept:

The determinant of a \( 2 \times 2 \) matrix \( \begin{vmatrix} a & b
c & d \end{vmatrix} \) is \( ad - bc \).
Trigonometric Identity: \( 2 \sin \theta \cos \theta = \sin 2\theta \).



Step 1: Expand the given determinant

\[ \begin{vmatrix} \cos x & \sin x \\ -\cos x & \sin x \end{vmatrix} = (\cos x)(\sin x) - (\sin x)(-\cos x) \] \[ = \cos x \sin x + \sin x \cos x \] \[ = 2 \sin x \cos x \]


Step 2: Equate to the given value


Using the identity \( 2 \sin x \cos x = \sin 2x \): \[ \sin 2x = 1 \]


Step 3: Solve for \( x \)


For \( \sin \theta = 1 \), the general solution is \( \theta = n\pi + (-1)^n \frac{\pi}{2} \).
For the principal value: \[ 2x = \frac{\pi}{2} \] \[ x = \frac{\pi}{4} \] Quick Tip: Tip 1: If you forget the \( \sin 2x \) identity, just substitute the options into the expanded expression \( 2\sin x \cos x \).
Tip 2: At \( x = \pi/4 \), \( \sin x = \cos x = 1/\sqrt{2} \), so \( 2(1/\sqrt{2})(1/\sqrt{2}) = 1 \).

Concept:

Since \( A \) and \( B \) are symmetric, \( A^T = A \) and \( B^T = B \).
A matrix \( C \) is symmetric if \( C^T = C \).
A matrix \( C \) is skew-symmetric if \( C^T = -C \).



Step 1: Let the resulting matrix be \( C \)

Define \( C = AB - BA \).
To check its property, we must find its transpose \( C^T \).


Step 2: Find the transpose using properties
\[ C^T = (AB - BA)^T \] \[ C^T = (AB)^T - (BA)^T \]
Apply the reversal law \( (XY)^T = Y^T X^T \): \[ C^T = B^T A^T - A^T B^T \]


Step 3: Substitute the given conditions

Since \( A^T = A \) and \( B^T = B \): \[ C^T = BA - AB \]


Step 4: Compare \( C^T \) with \( C \)

Factor out a negative sign: \[ C^T = -(AB - BA) \] \[ C^T = -C \]
Since \( C^T = -C \), the matrix is skew-symmetric. Quick Tip: Tip 1: For any symmetric matrices \( A, B \):
Tip 2: \( AB + BA \) is always symmetric.
Tip 3: \( AB - BA \) is always skew-symmetric.

Concept:

For a continuous and monotonic function, the absolute maximum and minimum values in a closed interval \( [a, b] \) occur at the endpoints or at critical points.
An exponential function \( e^{kx} \) is strictly monotonic.
If \( k < 0 \), the function is strictly decreasing.



Step 1: Determine the nature of the function by finding its derivative

The given function is: \[ f(x) = e^{-x} \]
Differentiating with respect to \( x \): \[ f'(x) = \frac{d}{dx}(e^{-x}) = -e^{-x} \]
Since \( e^{-x} > 0 \) for all real \( x \), we have: \[ f'(x) = -e^{-x} < 0 for all x \in [0, 3] \]


Step 2: Evaluate the function at the boundary points of the interval

Because \( f'(x) < 0 \), the function \( f(x) \) is strictly decreasing on the interval \( [0, 3] \).

This implies the maximum value is at the left endpoint and the minimum value is at the right endpoint.


Value at the left endpoint \( x = 0 \): \[ f(0) = e^{-(0)} = 1 \]
Value at the right endpoint \( x = 3 \): \[ f(3) = e^{-3} \]


Step 3: Identify the least value

Since the function is strictly decreasing: \[ f(3) < f(0) \implies e^{-3} < 1 \]
Thus, the least value of the function in the interval \( [0, 3] \) is \( e^{-3} \). Quick Tip: For any strictly decreasing function in \( [a, b] \), the least value is always \( f(b) \).
For any strictly increasing function in \( [a, b] \), the least value is always \( f(a) \).

Concept:

Integration by substitution: If an integral has the form \( \int f(g(x))g'(x) dx \), we substitute \( u = g(x) \).
The derivative of the denominator often appears in the numerator of log-type integrals.



Step 1: Choose an appropriate substitution for the integral

Let the integral be \( I \): \[ I = \int \frac{3ax}{b^2 + c^2x^2} dx \]
Put \( t = b^2 + c^2x^2 \).


Differentiating with respect to \( x \): \[ \frac{dt}{dx} = 0 + c^2(2x) = 2c^2x \] \[ dt = 2c^2x dx \implies x dx = \frac{1}{2c^2} dt \]


Step 2: Perform the substitution and integrate

Substitute \( t \) and \( x dx \) back into the integral: \[ I = \int \frac{3a}{t} \cdot \left( \frac{1}{2c^2} \right) dt \]
Take the constants outside the integral: \[ I = \frac{3a}{2c^2} \int \frac{1}{t} dt \] \[ I = \frac{3a}{2c^2} \log |t| + K \]


Step 3: Substitute back the original variable and compare coefficients

Replacing \( t \) with its expression in \( x \): \[ I = \frac{3a}{2c^2} \log |b^2 + c^2x^2| + K \]
Comparing this result with the given form \( A \log |b^2 + c^2x^2| + K \): \[ A = \frac{3a}{2c^2} \] Quick Tip: Use the shortcut: \( \int \frac{f'(x)}{f(x)} dx = \log |f(x)| + C \).
Adjust the constant factor in the numerator to match the derivative of the denominator exactly.

Concept:



Property of Definite Integrals: \( \int_{-a}^{a} f(x) dx = 0 \) if \( f(x) \) is an odd function.
A function is odd if \( f(-x) = -f(x) \).
A function is even if \( f(-x) = f(x) \).



Step 1: Define the integrand function and test for symmetry


Let the integrand be \( f(x) \):

\[ f(x) = \frac{x^3}{x^2 + 2|x| + 1} \]

To check if the function is odd or even, replace \( x \) with \( -x \):

\[ f(-x) = \frac{(-x)^3}{(-x)^2 + 2|-x| + 1} \]


Step 2: Simplify the expression for \( f(-x) \)


Using the properties \( (-x)^3 = -x^3 \), \( (-x)^2 = x^2 \), and \( |-x| = |x| \):

\[ f(-x) = \frac{-x^3}{x^2 + 2|x| + 1} \] \[ f(-x) = - \left( \frac{x^3}{x^2 + 2|x| + 1} \right) \] \[ f(-x) = -f(x) \]


This proves that \( f(x) \) is an odd function.



Step 3: Apply the definite integral property


Since the function is odd and the limits of integration are symmetric about the origin (\( -1 \) to \( 1 \)):

\[ \int_{-1}^{1} f(x) dx = 0 \]

Therefore:

\[ \int_{-1}^{1} \frac{x^3}{x^2 + 2|x| + 1} dx = 0 \] Quick Tip: Always check for odd functions when the integration limits are of the form \( [-a, a] \).
Odd function times even function is always an odd function. Here, \( x^3 \) is odd and the denominator is even.

Concept:

The area bounded by a curve \( y = f(x) \), the x-axis, and lines \( x=a, x=b \) is given by \( \int_{a}^{b} |y| dx \).
Absolute value function definition: \( |x| = x \) for \( x \geq 0 \) and \( |x| = -x \) for \( x < 0 \).



Step 1: Redefine the function to handle the absolute value

The given curve is \( y = x|x| \).

For \( x \geq 0 \), \( y = x(x) = x^2 \).

For \( x < 0 \), \( y = x(-x) = -x^2 \).


Thus: \[ y = \begin{cases} -x^2 & if x < 0
x^2 & if x \geq 0 \end{cases} \]


Step 2: Set up the integral for area

The area required is the integral of the absolute value of \( y \) from \( -1 \) to \( 1 \): \[ Area = \int_{-1}^{1} |x|x|| dx \]
Since \( |x|x|| = x^2 \) for all real \( x \) (because \( x^2 \) is always non-negative): \[ Area = \int_{-1}^{1} x^2 dx \]


Step 3: Evaluate the definite integral

Since \( x^2 \) is an even function, we can simplify the integral: \[ Area = 2 \int_{0}^{1} x^2 dx \] \[ Area = 2 \left[ \frac{x^3}{3} \right]_{0}^{1} \] \[ Area = 2 \left( \frac{1^3}{3} - \frac{0^3}{3} \right) \] \[ Area = 2 \left( \frac{1}{3} \right) = \frac{2}{3} \] Quick Tip: Area is a physical quantity and can never be zero or negative.
In area problems, if you get zero, you likely forgot to take the absolute value of the function before integrating.

Concept:

A first-order linear differential equation in \( x \) has the standard form:
\[ \frac{dx}{dy} + P_1(y)x = Q_1(y) \]
The Integrating Factor (I.F.) for such an equation is defined as:
\[ I.F. = e^{\int P_1(y) dy} \]



Step 1: Convert the given equation to standard form

The given equation is: \[ R \frac{dx}{dy} + Px = Q \]
To get it into standard form, we divide the entire equation by \( R \) (assuming \( R \neq 0 \)): \[ \frac{dx}{dy} + \left( \frac{P}{R} \right) x = \frac{Q}{R} \]


Step 2: Identify the coefficient of \( x \)

Comparing with the standard form \( \frac{dx}{dy} + P_1 x = Q_1 \), we find: \[ P_1 = \frac{P}{R} \]
Note that \( P \) and \( R \) are given as functions of \( y \), so \( P_1 \) is also a function of \( y \).


Step 3: Calculate the Integrating Factor

Using the formula for I.F.: \[ I.F. = e^{\int P_1 dy} \] \[ I.F. = e^{\int \frac{P}{R} dy} \]
This matches option (C). Quick Tip: Always ensure the coefficient of the derivative term (\( \frac{dx}{dy} \) or \( \frac{dy}{dx} \)) is 1 before identifying \( P \).
The integral in the exponent must be with respect to the independent variable (\( y \) in this case).

Concept:

Order: The order of a differential equation is the order of the highest derivative present in the equation.
Degree: The degree is the power of the highest order derivative, provided the equation is a polynomial in its derivatives.



Step 1: Expand the given differential equation

The given equation is: \[ \frac{d}{dx}(y')^3 + (y')^3 = 1 \]
We need to perform the differentiation in the first term with respect to \( x \).
Using the chain rule: \[ \frac{d}{dx} \left( \frac{dy}{dx} \right)^3 = 3 \left( \frac{dy}{dx} \right)^2 \cdot \frac{d}{dx} \left( \frac{dy}{dx} \right) \] \[ = 3(y')^2 \cdot y'' \]


Step 2: Rewrite the full equation

Substituting the derivative back into the original equation: \[ 3(y')^2 y'' + (y')^3 = 1 \]
Or in standard notation: \[ 3 \left( \frac{dy}{dx} \right)^2 \left( \frac{d^2y}{dx^2} \right) + \left( \frac{dy}{dx} \right)^3 = 1 \]


Step 3: Identify the order and degree

The highest derivative present in the equation is the second derivative, \( y'' \) or \( \frac{d^2y}{dx^2} \).
Therefore, the order is 2.


The highest order derivative term is \( 3(y')^2 y'' \).
The exponent (power) of this term \( (y'') \) is 1.
Therefore, the degree is 1. Quick Tip: Always perform indicated operations like \( \frac{d}{dx} \) before determining the order and degree.
Ensure all derivatives are free from radicals or fractional exponents to correctly identify the degree.

Concept:

Two non-zero vectors \( \vec{a} \) and \( \vec{b} \) are perpendicular if and only if their dot product (scalar product) is zero.
If \( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \) and \( \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \), then \( \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 \).



Step 1: Set up the dot product equation

Let \( \vec{a} = \hat{i} + 2\hat{j} + 3\hat{k} \) and \( \vec{b} = 2\hat{i} - p\hat{j} + \hat{k} \).
Since the vectors are perpendicular: \[ \vec{a} \cdot \vec{b} = 0 \]


Step 2: Calculate the dot product

Multiply the corresponding components of the vectors: \[ (1)(2) + (2)(-p) + (3)(1) = 0 \] \[ 2 - 2p + 3 = 0 \]


Step 3: Solve for \( p \)

Combine the numerical terms: \[ 5 - 2p = 0 \]
Isolate \( p \): \[ 2p = 5 \] \[ p = \frac{5}{2} \] Quick Tip: Perpendicularity check: \( \vec{a} \cdot \vec{b} = 0 \).
Parallelism check: \( \vec{a} \times \vec{b} = 0 \) or components are proportional.

Concept:

Three points \( A, B, C \) are collinear if the vectors \( \vec{AB} \) and \( \vec{BC} \) (or \( \vec{AC} \)) are parallel.
Parallel vectors have proportional components: if \( \vec{u} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k} \) and \( \vec{v} = x_2\hat{i} + y_2\hat{j} + z_2\hat{k} \), then \( \frac{x_1}{x_2} = \frac{y_1}{y_2} = \frac{z_1}{z_2} \).



Step 1: Find the vectors connecting the points

Let the points be \( A(-\hat{i} - \hat{j} + 2\hat{k}) \), \( B(2\hat{i} + m\hat{j} + 5\hat{k}) \), and \( C(3\hat{i} + 11\hat{j} + 6\hat{k}) \).


Vector \( \vec{AB} = \vec{OB} - \vec{OA} \): \[ \vec{AB} = (2 - (-1))\hat{i} + (m - (-1))\hat{j} + (5 - 2)\hat{k} \] \[ \vec{AB} = 3\hat{i} + (m + 1)\hat{j} + 3\hat{k} \]


Vector \( \vec{BC} = \vec{OC} - \vec{OB} \): \[ \vec{BC} = (3 - 2)\hat{i} + (11 - m)\hat{j} + (6 - 5)\hat{k} \] \[ \vec{BC} = 1\hat{i} + (11 - m)\hat{j} + 1\hat{k} \]


Step 2: Apply the condition for collinearity

Since the points are collinear, \( \vec{AB} = \lambda \vec{BC} \).
Equating the ratios of the components: \[ \frac{3}{1} = \frac{m + 1}{11 - m} = \frac{3}{1} \]


Step 3: Solve for \( m \)

From the first equality: \[ 3 = \frac{m + 1}{11 - m} \]
Cross-multiply: \[ 3(11 - m) = m + 1 \] \[ 33 - 3m = m + 1 \]


Group the terms: \[ 33 - 1 = m + 3m \] \[ 32 = 4m \] \[ m = 8 \] Quick Tip: For collinearity of \( A, B, C \), the area of triangle \( ABC \) must be zero.
Alternatively, checking the ratio of the change in components is often the fastest method.

Concept:

Conditional Probability: The probability of an event \( A \) given that event \( B \) has occurred is \( P(A|B) = \frac{P(A \cap B)}{P(B)} \), where \( P(B) > 0 \).
For an unbiased die, the sample space is \( S = \{1, 2, 3, 4, 5, 6\} \).



Step 1: Verify the Assertion (A)

Let \( A \) be the event of getting a prime number. \[ A = \{2, 3, 5\} \implies n(A) = 3 \]

Let \( B \) be the event that the number appearing on the die is odd. \[ B = \{1, 3, 5\} \implies n(B) = 3 \]


The intersection \( A \cap B \) is the set of numbers that are both prime and odd: \[ A \cap B = \{3, 5\} \implies n(A \cap B) = 2 \]


Calculating the conditional probability: \[ P(A|B) = \frac{n(A \cap B)}{n(B)} = \frac{2}{3} \]

Thus, Assertion (A) is True.


Step 2: Verify the Reason (R)

The formula provided in the Reason is: \[ P(A|B) = \frac{P(A \cup B)}{P(B)} \]

This formula is incorrect. The correct formula for conditional probability is: \[ P(A|B) = \frac{P(A \cap B)}{P(B)}, \quad P(B) > 0 \]

Therefore, Reason (R) is False.


Step 3: Check the relationship between A and R

Assertion (A) is True, as \[ P(A|B) = \frac{2}{3}. \]

However, Reason (R) is False because conditional probability is calculated using the intersection \( A \cap B \), not the union \( A \cup B \).

Therefore, Assertion (A) is true and Reason (R) is false. Quick Tip: Tip 1: In conditional probability, the given condition reduces the size of the sample space. Tip 2: Remember that \( P(A|B) \) involves the intersection \( A \cap B \), not the union \( A \cup B \).

Concept:

The direction of a line is determined by the vector \( \vec{b} \) in the form \( \vec{r} = \vec{a} + \lambda\vec{b} \).
Two lines are perpendicular if the dot product of their direction vectors is zero.



Step 1: Find the direction vectors of the lines in Assertion (A)

For the first line: \( x = py + q \implies \frac{x - q}{p} = y \) and \( z = ry + s \implies \frac{z - s}{r} = y \).
The symmetric form is: \[ \frac{x - q}{p} = \frac{y - 0}{1} = \frac{z - s}{r} \]
The direction vector is \( \vec{b}_1 = p\hat{i} + 1\hat{j} + r\hat{k} \).


For the second line: \( x = p'y + q' \implies \frac{x - q'}{p'} = y \) and \( z = r'y + s' \implies \frac{z - s'}{r'} = y \).
The symmetric form is: \[ \frac{x - q'}{p'} = \frac{y - 0}{1} = \frac{z - s'}{r'} \]
The direction vector is \( \vec{b}_2 = p'\hat{i} + 1\hat{j} + r'\hat{k} \).


Step 2: Apply the perpendicularity condition

The lines are perpendicular if \( \vec{b}_1 \cdot \vec{b}_2 = 0 \). \[ (p\hat{i} + \hat{j} + r\hat{k}) \cdot (p'\hat{i} + \hat{j} + r'\hat{k}) = 0 \] \[ (p)(p') + (1)(1) + (r)(r') = 0 \] \[ pp' + rr' + 1 = 0 \]
Thus, Assertion (A) is True.


Step 3: Evaluate the Reason (R) and the link

Reason (R) states the general vector condition for perpendicularity, \( \vec{b}_1 \cdot \vec{b}_2 = 0 \).
This is True.
Since we derived the result in Step 2 directly from this principle, the Reason correctly explains the Assertion. Quick Tip: When a line is given as \( x = ay+b, z = cy+d \), the direction ratios are \( (a, 1, c) \).
If denominators in symmetric form are \( l, m, n \), the direction vector is \( l\hat{i} + m\hat{j} + n\hat{k} \).

Concept:

Conversion to tangent form: To simplify an expression involving \( \tan^{-1} \), we try to express the inner term as \( \tan \theta \).
Trigonometric Identity: \( \tan \left( \frac{\pi}{4} - \theta \right) = \frac{1 - \tan \theta}{1 + \tan \theta} \).



Step 1: Divide numerator and denominator by \( \cos 2x \)

Divide both the numerator and the denominator inside the brackets by \( \cos 2x \): \[ \frac{\frac{\cos 2x}{\cos 2x} - \frac{\sin 2x}{\cos 2x}}{\frac{\cos 2x}{\cos 2x} + \frac{\sin 2x}{\cos 2x}} = \frac{1 - \tan 2x}{1 + \tan 2x} \]


Step 2: Rewrite the expression using tangent identity

We know that \( \tan \frac{\pi}{4} = 1 \). The expression can be written as: \[ \frac{\tan \frac{\pi}{4} - \tan 2x}{1 + \tan \frac{\pi}{4} \tan 2x} \]
Using the formula \( \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \), we get: \[ = \tan \left( \frac{\pi}{4} - 2x \right) \]


Step 3: Simplify the inverse tangent function

Substitute the result back into the original expression: \[ \tan^{-1} \left[ \tan \left( \frac{\pi}{4} - 2x \right) \right] \]
Check the range: Given \( 0 < x < \frac{\pi}{4} \), then \( 0 < 2x < \frac{\pi}{2} \).


Subtracting from \( \frac{\pi}{4} \): \[ \frac{\pi}{4} - \frac{\pi}{2} < \frac{\pi}{4} - 2x < \frac{\pi}{4} - 0 \] \[ -\frac{\pi}{4} < \frac{\pi}{4} - 2x < \frac{\pi}{4} \]
Since this is within the principal value branch \( (-\frac{\pi}{2}, \frac{\pi}{2}) \): \[ = \frac{\pi}{4} - 2x \] Quick Tip: When you see \( \cos \theta \pm \sin \theta \), dividing by \( \cos \theta \) is almost always the first step to convert it to a tangent form.
Always verify the range of the argument for inverse functions to ensure the result is correct.

Concept:

Principal values:
\( \sin^{-1}(1) = \frac{\pi}{2} \) since \( \sin \frac{\pi}{2} = 1 \).
\( \cos^{-1}(-x) = \pi - \cos^{-1}(x) \).



Step 1: Determine the principal values of inverse terms

For the first term: \[ \sin^{-1}(1) = \frac{\pi}{2} \]


For the second term: \[ \cos^{-1} \left( -\frac{1}{2} \right) = \pi - \cos^{-1} \left( \frac{1}{2} \right) \]
Since \( \cos \frac{\pi}{3} = \frac{1}{2} \): \[ \cos^{-1} \left( -\frac{1}{2} \right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]


Step 2: Substitute values into the expression

The expression becomes: \[ \tan \left( \frac{\pi}{2} - \frac{2\pi}{3} \right) \]


Step 3: Simplify and calculate the final value

Find the common denominator for the angle: \[ \frac{\pi}{2} - \frac{2\pi}{3} = \frac{3\pi - 4\pi}{6} = -\frac{\pi}{6} \]
Now find the tangent: \[ \tan \left( -\frac{\pi}{6} \right) = -\tan \left( \frac{\pi}{6} \right) \] \[ = -\frac{1}{\sqrt{3}} \] Quick Tip: Remember that \( \cos^{-1} \) of a negative value always lies in the second quadrant \( (\frac{\pi}{2}, \pi] \).
\( \tan(-\theta) = -\tan \theta \), a property of odd functions.

Concept:

Continuity Check: A function \( f(x) \) is continuous at \( x = a \) if \( LHL = RHL = f(a) \).
Modulus property: \( |x - 3| = -(x - 3) \) when \( x < 3 \).



Step 1: Calculate the Left Hand Limit (LHL) at \( x = 3 \)

For \( x < 3 \), \( |x - 3| = -(x - 3) \). \[ LHL = \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} \frac{-(x - 3)}{2(x - 3)} \]
Cancelling the common term \( (x - 3) \): \[ LHL = \lim_{x \to 3^-} \left( -\frac{1}{2} \right) = -\frac{1}{2} \]


Step 2: Calculate the Right Hand Limit (RHL) at \( x = 3 \)

For \( x \geq 3 \), the function is defined as \( \frac{x - 6}{6} \). \[ RHL = \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} \frac{x - 6}{6} \] \[ RHL = \frac{3 - 6}{6} = \frac{-3}{6} = -\frac{1}{2} \]


Step 3: Compare LHL, RHL, and the function value \( f(3) \)

From the definition, \( f(3) \) is calculated using the \( x \geq 3 \) case: \[ f(3) = \frac{3 - 6}{6} = -\frac{1}{2} \]
Since: \[ LHL = RHL = f(3) = -\frac{1}{2} \]
The function \( f(x) \) is continuous at \( x = 3 \). Quick Tip: Modulus functions create corners in graphs but don't always create discontinuities.
When a function involves a fraction like \( \frac{|x-a|}{x-a} \), it's essentially a jump function with values \( \pm 1 \) (or scaled versions thereof).

Concept:

Implicit Differentiation: Differentiate every term with respect to \( x \), treating \( y \) as a function of \( x \) (using the chain rule).
Product Rule: \( \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \).



Step 1: Differentiate the equation with respect to \( x \)

Given: \( \sqrt{3}x^2 + \sqrt{3}y^2 = 4xy \)
Differentiating both sides: \[ \sqrt{3}(2x) + \sqrt{3}\left( 2y \frac{dy}{dx} \right) = 4 \left[ x \frac{dy}{dx} + y(1) \right] \] \[ 2\sqrt{3}x + 2\sqrt{3}y \frac{dy}{dx} = 4x \frac{dy}{dx} + 4y \]


Step 2: Rearrange to solve for \( \frac{dy}{dx} \)

Gather all \( \frac{dy}{dx} \) terms on one side: \[ 2\sqrt{3}y \frac{dy}{dx} - 4x \frac{dy}{dx} = 4y - 2\sqrt{3}x \]
Factor out the derivative: \[ \frac{dy}{dx} (2\sqrt{3}y - 4x) = 4y - 2\sqrt{3}x \] \[ \frac{dy}{dx} = \frac{4y - 2\sqrt{3}x}{2\sqrt{3}y - 4x} \]


Step 3: Substitute the coordinates \( \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \)
\[ \frac{dy}{dx} = \frac{4(\frac{\sqrt{3}}{2}) - 2\sqrt{3}(\frac{1}{2})}{2\sqrt{3}(\frac{\sqrt{3}}{2}) - 4(\frac{1}{2})} \]
Simplify numerator: \( 2\sqrt{3} - \sqrt{3} = \sqrt{3} \)
Simplify denominator: \( 3 - 2 = 1 \)

\[ \frac{dy}{dx} = \frac{\sqrt{3}}{1} = \sqrt{3} \] Quick Tip: You can often plug in the coordinates right after differentiation instead of rearranging to save time.
Ensure you apply the chain rule correctly to every term containing \( y \).

Concept:

Double Angle Formulas: \( 1 + \cos 2x = 2 \cos^2 x \).
Relation between Sine and Cosine: \( \cos x = \sin \left( \frac{\pi}{2} - x \right) \).



Step 1: Substitute the identity for \( 1 + \cos 2x \)
\[ \sqrt{\frac{1 + \cos 2x}{2}} = \sqrt{\frac{2 \cos^2 x}{2}} \] \[ = \sqrt{\cos^2 x} = |\cos x| \]


Step 2: Determine the sign based on the given interval

Given \( 0 < x < \frac{\pi}{2} \), \( x \) is in the first quadrant.
In the first quadrant, \( \cos x \) is positive. \[ |\cos x| = \cos x \]


Step 3: Apply the inverse sine transformation

The expression becomes: \[ \sin^{-1}(\cos x) \]
Convert cosine to sine to match the inverse function: \[ \sin^{-1} \left[ \sin \left( \frac{\pi}{2} - x \right) \right] \]
Check the range: Since \( 0 < x < \frac{\pi}{2} \), then \( 0 < \frac{\pi}{2} - x < \frac{\pi}{2} \).
This is within the principal range of \( \sin^{-1} \).

\[ = \frac{\pi}{2} - x \] Quick Tip: Memorizing \( \sin^{-1}(\cos x) = \frac{\pi}{2} - x \) for small \( x \) saves time in calculus.
Always pay attention to the interval to correctly resolve square roots of squares.

Concept:

Inverse function values:
\( \sin^{-1}(-1) = -\frac{\pi}{2} \).
\( \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3} \).



Step 1: Find the values of inverse functions

Let \( \alpha = \sin^{-1}(-1) \). Since \( \sin(-\frac{\pi}{2}) = -1 \): \[ \alpha = -\frac{\pi}{2} \]


Let \( \beta = \tan^{-1}(-\sqrt{3}) \). Since \( \tan(-\frac{\pi}{3}) = -\sqrt{3} \): \[ \beta = -\frac{\pi}{3} \]


Step 2: Substitute values into the expression
\[ \cos [ \alpha - \beta ] = \cos \left[ -\frac{\pi}{2} - \left( -\frac{\pi}{3} \right) \right] \] \[ = \cos \left( -\frac{\pi}{2} + \frac{\pi}{3} \right) \]


Step 3: Simplify and calculate the result
\[ \cos \left( \frac{-3\pi + 2\pi}{6} \right) = \cos \left( -\frac{\pi}{6} \right) \]
Since \( \cos(-\theta) = \cos \theta \): \[ = \cos \frac{\pi}{6} \] \[ = \frac{\sqrt{3}}{2} \] Quick Tip: Note the signs carefully when subtracting negative angles. \( \sin^{-1} \) and \( \tan^{-1} \) of negative values result in negative angles in the fourth quadrant \( (-\frac{\pi}{2}, 0) \).

Concept:

Vector area of a triangle: The area of a triangle with vertices \( A, B, C \) is given by:
\[ Area = \frac{1}{2} | \vec{AB} \times \vec{AC} | \]
Position vectors: \( \vec{AB} = P.V. of B - P.V. of A \).



Step 1: Determine the vectors \( \vec{AB} \) and \( \vec{AC} \)
\[ \vec{AB} = (2 - 1)\hat{i} + (-1 - 2)\hat{j} + (4 - 3)\hat{k} = \hat{i} - 3\hat{j} + \hat{k} \] \[ \vec{AC} = (4 - 1)\hat{i} + (5 - 2)\hat{j} + (-1 - 3)\hat{k} = 3\hat{i} + 3\hat{j} - 4\hat{k} \]


Step 2: Calculate the cross product \( \vec{AB} \times \vec{AC} \)
\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 1 \\ 3 & 3 & -4 \end{vmatrix} \] \[ = \hat{i} [(-3)(-4) - (3)(1)] - \hat{j} [(1)(-4) - (3)(1)] + \hat{k} [(1)(3) - (3)(-3)] \] \[ = \hat{i} [12 - 3] - \hat{j} [-4 - 3] + \hat{k} [3 + 9] \] \[ = 9\hat{i} + 7\hat{j} + 12\hat{k} \]


Step 3: Find the magnitude and calculate area
\[ | \vec{AB} \times \vec{AC} | = \sqrt{9^2 + 7^2 + 12^2} \] \[ = \sqrt{81 + 49 + 144} = \sqrt{274} \]

\[ Area = \frac{1}{2} \sqrt{274} sq. units \] Quick Tip: You can use any two adjacent sides (e.g., \( \vec{BA} \) and \( \vec{BC} \)) to get the same area.
Double-check your determinant calculation; cross-products are common sources of small errors.

Concept:

Parallelogram Law: If \( \vec{a} \) and \( \vec{b} \) are adjacent sides, then the diagonals are \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \).



Step 1: Find the first diagonal vector \( \vec{d_1} \)
\[ \vec{d_1} = \vec{a} + \vec{b} \] \[ = (3\hat{i} - 2\hat{j} + 2\hat{k}) + (\hat{i} + 0\hat{j} + 2\hat{k}) \] \[ = 4\hat{i} - 2\hat{j} + 4\hat{k} \]


Step 2: Find the second diagonal vector \( \vec{d_2} \)
\[ \vec{d_2} = \vec{a} - \vec{b} \] \[ = (3\hat{i} - 2\hat{j} + 2\hat{k}) - (\hat{i} + 0\hat{j} + 2\hat{k}) \] \[ = 2\hat{i} - 2\hat{j} + 0\hat{k} = 2\hat{i} - 2\hat{j} \]


Step 3: Calculate the lengths of the diagonals

Length of \( \vec{d_1} \): \[ | \vec{d_1} | = \sqrt{4^2 + (-2)^2 + 4^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6 \]


Length of \( \vec{d_2} \): \[ | \vec{d_2} | = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \] Quick Tip: One diagonal is the sum, representing the vector from origin to opposite vertex.
The other diagonal is the difference, representing the vector connecting the endpoints of side vectors.

Concept:

Integration by Parts: \( \int u \cdot dv = uv - \int v \cdot du \).
For logarithmic functions, we take the log term as \( u \) and \( 1 \) as \( dv \).
Substitution method and algebraic adjustment for rational integrals.




Step 1: Apply Integration by Parts

Let \( I = \int_{0}^{1} \log(1 + x^2) \cdot 1 \, dx \).
Taking \( u = \log(1 + x^2) \) and \( dv = dx \), we have \( du = \frac{2x}{1 + x^2}dx \) and \( v = x \). \[ I = \left[ x \log(1 + x^2) \right]_{0}^{1} - \int_{0}^{1} x \cdot \frac{2x}{1 + x^2} \, dx \]



Step 2: Evaluate the boundary terms and simplify the remaining integral

Evaluating the first part: \[ \left[ 1 \cdot \log(1 + 1^2) \right] - \left[ 0 \cdot \log(1 + 0^2) \right] = \log 2 - 0 = \log 2 \]
Now the integral part becomes: \[ -2 \int_{0}^{1} \frac{x^2}{1 + x^2} \, dx \]



Step 3: Solve the integral of the rational function

Use algebraic manipulation: \( \frac{x^2}{1 + x^2} = \frac{x^2 + 1 - 1}{1 + x^2} = 1 - \frac{1}{1 + x^2} \). \[ -2 \int_{0}^{1} \left( 1 - \frac{1}{1 + x^2} \right) \, dx = -2 \left[ x - \tan^{-1}x \right]_{0}^{1} \] \[ = -2 \left[ (1 - \tan^{-1}1) - (0 - \tan^{-1}0) \right] \] \[ = -2 \left[ 1 - \frac{\pi}{4} \right] = -2 + \frac{\pi}{2} \]



Step 4: Combine the results for the final answer
\[ I = \log 2 - 2 + \frac{\pi}{2} \] Quick Tip: Always remember to evaluate limits carefully for the term \( uv \) in integration by parts.
When the numerator and denominator of a rational function have the same degree, divide or use the \( +1, -1 \) trick.

Concept:



Law of Total Probability: If \( E_1, E_2 \) are mutually exclusive events, then \( P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2) \).




Step 1: Define the events and find probabilities of bag selection


Let \( E_1 \) be the event that Bag I is chosen (die shows \( \{1, 2\} \)).
Let \( E_2 \) be the event that Bag II is chosen (die shows \( \{3, 4, 5, 6\} \)). \[ P(E_1) = \frac{2}{6} = \frac{1}{3} \] \[ P(E_2) = \frac{4}{6} = \frac{2}{3} \]



Step 2: Find the conditional probabilities of drawing a red ball


Let \( A \) be the event that the ball drawn is red.
Bag I has 3 red and 4 white balls (Total 7). \[ P(A|E_1) = \frac{3}{7} \]
Bag II has 8 red and 6 white balls (Total 14). \[ P(A|E_2) = \frac{8}{14} = \frac{4}{7} \]



Step 3: Calculate the total probability of drawing a red ball


Using the Total Probability Theorem:

\[ P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2) \] \[ P(A) = \left( \frac{1}{3} \right) \left( \frac{3}{7} \right) + \left( \frac{2}{3} \right) \left( \frac{4}{7} \right) \] \[ P(A) = \frac{3}{21} + \frac{8}{21} = \frac{11}{21} \] Quick Tip: Draw a tree diagram to visualize the branching of choices for multi-stage probability problems.
Always check if the fractions can be simplified before adding to make calculations easier.

Concept:

\( P(atleast one) = P(X \cup Y) = a \).
\( P(exactly one) = P(X \cup Y) - P(X \cap Y) = b \).
Complement rule: \( P(X') = 1 - P(X) \).




Step 1: Find the probability of intersection

From the given information: \[ P(X \cup Y) = a \] \[ P(X \cup Y) - P(X \cap Y) = b \]
Substituting \( a \) in the second equation: \[ a - P(X \cap Y) = b \] \[ P(X \cap Y) = a - b \]



Step 2: Express the sum of individual probabilities

Using the addition theorem: \[ P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) \] \[ a = P(X) + P(Y) - (a - b) \] \[ P(X) + P(Y) = a + a - b = 2a - b \]



Step 3: Evaluate the target expression

We need to find \( P(X') + P(Y') \): \[ P(X') + P(Y') = (1 - P(X)) + (1 - P(Y)) \] \[ = 2 - (P(X) + P(Y)) \]
Substitute the expression from Step 2: \[ = 2 - (2a - b) = 2 - 2a + b \]
Hence proved. Quick Tip: "At least one" is the union \( A \cup B \).
"Exactly one" can also be written as \( P(A) + P(B) - 2P(A \cap B) \).

Concept:

Rationalization of the integrand to remove the square root from the numerator.
Integration of square root functions using standard formulas.




Step 1: Rationalize the numerator

Multiply the numerator and denominator by \( \sqrt{x + 2} \): \[ \int \sqrt{\frac{x + 2}{x - 2} \cdot \frac{x + 2}{x + 2}} dx = \int \frac{x + 2}{\sqrt{x^2 - 4}} dx \]



Step 2: Split the integral
\[ I = \int \frac{x}{\sqrt{x^2 - 4}} dx + \int \frac{2}{\sqrt{x^2 - 4}} dx \]
Let these be \( I_1 \) and \( I_2 \).



Step 3: Evaluate each part

For \( I_1 = \int \frac{x}{\sqrt{x^2 - 4}} dx \), put \( x^2 - 4 = t \), so \( 2x \, dx = dt \): \[ I_1 = \frac{1}{2} \int \frac{dt}{\sqrt{t}} = \frac{1}{2} [2\sqrt{t}] = \sqrt{x^2 - 4} \]
For \( I_2 = \int \frac{2}{\sqrt{x^2 - 4}} dx \), use the formula \( \int \frac{dx}{\sqrt{x^2 - a^2}} = \log |x + \sqrt{x^2 - a^2}| \): \[ I_2 = 2 \log |x + \sqrt{x^2 - 4}| \]



Step 4: Combine the terms
\[ I = \sqrt{x^2 - 4} + 2 \log |x + \sqrt{x^2 - 4}| + C \] Quick Tip: Rationalizing the numerator is a powerful technique for integrals of the form \( \sqrt{\frac{L_1}{L_2}} \).
Always check for the direct derivative of the term inside the square root in the numerator.

Concept:



Partial Fractions: Breaking down complex rational functions.
Note: Here we only use partial fractions on \( x^2 \) algebraically, not for integration directly yet.




Step 1: Apply partial fractions algebraically


Let \( x^2 = t \). Then we consider \( \frac{t}{(t + 9)(t + 16)} \). \[ \frac{t}{(t + 9)(t + 16)} = \frac{A}{t + 9} + \frac{B}{t + 16} \] \[ t = A(t + 16) + B(t + 9) \]
Putting \( t = -9 \): \( -9 = A(7) \implies A = -9/7 \).
Putting \( t = -16 \): \( -16 = B(-7) \implies B = 16/7 \).



Step 2: Rewrite the integrand with \( x \)

\[ \frac{x^2}{(x^2 + 9)(x^2 + 16)} = \frac{1}{7} \left[ \frac{16}{x^2 + 16} - \frac{9}{x^2 + 9} \right] \]



Step 3: Integrate each term using standard formulas


Using \( \int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) \):

\[ I = \frac{1}{7} \left[ 16 \cdot \frac{1}{4} \tan^{-1} \left( \frac{x}{4} \right) - 9 \cdot \frac{1}{3} \tan^{-1} \left( \frac{x}{3} \right) \right] \] \[ I = \frac{1}{7} \left[ 4 \tan^{-1} \left( \frac{x}{4} \right) - 3 \tan^{-1} \left( \frac{x}{3} \right) \right] + C \] Quick Tip: When the numerator and denominator contain only \( x^2 \), replace \( x^2 \) with \( t \) to simplify partial fraction working, then substitute \( x^2 \) back before integrating.

Concept:

Definite integral properties for even functions: \( \int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx \).
Half-angle identity: \( 1 + \cos 2x = 2 \cos^2 x \).




Step 1: Evaluate \( I_1 \)

The function \( \frac{1}{1 + \cos 2x} \) is even. \[ I_1 = 2 \int_{0}^{\pi/4} \frac{dx}{2 \cos^2 x} = \int_{0}^{\pi/4} \sec^2 x \, dx \] \[ I_1 = \left[ \tan x \right]_{0}^{\pi/4} = \tan \frac{\pi}{4} - \tan 0 = 1 \]



Step 2: Evaluate \( I_2 \)

The function \( |x| \) is even. \[ I_2 = 2 \int_{0}^{1/2} x \, dx \] \[ I_2 = 2 \left[ \frac{x^2}{2} \right]_{0}^{1/2} = \left[ x^2 \right]_{0}^{1/2} \] \[ I_2 = (1/2)^2 - 0 = \frac{1}{4} \]



Step 3: Verify the given relationship

Calculate \( I_1 - 4I_2 \): \[ 1 - 4(1/4) = 1 - 1 = 0 \]
Hence shown. Quick Tip: Checking for even/odd functions simplifies integrals over symmetric intervals significantly.
Always simplify the trigonometric denominator first before integrating.

Concept:



Homogeneous Differential Equations: Equations where the total degree of every term is the same.
Method of substitution: Put \( y = vx \).




Step 1: Identify as homogeneous and substitute


Rearranging the equation:

\[ \frac{dy}{dx} = \frac{y^2 - x^2}{2xy} \]
Put \( y = vx \), then \( \frac{dy}{dx} = v + x \frac{dv}{dx} \). \[ v + x \frac{dv}{dx} = \frac{v^2 x^2 - x^2}{2x(vx)} = \frac{v^2 - 1}{2v} \]



Step 2: Separate the variables

\[ x \frac{dv}{dx} = \frac{v^2 - 1}{2v} - v = \frac{v^2 - 1 - 2v^2}{2v} = -\frac{1 + v^2}{2v} \] \[ \frac{2v}{1 + v^2} dv = -\frac{1}{x} dx \]



Step 3: Integrate and substitute back

\[ \int \frac{2v}{1 + v^2} dv = -\int \frac{dx}{x} \] \[ \log(1 + v^2) = -\log x + \log C = \log(\frac{C}{x}) \] \[ 1 + v^2 = \frac{C}{x} \implies 1 + \frac{y^2}{x^2} = \frac{C}{x} \] \[ \frac{x^2 + y^2}{x^2} = \frac{C}{x} \implies x^2 + y^2 = Cx \] Quick Tip: A differential equation \( Mdx + Ndy = 0 \) is homogeneous if \( M \) and \( N \) are of the same degree.
For homogeneous equations, the substitution \( y = vx \) always reduces it to variable separable form.

Concept:

Variable Separable Method: Grouping \( y \) terms with \( dy \) and \( x \) terms with \( dx \).
Particular Solution involves finding the integration constant \( C \) using boundary conditions.




Step 1: Separate the variables
\[ (1 + e^{2x})dy = -(1 + y^2)e^x dx \] \[ \frac{dy}{1 + y^2} = -\frac{e^x}{1 + (e^x)^2} dx \]



Step 2: Integrate both sides
\[ \int \frac{dy}{1 + y^2} = -\int \frac{e^x}{1 + (e^x)^2} dx \]
For the RHS, put \( e^x = t \), so \( e^x dx = dt \). \[ \tan^{-1} y = -\int \frac{dt}{1 + t^2} = -\tan^{-1} t + C \] \[ \tan^{-1} y = -\tan^{-1} e^x + C \implies \tan^{-1} y + \tan^{-1} e^x = C \]



Step 3: Use boundary condition to find \( C \)

Given \( y(1) = 0 \), substitute \( x = 1, y = 0 \): \[ \tan^{-1} 0 + \tan^{-1} e^1 = C \] \[ 0 + \tan^{-1} e = C \implies C = \tan^{-1} e \]



Step 4: Write the particular solution
\[ \tan^{-1} y + \tan^{-1} e^x = \tan^{-1} e \] Quick Tip: Always check if simple variable separation is possible before attempting more complex methods.
Ensure substitution is correctly substituted back before applying initial conditions.

Concept:

Graphical Method: Represent the linear inequalities as half-planes on a coordinate system and find the intersection to determine the feasible region.
Corner Point Theorem: The optimal value of the objective function occurs at one of the vertices of the feasible region.



Step 1: Plot the boundary lines and determine the feasible region

Convert inequalities to equations to find boundary lines:

Line 1: \( x + y = 7 \). Intersects axes at \( (7, 0) \) and \( (0, 7) \). The region \( x + y \leq 7 \) is towards the origin.
Line 2: \( 2x - 3y + 6 = 0 \implies 3y = 2x + 6 \implies y = \frac{2}{3}x + 2 \). Intersects axes at \( (0, 2) \) and \( (-3, 0) \). Testing origin \( (0, 0) \): \( 6 \geq 0 \) (True), so the region is towards the origin side.
Non-negative constraints: \( x \geq 0, y \geq 0 \) restrict the region to the first quadrant.







Step 2: Identify corner points

Find the intersection of \( x + y = 7 \) and \( 2x - 3y = -6 \): \[ x = 7 - y \implies 2(7 - y) - 3y = -6 \implies 14 - 5y = -6 \implies 5y = 20 \implies y = 4, x = 3 \]
The corner points of the feasible region are \( O(0, 0), A(7, 0), B(3, 4), \) and \( C(0, 2) \).


Step 3: Evaluate \( Z = 13x - 15y \) at corner points


At \( O(0, 0) \): \( Z = 13(0) - 15(0) = 0 \)
At \( A(7, 0) \): \( Z = 13(7) - 15(0) = 91 \)
At \( B(3, 4) \): \( Z = 13(3) - 15(4) = 39 - 60 = -21 \)
At \( C(0, 2) \): \( Z = 13(0) - 15(2) = -30 \)



The minimum value of \( Z \) is \( -30 \) at the point \( (0, 2) \). Quick Tip: For minimization, look for the vertex furthest in the direction opposite to the gradient of \( Z \).
Always test the origin \( (0,0) \) in inequalities to quickly determine which side of the line to shade.

Concept:

Direction Ratios (DRs): DRs of line through \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) are \( (x_2-x_1, y_2-y_1, z_2-z_1) \).
Parallel Lines: Two lines are parallel if their DRs are proportional.
Distance between parallel lines: \( d = \frac{|\vec{CP} \times \vec{d}|}{|\vec{d}|} \), where \( P \) is on line \( AB \), \( C \) is on line \( CD \), and \( \vec{d} \) is the direction vector.



Step 1: Determine the direction ratios of both lines

For line \( AB \): \( DRs = (2 - 0, 3 - 4, -1 - 1) = (2, -1, -2) \).

For line \( CD \): \( DRs = (2 - 4, 6 - 5, 2 - 0) = (-2, 1, 2) \).


Comparing the DRs: \[ \frac{2}{-2} = \frac{-1}{1} = \frac{-2}{2} = -1 \]
Since the DRs are proportional, the lines \( AB \) and \( CD \) are parallel.


Step 2: Set up vector quantities for distance calculation

Let point \( P = A(0, 4, 1) \) on line \( AB \).

Let point \( C = (4, 5, 0) \) on line \( CD \).

Vector \( \vec{CP} = (0 - 4)\hat{i} + (4 - 5)\hat{j} + (1 - 0)\hat{k} = -4\hat{i} - \hat{j} + \hat{k} \).

Direction vector of line \( CD \), \( \vec{b} = -2\hat{i} + \hat{j} + 2\hat{k} \).


Step 3: Calculate the distance

Cross product \( \vec{CP} \times \vec{b} \): \[ \vec{CP} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & -1 & 1 \\ -2 & 1 & 2 \end{vmatrix} = \hat{i}(-2 - 1) - \hat{j}(-8 + 2) + \hat{k}(-4 - 2) = -3\hat{i} + 6\hat{j} - 6\hat{k} \]
Magnitude \( |\vec{CP} \times \vec{b}| = \sqrt{(-3)^2 + 6^2 + (-6)^2} = \sqrt{9 + 36 + 36} = \sqrt{81} = 9 \).

Magnitude \( |\vec{b}| = \sqrt{(-2)^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = 3 \).

\[ Distance = \frac{|\vec{CP} \times \vec{b}|}{|\vec{b}|} = \frac{9}{3} = 3 units \] Quick Tip: Parallel lines have direction ratios \( a, b, c \) and \( ka, kb, kc \).
Distance can also be found using the projection method: \( \sqrt{|\vec{CP}|^2 - (proj_{\vec{b}}\vec{CP})^2} \).

Concept:



Equivalence Relation: A relation is an equivalence relation if it is reflexive, symmetric, and transitive.
Divisibility: \( a \) is divisible by \( p \) if \( a = kp \) for some integer \( k \).



Step 1: Check for Reflexivity


For any \( x \in Z \), \( |x - x| = 0 \).


Since \( 0 = 0 \cdot p \), where \( 0 \) is an integer, \( 0 \) is divisible by \( p \).


Thus, \( (x, x) \in R \). \( R \) is reflexive.



Step 2: Check for Symmetry


Let \( (x, y) \in R \). Then \( |x - y| \) is divisible by \( p \).


Since \( |x - y| = |y - x| \), it follows that \( |y - x| \) is also divisible by \( p \).


Thus, \( (y, x) \in R \). \( R \) is symmetric.



Step 3: Check for Transitivity


Let \( (x, y) \in R \) and \( (y, z) \in R \).


Then \( |x - y| = k_1 p \) and \( |y - z| = k_2 p \) for integers \( k_1, k_2 \geq 0 \).


This implies \( x - y = \pm k_1 p \) and \( y - z = \pm k_2 p \).

Adding the equations: \( x - z = (x - y) + (y - z) = (\pm k_1 \pm k_2) p \).


Let \( K = \pm k_1 \pm k_2 \), which is an integer. Then \( x - z = Kp \).


Taking magnitude: \( |x - z| = |K| p \), which is divisible by \( p \).


Thus, \( (x, z) \in R \). \( R \) is transitive.



Since \( R \) is reflexive, symmetric, and transitive, it is an equivalence relation. Quick Tip: Modulo arithmetic rules: If \( a \equiv b \pmod{p} \) and \( b \equiv c \pmod{p} \), then \( a \equiv c \pmod{p} \).
0 is divisible by every non-zero integer.

Concept:

One-one (Injective): \( f(x_1) = f(x_2) \implies x_1 = x_2 \).
Onto (Surjective): For every \( y \) in the codomain, there exists an \( x \) in the domain such that \( f(x) = y \).



Step 1: Verify if the function is one-one

Let \( f(x_1) = f(x_2) \): \[ \frac{3x_1 + 2}{5x_1 - 3} = \frac{3x_2 + 2}{5x_2 - 3} \]
Cross-multiplying: \[ (3x_1 + 2)(5x_2 - 3) = (3x_2 + 2)(5x_1 - 3) \] \[ 15x_1x_2 - 9x_1 + 10x_2 - 6 = 15x_1x_2 - 9x_2 + 10x_1 - 6 \] \[ -9x_1 + 10x_2 = -9x_2 + 10x_1 \] \[ 19x_2 = 19x_1 \implies x_1 = x_2 \]
Thus, \( f \) is one-one.


Step 2: Verify if the function is onto

Let \( y = f(x) = \frac{3x + 2}{5x - 3} \). We need to express \( x \) in terms of \( y \): \[ y(5x - 3) = 3x + 2 \] \[ 5xy - 3y = 3x + 2 \] \[ 5xy - 3x = 3y + 2 \implies x(5y - 3) = 3y + 2 \] \[ x = \frac{3y + 2}{5y - 3} \]
Since the codomain is \( R - \{3/5\} \), \( y \neq 3/5 \), so \( 5y - 3 \neq 0 \).

For any \( y \in R - \{3/5\} \), \( x \) is a real number.

Also, if \( x = 3/5 \), then \( 3(3/5) + 2 = y(5(3/5) - 3) \implies 1.8 + 2 = y(0) \), which is impossible. So \( x \in R - \{3/5\} \).

Thus, every \( y \) has a pre-image, and the function is onto. Quick Tip: For rational functions \( \frac{ax+b}{cx+d} \), if \( ad - bc \neq 0 \), the function is always one-one.
The horizontal asymptote \( y = a/c \) is always excluded from the range of such functions.

Concept:



Matrix Inverse: \( A^{-1} = \frac{1}{|A|} adj(A) \).
System Solution: \( X = A^{-1}B \).



Step 1: Calculate the determinant \( |A| \)

\[ |A| = 0(-1 - 2) - 2(0 - (-2)) + 1(2 - (-1)) \] \[ |A| = 0 - 2(2) + 1(3) = -4 + 3 = -1 \]
Since \( |A| \neq 0 \), \( A^{-1} \) exists.


Step 2: Find the adjoint of \( A \)


Cofactors:

\( A_{11}=-1, A_{12}=-2, A_{13}=3, A_{21}=-1, A_{22}=-1, A_{23}=2, A_{31}=-3, A_{32}=-2, A_{33}=4 \) \[ adj(A) = \begin{bmatrix} -1 & -2 & 3 \\ -1 & -1 & 2 \\ -3 & -2 & 4 \end{bmatrix}^T = \begin{bmatrix} -1 & -1 & -3 \\ -2 & -1 & -2 \\ 3 & 2 & 4 \end{bmatrix} \] \[ A^{-1} = \frac{1}{-1} \begin{bmatrix} -1 & -1 & -3 \\ -2 & -1 & -2 \\ 3 & 2 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 3 \\ 2 & 1 & 2 \\ -3 & -2 & -4 \end{bmatrix} \]


Step 3: Solve the system


Note the system matrix \( C = \begin{bmatrix} 0 & -2 & 1 \\ 2 & -1 & -1 \\ 1 & -2 & 0 \end{bmatrix} \). This is \( A^T \).

Since \( (A^T)^{-1} = (A^{-1})^T \), we use the transpose of \( A^{-1} \). \[ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 & 2 & -3 \\ 1 & 1 & -2 \\ 3 & 2 & -4 \end{bmatrix} \begin{bmatrix} 7 \\ 8 \\ 10 \end{bmatrix} = \begin{bmatrix} 7 + 16 - 30 \\ 7 + 8 - 20 \\ 21 + 16 - 40 \end{bmatrix} = \begin{bmatrix} -7 \\ -5 \\ -3 \end{bmatrix} \]
Solution: \( x = -7, y = -5, z = -3 \). Quick Tip: Always check if the coefficients of the system match the matrix \( A \) or its transpose \( A^T \).
\( (A^T)^{-1} = (A^{-1})^T \).

Concept:

Singular Matrix: A matrix whose determinant is zero.
Trigonometric Identities: \( \sin 3x = 3\sin x - 4\sin^3 x, \cos 2x = 1 - 2\sin^2 x \).



Step 1: Expand the determinant and set it to zero
\[ 3(8 - 7\cos 2x) - (-1)(-14 + 11\cos 2x) + \sin 3x(-49 + 44) = 0 \] \[ 24 - 21\cos 2x - 14 + 11\cos 2x - 5\sin 3x = 0 \] \[ 10 - 10\cos 2x - 5\sin 3x = 0 \implies 2 - 2\cos 2x - \sin 3x = 0 \]


Step 2: Substitute identities and simplify
\[ 2 - 2(1 - 2\sin^2 x) - (3\sin x - 4\sin^3 x) = 0 \] \[ 2 - 2 + 4\sin^2 x - 3\sin x + 4\sin^3 x = 0 \] \[ 4\sin^3 x + 4\sin^2 x - 3\sin x = 0 \]


Step 3: Solve the equation
\[ \sin x (4\sin^2 x + 4\sin x - 3) = 0 \]

\( \sin x = 0 \implies x = 0 \in [0, \pi/2] \).
\( 4\sin^2 x + 4\sin x - 3 = 0 \implies (2\sin x - 1)(2\sin x + 3) = 0 \).
\( \sin x = 1/2 \implies x = \pi/6 \in [0, \pi/2] \).
\( \sin x = -3/2 \) (Impossible).



The values of \( x \) are \( 0 \) and \( \pi/6 \). Quick Tip: For singular matrices, \( |A| = 0 \).
When solving trigonometric equations, always filter solutions based on the given interval \( [0, \pi/2] \).

Concept:



Parametric Differentiation: \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \).
Second Order Parametric Differentiation: \( \frac{d^2y}{dx^2} = \frac{d}{dt} \left( \frac{dy}{dx} \right) \cdot \frac{dt}{dx} \).



Step 1: Find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \)

\[ \frac{dx}{dt} = a(\cos t - (\cos t - t \sin t)) = a t \sin t \]

\[ \frac{dy}{dt} = b(-\sin t + (\sin t + t \cos t)) = b t \cos t \]



Step 2: Calculate \( \frac{dy}{dx} \)

\[ \frac{dy}{dx} = \frac{b t \cos t}{a t \sin t} = \frac{b}{a} \cot t \]


Step 3: Find the second derivative

\[ \frac{d^2y}{dx^2} = \frac{d}{dt} \left( \frac{b}{a} \cot t \right) \frac{dt}{dx} \] \[ \frac{d^2y}{dx^2} = -\frac{b}{a} \csc^2 t \cdot \frac{1}{at \sin t} \] \[ \frac{d^2y}{dx^2} = -\frac{b}{a^2 t \sin^3 t} \] Quick Tip: Don't forget the \( \frac{dt}{dx} \) factor when finding the second derivative of parametric functions.
Product rule is essential when differentiating terms like \( t \cos t \) or \( t \sin t \).

Concept:

A circle with its center at the origin \( (0, 0) \) and radius \( r \) is represented by the equation \( x^2 + y^2 = r^2 \).
Concentric circles are circles that share the same center but have different radii.



Step 1: Identify the center and radius for the boundary circle \( C_1 \)

The equation for \( C_1 \) is: \[ x^2 + y^2 = 64 \]
Comparing this with the standard form \( x^2 + y^2 = r^2 \): \[ r_1^2 = 64 \implies r_1 = 8 units \]
The center is at the origin \( (0, 0) \).


Step 2: Identify the center and radius for the pond circle \( C_2 \)

The equation for \( C_2 \) is: \[ x^2 + y^2 = 4 \]
Comparing this with the standard form: \[ r_2^2 = 4 \implies r_2 = 2 units \]
The center is also at the origin \( (0, 0) \).


Step 3: Draw the diagram showing concentric circles

Since both circles have the same center, they are concentric.
The pond \( C_2 \) is located inside the roundabout boundary \( C_1 \).



Quick Tip: For any circle \( x^2 + y^2 = a^2 \), the x-intercepts are \( (\pm a, 0) \) and the y-intercepts are \( (0, \pm a) \).
Concentric circles never intersect because their radii are different while their center is the same.

Concept:

To express \( y \) as a function of \( x \), we isolate the \( y \) variable on one side of the equation.
A circle \( x^2 + y^2 = r^2 \) can be split into two functions: \( y = \sqrt{r^2 - x^2} \) (upper semicircle) and \( y = -\sqrt{r^2 - x^2} \) (lower semicircle).



Step 1: Derive the function for the boundary circle \( C_1 \)

The equation for \( C_1 \) is: \[ x^2 + y^2 = 64 \]
Subtract \( x^2 \) from both sides: \[ y^2 = 64 - x^2 \]
Taking the square root: \[ y = \pm \sqrt{64 - x^2} \]
Considering the upper part of the circle to represent it as a function: \[ f_1(x) = \sqrt{64 - x^2} \]


Step 2: Derive the function for the pond circle \( C_2 \)

The equation for \( C_2 \) is: \[ x^2 + y^2 = 4 \]
Subtract \( x^2 \) from both sides: \[ y^2 = 4 - x^2 \]
Taking the square root: \[ y = \pm \sqrt{4 - x^2} \]
Considering the upper part: \[ f_2(x) = \sqrt{4 - x^2} \] Quick Tip: Mathematically, a full circle is a relation and fails the vertical line test. We use the positive root to represent the upper boundary as a function.
The domain for \( C_1 \) is \( [-8, 8] \) and for \( C_2 \) is \( [-2, 2] \).

Concept:



The total area of a circle \( x^2 + y^2 = r^2 \) is four times the area in the first quadrant.
Area in the first quadrant = \( \int_{0}^{r} y \, dx \).
Standard Integral: \( \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C \).



Step 1: Set up the integral for the total area of circle \( C_1 \)


The radius is \( r = 8 \). The total area \( A_1 \) is given by: \[ A_1 = 4 \int_{0}^{8} \sqrt{64 - x^2} \, dx \]


Step 2: Apply the integration formula


Using the standard formula with \( a = 8 \): \[ A_1 = 4 \left[ \frac{x}{2}\sqrt{64 - x^2} + \frac{64}{2}\sin^{-1}\left(\frac{x}{8}\right) \right]_{0}^{8} \]


Step 3: Evaluate the definite integral with limits


Substituting the upper limit \( x = 8 \):

\[ \left( \frac{8}{2}\sqrt{64 - 64} + 32\sin^{-1}(1) \right) = 0 + 32 \cdot \frac{\pi}{2} = 16\pi \]


Substituting the lower limit \( x = 0 \):

\[ \left( \frac{0}{2}\sqrt{64 - 0} + 32\sin^{-1}(0) \right) = 0 + 0 = 0 \]


Step 4: Final multiplication
\[ A_1 = 4 \times (16\pi - 0) \] \[ A_1 = 64\pi sq. units \] Quick Tip: Always verify integration results for circles using the geometric formula \( \pi r^2 \).
The factor 4 accounts for symmetry across the x and y axes.

Concept:

The pond is bounded by \( C_2 : x^2 + y^2 = 4 \).
It is a circle with center at origin and radius \( r = 2 \).
Area = \( 4 \times \int_{0}^{r} \sqrt{r^2 - x^2} \, dx \).



Step 1: Set up the integral for circle \( C_2 \)

With radius \( r = 2 \): \[ A_2 = 4 \int_{0}^{2} \sqrt{4 - x^2} \, dx \]


Step 2: Apply the integration formula

Using the standard formula with \( a = 2 \): \[ A_2 = 4 \left[ \frac{x}{2}\sqrt{4 - x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right) \right]_{0}^{2} \]


Step 3: Evaluate the definite integral

Substituting the upper limit \( x = 2 \): \[ \left( \frac{2}{2}\sqrt{4 - 4} + 2\sin^{-1}(1) \right) = 0 + 2 \cdot \frac{\pi}{2} = \pi \]
Substituting the lower limit \( x = 0 \): \[ \left( 0 + 2\sin^{-1}(0) \right) = 0 \]


Step 4: Final multiplication
\[ A_2 = 4 \times (\pi - 0) \] \[ A_2 = 4\pi sq. units \] Quick Tip: Integration is the standard tool for finding areas of regions with curved boundaries.
For a circle of radius \( r \), the integration result \( \int_{0}^{r} \sqrt{r^2 - x^2} \, dx \) always yields \( \frac{\pi r^2}{4} \).

Concept:



The rate of decrease in the number of subscribers is constant with respect to the increase in fee.
If \( k \) subscribers discontinue for every Rs 1 increase, then for an increase of Rs \( x \), \( k \cdot x \) subscribers will discontinue.



Step 1: Identify the rate of discontinuation


According to the problem, for every Rs 1 increase in the annual subscription fee, the number of subscribers that discontinue is 10.

\[ Rate of discontinuation = 10 subscribers / Rs 1 increase \]



Step 2: Calculate the total discontinued subscribers for Rs x increase


If the fee is increased by Rs \( x \), the total number of subscribers who will discontinue is given by the product of the rate and the amount of increase.

\[ Number of discontinued subscribers = 10 \times x = 10x \] Quick Tip: Always establish the relationship between the independent variable (fee increase) and the dependent variable (number of subscribers).
Linear relationships are of the form \( y = mx + c \). Here, the change in \( y \) is \( mx \).

Concept:

Total Revenue (\( R \)) = (Number of Subscribers) \(\times\) (Subscription Fee per subscriber).
New Fee = Original Fee + Increase (\( x \)).
New Number of Subscribers = Original Number - Discontinued Subscribers.



Step 1: Determine the new fee per subscriber

Original annual subscription fee = Rs 300.
Increase in fee = Rs \( x \). \[ New Fee = (300 + x) \]


Step 2: Determine the new number of subscribers

Original number of subscribers = 5000.
Number of subscribers who discontinue = \( 10x \). \[ Remaining Subscribers = (5000 - 10x) \]


Step 3: Formulate the revenue function \( R(x) \)
\[ R(x) = New Fee \times Remaining Subscribers \] \[ R(x) = (300 + x)(5000 - 10x) \] \[ R(x) = 1500000 - 3000x + 5000x - 10x^2 \] \[ R(x) = -10x^2 + 2000x + 1500000 \] Quick Tip: Revenue functions in such problems are usually quadratic.
Ensure units are consistent; here both the original fee and the increase are in Rupees.

Concept:

To maximize a function, find its first derivative and set it to zero (\( R'(x) = 0 \)) to find critical points.
Use the second derivative test (\( R''(x) < 0 \)) to confirm a local maximum.



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

From the previous part: \[ R(x) = -10x^2 + 2000x + 1500000 \]
Differentiating with respect to \( x \): \[ R'(x) = \frac{d}{dx}(-10x^2 + 2000x + 1500000) \] \[ R'(x) = -20x + 2000 \]


Step 2: Solve for the critical point

Set \( R'(x) = 0 \) to find the value of \( x \): \[ -20x + 2000 = 0 \] \[ 20x = 2000 \] \[ x = \frac{2000}{20} = 100 \]


Step 3: Verify the maximum using the second derivative

Differentiating \( R'(x) \) again: \[ R''(x) = \frac{d}{dx}(-20x + 2000) = -20 \]
Since \( R''(100) = -20 \), which is less than 0, the function \( R(x) \) is maximized at \( x = 100 \). Quick Tip: For a quadratic function \( ax^2 + bx + c \), the maximum/minimum occurs at \( x = -b/(2a) \).
Here, \( x = -2000 / (2 \times -10) = 100 \).

Concept:



A function is increasing on an interval if its first derivative is positive (\( R'(x) > 0 \)).
A function is decreasing on an interval if its first derivative is negative (\( R'(x) < 0 \)).



Step 1: Identify the first derivative and the boundary of the domain


We found \( R'(x) = -20x + 2000 \).


The total subscribers cannot be negative: \( 5000 - 10x \geq 0 \implies x \leq 500 \).


Thus, the practical domain for the increase \( x \) is \( [0, 500] \). The question specifies a broader range (0, 5000), but we analyze based on the derivative sign.



Step 2: Find the interval where \( R(x) \) is increasing


Set \( R'(x) > 0 \):

\[ -20x + 2000 > 0 \] \[ 2000 > 20x \implies x < 100 \]
So, \( R(x) \) is increasing in the interval \( (0, 100) \).



Step 3: Find the interval where \( R(x) \) is decreasing


Set \( R'(x) < 0 \):

\[ -20x + 2000 < 0 \] \[ 2000 < 20x \implies x > 100 \]


So, \( R(x) \) is decreasing in the interval \( (100, 5000) \). Note: realistically, it stops at \( x=500 \). Quick Tip: The turning point of the parabola marks the boundary between the increasing and decreasing intervals.
Since the parabola opens downwards (\( a = -10 \)), it increases before the peak and decreases after.

Concept:



The possible outcomes for a single ticket holder correspond to the various prize tiers available or winning no prize at all.



Step 1: List the available prize amounts


From the problem description, the different prize amounts are:



First Prize: Rs 3,00,000
Second Prize: Rs 2,00,000
Third Prize: Rs 50,000



Step 2: Account for the possibility of not winning


Since there are only \( 1 + 2 + 3 = 6 \) winning tickets out of 1,00,000, it is possible for a person to not win any prize. In this case, the amount won is Rs 0.



Step 3: Combine all possible amounts


The set of possible winning amounts is:

\[ \{Rs 3,00,000, Rs 2,00,000, Rs 50,000, Rs 0\} \] Quick Tip: In discrete probability distributions, always include the zero outcome if the number of prizes is less than the total number of entries.

Concept:



"At least Rs 2,00,000" means the person wins either a first prize (Rs 3,00,000) or a second prize (Rs 2,00,000).
Probability \( P(E) = \frac{Number of favorable outcomes}{Total number of outcomes} \).



Step 1: Identify the total number of outcomes


The total number of tickets sold is:

\[ N = 1,00,000 \]


Step 2: Identify the number of favorable outcomes


Number of first prize tickets = 1.
Number of second prize tickets = 2.
Total winning tickets for at least Rs 2,00,000: \[ n = 1 + 2 = 3 \]


Step 3: Calculate the probability

\[ P(at least Rs 2,00,000) = \frac{n}{N} \] \[ P = \frac{3}{1,00,000} \] \[ P = 0.00003 \] Quick Tip: When a question says "at least", sum the counts of all tiers meeting or exceeding that value.
Ensure you count all individual prizes (e.g., two second prizes count as 2 successful outcomes).

Concept:



The event "not winning any amount" is the complement of "winning any prize".
\( P(Not Win) = 1 - P(Win any prize) \).



Step 1: Calculate the total number of winning tickets


Number of 1st prizes = 1.
Number of 2nd prizes = 2.
Number of 3rd prizes = 3.
Total winning tickets: \[ Win = 1 + 2 + 3 = 6 \]


Step 2: Calculate the number of non-winning tickets


Total tickets = 1,00,000.
Non-winning tickets: \[ Non-Win = 1,00,000 - 6 = 99,994 \]


Step 3: Calculate the probability

\[ P(Not Win) = \frac{99,994}{1,00,000} \] \[ P = 0.99994 \]
Alternatively, using the complement: \[ P = 1 - \frac{6}{1,00,000} = 1 - 0.00006 = 0.99994 \] Quick Tip: Using the complement rule \( 1 - P(E) \) is often safer than counting large numbers of non-favorable outcomes.

Concept:

This involves two independent events, \( A \) (winning any prize in the first jackpot) and \( B \) (winning the second jackpot).
Probability of exactly one = \( P(A \cap B') + P(A' \cap B) \).
Since events are independent, \( P(A \cap B') = P(A) \times P(B') \).



Step 1: Find individual probabilities of winning

For Jackpot 1 (event \( A \)): Total winning tickets = 6. \[ P(A) = \frac{6}{1,00,000} \]
For Jackpot 2 (event \( B \)): \[ P(B) = \frac{1}{1,00,000} \]


Step 2: Find probabilities of not winning
\[ P(A') = 1 - \frac{6}{1,00,000} = \frac{99,994}{1,00,000} \] \[ P(B') = 1 - \frac{1}{1,00,000} = \frac{99,999}{1,00,000} \]


Step 3: Apply the formula for exactly one success
\[ P(Exactly one) = P(A)P(B') + P(A')P(B) \] \[ P = \left( \frac{6}{10^5} \times \frac{99,999}{10^5} \right) + \left( \frac{99,994}{10^5} \times \frac{1}{10^5} \right) \] \[ P = \frac{599,994 + 99,994}{10^{10}} = \frac{699,988}{10^{10}} \] \[ P \approx \frac{7}{10^5} = 0.00007 (approximately) \] Quick Tip: For very small probabilities, \( P(A)P(B') + P(A')P(B) \) is approximately \( P(A) + P(B) \). Always treat separate lotteries/jackpots as independent events unless specified otherwise.