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

Concept:

The dot product of two vectors \( \vec{u} \) and \( \vec{v} \) follows the distributive law: \( (\vec{u} + \vec{v}) \cdot (\vec{u} - \vec{v}) = |\vec{u}|^2 - |\vec{v}|^2 \).
Magnitude of a vector is always a non-negative scalar value.



Step 1: Expand the given dot product equation

Given: \[ (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 198 \]
Using the property \( (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = \vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} - \vec{b} \cdot \vec{b} \): \[ |\vec{a}|^2 - |\vec{b}|^2 = 198 \quad \dots (i) \]


Step 2: Substitute the relationship between the magnitudes

It is given that \( |\vec{a}| = 10|\vec{b}| \).
Substitute this into equation (i): \[ (10|\vec{b}|)^2 - |\vec{b}|^2 = 198 \] \[ 100|\vec{b}|^2 - |\vec{b}|^2 = 198 \]


Step 3: Solve for \( |\vec{b}| \)
\[ 99|\vec{b}|^2 = 198 \]
Dividing both sides by 99: \[ |\vec{b}|^2 = \frac{198}{99} = 2 \]
Taking the square root (magnitude is positive): \[ |\vec{b}| = \sqrt{2} \] Quick Tip: Remember the algebraic identity \( (x+y)(x-y) = x^2 - y^2 \) directly applies to the dot product of the sum and difference of two vectors.
Always check if the question asks for \( |\vec{a}| \) or \( |\vec{b}| \) before selecting the final option.

Concept:



Direction cosines represent the unit vector along a line.
The cosine of the angle between two lines with direction cosines \( (l_1, m_1, n_1) \) and \( (l_2, m_2, n_2) \) is the dot product of their unit vectors.



Step 1: Identify the standard formula for the angle between lines


The angle \( \theta \) between two lines having direction cosines \( (l_1, m_1, n_1) \) and \( (l_2, m_2, n_2) \) is given by:

\[ \cos \theta = l_1l_2 + m_1m_2 + n_1n_2 \]



Step 2: Apply the condition for the acute angle


Since the question specifies that \( \theta \) is an acute angle, \( \cos \theta \) must be non-negative.


Therefore, we take the absolute value of the expression to ensure the result corresponds to the acute angle:

\[ \cos \theta = | l_1l_2 + m_1m_2 + n_1n_2 | \] Quick Tip: For lines, the angle is usually considered acute (\( 0 \le \theta \le \pi/2 \)), which is why the modulus sign is used in the formula.
If the direction ratios were given instead, the formula would be \( \cos \theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}} \).

Concept:

A line perpendicular to two lines is parallel to the cross product of the vectors along those two lines.
If two lines have direction ratios \( (a_1, b_1, c_1) \) and \( (a_2, b_2, c_2) \), the perpendicular line has direction ratios proportional to \( (b_1c_2 - b_2c_1, c_1a_2 - c_2a_1, a_1b_2 - a_2b_1) \).



Step 1: Represent the direction ratios as vectors

Let the vectors along the given lines be: \[ \vec{v_1} = 1\hat{i} + 0\hat{j} + 0\hat{k} = \hat{i} \] \[ \vec{v_2} = 0\hat{i} - 1\hat{j} + 0\hat{k} = -\hat{j} \]


Step 2: Find the cross product of the vectors

The vector \( \vec{v} \) perpendicular to both \( \vec{v_1} \) and \( \vec{v_2} \) is given by: \[ \vec{v} = \vec{v_1} \times \vec{v_2} = \hat{i} \times (-\hat{j}) \]
Using the cyclic cross product rules (\( \hat{i} \times \hat{j} = \hat{k} \)): \[ \vec{v} = -(\hat{i} \times \hat{j}) = -\hat{k} \]


Step 3: Determine the direction ratios

The vector \( -\hat{k} \) can be written in components as: \[ 0\hat{i} + 0\hat{j} - 1\hat{k} \]
Thus, the direction ratios are \( <0, 0, -1> \). Quick Tip: The direction ratios \( <1, 0, 0> \) and \( <0, -1, 0> \) correspond to the x-axis and negative y-axis respectively.
A line perpendicular to both the x and y axes must lie along the z-axis, which has DRs like \( <0, 0, 1> \) or \( <0, 0, -1> \).

Concept:

For independent events \( E \) and \( F \), \( P(E \cap F) = P(E) \cdot P(F) \).
For independent events, \( P(E|F) = P(E) \) and \( P(F|E) = P(F) \).
Addition theorem: \( P(E \cup F) = P(E) + P(F) - P(E \cap F) \).



Step 1: Find \( P(F) \) using the union formula

Since \( E \) and \( F \) are independent: \[ P(E \cup F) = P(E) + P(F) - P(E)P(F) \]
Substitute the given values: \[ \frac{1}{2} = \frac{3}{10} + P(F) - \frac{3}{10}P(F) \] \[ \frac{1}{2} - \frac{3}{10} = \left( 1 - \frac{3}{10} \right)P(F) \] \[ \frac{5 - 3}{10} = \frac{7}{10}P(F) \] \[ \frac{2}{10} = \frac{7}{10}P(F) \implies P(F) = \frac{2}{7} \]


Step 2: Calculate the required expression using independence properties

Given \( E \) and \( F \) are independent: \[ P(E|F) = P(E) = \frac{3}{10} \] \[ P(F|E) = P(F) = \frac{2}{7} \]


Step 3: Evaluate the difference
\[ P(E|F) - P(F|E) = \frac{3}{10} - \frac{2}{7} \]
Taking the LCM of 10 and 7 which is 70: \[ \frac{3(7) - 2(10)}{70} = \frac{21 - 20}{70} = \frac{1}{70} \] Quick Tip: Always use the property \( P(A|B) = P(A) \) for independent events to save time.
The union of independent events can also be expressed as \( P(E \cup F) = 1 - P(E^c)P(F^c) \).

Concept:



The domain of the function \( f(y) = \cos^{-1} y \) is the set of all real numbers \( y \) such that \( -1 \le y \le 1 \).



Step 1: Set up the inequality for the domain


For the expression \( \cos^{-1} (4x + 1) \) to be defined, the argument must lie within the range of the cosine function:

\[ -1 \le 4x + 1 \le 1 \]


Step 2: Solve the double inequality for \( x \)


Subtract 1 from all parts of the inequality:

\[ -1 - 1 \le 4x \le 1 - 1 \] \[ -2 \le 4x \le 0 \]
Now, divide the entire inequality by 4: \[ -\frac{2}{4} \le x \le \frac{0}{4} \] \[ -\frac{1}{2} \le x \le 0 \]


Step 3: State the final domain in interval notation


The range of values of \( x \) for which the function is defined is \( [-\frac{1}{2}, 0] \). Quick Tip: Domain problems for inverse trigonometric functions always start by setting the internal expression between \(-1\) and \(1\).
Watch out for the signs when dividing by a negative number; however, here the divisor 4 is positive, so the inequality signs remain unchanged.

Concept:

For an invertible matrix \( A \), if we are given a polynomial equation like \( A^2 + pA + qI = 0 \), we can find the expression for \( A^{-1} \) by multiplying the entire equation by \( A^{-1} \).
Multiplying \( A \) by \( A^{-1} \) results in the Identity matrix \( I \).



Step 1: Rearrange the given matrix equation

The given equation is: \[ A^2 = 4A + 3I \]
We can rewrite this as: \[ A^2 - 4A = 3I \]


Step 2: Multiply the equation by \( A^{-1} \)

Since the identity matrix term is present, we assume \( A \) is invertible. Multiplying both sides by \( A^{-1} \): \[ A^{-1}(A^2 - 4A) = A^{-1}(3I) \] \[ A^{-1}A \cdot A - 4(A^{-1}A) = 3(A^{-1}I) \] \[ I \cdot A - 4I = 3A^{-1} \] \[ A - 4I = 3A^{-1} \]


Step 3: Express \( A^{-1} \) in terms of \( A \) and \( I \)

Dividing the equation by 3: \[ A^{-1} = \frac{1}{3}A - \frac{4}{3}I \]


Step 4: Compare with the given form to find \( x \) and \( y \)

The given form is \( A^{-1} = xA + yI \). Comparing coefficients: \[ x = \frac{1}{3}, \quad y = -\frac{4}{3} \]
Now, calculate \( x + y \): \[ x + y = \frac{1}{3} + \left( -\frac{4}{3} \right) = \frac{1 - 4}{3} = \frac{-3}{3} = -1 \] Quick Tip: To find the inverse using a characteristic equation, isolate the \( I \) term and multiply by \( A^{-1} \).
Be careful with the signs of coefficients when comparing with the target expression.

Concept:

A matrix \( M \) is symmetric if \( M' = M \).
A matrix \( M \) is skew-symmetric if \( M' = -M \).
Properties of Transpose: \( (A \pm B)' = A' \pm B' \) and \( (AB)' = B'A' \).



Step 1: Use the property of skew-symmetric matrices

Since \( A \) and \( B \) are skew-symmetric: \[ A' = -A \] \[ B' = -B \]


Step 2: Let \( P \) be the given matrix and find its transpose

Let \( P = AB' + BA' \).
Taking the transpose of both sides: \[ P' = (AB' + BA')' \]
Using the property \( (X+Y)' = X' + Y' \): \[ P' = (AB')' + (BA')' \]


Step 3: Apply the reversal law of transpose

Using \( (XY)' = Y'X' \): \[ P' = (B')'A' + (A')'B' \]
Since \( (M')' = M \): \[ P' = BA' + AB' \]


Step 4: Determine the nature of the matrix

We have \( P' = BA' + AB' \).
Since addition is commutative (\( X + Y = Y + X \)): \[ P' = AB' + BA' \] \[ P' = P \]
Since the transpose of the matrix is equal to the matrix itself, it is a symmetric matrix. Quick Tip: If you need to check symmetry, always compute the transpose of the whole expression.
Note that for skew-symmetric matrices, substituting \( A' = -A \) and \( B' = -B \) directly into \( AB' + BA' \) gives \( A(-B) + B(-A) = -AB - BA = -(AB + BA) \). Transposing this gives the same result.

Concept:

For matrix multiplication \( A_{m \times n} \cdot B_{p \times q} \) to be defined, the number of columns in \( A \) must equal the number of rows in \( B \), i.e., \( n = p \).
The resulting matrix will have an order of \( m \times q \).



Step 1: Determine the order of the known matrices

Let the matrix on the left be \( A = [2 \quad 1] \). It has 1 row and 2 columns.
Order of \( A = 1 \times 2 \).
The matrix on the right is \( C = [3 \quad 4 \quad 5] \). It has 1 row and 3 columns.
Order of \( C = 1 \times 3 \).


Step 2: Apply the multiplication rule to find the number of rows in X

Let the order of matrix \( X \) be \( m \times n \).
Since the product \( A \cdot X \) is defined, the number of columns in \( A \) must equal the number of rows in \( X \). \[ 2 = m \]


Step 3: Apply the product order rule to find the number of columns in X

The order of the product \( AX \) is given by (rows of \( A \)) \( \times \) (columns of \( X \)).
The order of \( AX \) is \( 1 \times n \).
Comparing this with the order of the result matrix \( C \) (\( 1 \times 3 \)): \[ n = 3 \]


Step 4: State the final order

The order of matrix \( X \) is \( m \times n = 2 \times 3 \). Quick Tip: Always write down the dimensions in the form \( (1 \times 2) \times (m \times n) = (1 \times 3) \).
Match the "inner" numbers for existence and "outer" numbers for the final result.

Concept:

The binomial expansion applies to matrices if they commute (e.g., \( I \) and \( A \)).
\( (I - A)^3 = I^3 - 3I^2A + 3IA^2 - A^3 \).
Idempotent matrix property: If \( A^2 = A \), then \( A^k = A \) for all integers \( k \ge 1 \).



Step 1: Expand the cubic expression

Using the algebraic identity for \( (a-b)^3 \): \[ (I - A)^3 = I^3 - 3I^2A + 3IA^2 - A^3 \]
Since \( I^k = I \) and \( IA = AI = A \): \[ (I - A)^3 = I - 3A + 3A^2 - A^3 \]


Step 2: Simplify the higher powers of A

Given \( A^2 = A \).
Then: \[ A^3 = A^2 \cdot A = A \cdot A = A^2 = A \]
Substituting \( A^2 = A \) and \( A^3 = A \) into the expansion: \[ (I - A)^3 = I - 3A + 3A - A \]


Step 3: Calculate the final simplified matrix expression
\[ (I - A)^3 = I + (-3A + 3A) - A \] \[ (I - A)^3 = I - A \]


Step 4: Compare with the given form to find \( x \)

The given equation is \( (I - A)^3 = xA + I \).
Equating the simplified expression to the given form: \[ I - A = xA + I \]
Subtracting \( I \) from both sides: \[ -A = xA \]
By comparing the coefficients of \( A \): \[ x = -1 \] Quick Tip: For idempotent matrices where \( A^2 = A \), any power of the matrix simplifies back to the matrix itself.
Be careful with the signs in the binomial expansion of \( (I - A)^3 \); the signs alternate: \( +, -, +, - \).

Concept:



The fundamental property of adjoint: \( A(adj A) = |A|I \).
Determinant of a scalar multiple: If \( A \) is a square matrix of order \( n \), then \( |kA| = k^n |A| \).



Step 1: Find the determinant \( |A| \) from the given matrix product


The given product is: \[ A(adj A) = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix} \]


Factoring out 3 from the scalar matrix:

\[ A(adj A) = 3 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = 3I \]
Using the property \( A(adj A) = |A|I \), we find: \[ |A| = 3 \]


Step 2: Identify the order of the matrix


The identity matrix \( I \) in the result is of order \( 3 \times 3 \).
Thus, the order of matrix \( A \) is \( n = 3 \).



Step 3: Calculate \( |2A| \)


Using the property \( |kA| = k^n |A| \), where \( k = 2 \), \( n = 3 \), and \( |A| = 3 \): \[ |2A| = 2^3 \cdot |A| \] \[ |2A| = 8 \cdot 3 \] \[ |2A| = 24 \] Quick Tip: Remember to raise the scalar to the power of the matrix order when pulling it out of a determinant.
A common mistake is simply multiplying the determinant by the scalar without considering the order.

Concept:

A function \( f(x) \) is continuous at \( x = c \) if the limit exists and equals the value of the function: \( \lim_{x \to c} f(x) = f(c) \).
Product of a function tending to zero and a bounded function results in a limit of zero.



Step 1: Find the value of the function at \( x = 0 \)

The function definition for \( x = 0 \) is \( f(x) = k(x + 1) \). \[ f(0) = k(0 + 1) = k \]


Step 2: Calculate the limit of the function as \( x \to 0 \)

We need to find \( \lim_{x \to 0} x^2 \sin \frac{1}{x} \).
We know that for any real value of \( \frac{1}{x} \), the sine function is bounded: \[ -1 \le \sin \frac{1}{x} \le 1 \]
Multiplying throughout by \( x^2 \) (which is non-negative): \[ -x^2 \le x^2 \sin \frac{1}{x} \le x^2 \]


Step 3: Apply the Squeeze Theorem

As \( x \to 0 \), both \( -x^2 \to 0 \) and \( x^2 \to 0 \).
By Squeeze Theorem: \[ \lim_{x \to 0} x^2 \sin \frac{1}{x} = 0 \]


Step 4: Equate limit and value for continuity

For \( f(x) \) to be continuous at \( x = 0 \): \[ \lim_{x \to 0} f(x) = f(0) \] \[ 0 = k \] Quick Tip: Functions of the form \( x^n \sin(1/x) \) or \( x^n \cos(1/x) \) always have a limit of 0 at \( x=0 \) if \( n > 0 \) because the oscillation is suppressed by the power of \( x \).

Concept:

Derivatives of inverse trigonometric functions.
Chain rule of differentiation or implicit differentiation.
Relationship between trigonometric and inverse trigonometric functions.



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

Given: \( y = 2 \cos^{-1} x \).
Differentiating both sides w.r.t. \( x \): \[ \frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\cos^{-1} x) \]
Using the standard derivative \( \frac{d}{dx}(\cos^{-1} x) = \frac{-1}{\sqrt{1 - x^2}} \): \[ \frac{dy}{dx} = \frac{-2}{\sqrt{1 - x^2}} \quad \dots (i) \]


Step 2: Convert the expression in terms of \( y \)

From the original equation: \( \cos^{-1} x = \frac{y}{2} \implies x = \cos \frac{y}{2} \).
Substitute this value into equation (i): \[ \frac{dy}{dx} = \frac{-2}{\sqrt{1 - \cos^2 \frac{y}{2}}} \]


Step 3: Simplify using trigonometric identity

Since \( 1 - \cos^2 \theta = \sin^2 \theta \): \[ \frac{dy}{dx} = \frac{-2}{\sqrt{\sin^2 \frac{y}{2}}} \] \[ \frac{dy}{dx} = \frac{-2}{\sin \frac{y}{2}} \] \[ \frac{dy}{dx} = -2 \csc \frac{y}{2} \] Quick Tip: When options are given in terms of the dependent variable \( y \), look to rearrange the initial equation to substitute back the trigonometric equivalent of \( x \).

Concept:

Volume of a sphere: \( V = \frac{4}{3}\pi r^3 \).
Relation between radius and diameter: \( D = 2r \) or \( r = \frac{D}{2} \).
Rate of change is the derivative of one quantity with respect to another.



Step 1: Express Volume in terms of Diameter
\[ V = \frac{4}{3}\pi \left( \frac{D}{2} \right)^3 \] \[ V = \frac{4}{3}\pi \cdot \frac{D^3}{8} \] \[ V = \frac{\pi D^3}{6} \]


Step 2: Differentiate Volume with respect to Diameter

Rate of change \( \frac{dV}{dD} \) is: \[ \frac{dV}{dD} = \frac{d}{dD} \left( \frac{\pi D^3}{6} \right) \] \[ \frac{dV}{dD} = \frac{\pi}{6} \cdot 3D^2 = \frac{\pi D^2}{2} \]


Step 3: Substitute the given radius value

Given radius \( r = 5 cm \), so diameter \( D = 10 cm \). \[ \left. \frac{dV}{dD} \right|_{D=10} = \frac{\pi (10)^2}{2} \] \[ \frac{dV}{dD} = \frac{100\pi}{2} = 50\pi cm^3/cm \] Quick Tip: Be very careful to differentiate with respect to what is asked (diameter vs radius). If differentiated w.r.t radius, the answer would be \( 4\pi r^2 = 100\pi \), which is a common distractor.

Concept:

Method of substitution: Use when a part of the integrand is the derivative of another part.
Standard Integral Formula: \( \int \frac{dt}{\sqrt{t^2 + a^2}} = \log|t + \sqrt{t^2 + a^2}| + C \).



Step 1: Use substitution method

Let \( \sin x = t \).
Then, differentiating with respect to \( x \): \[ \cos x \, dx = dt \]


Step 2: Transform the integral

Substitute \( t \) and \( dt \) into the original integral: \[ I = \int \frac{dt}{\sqrt{t^2 + 1}} \]


Step 3: Apply the standard integral formula

Using \( \int \frac{dt}{\sqrt{t^2 + a^2}} = \log|t + \sqrt{t^2 + a^2}| + C where a=1 \): \[ I = \log|t + \sqrt{t^2 + 1}| + C \]


Step 4: Substitute back the original variable

Replace \( t \) with \( \sin x \): \[ I = \log|\sin x + \sqrt{\sin^2 x + 1}| + C \] Quick Tip: Recognize derivatives of basic trig functions immediately. Since \( \cos x \) is in the numerator, substituting for \( \sin x \) is the standard approach for this family of integrals.

Concept:

Property of even functions: If \( f(x) = f(-x) \), then \( \int_{-a}^{a} f(x) dx = 2\int_{0}^{a} f(x) dx = 2\int_{-a}^{0} f(x) dx \).
Absolute value definition: \( |x| = x \) for \( x \ge 0 \) and \( |x| = -x \) for \( x < 0 \).



Step 1: Check the parity of the function

Let \( f(x) = 1 - |x| \).
Calculate \( f(-x) \): \[ f(-x) = 1 - |-x| = 1 - |x| = f(x) \]
Thus, the function is an even function.


Step 2: Apply integral properties for even functions

Since \( f(x) \) is even: \[ \int_{-1}^{1} (1 - |x|) dx = 2 \int_{-1}^{0} (1 - |x|) dx \]


Step 3: Define the absolute value in the relevant interval

In the interval \( [-1, 0] \), the value of \( x \) is negative or zero (\( x \le 0 \)).
Therefore, \( |x| = -x \).
Substitute this into the integral: \[ 1 - |x| = 1 - (-x) = 1 + x \]


Step 4: Write the simplified integral
\[ \int_{-1}^{1} (1 - |x|) dx = 2 \int_{-1}^{0} (1 + x) dx \] Quick Tip: Symmetry is your best friend in definite integration. Always check if the function is even or odd when the limits are of the form \( [-a, a] \).

Concept:

The area of a region bounded by a curve \( x = f(y) \) and the y-axis from \( y = a \) to \( y = b \) is given by \( \int_{a}^{b} x \, dy \).
Symmetry of the circle can be used to simplify the integration.



Step 1: Identify the equation of the boundary and limits

The given circle is \( x^2 + y^2 = 9 \).
The shaded region is bounded below by the line \( y = 1 \) and above by the top of the circle, where \( y = 3 \) (since the radius is 3).
Thus, the limits of integration for \( y \) are from 1 to 3.


Step 2: Express \( x \) in terms of \( y \)

From the circle equation: \[ x^2 = 9 - y^2 \] \[ x = \pm \sqrt{9 - y^2} \]
For any height \( y \), the width of the shaded region extends from \( x = -\sqrt{9 - y^2} \) to \( x = \sqrt{9 - y^2} \).


Step 3: Set up the integral for the area

The total width at any height \( y \) is: \[ Width = \sqrt{9 - y^2} - (-\sqrt{9 - y^2}) = 2\sqrt{9 - y^2} \]
The area \( A \) is the integral of this width with respect to \( y \) from 1 to 3: \[ A = \int_{1}^{3} 2\sqrt{9 - y^2} dy \] \[ A = 2 \int_{1}^{3} \sqrt{9 - y^2} dy \] Quick Tip: When integrating with respect to the y-axis, the function must be \( x = f(y) \).
Always check for symmetry; here the region is symmetric about the y-axis, hence the factor of 2.

Concept:

A differential equation \( \frac{dy}{dx} = F(x, y) \) is said to be homogeneous if \( F(x, y) \) is a homogeneous function of degree zero.
A function \( F(x, y) \) is homogeneous of degree \( n \) if \( F(\lambda x, \lambda y) = \lambda^n F(x, y) \).



Step 1: Test function (i)
\( F(x, y) = 3x + 2y \) \[ F(\lambda x, \lambda y) = 3(\lambda x) + 2(\lambda y) = \lambda(3x + 2y) = \lambda^1 F(x, y) \]
This is a homogeneous function of degree 1. Not degree 0.


Step 2: Test function (ii)
\( F(x, y) = \sin \frac{y}{x} + \log \frac{y}{x} \) \[ F(\lambda x, \lambda y) = \sin \frac{\lambda y}{\lambda x} + \log \frac{\lambda y}{\lambda x} = \sin \frac{y}{x} + \log \frac{y}{x} = \lambda^0 F(x, y) \]
This is homogeneous of degree 0.


Step 3: Test function (iii)
\( F(x, y) = e^{y/x} + 1 \) \[ F(\lambda x, \lambda y) = e^{\lambda y / \lambda x} + 1 = e^{y/x} + 1 = \lambda^0 F(x, y) \]
This is homogeneous of degree 0.


Step 4: Test function (iv)
\( F(x, y) = \sqrt{x^2 + y^2} - y \) \[ F(\lambda x, \lambda y) = \sqrt{(\lambda x)^2 + (\lambda y)^2} - \lambda y = \lambda\sqrt{x^2 + y^2} - \lambda y = \lambda^1 F(x, y) \]
This is a homogeneous function of degree 1. Not degree 0. Quick Tip: A quick shortcut: If every term in the function can be written purely as a function of \( (y/x) \) or \( (x/y) \), it is homogeneous of degree 0.

Concept:



Dot product definition: \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \), where \( \theta \) is the angle between the vectors.
Triangle Inequality: \( |\vec{a} + \vec{b}| \le |\vec{a}| + |\vec{b}| \).
Cross product magnitude: \( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \).



Step 1: Evaluate statement (A)


We know \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \).
Since the range of \( \cos \theta \) is \( [-1, 1] \), we have: \[ \cos \theta \le 1 \]


Multiplying by the non-negative scalar \( |\vec{a}| |\vec{b}| \):

\[ |\vec{a}| |\vec{b}| \cos \theta \le |\vec{a}| |\vec{b}| \] \[ \vec{a} \cdot \vec{b} \le |\vec{a}| |\vec{b}| \]
This is essentially a form of the Cauchy-Schwarz inequality and is always true.



Step 2: Evaluate statements (B), (C), and (D)


(B) \( |\vec{a} + \vec{b}| \ge |\vec{a}| + |\vec{b}| \) contradicts the Triangle Inequality (\( \le \)). It is only true (as equality) when vectors are in the same direction.


(C) \( |\vec{a} - \vec{b}| = |\vec{a}| - |\vec{b}| \) is false in general.


(D) \( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \). Since \( \sin \theta \le 1 \), the magnitude is always \( \le |\vec{a}| |\vec{b}| \). Thus (D) is false. Quick Tip: Geometric inequalities like Triangle Inequality and Cauchy-Schwarz are fundamental to vector algebra. Remember: Dot product \( \le \) product of magnitudes; Cross product \( \le \) product of magnitudes.

Concept:

General solution contains an arbitrary constant \( C \).
Particular solutions are obtained by assigning specific values to \( C \).
Range of exponential functions: \( e^k > 0 \) for all real \( k \).



Step 1: Solve the differential equation to check Reason (R)
\[ \frac{dy}{dx} = e^x \cdot e^y \]
Separating variables: \[ e^{-y} dy = e^x dx \]
Integrating both sides: \[ \int e^{-y} dy = \int e^x dx \] \[ -e^{-y} = e^x + K \] \[ e^x + e^{-y} = -K \]
Let \( -K = C \), then \( e^x + e^{-y} = C \).
Thus, Reason (R) is true.


Step 2: Analyze the Assertion (A)

The assertion claims \( e^x + e^{-y} = -2 \) can be a particular solution.
From a purely algebraic standpoint, if we differentiate \( e^x + e^{-y} = -2 \), we get: \[ e^x - e^{-y} \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{e^x}{e^{-y}} = e^{x+y} \]
However, for real variables \( x \) and \( y \), \( e^x > 0 \) and \( e^{-y} > 0 \).
The sum of two positive numbers must be positive: \( e^x + e^{-y} > 0 \).
Therefore, \( e^x + e^{-y} = -2 \) has no real solutions for \( x \) and \( y \).
In the context of real-valued functions, it cannot be a solution. Hence, Assertion (A) is false. Quick Tip: Always check the domain and range of your solutions. While a constant \( C \) can be any real number algebraically, it must be within the range of the function on the other side of the equation for a real solution to exist.

Concept:



Collinear vectors: Two vectors are collinear if one is a scalar multiple of the other (\( \vec{b} = \lambda \vec{a} \)).
Dot product of parallel/anti-parallel vectors: \( \vec{a} \cdot (k\vec{a}) = k|\vec{a}|^2 \).
Orthogonal vectors: Two non-zero vectors are perpendicular if their dot product is 0.



Step 1: Evaluate Assertion (A)


Two vectors \( \vec{u} \) and \( \vec{v} \) are collinear if \( \vec{v} = \lambda \vec{u} \) for some scalar \( \lambda \).


Here, \( (-2\vec{a}) \) is a scalar multiple of \( \vec{a} \) with \( \lambda = -2 \).


Therefore, they are collinear (specifically, they are anti-parallel).
Assertion (A) is true.



Step 2: Evaluate Reason (R)


Calculate the dot product:

\[ \vec{a} \cdot (-2\vec{a}) = -2(\vec{a} \cdot \vec{a}) = -2|\vec{a}|^2 \]
Since \( \vec{a} \neq \vec{0} \), \( |\vec{a}|^2 > 0 \).


Thus, \( -2|\vec{a}|^2 \neq 0 \).
Reason (R) is false. Quick Tip: Collinear vectors have an angle of \( 0^\circ \) or \( 180^\circ \) between them. Their dot product is zero only if at least one vector is the zero vector.

Concept:



Standard form of a line: \( \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \), where \( \langle a, b, c \rangle \) are direction ratios.
Condition for perpendicularity: Two lines with direction ratios \( \langle a_1, b_1, c_1 \rangle \) and \( \langle a_2, b_2, c_2 \rangle \) are perpendicular if \( a_1a_2 + b_1b_2 + c_1c_2 = 0 \).



Step 1: Convert the equations of lines to standard form


Line 1: \( \frac{x - 3}{1} = \frac{1 - y}{1} = \frac{z + 2}{p} \)
Rewriting the middle term to match standard form: \( \frac{1 - y}{1} = \frac{-(y - 1)}{1} = \frac{y - 1}{-1} \)


Standard Form \( L_1 \): \( \frac{x - 3}{1} = \frac{y - 1}{-1} = \frac{z + 2}{p} \)
Direction ratios \( \langle a_1, b_1, c_1 \rangle = \langle 1, -1, p \rangle \)


Step 2: Convert the second line to standard form


Line 2: \( \frac{2 - x}{3} = \frac{y + 1}{5} = \frac{z + 56}{2p} \)
Rewriting the first term: \( \frac{2 - x}{3} = \frac{-(x - 2)}{3} = \frac{x - 2}{-3} \)


Standard Form \( L_2 \): \( \frac{x - 2}{-3} = \frac{y + 1}{5} = \frac{z + 56}{2p} \)
Direction ratios \( \langle a_2, b_2, c_2 \rangle = \langle -3, 5, 2p \rangle \)


Step 3: Apply the perpendicularity condition


Since the lines are perpendicular: \[ a_1a_2 + b_1b_2 + c_1c_2 = 0 \] \[ (1)(-3) + (-1)(5) + (p)(2p) = 0 \] \[ -3 - 5 + 2p^2 = 0 \] \[ 2p^2 = 8 \] \[ p^2 = 4 \] \[ p = \pm 2 \] Quick Tip: Always check if the coefficients of \( x, y, z \) in the numerator are exactly 1 before identifying direction ratios.
The sign of terms like \( (1-y) \) or \( (2-x) \) must be factored out to the denominator.

Concept:



Vector equation of a line through point \( \vec{a} \) parallel to vector \( \vec{b} \) is \( \vec{r} = \vec{a} + k\vec{b} \).
A line perpendicular to two vectors \( \vec{b_1} \) and \( \vec{b_2} \) is parallel to their cross product \( \vec{b_1} \times \vec{b_2} \).



Step 1: Identify the direction vectors of the given lines


From the given equations:


Direction of line 1: \( \vec{b_1} = 3\hat{i} + 4\hat{j} + 2\hat{k} \)
Direction of line 2: \( \vec{b_2} = \hat{i} - \hat{j} + \hat{k} \)


Step 2: Find the direction of the required line


The required line is perpendicular to both \( \vec{b_1} \) and \( \vec{b_2} \).


Direction vector \( \vec{b} = \vec{b_1} \times \vec{b_2} \):

\[ \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 2 \\ 1 & -1 & 1 \end{vmatrix} \] \[ \vec{b} = \hat{i}(4 + 2) - \hat{j}(3 - 2) + \hat{k}(-3 - 4) \] \[ \vec{b} = 6\hat{i} - \hat{j} - 7\hat{k} \]


Step 3: Write the vector equation


The line passes through the origin \( \vec{a} = \vec{0} \).
The equation is: \[ \vec{r} = \vec{0} + \gamma(6\hat{i} - \hat{j} - 7\hat{k}) \] \[ \vec{r} = \gamma(6\hat{i} - \hat{j} - 7\hat{k}) \] Quick Tip: For lines through the origin, the position vector part is zero.
When performing a cross product, double-check the signs of the \( \hat{j} \) component cofactor.

Concept:

Product of functions with inverse trigonometric exponents: \( \sin^{-1} t + \cos^{-1} t = \frac{\pi}{2} \).
Differentiation of parametric functions or implicit functions.



Step 1: Establish a relation between x and y

Multiply \( x \) and \( y \): \[ x \cdot y = e^{\sin^{-1} t} \cdot e^{\cos^{-1} t} \] \[ xy = e^{\sin^{-1} t + \cos^{-1} t} \]
Since \( \sin^{-1} t + \cos^{-1} t = \frac{\pi}{2} \): \[ xy = e^{\pi/2} \]


Step 2: Differentiate implicitly with respect to x
\[ \frac{d}{dx}(xy) = \frac{d}{dx}(e^{\pi/2}) \]
Using the product rule: \[ y + x\frac{dy}{dx} = 0 \] \[ \frac{dy}{dx} = -\frac{y}{x} \]


Step 3: Find values of x and y at the given point

At \( t = \frac{1}{\sqrt{2}} \): \[ \sin^{-1} \left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}, \quad \cos^{-1} \left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4} \] \[ x = e^{\pi/4} \] \[ y = e^{\pi/4} \]


Step 4: Calculate the final derivative
\[ \left. \frac{dy}{dx} \right|_{t=1/\sqrt{2}} = -\frac{e^{\pi/4}}{e^{\pi/4}} = -1 \] Quick Tip: Using the identity \( \sin^{-1} t + \cos^{-1} t = \pi/2 \) turns a complex parametric differentiation problem into a simple implicit one.
Always look for algebraic simplifications before starting the calculus.

Concept:

A function \( f(x) \) is increasing if its first derivative \( f'(x) > 0 \).
Product rule: \( \frac{d}{dx}(uv) = u'v + uv' \).



Step 1: Find the first derivative of the function
\( f(x) = x \log x \) \[ f'(x) = \frac{d}{dx}(x) \cdot \log x + x \cdot \frac{d}{dx}(\log x) \] \[ f'(x) = 1 \cdot \log x + x \cdot \frac{1}{x} \] \[ f'(x) = \log x + 1 \]


Step 2: Set the condition for an increasing function

For \( f(x) \) to be increasing: \[ f'(x) > 0 \] \[ \log x + 1 > 0 \] \[ \log x > -1 \]


Step 3: Solve the logarithmic inequality

Taking anti-log (base \( e \)) on both sides: \[ x > e^{-1} \] \[ x > \frac{1}{e} \]


Step 4: Write the result in interval notation

Given domain is \( (0, \infty) \).
The sub-interval where the function increases is \( \left( \frac{1}{e}, \infty \right) \). Quick Tip: The base of the logarithm in calculus is always \( e \) unless specified otherwise.
Remember that \( \log x \) is only defined for \( x > 0 \).

Concept:

Solve from the innermost bracket outwards.
Standard values: \( \tan^{-1} 1 = \frac{\pi}{4} \).
Trigonometric relations: \( \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \).



Step 1: Evaluate the innermost term

Inside the brackets we have \( \tan^{-1} 1 \).
We know that \( \tan \frac{\pi}{4} = 1 \), so: \[ \tan^{-1} 1 = \frac{\pi}{4} \]


Step 2: Evaluate the cosine term

The expression becomes \( \cos(\frac{\pi}{4}) \). \[ \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \]


Step 3: Evaluate the cotangent inverse term

Now the expression inside the sine is \( \cot^{-1} [\sqrt{2} \cdot \frac{1}{\sqrt{2}}] \). \[ \cot^{-1} (1) \]
Since \( \cot \frac{\pi}{4} = 1 \): \[ \cot^{-1} (1) = \frac{\pi}{4} \]


Step 4: Find the final value

The whole expression reduces to: \[ \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \] Quick Tip: Inverse functions of the same value often cancel out if the scaling factors lead back to 1.
Always verify if the value lies in the principal range of the inverse trigonometric function.

Concept:



Symmetric relation: If \( (a, b) \in R \), then \( (b, a) \) must be in \( R \).
Equivalence relation: A relation that is Reflexive, Symmetric, and Transitive.



Step 1: Check for symmetry


Given \( R = \{(1, 1), (3, 3), (1, 2)\} \).


Here, \( (1, 2) \in R \). For \( R \) to be symmetric, \( (2, 1) \) must be in \( R \).


Since \( (2, 1) \notin R \), the relation is not symmetric.



Step 2: Determine elements for Reflexivity


For \( R \cup R_1 \) to be reflexive on \( \{1, 2, 3\} \), it must contain \( (1, 1), (2, 2), (3, 3) \).

\( R \) already has \( (1, 1) \) and \( (3, 3) \).


So, we must add \( (2, 2) \).



Step 3: Determine elements for Symmetry


As found in Step 1, we must add \( (2, 1) \) to ensure symmetry for the existing \( (1, 2) \).



Step 4: Check for Transitivity


The current set is \( \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\} \).


Checking pairs: \( (1, 2) \) and \( (2, 1) \implies (1, 1) \) (present). \( (2, 1) \) and \( (1, 2) \implies (2, 2) \) (present).


The relation is now transitive.


Step 5: Identify \( R_1 \)


The elements added were \( (2, 2) \) and \( (2, 1) \).
Thus, \( R_1 = \{(2, 2), (2, 1)\} \). Quick Tip: Smallest equivalence relation is always the diagonal set \( \{(x,x) : x \in A\} \).
Start with reflexivity, then add for symmetry, and finally check for transitivity.

Concept:

Magnitude identity: \( |\vec{u}|^2 = \vec{u} \cdot \vec{u} \).
Square of sum/diff: \( |\vec{a} \pm k\vec{b}|^2 = |\vec{a}|^2 + k^2|\vec{b}|^2 \pm 2k(\vec{a} \cdot \vec{b}) \).
Dot product: \( \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta \).



Step 1: Square both sides of the given equation
\[ |\vec{a} + 2\vec{b}|^2 = |2\vec{a} - \vec{b}|^2 \]
Expanding using dot product properties: \[ |\vec{a}|^2 + 4|\vec{b}|^2 + 4(\vec{a} \cdot \vec{b}) = 4|\vec{a}|^2 + |\vec{b}|^2 - 4(\vec{a} \cdot \vec{b}) \]


Step 2: Substitute values for unit vectors

Since \( \vec{a} \) and \( \vec{b} \) are unit vectors, \( |\vec{a}| = 1 \) and \( |\vec{b}| = 1 \). \[ 1^2 + 4(1^2) + 4(\vec{a} \cdot \vec{b}) = 4(1^2) + 1^2 - 4(\vec{a} \cdot \vec{b}) \] \[ 1 + 4 + 4(\vec{a} \cdot \vec{b}) = 4 + 1 - 4(\vec{a} \cdot \vec{b}) \] \[ 5 + 4(\vec{a} \cdot \vec{b}) = 5 - 4(\vec{a} \cdot \vec{b}) \]


Step 3: Solve for the dot product
\[ 8(\vec{a} \cdot \vec{b}) = 0 \] \[ \vec{a} \cdot \vec{b} = 0 \]


Step 4: Calculate the angle
\[ |\vec{a}||\vec{b}|\cos\theta = 0 \] \[ (1)(1)\cos\theta = 0 \implies \cos\theta = 0 \] \[ \theta = \frac{\pi}{2} or 90^\circ \] Quick Tip: If magnitudes of sum and difference linear combinations are equal with swapped coefficients, the vectors are usually orthogonal.
Squaring is the standard technique to remove absolute magnitude signs in vector equations.

Concept:



This problem follows a Binomial Distribution because there are a fixed number of independent trials (\( n = 3 \)) with only two possible outcomes (success or failure).
The probability of \( r \) successes in \( n \) trials is given by \( P(X = r) = \binom{n}{r} p^r q^{n-r} \), where \( p \) is the probability of success and \( q = 1 - p \) is the probability of failure.



Step 1: Identify the parameters of the binomial distribution


Given the success rate is 90%, we have:

\[ p = 0.9 = \frac{9}{10} \]
Probability of failure: \[ q = 1 - p = 1 - 0.9 = 0.1 = \frac{1}{10} \]
Number of surgeries (trials): \[ n = 3 \]


Step 2: Apply the binomial formula for exactly one success


We need to find \( P(X = 1) \):

\[ P(X = 1) = \binom{3}{1} (p)^1 (q)^{3-1} \] \[ P(X = 1) = 3 \cdot (0.9)^1 \cdot (0.1)^2 \]


Step 3: Calculate the final probability

\[ P(X = 1) = 3 \cdot 0.9 \cdot 0.01 \] \[ P(X = 1) = 2.7 \cdot 0.01 = 0.027 \]
Thus, the probability that exactly one surgery is successful is 0.027. Quick Tip: For independent events, you can also think of this as: \( P(SFF) + P(FSF) + P(FFS) \).
Always ensure that \( p + q = 1 \).

Concept:



"At most two" means the number of successful surgeries can be 0, 1, or 2.
Complementary probability: \( P(X \le 2) = 1 - P(X > 2) = 1 - P(X = 3) \).



Step 1: Define the parameters


From the previous part:

\[ n = 3, \quad p = 0.9, \quad q = 0.1 \]


Step 2: Calculate the probability of all surgeries being successful


Using the binomial formula for \( r = 3 \):

\[ P(X = 3) = \binom{3}{3} (p)^3 (q)^0 \] \[ P(X = 3) = 1 \cdot (0.9)^3 \cdot 1 \] \[ P(X = 3) = 0.729 \]


Step 3: Use the complement rule to find the final probability

\[ P(at most 2 successes) = 1 - P(X = 3) \] \[ P(X \le 2) = 1 - 0.729 \] \[ P(X \le 2) = 0.271 \]
Thus, the probability that at most two surgeries are successful is 0.271. Quick Tip: When asked for "at most" or "at least", check if calculating the complement is faster than summing individual probabilities.

Concept:

When an integrand contains different fractional powers of \( x \), use a substitution \( x = t^n \) where \( n \) is the LCM of the denominators of the fractional powers.
After substitution, perform polynomial division if the degree of the numerator is greater than or equal to the degree of the denominator.



Step 1: Choose an appropriate substitution

The fractional powers are \( \frac{1}{2} \) and \( \frac{1}{3} \). The LCM of 2 and 3 is 6.
Let \( x = t^6 \).
Then \( dx = 6t^5 \, dt \).
Also, \( x^{1/2} = (t^6)^{1/2} = t^3 \) and \( x^{1/3} = (t^6)^{1/3} = t^2 \).


Step 2: Rewrite the integral in terms of \( t \)
\[ I = \int \frac{6t^5}{t^3 + t^2} dt \] \[ I = \int \frac{6t^5}{t^2(t + 1)} dt = \int \frac{6t^3}{t + 1} dt \]


Step 3: Perform polynomial division

Divide \( t^3 \) by \( t + 1 \): \[ \frac{t^3}{t + 1} = t^2 - t + 1 - \frac{1}{t + 1} \]
So, the integral becomes: \[ I = 6 \int \left( t^2 - t + 1 - \frac{1}{t + 1} \right) dt \]


Step 4: Integrate and back-substitute
\[ I = 6 \left[ \frac{t^3}{3} - \frac{t^2}{2} + t - \log|t + 1| \right] + C \] \[ I = 2t^3 - 3t^2 + 6t - 6\log|t + 1| + C \]
Since \( x = t^6 \implies t = x^{1/6} \): \[ I = 2\sqrt{x} - 3x^{1/3} + 6x^{1/6} - 6\log|x^{1/6} + 1| + C \] Quick Tip: Always simplify the powers of \( t \) in the denominator before dividing.
The substitution \( x = t^{LCM} \) is a standard technique for eliminating radicals.

Concept:

Use inverse trigonometric identities to simplify the integrand first.
Use Integration by Parts: \( \int u \, v \, dx = u \int v \, dx - \int (u' \int v \, dx) \, dx \).
Identity: \( \tan^{-1} \left( \frac{a - b}{1 + ab} \right) = \tan^{-1} a - \tan^{-1} b \).



Step 1: Simplify the integrand using identities

The integrand is \( \tan^{-1} \left( \frac{1 - x}{1 + (1)(x)} \right) \).
Using the identity \( \tan^{-1} \left( \frac{A - B}{1 + AB} \right) = \tan^{-1} A - \tan^{-1} B \): \[ \tan^{-1} \left( \frac{1 - x}{1 + x} \right) = \tan^{-1} 1 - \tan^{-1} x = \frac{\pi}{4} - \tan^{-1} x \]


Step 2: Set up the integral and apply linearity
\[ I = \int \left( \frac{\pi}{4} - \tan^{-1} x \right) dx \] \[ I = \frac{\pi}{4}x - \int 1 \cdot \tan^{-1} x \, dx \quad \dots (i) \]


Step 3: Integrate \( \tan^{-1} x \) by parts

Let \( u = \tan^{-1} x \) and \( dv = dx \).
Then \( du = \frac{1}{1 + x^2} dx \) and \( v = x \). \[ \int \tan^{-1} x \, dx = x \tan^{-1} x - \int \frac{x}{1 + x^2} dx \]
Multiply and divide the second term by 2 to get the derivative of the denominator: \[ \int \tan^{-1} x \, dx = x \tan^{-1} x - \frac{1}{2} \log|1 + x^2| \]


Step 4: Substitute back to find the final result

Substitute the result from Step 3 into equation (i): \[ I = \frac{\pi}{4}x - \left( x \tan^{-1} x - \frac{1}{2} \log(1 + x^2) \right) + C \] \[ I = x \left( \frac{\pi}{4} - \tan^{-1} x \right) + \frac{1}{2} \log(1 + x^2) + C \] Quick Tip: Always look for inverse trig identities before integrating. It is much easier than substituting \( x = \tan \theta \).
Remember that \( \int \tan^{-1} x \, dx = x \tan^{-1} x - \frac{1}{2}\ln(1+x^2) + C \) is a useful result to memorize.

Concept:

Use the definite integral property: \( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx \).
Trigonometric substitution for \( \sin 2x \) using \( \tan x \).



Step 1: Apply the property \( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx \)

Let \( I = \int_{0}^{\pi/2} \frac{\sin^2 x}{1 + \sin 2x} dx \quad \dots (i) \)
Replacing \( x \) by \( \frac{\pi}{2} - x \): \[ I = \int_{0}^{\pi/2} \frac{\sin^2(\pi/2 - x)}{1 + \sin(2(\pi/2 - x))} dx \]
Since \( \sin(\pi/2 - x) = \cos x \) and \( \sin(\pi - 2x) = \sin 2x \): \[ I = \int_{0}^{\pi/2} \frac{\cos^2 x}{1 + \sin 2x} dx \quad \dots (ii) \]


Step 2: Add equations (i) and (ii)
\[ 2I = \int_{0}^{\pi/2} \frac{\sin^2 x + \cos^2 x}{1 + \sin 2x} dx \] \[ 2I = \int_{0}^{\pi/2} \frac{1}{1 + \sin 2x} dx \]


Step 3: Convert to tan form and solve

Using \( \sin 2x = \frac{2 \tan x}{1 + \tan^2 x} \): \[ 2I = \int_{0}^{\pi/2} \frac{1}{1 + \frac{2 \tan x}{1 + \tan^2 x}} dx = \int_{0}^{\pi/2} \frac{\sec^2 x}{(1 + \tan x)^2} dx \]
Let \( \tan x = t \), then \( \sec^2 x \, dx = dt \).
Limits: when \( x = 0, t = 0 \); when \( x = \pi/2, t \to \infty \). \[ 2I = \int_{0}^{\infty} \frac{dt}{(1 + t)^2} = \left[ -\frac{1}{1 + t} \right]_{0}^{\infty} \] \[ 2I = (0) - (-1) = 1 \]


Step 4: Determine final value
\[ 2I = 1 \implies I = \frac{1}{2} \] Quick Tip: The substitution \( x \to a-x \) often helps simplify symmetric trigonometric denominators involving \( \sin 2x \) or \( \sin x + \cos x \).

Concept:

Use substitution when the numerator is the derivative of a part of the denominator.
Use Partial Fractions to decompose the rational expression.



Step 1: Substitute for the trigonometric function

Let \( \sin x = t \).
Then \( \cos x \, dx = dt \).
The integral becomes: \[ I = \int \frac{dt}{(2 + t)(4 + t)} \]


Step 2: Decompose using partial fractions

Let \( \frac{1}{(t + 2)(t + 4)} = \frac{A}{t + 2} + \frac{B}{t + 4} \) \[ 1 = A(t + 4) + B(t + 2) \]
Put \( t = -2 \implies 1 = 2A \implies A = 1/2 \).
Put \( t = -4 \implies 1 = -2B \implies B = -1/2 \). \[ \frac{1}{(t + 2)(t + 4)} = \frac{1}{2} \left[ \frac{1}{t + 2} - \frac{1}{t + 4} \right] \]


Step 3: Integrate with respect to \( t \)
\[ I = \frac{1}{2} \int \left( \frac{1}{t + 2} - \frac{1}{t + 4} \right) dt \] \[ I = \frac{1}{2} (\log|t + 2| - \log|t + 4|) + C \] \[ I = \frac{1}{2} \log \left| \frac{t + 2}{t + 4} \right| + C \]


Step 4: Back-substitute and state the final result
\[ I = \frac{1}{2} \log \left| \frac{\sin x + 2}{\sin x + 4} \right| + C \] Quick Tip: When partial fractions have terms like \( \frac{1}{(x+a)(x+b)} \), the shortcut is \( \frac{1}{b-a} [ \frac{1}{x+a} - \frac{1}{x+b} ] \).

Concept:

For integrals of the form \( \int \frac{px+q}{ax^2+bx+c} dx \), express the numerator as \( A \cdot \frac{d}{dx}(denominator) + B \).
Split into two parts: a logarithmic part and an inverse tangent part (by completing the square).



Step 1: Express numerator in terms of denominator's derivative

Derivative of \( x^2 + 4x + 5 \) is \( 2x + 4 \).
Let \( x + 3 = A(2x + 4) + B \).
Comparing coefficients of \( x \): \( 1 = 2A \implies A = 1/2 \).
Comparing constants: \( 3 = 4A + B \implies 3 = 2 + B \implies B = 1 \).
So, \( x + 3 = \frac{1}{2}(2x + 4) + 1 \).


Step 2: Split the integral
\[ I = \int \frac{\frac{1}{2}(2x + 4) + 1}{x^2 + 4x + 5} dx \] \[ I = \frac{1}{2} \int \frac{2x + 4}{x^2 + 4x + 5} dx + \int \frac{1}{x^2 + 4x + 5} dx \]


Step 3: Integrate the two parts

Part 1: \( \frac{1}{2} \log|x^2 + 4x + 5| \)
Part 2: Complete the square: \( x^2 + 4x + 5 = (x + 2)^2 + 1 \). \[ \int \frac{1}{(x + 2)^2 + 1^2} dx = \tan^{-1}(x + 2) \]


Step 4: Combine the results
\[ I = \frac{1}{2} \log(x^2 + 4x + 5) + \tan^{-1}(x + 2) + C \]
(Note: \( x^2 + 4x + 5 > 0 \) for all \( x \), so modulus is not strictly necessary). Quick Tip: Always check if the numerator is directly the derivative of the denominator first. If not, the method of "A and B" coefficients is reliable.

Concept:

This is a Homogeneous Differential Equation because each term has the same total degree (degree 2).
Substitution: Let \( y = vx \), then \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).



Step 1: Rewrite the equation for \( dy/dx \)
\[ 2xy dy = -(x^2 - y^2) dx \] \[ \frac{dy}{dx} = \frac{y^2 - x^2}{2xy} \]


Step 2: Apply the homogeneous substitution

Let \( y = vx \implies \frac{dy}{dx} = v + x \frac{dv}{dx} \). \[ v + x \frac{dv}{dx} = \frac{v^2 x^2 - x^2}{2x(vx)} = \frac{x^2(v^2 - 1)}{x^2(2v)} \] \[ v + x \frac{dv}{dx} = \frac{v^2 - 1}{2v} \]


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


Step 4: Integrate and back-substitute
\[ \int \frac{2v}{v^2 + 1} dv = -\int \frac{1}{x} dx \] \[ \log|v^2 + 1| = -\log|x| + \log C \] \[ \log|v^2 + 1| + \log|x| = \log C \implies x(v^2 + 1) = C \]
Substitute \( v = y/x \): \[ x \left( \frac{y^2}{x^2} + 1 \right) = C \implies \frac{x(y^2 + x^2)}{x^2} = C \implies x^2 + y^2 = Cx \]
This is the general solution. Quick Tip: For homogeneous equations, the resulting separable equation will always involve \( x \frac{dv}{dx} \).
The solution \( x^2 + y^2 = Cx \) represents a family of circles.

Concept:

This is a Variable Separable Differential Equation.
Group all \( x \) terms with \( dx \) and all \( y \) terms with \( dy \).
Find the particular solution by substituting the initial values into the general solution.



Step 1: Separate the variables

Divide the whole equation by \( \cos x \cos y \): \[ \frac{\sin x \cos y}{\cos x \cos y} dx + \frac{\cos x \sin y}{\cos x \cos y} dy = 0 \] \[ \tan x \, dx + \tan y \, dy = 0 \]


Step 2: Integrate to find the general solution
\[ \int \tan x \, dx + \int \tan y \, dy = 0 \] \[ \log|\sec x| + \log|\sec y| = \log C \] \[ \log|\sec x \cdot \sec y| = \log C \implies \sec x \sec y = C \]


Step 3: Find the constant \( C \) using initial conditions

Given \( y = \pi/4 \) when \( x = 0 \): \[ \sec(0) \cdot \sec(\pi/4) = C \] \[ 1 \cdot \sqrt{2} = C \implies C = \sqrt{2} \]


Step 4: Write the particular solution
\[ \sec x \sec y = \sqrt{2} \]
Or in cosine terms: \[ \frac{1}{\cos x} \cdot \frac{1}{\cos y} = \sqrt{2} \implies \cos x \cos y = \frac{1}{\sqrt{2}} \] Quick Tip: For separable equations with trigonometric terms, using log forms for integrals often simplifies finding the constant \( C \).
Remember that \( \int \tan \theta \, d\theta = \log|\sec \theta| \) or \( -\log|\cos \theta| \).

Concept:

Logarithmic differentiation is used when a function is of the form \( [f(x)]^{g(x)} \).
Property: \( \log(a^b) = b \log a \).
Product rule: \( \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx} \).
Chain rule: \( \frac{d}{dx}(\log u) = \frac{1}{u} \frac{du}{dx} \).



Step 1: Apply logarithms to both sides of the equation

Given: \[ (\sin x)^y = y^{\cos x} \]
Taking natural logarithm (\( \ln \)) on both sides: \[ \ln [(\sin x)^y] = \ln [y^{\cos x}] \]
Using the property \( \ln(a^b) = b \ln a \): \[ y \ln(\sin x) = \cos x \ln(y) \]


Step 2: Differentiate implicitly with respect to \( x \)

Applying the product rule on both sides: \[ \frac{d}{dx}[y] \cdot \ln(\sin x) + y \cdot \frac{d}{dx}[\ln(\sin x)] = \frac{d}{dx}[\cos x] \cdot \ln(y) + \cos x \cdot \frac{d}{dx}[\ln(y)] \] \[ \frac{dy}{dx} \ln(\sin x) + y \cdot \frac{1}{\sin x} \cdot \cos x = -\sin x \ln(y) + \cos x \cdot \frac{1}{y} \frac{dy}{dx} \] \[ \frac{dy}{dx} \ln(\sin x) + y \cot x = -\sin x \ln(y) + \frac{\cos x}{y} \frac{dy}{dx} \]


Step 3: Rearrange terms to isolate \( \frac{dy}{dx} \)

Group all terms containing \( \frac{dy}{dx} \) on the left side: \[ \frac{dy}{dx} \ln(\sin x) - \frac{\cos x}{y} \frac{dy}{dx} = -\sin x \ln(y) - y \cot x \]
Factor out \( \frac{dy}{dx} \): \[ \frac{dy}{dx} \left[ \ln(\sin x) - \frac{\cos x}{y} \right] = - [\sin x \ln(y) + y \cot x] \]


Step 4: Solve for \( \frac{dy}{dx} \) and simplify
\[ \frac{dy}{dx} \left[ \frac{y \ln(\sin x) - \cos x}{y} \right] = - [\sin x \ln(y) + y \cot x] \] \[ \frac{dy}{dx} = \frac{-y [\sin x \ln(y) + y \cot x]}{y \ln(\sin x) - \cos x} \]
Multiplying the numerator and denominator by \( -1 \) to clean the signs: \[ \frac{dy}{dx} = \frac{y [y \cot x + \sin x \ln(y)]}{\cos x - y \ln(\sin x)} \] Quick Tip: When differentiating expressions with variables in exponents, always take logs first.
Be careful with the chain rule for terms like \( \ln(\sin x) \), which requires an extra \( \cos x \) factor.

Concept:

Vector form of a line: \( \vec{r} = \vec{a} + \lambda \vec{b} \), where \( \vec{a} \) is a point and \( \vec{b} \) is the direction.
Intersection: Equate components of the general points on both lines and solve for parameters.



Step 1: Find vector forms of the lines

For \( l_1 \): Point \( \vec{a_1} = -3\hat{i} + \hat{j} + 5\hat{k} \), Direction \( \vec{b_1} = -3\hat{i} + \hat{j} + 5\hat{k} \). \[ \vec{r} = (-3\hat{i} + \hat{j} + 5\hat{k}) + \lambda(-3\hat{i} + \hat{j} + 5\hat{k}) \]
For \( l_2 \): Standard form is \( \frac{x + 1}{-1} = \frac{y - 2}{2} = \frac{z - 5}{5} \).
Point \( \vec{a_2} = -\hat{i} + 2\hat{j} + 5\hat{k} \), Direction \( \vec{b_2} = -\hat{i} + 2\hat{j} + 5\hat{k} \). \[ \vec{r} = (-\hat{i} + 2\hat{j} + 5\hat{k}) + \mu(-\hat{i} + 2\hat{j} + 5\hat{k}) \]


Step 2: Set up equations for intersection

Equating general points: \[ -3\lambda - 3 = -\mu - 1 \implies 3\lambda - \mu = -2 \quad \dots(1) \] \[ \lambda + 1 = 2\mu + 2 \implies \lambda - 2\mu = 1 \quad \dots(2) \] \[ 5\lambda + 5 = 5\mu + 5 \implies \lambda = \mu \quad \dots(3) \]


Step 3: Solve the system

Substituting \( \lambda = \mu \) into (2): \[ \lambda - 2\lambda = 1 \implies -\lambda = 1 \implies \lambda = -1, \mu = -1 \]
Check in equation (1): \[ 3(-1) - (-1) = -3 + 1 = -2 \].
LHS = RHS.


Step 4: Conclusion

Since the values of \( \lambda \) and \( \mu \) satisfy all three equations, the lines intersect.
Substituting \( \lambda = -1 \) in \( l_1 \): \[ Point = (-3(-1)-3, -1+1, 5(-1)+5) = (0, 0, 0) \]
The lines intersect at the origin. Quick Tip: Always normalize the sign of \( y \) or \( z \) in Cartesian form before reading direction ratios.
Lines intersect if the system of three equations for two parameters is consistent.

Concept:

Distance between two parallel lines \( \vec{r} = \vec{a_1} + \lambda \vec{b} \) and \( \vec{r} = \vec{a_2} + \mu \vec{b} \) is \( d = \frac{|(\vec{a_2} - \vec{a_1}) \times \vec{b}|}{|\vec{b}|} \).
For a square, adjacent sides are perpendicular, so the dot product of their direction vectors is 0.



Step 1: Find the side length of the square

Here \( \vec{a_2} - \vec{a_1} = (3-1)\hat{i} + (3-2)\hat{j} + (-5+4)\hat{k} = 2\hat{i} + \hat{j} - \hat{k} \).
Direction vector \( \vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k} \). \[ (\vec{a_2} - \vec{a_1}) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 2 & 3 & 6 \end{vmatrix} = \hat{i}(6+3) - \hat{j}(12+2) + \hat{k}(6-2) = 9\hat{i} - 14\hat{j} + 4\hat{k} \] \[ Side s = \frac{\sqrt{9^2 + (-14)^2 + 4^2}}{\sqrt{2^2 + 3^2 + 6^2}} = \frac{\sqrt{81 + 196 + 16}}{\sqrt{4 + 9 + 36}} = \frac{\sqrt{293}}{7} \]


Step 2: Calculate the area
\[ Area = s^2 = \left( \frac{\sqrt{293}}{7} \right)^2 = \frac{293}{49} sq units \]


Step 3: Find the value of \( p \)

The adjacent sides are perpendicular.
Direction ratio of first side: \( \langle 2, 3, 6 \rangle \).
Direction ratio of other side: \( \langle -3, 6, p \rangle \). \[ (2)(-3) + (3)(6) + (6)(p) = 0 \] \[ -6 + 18 + 6p = 0 \implies 12 + 6p = 0 \implies p = -2 \] Quick Tip: Side of a square equals the distance between its parallel lines.
The dot product of direction ratios of perpendicular lines is always zero.

Concept:

A function is one-one if \( f(x_1) = f(x_2) \implies x_1 = x_2 \).
A function is onto if its codomain equals its range.



Step 1: Prove the function is one-one

Let \( f(x_1) = f(x_2) \) for \( x_1, x_2 \in R_+ \). \[ 4x_1^2 + 12x_1 + 15 = 4x_2^2 + 12x_2 + 15 \] \[ 4(x_1^2 - x_2^2) + 12(x_1 - x_2) = 0 \] \[ 4(x_1 - x_2)(x_1 + x_2) + 12(x_1 - x_2) = 0 \] \[ 4(x_1 - x_2)[x_1 + x_2 + 3] = 0 \]
Since \( x_1, x_2 \ge 0 \), the term \( x_1 + x_2 + 3 \ge 3 \) (never zero).
Thus, \( x_1 - x_2 = 0 \implies x_1 = x_2 \).
So, \( f \) is one-one.


Step 2: Find set \( A \) for the onto condition

Set \( A \) must be the range of \( f \) on \( [0, \infty) \).
For \( x = 0 \), \( f(0) = 15 \).
Since \( f'(x) = 8x + 12 > 0 \) for \( x \ge 0 \), the function is strictly increasing.
As \( x \to \infty \), \( f(x) \to \infty \).
Thus, the range is \( [15, \infty) \).
So, Set \( A = [15, \infty) \).


Step 3: Check if \( f(a) = 7 \) for \( a \in R_+ \)
\[ 4a^2 + 12a + 15 = 7 \] \[ 4a^2 + 12a + 8 = 0 \] \[ a^2 + 3a + 2 = 0 \implies (a + 1)(a + 2) = 0 \] \[ a = -1 or a = -2 \]
Neither value is in \( R_+ = [0, \infty) \).
Therefore, no such \( a \) exists. Quick Tip: Use calculus (\( f'(x) \)) to quickly check one-one nature: if \( f'(x) \) is always positive or always negative on the domain, it is one-one.
The range of a continuous increasing function on \( [a, b] \) is \( [f(a), f(b)] \).

Concept:



Convert inequalities to equations to draw boundary lines.
Find the feasible region and its corner points.
Evaluate the objective function \( Z \) at each corner point.



Step 1: Draw the constraint lines


(i) \( x = 20 \) (vertical line).


(ii) \( 2x + 3y = 120 \). Points: \( (60, 0) \) and \( (0, 40) \).


(iii) \( 2x + y = 60 \). Points: \( (30, 0) \) and \( (0, 60) \).



Step 2: Determine corner points of the feasible region


The region satisfies \( x \le 20 \), \( y \le 40 \), etc., and stays in the 1st quadrant.


Points of intersection:


Lines (ii) and (iii): \( 2x+3y=120 \) and \( 2x+y=60 \). Subtraction: \( 2y=60 \implies y=30, x=15 \). Point \( (15, 30) \).
Lines (i) and (iii): \( x=20, 2(20)+y=60 \implies y=20 \). Point \( (20, 20) \).
Vertices: \( O(0,0), A(20,0), B(20,20), C(15,30), D(0,40) \).


Step 3: Evaluate \( Z = 8x + 8y \)




At \( O(0,0) \), \( Z = 0 \).
At \( A(20,0) \), \( Z = 160 \).
At \( B(20,20) \), \( Z = 160 + 160 = 320 \).
At \( C(15,30) \), \( Z = 120 + 240 = 360 \).
At \( D(0,40) \), \( Z = 320 \).


Step 4: Conclusion

The maximum value of \( Z \) is 360 at the point (15, 30). Quick Tip: Always shade the common area to identify the feasible region clearly.
The optimal solution in a linear problem always lies at a corner point.

Concept:



Let \( x \) be the number of Rs 1,000 vouchers and \( y \) be the number of Rs 500 vouchers.
System \( AX = B \) is solved as \( X = A^{-1}B \).



Step 1: Formulate the equations


Total vouchers: \( x + y = 60 \)
Given condition: \( x + 3y = 100 \)
Matrix form \( \begin{bmatrix} 1 & 1 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 60 \\ 100 \end{bmatrix} \).



Step 2: Find \( A^{-1} \)

\( |A| = 3 - 1 = 2 \). \( adj A = \begin{bmatrix} 3 & -1 \\ -1 & 1 \end{bmatrix} \). \( A^{-1} = \frac{1}{2} \begin{bmatrix} 3 & -1 \\ -1 & 1 \end{bmatrix} \).


Step 3: Solve for \( x \) and \( y \)

\( \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 3 & -1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 60 \\ 100 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 180 - 100 \\ -60 + 100 \end{bmatrix} \) \[ \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 80 \\ 40 \end{bmatrix} = \begin{bmatrix} 40 \\ 20 \end{bmatrix} \]
So, there are 40 vouchers of Rs 1,000 and 20 vouchers of Rs 500. Quick Tip: Always double check the word problem wording: "3 times the number of Rs 500 vouchers" means \( 3y \).
Verify by substitution: \( 40 + 20 = 60 \) and \( 40 + 3(20) = 100 \). Correct.

Concept:

If \( PQ - RS \) is a null matrix, then \( PQ = RS \).
Therefore, \( S = R^{-1}PQ \), provided \( R^{-1} \) exists.



Step 1: Calculate \( PQ \)

Given, \[ P = \begin{bmatrix} 2 & -1\\ 3 & 4 \end{bmatrix}, \qquad Q = \begin{bmatrix} 5 & 2\\ 7 & 4 \end{bmatrix} \]

Therefore, \[ PQ = \begin{bmatrix} 2 & -1\\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & 2\\ 7 & 4 \end{bmatrix} \]
\[ = \begin{bmatrix} 2(5)+(-1)(7) & 2(2)+(-1)(4)\\ 3(5)+4(7) & 3(2)+4(4) \end{bmatrix} \]
\[ = \begin{bmatrix} 10-7 & 4-4\\ 15+28 & 6+16 \end{bmatrix} = \begin{bmatrix} 3 & 0\\ 43 & 22 \end{bmatrix}. \]


Step 2: Find \( R^{-1} \)

Given, \[ R = \begin{bmatrix} 2 & 5\\ 3 & 8 \end{bmatrix} \]

The determinant of \( R \) is \[ |R| = (2)(8)-(5)(3) \]
\[ =16-15=1 \neq 0. \]

Therefore, \( R^{-1} \) exists.

Using \[ R^{-1}=\frac{1}{|R|} \begin{bmatrix} d & -b\\ -c & a \end{bmatrix}, \]

we get \[ R^{-1} = \begin{bmatrix} 8 & -5\\ -3 & 2 \end{bmatrix}. \]


Step 3: Find the matrix \( S \)

Since \[ PQ-RS=O, \]
we have \[ PQ=RS. \]

Multiplying both sides by \(R^{-1}\) from the left, \[ S=R^{-1}PQ. \]

Therefore, \[ S= \begin{bmatrix} 8 & -5\\ -3 & 2 \end{bmatrix} \begin{bmatrix} 3 & 0\\ 43 & 22 \end{bmatrix} \]
\[ = \begin{bmatrix} 8(3)+(-5)(43) & 8(0)+(-5)(22)\\ (-3)(3)+2(43) & (-3)(0)+2(22) \end{bmatrix} \]
\[ = \begin{bmatrix} 24-215 & -110\\ -9+86 & 44 \end{bmatrix} \]
\[ = \boxed{ \begin{bmatrix} -191 & -110\\ 77 & 44 \end{bmatrix}} \]

Hence, the required matrix is \[ \boxed{ S= \begin{bmatrix} -191 & -110\\ 77 & 44 \end{bmatrix}} \] Quick Tip: Tip 1: If \( PQ-RS=O \), first rearrange it as \( PQ=RS \). Tip 2: To find \( S \), multiply \( PQ \) by \( R^{-1} \) from the left.

Concept:

The equation of a straight line passing through two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the two-point form.
Formula: \( y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \).



Step 1: Identify the coordinates of points A and B

The coordinates of vertex A are \( (x_1, y_1) = (0, 4) \).
The coordinates of vertex B are \( (x_2, y_2) = (-2, 0) \).


Step 2: Calculate the slope of line AB

Slope \( m = \frac{y_2 - y_1}{x_2 - x_1} \): \[ m = \frac{0 - 4}{-2 - 0} = \frac{-4}{-2} = 2 \]


Step 3: Apply the point-slope form to find the equation

Using point A(0, 4) and slope \( m = 2 \): \[ y - 4 = 2(x - 0) \] \[ y - 4 = 2x \]
Rearranging to standard form: \[ 2x - y + 4 = 0 \quad or \quad y = 2x + 4 \] Quick Tip: Since point A(0, 4) lies on the y-axis, the y-intercept \( c \) is 4.
Using the slope-intercept form \( y = mx + c \) directly gives \( y = 2x + 4 \) after finding the slope.

Concept:

The equation of a line can be found using the intercept form if the x-intercept and y-intercept are known.
Formula: \( \frac{x}{a} + \frac{y}{b} = 1 \), where \( a \) is the x-intercept and \( b \) is the y-intercept.



Step 1: Identify the intercepts from the coordinates

Line AC passes through A(0, 4) and C(3, 0).
The y-intercept \( b \) is the y-coordinate where \( x = 0 \), so \( b = 4 \).
The x-intercept \( a \) is the x-coordinate where \( y = 0 \), so \( a = 3 \).


Step 2: Substitute values into the intercept form
\[ \frac{x}{3} + \frac{y}{4} = 1 \]


Step 3: Simplify the equation

Multiplying the entire equation by the LCM of 3 and 4, which is 12: \[ 12 \left( \frac{x}{3} \right) + 12 \left( \frac{y}{4} \right) = 12(1) \] \[ 4x + 3y = 12 \]
Alternatively, in slope-intercept form: \[ 3y = -4x + 12 \implies y = -\frac{4}{3}x + 4 \] Quick Tip: The intercept form is the fastest way to write the equation when points lie on the axes.
Always verify by plugging the coordinates of one point back into your final equation.

Concept:

The area under a curve \( y = f(x) \) from \( x = a \) to \( x = b \) is given by \( \int_{a}^{b} y \, dx \).
Region OAC is the area bounded by the line AC, the x-axis, and the y-axis.



Step 1: Identify the function and the limits of integration

From part (ii), the equation of line AC is \( y = -\frac{4}{3}x + 4 \).
The region OAC starts at the y-axis (\( x = 0 \)) and ends at vertex C (\( x = 3 \)).
Therefore, the limits are from \( x = 0 \) to \( x = 3 \).


Step 2: Set up the definite integral for the area
\[ Area(OAC) = \int_{0}^{3} y \, dx \] \[ Area(OAC) = \int_{0}^{3} \left( -\frac{4}{3}x + 4 \right) dx \]


Step 3: Perform the integration
\[ Area(OAC) = \left[ -\frac{4}{3} \cdot \frac{x^2}{2} + 4x \right]_{0}^{3} \] \[ Area(OAC) = \left[ -\frac{2}{3}x^2 + 4x \right]_{0}^{3} \]


Step 4: Evaluate the definite integral
\[ Area(OAC) = \left( -\frac{2}{3}(3^2) + 4(3) \right) - \left( -\frac{2}{3}(0)^2 + 4(0) \right) \] \[ Area(OAC) = \left( -\frac{2}{3} \cdot 9 + 12 \right) - 0 \] \[ Area(OAC) = -6 + 12 = 6 sq units \] Quick Tip: Since OAC is a right-angled triangle, you can verify the result using the basic formula: Area \( = \frac{1}{2} \times base \times height = \frac{1}{2} \times 3 \times 4 = 6 \).

Concept:

The area of a region bounded by a curve \( y = f(x) \) and the x-axis is calculated using \( \int_{a}^{b} |y| \, dx \).
For regions in the second quadrant where \( x \) is negative, ensure limits are applied from left to right.



Step 1: Identify the function and limits for region AOB

From part (i), the equation of line AB is \( y = 2x + 4 \).
The region AOB is bounded by the line AB, the x-axis, and the y-axis.
The limits on the x-axis are from vertex B (\( x = -2 \)) to the origin (\( x = 0 \)).


Step 2: Set up the definite integral
\[ Area(AOB) = \int_{-2}^{0} y \, dx \] \[ Area(AOB) = \int_{-2}^{0} (2x + 4) dx \]


Step 3: Perform the integration
\[ Area(AOB) = \left[ \frac{2x^2}{2} + 4x \right]_{-2}^{0} \] \[ Area(AOB) = [ x^2 + 4x ]_{-2}^{0} \]


Step 4: Evaluate the limits to find the area
\[ Area(AOB) = (0^2 + 4(0)) - ((-2)^2 + 4(-2)) \] \[ Area(AOB) = 0 - (4 - 8) \] \[ Area(AOB) = 0 - (-4) = 4 sq units \] Quick Tip: Area is a scalar quantity and must always be positive. If your integration results in a negative value, check your limits or take the absolute value.

Concept:

Pythagorean Theorem: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides (\( c^2 = a^2 + b^2 \)).
The total width of the road is the sum of the distances from the point on the road to the bases of the two poles.



Step 1: Identify the geometry of the situation

Let the height of the larger pole be \( AB = 22 m \) and the smaller pole be \( CD = 16 m \).
The road width is \( BD = 20 m \).
Point R is on the road at a distance \( x \) from the smaller pole (CD), so \( RD = x \).
The distance from the larger pole (AB) is \( BR = BD - RD = 20 - x \).


Step 2: Find the length of ladder \( l_1 \) using Pythagoras

In right-angled triangle \( \triangle ABR \): \[ l_1^2 = AB^2 + BR^2 \] \[ l_1^2 = 22^2 + (20 - x)^2 \] \[ l_1 = \sqrt{484 + (20 - x)^2} \]


Step 3: Find the length of ladder \( l_2 \) using Pythagoras

In right-angled triangle \( \triangle CDR \): \[ l_2^2 = CD^2 + RD^2 \] \[ l_2^2 = 16^2 + x^2 \] \[ l_2 = \sqrt{256 + x^2} \]


Step 4: Combine to express \( p(x) \)
\[ p(x) = l_1 + l_2 \] \[ p(x) = \sqrt{484 + (20 - x)^2} + \sqrt{256 + x^2} \] Quick Tip: Always label your diagram points to avoid confusion between the distances to each pole.
Since point R is between the poles, the distances from it to the two poles must sum to the road width.

Concept:

Chain Rule for differentiation: \( \frac{d}{dx}[\sqrt{u}] = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx} \).



Step 1: Differentiate the first term of \( p(x) \)

Let \( p_1(x) = \sqrt{484 + (20 - x)^2} \).
Using the chain rule: \[ p_1'(x) = \frac{1}{2\sqrt{484 + (20 - x)^2}} \cdot \frac{d}{dx}[484 + (20 - x)^2] \] \[ p_1'(x) = \frac{1}{2\sqrt{484 + (20 - x)^2}} \cdot [0 + 2(20 - x)(-1)] \] \[ p_1'(x) = \frac{-(20 - x)}{\sqrt{484 + (20 - x)^2}} = \frac{x - 20}{\sqrt{484 + (20 - x)^2}} \]


Step 2: Differentiate the second term of \( p(x) \)

Let \( p_2(x) = \sqrt{256 + x^2} \).
Using the chain rule: \[ p_2'(x) = \frac{1}{2\sqrt{256 + x^2}} \cdot \frac{d}{dx}[256 + x^2] \] \[ p_2'(x) = \frac{1}{2\sqrt{256 + x^2}} \cdot [0 + 2x] \] \[ p_2'(x) = \frac{x}{\sqrt{256 + x^2}} \]


Step 3: Combine the derivatives
\[ p'(x) = p_1'(x) + p_2'(x) \] \[ p'(x) = \frac{x - 20}{\sqrt{484 + (20 - x)^2}} + \frac{x}{\sqrt{256 + x^2}} \] Quick Tip: Remember the derivative of \( (20-x)^2 \) involves an extra factor of \( -1 \) due to the chain rule on the internal term.

Concept:

To minimize a function \( f(x) \), find where \( f'(x) = 0 \) and check the second derivative \( f''(x) > 0 \).



Step 1: Formulate the function to minimize

Let \( S(x) = l_1^2 + l_2^2 \).
Substituting the expressions from part (i): \[ S(x) = [484 + (20 - x)^2] + [256 + x^2] \] \[ S(x) = 484 + 400 - 40x + x^2 + 256 + x^2 \] \[ S(x) = 2x^2 - 40x + 1140 \]


Step 2: Find the first derivative and solve for zero
\[ S'(x) = \frac{d}{dx}[2x^2 - 40x + 1140] \] \[ S'(x) = 4x - 40 \]
Setting \( S'(x) = 0 \): \[ 4x - 40 = 0 \implies x = 10 \]


Step 3: Confirm the minimum using the second derivative test
\[ S''(x) = \frac{d}{dx}[4x - 40] = 4 \]
Since \( S''(10) = 4 > 0 \), the function \( S(x) \) attains its minimum value at \( x = 10 \).
Thus, the sum of squares is minimum when point R is at a distance of 10 m from the smaller pole. Quick Tip: Minimizing the sum of squares is much simpler than minimizing the sum of lengths because it avoids square roots in the derivative.

Concept:



When the heights of the poles are equal, the setup becomes symmetric.



Step 1: Update the function with the new height


Now both poles have height 16 m.

\[ l_1^2 = 16^2 + (20 - x)^2 \] \[ l_2^2 = 16^2 + x^2 \]

The sum of squares \( S(x) \) is:

\[ S(x) = [256 + 400 - 40x + x^2] + [256 + x^2] \] \[ S(x) = 2x^2 - 40x + 912 \]


Step 2: Calculate the derivative and solve

\[ S'(x) = 4x - 40 \]
Setting \( S'(x) = 0 \): \[ 4x - 40 = 0 \implies x = 10 \]


Step 3: Determine the distance


Since \( x = 10 m \) and the total width is 20 m, the distance from the other pole is also \( 20 - 10 = 10 m \).


The ladders should be kept at the midpoint of the road (10 m from either pole). Quick Tip: In symmetric optimization problems, the extrema often occur at the geometric center of the constraints.

Concept:



The sum of probabilities of all mutually exclusive and exhaustive events is 1.



Step 1: Define the events


Let \( E_1 \) be the event that the student is a dropout.


Let \( E_2 \) be the event that the student is a regular student.



Step 2: Use given data to find probability


Given that 40% of the students are dropouts:

\[ P(E_1) = 40% = 0.40 = \frac{40}{100} = \frac{2}{5} \]


Step 3: Calculate the probability of a regular student


Since a student is either a dropout or a regular student:

\[ P(E_2) = 1 - P(E_1) \] \[ P(E_2) = 1 - 0.40 = 0.60 = \frac{3}{5} \]
The probability is 0.6. Quick Tip: Always identify the primary partitioning of the sample space before looking at conditional probabilities.

Concept:



If \( P(A) \) is the probability of an event happening, then \( 1 - P(A) \) is the probability of it not happening.



Step 1: Identify the conditional probability of qualifying


Let \( A \) be the event that a student qualifies the examination.


The problem states that of the dropouts (\( E_1 \)), 5% qualify:

\[ P(A|E_1) = 5% = 0.05 \]


Step 2: Find the probability of not qualifying


We need to find \( P(\bar{A}|E_1) \):

\[ P(\bar{A}|E_1) = 1 - P(A|E_1) \] \[ P(\bar{A}|E_1) = 1 - 0.05 = 0.95 \]
The probability is 0.95. Quick Tip: Read carefully: "selected from a group of dropout students" means the condition is already fixed, so it's a simple subtraction from 1.

Concept:

Bayes' Theorem: \( P(E_i|A) = \frac{P(E_i)P(A|E_i)}{\sum P(E_j)P(A|E_j)} \).



Step 1: List the known probabilities

Probability of being a dropout: \( P(E_1) = 0.40 \).
Probability of being a regular student: \( P(E_2) = 0.60 \).
Probability of qualifying given dropout: \( P(A|E_1) = 0.05 \).
Probability of qualifying given regular: \( P(A|E_2) = 0.10 \).


Step 2: Calculate total probability of qualifying
\[ P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2) \] \[ P(A) = (0.40 \times 0.05) + (0.60 \times 0.10) \] \[ P(A) = 0.02 + 0.06 = 0.08 \]


Step 3: Apply Bayes' Theorem to find \( P(Regular|Qualified) \)

The phrase "not a dropout" means the student is a regular student (\( E_2 \)). \[ P(E_2|A) = \frac{P(E_2)P(A|E_2)}{P(A)} \] \[ P(E_2|A) = \frac{0.60 \times 0.10}{0.08} = \frac{0.06}{0.08} \] \[ P(E_2|A) = \frac{6}{8} = 0.75 \]
The probability is 0.75. Quick Tip: Bayes' Theorem is essentially (Target Path Probability) / (Sum of all Path Probabilities).

Concept:

Total probability of the complementary event.



Step 1: Identify probabilities of not qualifying

Let \( \bar{A} \) be the event of not qualifying.
From dropouts: \( P(\bar{A}|E_1) = 0.95 \).
From regular students: \( P(\bar{A}|E_2) = 1 - 0.10 = 0.90 \).


Step 2: Calculate total probability of not qualifying
\[ P(\bar{A}) = P(E_1)P(\bar{A}|E_1) + P(E_2)P(\bar{A}|E_2) \] \[ P(\bar{A}) = (0.40 \times 0.95) + (0.60 \times 0.90) \] \[ P(\bar{A}) = 0.38 + 0.54 = 0.92 \]


Step 3: Apply Bayes' Theorem

We need to find \( P(E_2|\bar{A}) \): \[ P(E_2|\bar{A}) = \frac{P(E_2)P(\bar{A}|E_2)}{P(\bar{A})} \] \[ P(E_2|\bar{A}) = \frac{0.60 \times 0.90}{0.92} = \frac{0.54}{0.92} \] \[ P(E_2|\bar{A}) = \frac{54}{92} = \frac{27}{46} \approx 0.587 \]
The probability is 27/46. Quick Tip: Check if the denominators of parts (iii)(a) and (iii)(b) sum to 1. Here \( 0.08 + 0.92 = 1.0 \), which confirms total probability consistency.