CBSE Class 12 Mathematics Set 1- (65/2/1) Question Paper 2026 with Solution PDF

Concept:



A relation \(R\) on set \(A\) is reflexive if \((a, a) \in R\) for every \(a \in A\).

A relation \(R\) on set \(A\) is symmetric if \((a, b) \in R \implies (b, a) \in R\).

A relation \(R\) on set \(A\) is transitive if \((a, b) \in R\) and \((b, c) \in R \implies (a, c) \in R\).


Step 1: Checking for Reflexivity


For \(R\) to be reflexive on \(A = \{1, 2, 3\}\), the pairs \((1, 1)\), \((2, 2)\), and \((3, 3)\) must belong to \(R\).

In the given relation \(R = \{(1, 1), (2, 2), (1, 2)\}\), we see that \((3, 3) \notin R\).

Hence, the relation is not reflexive.


Step 2: Checking for Symmetry


For \(R\) to be symmetric, if \((1, 2) \in R\), then \((2, 1)\) must also be in \(R\).

Here, \((1, 2) \in R\) but \((2, 1) \notin R\).

Hence, the relation is not symmetric.


Step 3: Checking for Transitivity


We check all pairs \((a, b)\) and \((b, c)\) to see if \((a, c) \in R\).

For \((1, 1)\) and \((1, 2)\), we have \((1, 2) \in R\).

For \((1, 2)\) and \((2, 2)\), we have \((1, 2) \in R\).

There are no other pairs that violate the condition.

Hence, the relation is transitive only. Quick Tip: To check for transitivity, look for pairs where the second element of the first pair matches the first element of the second pair.
If you cannot find a counter-example \((a, b), (b, c) \in R\) but \((a, c) \notin R\), the relation is transitive.

Concept:

Matrix multiplication is not commutative in general, i.e., \(AB \neq BA\).

Algebraic identities in matrices must respect the order of multiplication.

Zero divisors exist in matrices; a product can be zero without either matrix being zero.


Step 1: Evaluating statement (i)
\((A + B)(A - B) = A(A - B) + B(A - B)\)
\(= A^2 - AB + BA - B^2\).

This is equal to \(A^2 - B^2\) only if \(AB = BA\).

Since \(AB \neq BA\) always, this statement is not always true.

Step 2: Evaluating statement (ii)
\(AB = BA\) is the commutative property.

Matrix multiplication is non-commutative in general.

Thus, this is not always true.

Step 3: Evaluating statement (iii)
\((A + B)^2 = (A + B)(A + B)\)
\(= A(A + B) + B(A + B)\)
\(= A^2 + AB + BA + B^2\).

This expansion is always valid for any square matrices \(A\) and \(B\).

Step 4: Evaluating statement (iv)

In matrices, \(AB = 0\) does not necessarily imply \(A = 0\) or \(B = 0\).

For example, let \(A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\) and \(B = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\).

Then \(AB = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\), but \(A \neq 0\) and \(B \neq 0\). Quick Tip: Remember that \((A+B)(A-B) = A^2 - B^2\) and \((A+B)^2 = A^2 + 2AB + B^2\) are ONLY true if matrices \(A\) and \(B\) commute (\(AB=BA\)).

Concept:

A square matrix \(A\) is symmetric if \(A = A^T\).

This implies the elements satisfy \(a_{ij} = a_{ji}\) for all \(i, j\).


Step 1: Identify relationships from symmetry

Since \(A\) is symmetric, \(A = A^T\):
\(\begin{bmatrix} 1 & a & b \\ -1 & 2 & c \\ 0 & 5 & 3 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 0 \\ a & 2 & 5 \\ b & c & 3 \end{bmatrix}\)

Step 2: Solve for variables \(a, b, c\)

Comparing corresponding elements:
\(a = -1\)
\(b = 0\)
\(c = 5\)

Step 3: Calculate the final expression

Substitute values into \(3a + b + c\):
\(3(-1) + 0 + 5\)
\(= -3 + 5\)
\(= 2\) Quick Tip: In a symmetric matrix, elements are mirrored across the main diagonal.
Simply equate \(a_{12} = a_{21}\), \(a_{13} = a_{31}\), and \(a_{23} = a_{32}\).

Concept:

\(A'\) denotes the transpose of matrix \(A\).

\(I\) is the identity matrix of the same order.

Matrix addition is performed element-wise.


Step 1: Determine the transpose \(A'\)

Given \(A = \begin{bmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{bmatrix}\).

The transpose Interchanges rows and columns:
\(A' = \begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix}\)

Step 2: Set up the equation \(A + A' = I\)
\(\begin{bmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{bmatrix} + \begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix} = \begin{bmatrix} 1 & 0
0 & 1 \end{bmatrix}\)
\(\begin{bmatrix} 2\cos x & 0 \\ 0 & 2\cos x \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

Step 3: Solve for \(x\)

Comparing the diagonal elements:
\(2\cos x = 1 \implies \cos x = \frac{1}{2}\).

For \(x \in [0, \pi/2]\), \(\cos x = \frac{1}{2}\) gives:
\(x = \frac{\pi}{3}\). Quick Tip: Adding a matrix to its transpose always results in a symmetric matrix.
The off-diagonal elements in this specific problem \((-\sin x + \sin x)\) automatically become zero.

Concept:



Inverse property of scalar multiplication: If \(k\) is a non-zero scalar and \(A\) is an invertible matrix, then \((kA)^{-1} = \frac{1}{k} A^{-1}\).


Step 1: Apply the scalar inverse property


In the given problem, the scalar is \(k = 3\).

Using the property \((kA)^{-1} = \frac{1}{k} A^{-1}\):
\((3A)^{-1} = \frac{1}{3} A^{-1}\).


Step 2: Verify the result


If we multiply \((3A)\) by \((\frac{1}{3}A^{-1})\):
\((3 \cdot \frac{1}{3}) (A \cdot A^{-1}) = 1 \cdot I = I\).

Since the product is the identity matrix, \(\frac{1}{3}A^{-1}\) is indeed the inverse.
Quick Tip: Do not confuse this with determinant properties.
For a matrix of order \(n\), \(|kA| = k^n |A|\), but for inverses, \((kA)^{-1} = \frac{1}{k} A^{-1}\) regardless of the order.

Concept:

Expansion of a \( 3 \times 3 \) determinant.

Formation and properties of quadratic equations.

Sum of roots of a quadratic equation \( Ax^2 + Bx + C = 0 \) is \( -\frac{B}{A} \).


Step 1: Expand the given determinant along the first row

Expand the determinant \( \Delta = -86 \):
\[ (-1) \begin{vmatrix} a & -1 \\ 4 & 2a \end{vmatrix} - (-2) \begin{vmatrix} -2 & -1 \\ 0 & 2a \end{vmatrix} + 5 \begin{vmatrix} -2 & a \\ 0 & 4 \end{vmatrix} = -86 \]

Evaluate the minor determinants:
\[ -1(2a^2 - (-4)) + 2(-4a - 0) + 5(-8 - 0) = -86 \]
\[ -1(2a^2 + 4) - 8a - 40 = -86 \]

Step 2: Simplify the equation to a standard quadratic form

Distribute the negative sign and combine constants:
\[ -2a^2 - 4 - 8a - 40 = -86 \]
\[ -2a^2 - 8a - 44 = -86 \]

Add 86 to both sides:
\[ -2a^2 - 8a + 42 = 0 \]

Divide by -2 to simplify coefficients:
\[ a^2 + 4a - 21 = 0 \]

Step 3: Find the sum of all possible values of \( a \)

The values of \( a \) are the roots of the quadratic equation \( a^2 + 4a - 21 = 0 \).

Using the sum of roots formula \( Sum = -\frac{b}{a} \):
\[ Sum = -\frac{4}{1} = -4 \] Quick Tip: Expand along the column or row with zeros to simplify calculations.
In questions asking for the 'sum' or 'product' of values, avoid solving for individual roots to save time.

Concept:

Differentiation of exponential functions \( \frac{d}{dx}(e^{f(x)}) = e^{f(x)} \cdot f'(x) \).

Implicit differentiation where \( y \) is treated as a function of \( x \).

Laws of exponents: \( \frac{e^A}{e^B} = e^{A-B} \).


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

The given equation is \( e^{-x} + e^{-y} = 2 \).

Differentiating both sides:
\[ \frac{d}{dx}(e^{-x}) + \frac{d}{dx}(e^{-y}) = \frac{d}{dx}(2) \]

Using the chain rule:
\[ e^{-x}(-1) + e^{-y}(-1)\frac{dy}{dx} = 0 \]
\[ -e^{-x} - e^{-y}\frac{dy}{dx} = 0 \]

Step 2: Isolate the derivative \( \frac{dy}{dx} \)

Move the term involving \( \frac{dy}{dx} \) to the right side:
\[ -e^{-x} = e^{-y}\frac{dy}{dx} \]

Divide both sides by \( e^{-y} \):
\[ \frac{dy}{dx} = \frac{-e^{-x}}{e^{-y}} \]

Step 3: Simplify the expression using exponent properties

Apply the quotient rule for exponents \( \frac{a^m}{a^n} = a^{m-n} \):
\[ \frac{dy}{dx} = -e^{-x - (-y)} \]
\[ \frac{dy}{dx} = -e^{-x + y} \]
\[ \frac{dy}{dx} = -e^{y - x} \] Quick Tip: When differentiating implicit functions, remember that every \( y \) term differentiated must be multiplied by \( \frac{dy}{dx} \).
Keep track of signs carefully when dealing with negative exponents.

Concept:

Critical points occur where \( f'(x) = 0 \).

Local maxima and minima are determined by checking the sign of \( f''(x) \).

Second Derivative Test: \( f''(x) < 0 \implies Max \), \( f''(x) > 0 \implies Min \).


Step 1: Find the first derivative and critical points

Given \( f(x) = x + x^{-1} \).

Differentiate with respect to \( x \):
\[ f'(x) = 1 - \frac{1}{x^2} \]

To find critical points, set \( f'(x) = 0 \):
\[ 1 - \frac{1}{x^2} = 0 \implies x^2 = 1 \implies x = 1, -1 \]

Step 2: Perform the Second Derivative Test

Differentiate \( f'(x) \) to get \( f''(x) \):
\[ f''(x) = \frac{d}{dx}(1 - x^{-2}) = 2x^{-3} = \frac{2}{x^3} \]

At \( x = 1 \):
\[ f''(1) = \frac{2}{1^3} = 2 > 0 \quad (Local Minimum at x=1) \]

At \( x = -1 \):
\[ f''(-1) = \frac{2}{(-1)^3} = -2 < 0 \quad (Local Maximum at x=-1) \]

Step 3: Calculate the extreme values

Local Minimum Value at \( x = 1 \):
\[ f(1) = 1 + \frac{1}{1} = 2 \]

Local Maximum Value at \( x = -1 \):
\[ f(-1) = -1 + \frac{1}{-1} = -2 \]

Thus, the local maximum value is \( -2 \). Quick Tip: Counter-intuitively, for some functions like this, the local maximum value can be smaller than the local minimum value.
Always verify using derivatives rather than assuming magnitudes.

Concept:

Standard integration formula: \( \int \frac{1}{1 + (kx)^2} dx = \frac{1}{k} \tan^{-1}(kx) \).

Fundamental Theorem of Calculus for definite integrals.

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


Step 1: Evaluate the definite integral

Rewrite the integrand: \( 1 + 4x^2 = 1 + (2x)^2 \).

The integral is:
\[ I = \int_{0}^{2a} \frac{1}{1 + (2x)^2} dx = \left[ \frac{1}{2} \tan^{-1}(2x) \right]_{0}^{2a} \]

Substituting the limits:
\[ I = \frac{1}{2} [\tan^{-1}(2(2a)) - \tan^{-1}(0)] \]
\[ I = \frac{1}{2} \tan^{-1}(4a) \]

Step 2: Solve for the variable \( a \)

Set the result equal to the given value \( \frac{\pi}{6} \):
\[ \frac{1}{2} \tan^{-1}(4a) = \frac{\pi}{6} \]

Multiply by 2:
\[ \tan^{-1}(4a) = \frac{\pi}{3} \]

Taking tangent of both sides:
\[ 4a = \tan\left(\frac{\pi}{3}\right) \]

Step 3: Final Calculation

Substitute the value \( \tan(\pi/3) = \sqrt{3} \):
\[ 4a = \sqrt{3} \]
\[ a = \frac{\sqrt{3}}{4} \] Quick Tip: Always remember the \( \frac{1}{k} \) factor when using the \( \tan^{-1} \) formula for terms like \( (kx)^2 \).
Double-check your inverse trig values for standard angles (\( 30^\circ, 45^\circ, 60^\circ \)).

Concept:



Area bounded by curves can be calculated using integration with respect to \( x \) or \( y \).

Area with respect to y-axis: \( \int x \, dy \).

Symmetry of curves (Parabola \( y = x^2 \) is symmetric about the y-axis).



Step 1: Analyze the region and boundaries


The curve is \( y = x^2 \), which means \( x = \pm \sqrt{y} \).

The horizontal boundary is \( y = 16 \).

The region starts at the origin \( (0,0) \) and goes up to \( y = 16 \).


Step 2: Set up integration with respect to \( y \)


Integration along the y-axis covers horizontal strips.


For one side (the first quadrant), the area is \( \int_{0}^{16} x \, dy \).


Substituting \( x = \sqrt{y} \):

\[ Area_{side} = \int_{0}^{16} \sqrt{y} \, dy \]


Step 3: Apply symmetry to find total area


Since the parabola is symmetric about the y-axis, the total area is double the area on one side.

Total Area = \( 2 \times \int_{0}^{16} \sqrt{y} \, dy \).

This matches option (D). Quick Tip: When a curve is symmetric about the y-axis, calculating area with respect to \( y \) and doubling is often the most direct method.
Visualizing the bounded region helps in determining the correct variable for integration.

Concept:

A first-order differential equation can be solved using the variable separable method if it can be written in the form \( f(y)dy = g(x)dx \).

The integral of \( x^n \) is \( \frac{x^{n+1}}{n+1} + C \).


Step 1: Separate the variables

The given differential equation is:
\[ \frac{dy}{dx} = \frac{\sqrt{y}}{\sqrt{x}} \]

Separating the terms of \( y \) on the left and \( x \) on the right:
\[ \frac{1}{\sqrt{y}} dy = \frac{1}{\sqrt{x}} dx \]

This can be written as:
\[ y^{-1/2} dy = x^{-1/2} dx \]

Step 2: Integrate both sides

Integrating both sides of the equation:
\[ \int y^{-1/2} dy = \int x^{-1/2} dx \]

Applying the power rule for integration:
\[ \frac{y^{1/2}}{1/2} = \frac{x^{1/2}}{1/2} + C' \]
\[ 2\sqrt{y} = 2\sqrt{x} + C' \]

Step 3: Simplify the general solution

Divide the entire equation by 2:
\[ \sqrt{y} = \sqrt{x} + \frac{C'}{2} \]

Let \( \frac{C'}{2} = C \), a constant:
\[ \sqrt{y} - \sqrt{x} = C \] Quick Tip: Always separate variables completely before integrating.
The constant of integration can be adjusted (e.g., \( C/2 \) becomes \( C \)) to match the form of the given options.

Concept:

A linear differential equation of the form \( \frac{dy}{dx} + P(x)y = Q(x) \) has an integrating factor (I.F.) given by \( e^{\int P(x) dx} \).

\( \int \frac{1}{x} dx = \log|x| \) and \( e^{\log f(x)} = f(x) \).


Step 1: Write the equation in standard form

The given equation is:
\[ 2x \frac{dy}{dx} - y = 3 \]

To get \( \frac{dy}{dx} \) with a coefficient of 1, divide the whole equation by \( 2x \):
\[ \frac{dy}{dx} - \frac{1}{2x} y = \frac{3}{2x} \]

Comparing this with \( \frac{dy}{dx} + P(x)y = Q(x) \), we get:
\[ P(x) = -\frac{1}{2x} \]

Step 2: Calculate the Integrating Factor

Using the formula \( I.F. = e^{\int P(x) dx} \):
\[ I.F. = e^{\int -\frac{1}{2x} dx} \]
\[ I.F. = e^{-\frac{1}{2} \int \frac{1}{x} dx} \]
\[ I.F. = e^{-\frac{1}{2} \log x} \]

Step 3: Simplify using logarithmic properties

Apply the property \( n \log a = \log a^n \):
\[ I.F. = e^{\log x^{-1/2}} \]

Since \( e^{\log f(x)} = f(x) \):
\[ I.F. = x^{-1/2} \]
\[ I.F. = \frac{1}{\sqrt{x}} \] Quick Tip: Before identifying \( P(x) \), ensure the coefficient of \( \frac{dy}{dx} \) is exactly 1.
Be careful with the negative sign in \( P(x) \), as it directly affects the exponent of the Integrating Factor.

Concept:

For any scalar \( \lambda \) and vector \( \vec{a} \), the magnitude of the product is \( |\lambda \vec{a}| = |\lambda| |\vec{a}| \).

The magnitude of a vector is always a non-negative real number.


Step 1: Express the magnitude function

Given \( |\vec{a}| = 5 \), the magnitude of \( \lambda \vec{a} \) is:
\[ f(\lambda) = |\lambda \vec{a}| = |\lambda| \cdot 5 \]

We are given the interval for \( \lambda \) as \( [-2, 1] \).

Step 2: Find the smallest value

The smallest value of \( |\lambda| \) in the range \( [-2, 1] \) is 0 (which occurs at \( \lambda = 0 \)).

Smallest value of \( |\lambda \vec{a}| = 0 \cdot 5 = 0 \).

Step 3: Find the greatest value

The greatest value of \( |\lambda| \) in the range \( [-2, 1] \) is \( |-2| = 2 \).

Greatest value of \( |\lambda \vec{a}| = 2 \cdot 5 = 10 \).

Step 4: Calculate the sum

Sum of greatest and smallest values:
\[ Sum = 10 + 0 = 10 \] Quick Tip: Magnitude is always absolute. Even if \( \lambda \) is negative, \( |\lambda \vec{a}| \) is positive.
In an interval spanning across zero, the minimum absolute value is always zero.

Concept:

If a vector makes angles \( \alpha, \beta, \gamma \) with the axes, its direction cosines are \( \cos \alpha, \cos \beta, \cos \gamma \).

Property of direction cosines: \( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \).

A vector \( \vec{v} \) of magnitude \( r \) is given by \( r(l\hat{i} + m\hat{j} + n\hat{k}) \).


Step 1: Determine the direction cosines

The vector is perpendicular to the \( z \)-axis, so \( \gamma = 90^\circ \).

This means \( n = \cos \gamma = \cos 90^\circ = 0 \).

It makes equal angles with \( x \) and \( y \) axes, so \( \alpha = \beta \).

This means \( l = m \).

Step 2: Solve for \( l \) and \( m \)

Using \( l^2 + m^2 + n^2 = 1 \):
\[ l^2 + l^2 + 0 = 1 \]
\[ 2l^2 = 1 \implies l = \frac{1}{\sqrt{2}} \]

So, \( l = \frac{1}{\sqrt{2}} \) and \( m = \frac{1}{\sqrt{2}} \).

Step 3: Construct the vector

Given magnitude \( r = 3 \):
\[ \vec{v} = 3 \left( \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} + 0\hat{k} \right) \]
\[ \vec{v} = \frac{3}{\sqrt{2}} \hat{i} + \frac{3}{\sqrt{2}} \hat{j} \]

Rationalizing the denominator (multiply by \( \sqrt{2}/\sqrt{2} \)):
\[ \vec{v} = \frac{3\sqrt{2}}{2} \hat{i} + \frac{3\sqrt{2}}{2} \hat{j} \] Quick Tip: Perpendicular to the z-axis implies the vector lies in the XY plane (z-component is 0).
Equal angles with X and Y axes in the XY plane always imply a \( 45^\circ \) inclination.

Concept:

Standard symmetric form of a line: \( \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \), where \( a, b, c \) are direction ratios.

Direction cosines \( l, m, n \) are given by \( \frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}} \).


Step 1: Convert to standard form

The given equations are:
\[ \frac{2x - 1}{4} = \frac{1 - y}{3} = \frac{-z}{6} \]

Adjust the numerators so coefficients of \( x, y, z \) are \( +1 \):
\[ \frac{2(x - 1/2)}{4} = \frac{-(y - 1)}{3} = \frac{z}{-6} \]

Divide denominators accordingly:
\[ \frac{x - 1/2}{2} = \frac{y - 1}{-3} = \frac{z}{-6} \]

Step 2: Identify direction ratios and magnitude

The direction ratios are \( (a, b, c) = (2, -3, -6) \).

Calculate the magnitude \( \sqrt{a^2 + b^2 + c^2} \):
\[ \sqrt{2^2 + (-3)^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]

Step 3: Calculate direction cosines

The direction cosines are:
\[ l = \frac{2}{7}, \quad m = \frac{-3}{7}, \quad n = \frac{-6}{7} \] Quick Tip: Always ensure \( x, y, z \) have a coefficient of \( +1 \) in the numerator before identifying direction ratios.
Direction cosines are simply direction ratios normalized by their magnitude.

Concept:

Linear Programming Problem (LPP) deals with optimization (maximization or minimization).

Constraints are linear inequalities or equations that restrict the variables.

The target function representing cost, profit, etc., is the central focus of optimization.


Step 1: Definition of components in LPP

In any LPP, we have variables (usually \( x, y \)) and a primary function that we want to either maximize (like profit) or minimize (like cost).

This function is of the form \( Z = ax + by \).

Step 2: Identify the specific term

The function \( Z = ax + by \), which represents the goal of the problem, is technically termed the "Objective Function".

The other terms like 'constraints' refer to the limitations, and 'feasible region' refers to the set of possible solutions. Quick Tip: Always identify the objective function first in an LPP; it is the only function containing the 'Z' variable.
Constraints are the 'subject to' inequalities, while the objective function is the 'maximize/minimize' equation.

Concept:

The equation of a line passing through intercepts \( (a, 0) \) and \( (0, b) \) is \( \frac{x}{a} + \frac{y}{b} = 1 \).

For a region shaded towards the origin (including \( (0,0) \)), the inequality for a line \( Ax + By = C \) (where \( C > 0 \)) is usually \( Ax + By \leq C \).


Step 1: Find the equation of the first boundary line

Observing the graph, the first line passes through \( (5, 0) \) and \( (0, 5) \).

Using the intercept form:
\[ \frac{x}{5} + \frac{y}{5} = 1 \implies x + y = 5 \]

Since the shaded region is towards the origin, the inequality is \( x + y \leq 5 \).

Step 2: Find the equation of the second boundary line

The second line passes through \( (9, 0) \) and \( (0, 3) \).

Using the intercept form:
\[ \frac{x}{9} + \frac{y}{3} = 1 \]

Multiply the entire equation by 9:
\[ x + 3y = 9 \]

Since the shaded region is towards the origin, the inequality is \( x + 3y \leq 9 \).

Step 3: Combine the constraints

The non-trivial constraints (excluding \( x, y \geq 0 \)) defining this feasible region are:
\( x + y \leq 5 \) and \( x + 3y \leq 9 \). Quick Tip: For lines with positive intercepts, a region shaded "below" or "towards the origin" always uses the \( \leq \) sign.
Quickly check the intercepts on the axes from the graph to write the line equations in seconds.

Concept:
Concept:

Conditional Probability: \( P(X/Y) = \frac{P(X \cap Y)}{P(Y)} \).

De Morgan's Law: \( A' \cap B' = (A \cup B)' \).

Complementary Event: \( P(E') = 1 - P(E) \).


Step 1: Apply the definition of conditional probability

By the definition of conditional probability:
\[ P(A'/B') = \frac{P(A' \cap B')}{P(B')} \]

Step 2: Use De Morgan's Law to simplify the numerator

According to De Morgan's Law, the intersection of complements is the complement of the union:
\[ A' \cap B' = (A \cup B)' \]

Therefore, the probability is:
\[ P(A' \cap B') = P((A \cup B)') \]

Step 3: Apply the complement rule

Using the rule \( P(E') = 1 - P(E) \):
\[ P((A \cup B)') = 1 - P(A \cup B) \]

Substituting this back into the expression from Step 1:
\[ P(A'/B') = \frac{1 - P(A \cup B)}{P(B')} \] Quick Tip: Complementary probabilities in conditional statements like \( P(A'/B) = 1 - P(A/B) \) only work if the condition (the part after the slash) remains identical.
When the condition itself is a complement (like \( B' \)), you must expand using the full formula.

Concept:

Dot Product: \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \).

Cross Product Magnitude: \( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \).

Trigonometric Identity: \( \sin^2 \theta + \cos^2 \theta = 1 \) and \( \tan \theta = \frac{\sin \theta}{\cos \theta} \).


Step 1: Verify Assertion (A)

Let \( |\vec{a}| = a \) and \( |\vec{b}| = b \).

LHS = \( (ab \sin \theta)^2 + (ab \cos \theta)^2 \)
\(= a^2b^2 \sin^2 \theta + a^2b^2 \cos^2 \theta \)
\(= a^2b^2 (\sin^2 \theta + \cos^2 \theta) \)
\(= a^2b^2 (1) = |\vec{a}|^2 |\vec{b}|^2 \).

Thus, Assertion (A) is true. This is known as Lagrange's Identity.

Step 2: Verify Reason (R)

RHS of Reason = \( (\vec{a} \cdot \vec{b}) \tan \theta \)
\(= (ab \cos \theta) \left( \frac{\sin \theta}{\cos \theta} \right) \)
\(= ab \sin \theta \).

LHS of Reason = \( |\vec{a} \times \vec{b}| = ab \sin \theta \).

Thus, Reason (R) is true.

Step 3: Check if R explains A

If we substitute \( |\vec{a} \times \vec{b}| = (\vec{a} \cdot \vec{b}) \tan \theta \) into the identity in (A):
\( (\vec{a} \cdot \vec{b})^2 \tan^2 \theta + (\vec{a} \cdot \vec{b})^2 = (\vec{a} \cdot \vec{b})^2 (1 + \tan^2 \theta) \)
\(= (\vec{a} \cdot \vec{b})^2 \sec^2 \theta = (ab \cos \theta)^2 \cdot \frac{1}{\cos^2 \theta} = a^2b^2 \).

Since the relation in R can be used to derive/prove A, it serves as a correct explanation. Quick Tip: Lagrange's Identity is one of the most important relationships between dot and cross products.
Whenever you see cross product squared and dot product squared added together, think of the Pythagorean identity \( \sin^2 + \cos^2 = 1 \).

Concept:



Direction Cosines \( (l, m, n) \) of a line must satisfy the identity \( l^2 + m^2 + n^2 = 1 \).

Maximum and minimum values of the cosine function are \( 1 \) and \( -1 \).


Step 1: Verify Assertion (A)


Let the given values be \( l = 1, m = 1, n = 1 \).


We check the sum of squares:

\( l^2 + m^2 + n^2 = 1^2 + 1^2 + 1^2 = 1 + 1 + 1 = 3 \).


Since \( 3 \neq 1 \), these values cannot be the direction cosines of any line.

Thus, Assertion (A) is false.


Step 2: Verify Reason (R)


The reason states that \( \cos \theta = 1 \) is possible for \( \theta = 0 \).


We know that the cosine of zero degrees is exactly 1.


Thus, Reason (R) is true.


Step 3: Conclusion


Since Assertion (A) is false and Reason (R) is true, we select option (D). Quick Tip: Direction cosines are essentially components of a unit vector; hence their sum of squares must always be 1.
Direction ratios can be \( <1, 1, 1> \), but direction cosines would then be \( <1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3}> \).

Concept:

A function \( f: A \to B \) is said to be onto (surjective) if every element in the codomain \( B \) has at least one pre-image in the domain \( A \).

This means the range of the function must be equal to its codomain.


Step 1: Express \( x \) in terms of \( y \)

Let \( y = f(x) \), where \( y \in R \) (Codomain).
\[ y = \frac{x - 2}{x - 3} \]

Multiplying both sides by \( (x - 3) \):
\[ y(x - 3) = x - 2 \]
\[ xy - 3y = x - 2 \]
\[ xy - x = 3y - 2 \]
\[ x(y - 1) = 3y - 2 \]
\[ x = \frac{3y - 2}{y - 1} \]

Step 2: Analyze the existence of pre-image

For \( x \) to be defined as a real number, the denominator \( (y - 1) \) must not be zero.

If \( y = 1 \), then there is no real value of \( x \) such that \( f(x) = 1 \).

Check if \( y = 1 \) is in the codomain:

The codomain is given as \( R \), which contains the element 1.

Step 3: Conclusion on onto property

Since \( 1 \in R \) (Codomain) but has no pre-image in the domain \( R - \{3\} \), the range of the function is \( R - \{1\} \).

As Range \( \neq \) Codomain, the function is not onto. Quick Tip: To check if a function is onto, solve for \( x \). If any value in the codomain makes the expression for \( x \) undefined, the function is not onto.
For rational functions of form \( \frac{ax+b}{cx+d} \), the range is always \( R - \{a/c\} \).

Concept:

A function \( f: A \to B \) is injective (one-to-one) if \( f(x_1) = f(x_2) \implies x_1 = x_2 \) for all \( x_1, x_2 \in A \).

For a mapping on a Cartesian product, we assume two pairs have equal images and prove the original pairs must be identical.


Step 1: Set up the equality of images

Let \( (x_1, y_1) \) and \( (x_2, y_2) \) be two elements in the domain \( Z \times Z \).

Suppose their images are equal:
\[ f(x_1, y_1) = f(x_2, y_2) \]

By the definition of the function:
\[ (2y_1, 3x_1) = (2y_2, 3x_2) \]

Step 2: Equate the corresponding components

For two ordered pairs to be equal, their corresponding components must be equal:
\[ 2y_1 = 2y_2 \implies y_1 = y_2 \]
\[ 3x_1 = 3x_2 \implies x_1 = x_2 \]

Step 3: Conclusion on injectivity

Since \( x_1 = x_2 \) and \( y_1 = y_2 \), it follows that:
\[ (x_1, y_1) = (x_2, y_2) \]

As equality of images implies equality of elements, the function \( f \) is injective. Quick Tip: When checking injectivity for vector-valued functions, treat each component separately.
If each coordinate function is independent and injective, the entire mapping is injective.

Concept:

Parametric differentiation: If \( x = f(t) \) and \( y = g(t) \), then \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \).

Chain rule for differentiation: \( \frac{d}{dt}[f(t)^n] = n f(t)^{n-1} f'(t) \).


Step 1: Differentiate \( x \) with respect to \( t \)

Given \( x = a \sin^3 t \):
\[ \frac{dx}{dt} = a \cdot 3 \sin^2 t \cdot \frac{d}{dt}(\sin t) \]
\[ \frac{dx}{dt} = 3a \sin^2 t \cos t \]

Step 2: Differentiate \( y \) with respect to \( t \)

Given \( y = b \cos^3 t \):
\[ \frac{dy}{dt} = b \cdot 3 \cos^2 t \cdot \frac{d}{dt}(\cos t) \]
\[ \frac{dy}{dt} = 3b \cos^2 t (-\sin t) \]
\[ \frac{dy}{dt} = -3b \cos^2 t \sin t \]

Step 3: Find the derivative \( \frac{dy}{dx} \)
\[ \frac{dy}{dx} = \frac{-3b \cos^2 t \sin t}{3a \sin^2 t \cos t} \]
\[ \frac{dy}{dx} = -\frac{b}{a} \cdot \frac{\cos t}{\sin t} \]
\[ \frac{dy}{dx} = -\frac{b}{a} \cot t \]

Step 4: Evaluate at \( t = \frac{\pi}{4} \)

At \( t = \frac{\pi}{4} \):
\[ \frac{dy}{dx} = -\frac{b}{a} \cot\left(\frac{\pi}{4}\right) \]

Since \( \cot(45^\circ) = 1 \):
\[ \frac{dy}{dx} = -\frac{b}{a} \] Quick Tip: Always keep the negative sign from the derivative of cosine visible throughout the calculation.
Simplifying the trigonometric ratio to \( \cot t \) makes the final substitution much faster.

Concept:

To find the absolute maximum of a function on a closed interval, we evaluate the function at critical points and at the endpoints of the interval.

Critical points are where \( f'(x) = 0 \).


Step 1: Find the derivative and critical points

Given \( f(x) = \cos x + \sin^2 x \).
\[ f'(x) = -\sin x + 2 \sin x \cos x \]
\[ f'(x) = \sin x (2 \cos x - 1) \]

Set \( f'(x) = 0 \):

1. \( \sin x = 0 \implies x = 0, \pi \) (Endpoints).

2. \( 2 \cos x - 1 = 0 \implies \cos x = \frac{1}{2} \implies x = \frac{\pi}{3} \).

Step 2: Evaluate the function at endpoints and critical points

- At \( x = 0 \):
\[ f(0) = \cos 0 + \sin^2 0 = 1 + 0 = 1 \]


- At \( x = \pi \):
\[ f(\pi) = \cos \pi + \sin^2 \pi = -1 + 0 = -1 \]


- At \( x = \frac{\pi}{3} \):
\[ f\left(\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) + \sin^2\left(\frac{\pi}{3}\right) = \frac{1}{2} + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{2} + \frac{3}{4} = \frac{5}{4} = 1.25 \]

Step 3: Determine the absolute maximum

Comparing the values: \( 1, -1, 1.25 \).

The greatest value is \( \frac{5}{4} \).

Hence, absolute maximum value is \( \frac{5}{4} \). Quick Tip: Always check the endpoints of the interval; local maxima from derivatives aren't always global maxima.
Converting the function to a single trig variable (using \( \sin^2 x = 1 - \cos^2 x \)) can sometimes simplify the calculation.

Concept:

Volume of a solid hemisphere, \( V = \frac{2}{3}\pi r^3 \).

Total Surface area of a solid hemisphere, \( S = 3\pi r^2 \).

Rate of change is the derivative with respect to time \( t \).


Step 1: Use the uniform rate of volume

Given that volume increases at a uniform rate, \( \frac{dV}{dt} = k \) (constant).

Since \( V = \frac{2}{3}\pi r^3 \):
\[ \frac{d}{dt}\left(\frac{2}{3}\pi r^3\right) = k \]
\[ \frac{2}{3}\pi (3r^2) \frac{dr}{dt} = k \]
\[ 2\pi r^2 \frac{dr}{dt} = k \implies \frac{dr}{dt} = \frac{k}{2\pi r^2} \]

Step 2: Find the rate of change of surface area

We need to find \( \frac{dS}{dt} \) where \( S = 3\pi r^2 \):
\[ \frac{dS}{dt} = \frac{d}{dt}(3\pi r^2) = 6\pi r \frac{dr}{dt} \]

Step 3: Substitute and prove

Substitute the expression for \( \frac{dr}{dt} \) into the surface area equation:
\[ \frac{dS}{dt} = 6\pi r \left( \frac{k}{2\pi r^2} \right) \]
\[ \frac{dS}{dt} = \frac{3k}{r} \]

Since \( 3k \) is a constant, we have:
\[ \frac{dS}{dt} \propto \frac{1}{r} \]

Hence, the rate of surface area varies inversely as its radius. Quick Tip: "Uniform rate" always implies the derivative with respect to time is constant.
Be careful to distinguish between Curved Surface Area (\( 2\pi r^2 \)) and Total Surface Area (\( 3\pi r^2 \)) for a solid hemisphere.

Concept:

The position vector of the midpoint of a line segment joining two points is the average of their position vectors.

Vector addition and the concept of relative position vectors: \( \vec{PQ} = \vec{Q} - \vec{P} \).


Step 1: Prove the vector expression for the median

Let the position vectors of vertices A, B, and C be \( \vec{a}, \vec{b}, \) and \( \vec{c} \) respectively.

By definition:
\( \vec{AB} = \vec{b} - \vec{a} \) and \( \vec{AC} = \vec{c} - \vec{a} \).

D is the midpoint of BC, so its position vector \( \vec{d} = \frac{\vec{b} + \vec{c}}{2} \).

The median vector \( \vec{AD} = \vec{d} - \vec{a} \):
\[ \vec{AD} = \frac{\vec{b} + \vec{c}}{2} - \vec{a} \]
\[ \vec{AD} = \frac{\vec{b} + \vec{c} - 2\vec{a}}{2} \]
\[ \vec{AD} = \frac{(\vec{b} - \vec{a}) + (\vec{c} - \vec{a})}{2} = \frac{\vec{AB} + \vec{AC}}{2} \] (Proved).

Step 2: Calculate the vector \( \vec{AD} \)
\[ \vec{AD} = \frac{(\hat{j} + \hat{k}) + (3\hat{i} - \hat{j} + 4\hat{k})}{2} \]
\[ \vec{AD} = \frac{3\hat{i} + 0\hat{j} + 5\hat{k}}{2} \]
\[ \vec{AD} = \frac{3}{2}\hat{i} + \frac{5}{2}\hat{k} \]

Step 3: Find the length (magnitude) of the median

Length \( |\vec{AD}| = \sqrt{\left(\frac{3}{2}\right)^2 + (0)^2 + \left(\frac{5}{2}\right)^2} \)
\[ |\vec{AD}| = \sqrt{\frac{9}{4} + \frac{25}{4}} \]
\[ |\vec{AD}| = \sqrt{\frac{34}{4}} = \frac{\sqrt{34}}{2} units \] Quick Tip: Think of the median as the vector diagonal of a parallelogram divided by two.
Midpoint formulas in vectors are algebraically identical to midpoint formulas in coordinate geometry.

Concept:

Any general point on a line in vector form can be written as \( (x, y, z) \) by grouping terms with and without \( \lambda \).

If a condition is given for coordinates, solve the linear equation in \( \lambda \).


Step 1: Find the general coordinates of any point on the line

The line is \( \vec{r} = (2\lambda)\hat{i} + (-1 - 2\lambda)\hat{j} + (3 + \lambda)\hat{k} \).

Coordinates of a general point \( P(x, y, z) \) are:
\[ x = 2\lambda \]
\[ y = -1 - 2\lambda \]
\[ z = 3 + \lambda \]

Step 2: Apply the sum condition

The problem states that the sum of coordinates is 3:
\[ x + y + z = 3 \]

Substitute the parametric expressions:
\[ (2\lambda) + (-1 - 2\lambda) + (3 + \lambda) = 3 \]

Step 3: Solve for \( \lambda \)

Combine like terms:
\[ 2\lambda - 2\lambda + \lambda - 1 + 3 = 3 \]
\[ \lambda + 2 = 3 \]
\[ \lambda = 1 \]

Step 4: Determine the coordinates of the point

Substitute \( \lambda = 1 \) back into the expressions for \( x, y, z \):
\[ x = 2(1) = 2 \]
\[ y = -1 - 2(1) = -3 \]
\[ z = 3 + 1 = 4 \]

Hence, the co-ordinates of the point are \( (2, -3, 4) \). Quick Tip: A vector line equation \( \vec{r} = \vec{a} + \lambda\vec{b} \) is just a collection of three parametric equations.
Always verify your final point by checking if the sum of its coordinates matches the given condition.

Concept:

Integration of the form \( \int \frac{px+q}{\sqrt{ax^2+bx+c}} \, dx \).

Express the numerator as \( px+q = A \cdot \frac{d}{dx}(quadratic) + B \).

Use substitution for the first part and completion of squares for the second part.


Step 1: Express the numerator in terms of the derivative of the quadratic

The derivative of the expression inside the square root is \( \frac{d}{dx}(9x - x^2) = 9 - 2x \).

Let \( x + 2 = A(9 - 2x) + B \).

Comparing coefficients of \( x \): \( 1 = -2A \implies A = -\frac{1}{2} \).

Comparing constants: \( 2 = 9A + B \implies 2 = 9(-\frac{1}{2}) + B \implies B = 2 + \frac{9}{2} = \frac{13}{2} \).

So, \( x + 2 = -\frac{1}{2}(9 - 2x) + \frac{13}{2} \).

Step 2: Split the integral into two parts
\[ I = \int \frac{-\frac{1}{2}(9 - 2x) + \frac{13}{2}}{\sqrt{9x - x^2}} \, dx \]
\[ I = -\frac{1}{2} \int \frac{9 - 2x}{\sqrt{9x - x^2}} \, dx + \frac{13}{2} \int \frac{1}{\sqrt{9x - x^2}} \, dx \]

Let \( I = I_1 + I_2 \).

Step 3: Evaluate the first integral \( I_1 \)

For \( I_1 = -\frac{1}{2} \int \frac{9 - 2x}{\sqrt{9x - x^2}} \, dx \), let \( 9x - x^2 = t \).

Then \( (9 - 2x) \, dx = dt \).
\[ I_1 = -\frac{1}{2} \int t^{-1/2} \, dt = -\frac{1}{2} [2t^{1/2}] = -\sqrt{9x - x^2} \]

Step 4: Evaluate the second integral \( I_2 \) by completing the square

For \( I_2 = \frac{13}{2} \int \frac{1}{\sqrt{9x - x^2}} \, dx \), consider the quadratic part:
\( 9x - x^2 = -(x^2 - 9x) = -[x^2 - 9x + (9/2)^2 - (9/2)^2] = (9/2)^2 - (x - 9/2)^2 \).
\[ I_2 = \frac{13}{2} \int \frac{dx}{\sqrt{(\frac{9}{2})^2 - (x - \frac{9}{2})^2}} \]

Using \( \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}(\frac{x}{a}) \):
\[ I_2 = \frac{13}{2} \sin^{-1}\left(\frac{x - 9/2}{9/2}\right) = \frac{13}{2} \sin^{-1}\left(\frac{2x - 9}{9}\right) \]

Step 5: Combine the results
\[ I = -\sqrt{9x - x^2} + \frac{13}{2} \sin^{-1}\left(\frac{2x - 9}{9}\right) + C \] Quick Tip: Always separate the integral into a 'substitution part' and a 'standard formula part'.
Check your values of A and B by expanding back to see if you get the original numerator.

Concept:

Use the property \( \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \).

Convert the cotangent function into sine and cosine.


Step 1: Simplify the integrand

Let \( I = \int_{\pi/12}^{5\pi/12} \frac{1}{1 + \sqrt{\frac{\cos x}{\sin x}}} \, dx = \int_{\pi/12}^{5\pi/12} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx \) --- (1)

Step 2: Apply the integration property

Sum of limits: \( a + b = \frac{\pi}{12} + \frac{5\pi}{12} = \frac{6\pi}{12} = \frac{\pi}{2} \).

Using \( \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(\frac{\pi}{2} - x) \, dx \):
\[ I = \int_{\pi/12}^{5\pi/12} \frac{\sqrt{\sin(\pi/2 - x)}}{\sqrt{\sin(\pi/2 - x)} + \sqrt{\cos(\pi/2 - x)}} \, dx \]
\[ I = \int_{\pi/12}^{5\pi/12} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} \, dx \] --- (2)

Step 3: Add equations (1) and (2)
\[ 2I = \int_{\pi/12}^{5\pi/12} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx \]
\[ 2I = \int_{\pi/12}^{5\pi/12} 1 \, dx = [x]_{\pi/12}^{5\pi/12} \]

Step 4: Solve for \( I \)
\[ 2I = \frac{5\pi}{12} - \frac{\pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3} \]
\[ I = \frac{\pi}{6} \] Quick Tip: Whenever the limits add up to \( \pi/2 \) in a trig integral, the answer is often half the difference of the limits.
The transformation from \( 1/(1+\sqrt{\cot x}) \) to \( \sqrt{\sin x}/(\sqrt{\sin x}+\sqrt{\cos x}) \) is a standard technique.

Concept:

Split the integral at points where the modulus argument changes sign (\( x = 0 \)).

Recall \( |x| = x \) for \( x \geq 0 \) and \( |x| = -x \) for \( x < 0 \).


Step 1: Split the integral at \( x = 0 \)
\[ I = \int_{-\pi/6}^{0} (\sin |x| + \cos |x|) \, dx + \int_{0}^{\pi/2} (\sin |x| + \cos |x|) \, dx \]

Step 2: Remove modulus using appropriate signs

For \( x \in [-\pi/6, 0] \), \( |x| = -x \).

For \( x \in [0, \pi/2] \), \( |x| = x \).
\[ I = \int_{-\pi/6}^{0} (\sin(-x) + \cos(-x)) \, dx + \int_{0}^{\pi/2} (\sin x + \cos x) \, dx \]

Since \( \sin(-x) = -\sin x \) and \( \cos(-x) = \cos x \):
\[ I = \int_{-\pi/6}^{0} (-\sin x + \cos x) \, dx + \int_{0}^{\pi/2} (\sin x + \cos x) \, dx \]

Step 3: Integrate each term
\[ I = [\cos x + \sin x]_{-\pi/6}^{0} + [-\cos x + \sin x]_{0}^{\pi/2} \]

Step 4: Evaluate limits

Part 1: \( (\cos 0 + \sin 0) - (\cos(-\pi/6) + \sin(-\pi/6)) = (1 + 0) - (\frac{\sqrt{3}}{2} - \frac{1}{2}) = 1 - \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{3 - \sqrt{3}}{2} \).

Part 2: \( (-\cos(\pi/2) + \sin(\pi/2)) - (-\cos 0 + \sin 0) = (0 + 1) - (-1 + 0) = 1 + 1 = 2 \).

Step 5: Final addition
\[ I = \frac{3 - \sqrt{3}}{2} + 2 = \frac{3 - \sqrt{3} + 4}{2} = \frac{7 - \sqrt{3}}{2} \] Quick Tip: Always handle absolute values by splitting the interval at zero.
Remember the parity of trig functions: \( \cos \) is even, \( \sin \) is odd.

Concept:

Integrating a rate of change function to find the original function.

Use algebraic manipulation to integrate functions involving \( e^x \).


Step 1: Set up the integral for \( F(x) \)
\[ F(x) = \int \frac{1}{e^x + 1} \, dx \]

Step 2: Use a trick to integrate

Multiply numerator and denominator by \( e^{-x} \):
\[ F(x) = \int \frac{e^{-x}}{1 + e^{-x}} \, dx \]

Alternatively, add and subtract \( e^x \) in numerator:
\[ \int \frac{1+e^x-e^x}{e^x+1} dx = \int (1 - \frac{e^x}{e^x+1}) dx \]

Step 3: Integrate

Using the second method:
\[ F(x) = x - \log(e^x + 1) + C \]

Step 4: Find the constant \( C \)

Given \( F(0) = \log(1/2) = -\log 2 \).
\[ -\log 2 = 0 - \log(e^0 + 1) + C \]
\[ -\log 2 = -\log(1 + 1) + C \]
\[ -\log 2 = -\log 2 + C \implies C = 0 \]

Step 5: State the final function
\[ F(x) = x - \log(e^x + 1) \] Quick Tip: For integrals like \( 1/(e^x+1) \), substituting \( t = e^x \) is another valid approach.
Remember that \( \log(1/2) = \log(2^{-1}) = -\log 2 \).

Concept:

This is a homogeneous differential equation because it involves the term \( y/x \).

Substitute \( y = vx \) and \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).


Step 1: Apply the substitution

Let \( y = vx \). Then \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).

The equation becomes:
\[ x \left(v + x \frac{dv}{dx}\right) = vx - x \sin^2 v \]

Divide both sides by \( x \):
\[ v + x \frac{dv}{dx} = v - \sin^2 v \]

Step 2: Separate the variables
\[ x \frac{dv}{dx} = -\sin^2 v \]
\[ \frac{dv}{\sin^2 v} = -\frac{dx}{x} \]
\[ \csc^2 v \, dv = -\frac{dx}{x} \]

Step 3: Integrate both sides
\[ \int \csc^2 v \, dv = -\int \frac{1}{x} \, dx \]
\[ -\cot v = -\log|x| + C \]
\[ \cot\left(\frac{y}{x}\right) = \log|x| + C \]

Step 4: Apply initial condition

Given \( y(1) = \pi/6 \). When \( x = 1 \), \( y = \pi/6 \).
\[ \cot\left(\frac{\pi/6}{1}\right) = \log|1| + C \]
\[ \cot\left(\frac{\pi}{6}\right) = 0 + C \]
\[ \sqrt{3} = C \]

Step 5: State the final solution
\[ \cot\left(\frac{y}{x}\right) = \log|x| + \sqrt{3} \] Quick Tip: In homogeneous equations, the \( v \) term usually cancels out immediately after substitution.
Check the domain of \( \cot \) to ensure your solution is valid for the given point.

Concept:

This is a linear differential equation in the form \( \frac{dx}{dy} + P(y)x = Q(y) \).

Calculate the Integrating Factor \( I.F. = e^{\int P(y) \, dy} \).


Step 1: Write the equation in standard form

Divide the whole equation by \( y \log y \):
\[ \frac{dx}{dy} + \frac{1}{y \log y} x = \frac{2}{y^2 \log y} \]

Here, \( P(y) = \frac{1}{y \log y} \) and \( Q(y) = \frac{2}{y^2 \log y} \).

Step 2: Calculate the Integrating Factor
\[ I.F. = e^{\int \frac{1}{y \log y} \, dy} \]

Let \( \log y = u \implies \frac{1}{y} \, dy = du \).
\[ \int \frac{1}{u} \, du = \log u = \log(\log y) \]
\[ I.F. = e^{\log(\log y)} = \log y \]

Step 3: Solve the differential equation

The solution is given by \( x \cdot (I.F.) = \int Q(y) \cdot (I.F.) \, dy \).
\[ x \log y = \int \frac{2}{y^2 \log y} \cdot \log y \, dy \]
\[ x \log y = \int \frac{2}{y^2} \, dy \]

Step 4: Integrate and simplify
\[ x \log y = 2 \left[-\frac{1}{y}\right] + C \]
\[ x \log y = C - \frac{2}{y} \] Quick Tip: When the derivative is \( dx/dy \), the equation is linear in \( x \).
The Integrating Factor \( e^{\log(something)} \) simply results in 'something'.

Concept:

Corner Point Method: Find the vertices of the feasible region and evaluate the objective function at each.

The region must satisfy all inequalities simultaneously in the first quadrant.


Step 1: Plot the lines and find the feasible region

Line 1: \( x + y = 50 \). Intercepts are \( (50, 0) \) and \( (0, 50) \).

Line 2: \( 2x + y = 80 \). Intercepts are \( (40, 0) \) and \( (0, 80) \).

Region \( x, y \geq 0 \) means first quadrant.

Region \( x + y \leq 50 \) is below Line 1.

Region \( 2x + y \leq 80 \) is below Line 2.

The common shaded region is a quadrilateral.

Step 2: Find the point of intersection of the two lines

Solve \( 2x + y = 80 \) and \( x + y = 50 \).

Subtracting the equations: \( (2x + y) - (x + y) = 80 - 50 \implies x = 30 \).

Substitute \( x = 30 \) in \( x + y = 50 \implies y = 20 \).

The intersection point is \( (30, 20) \).

Step 3: Identify the corner points of the feasible region

The corner points are:

1. \( O(0, 0) \)

2. \( A(40, 0) \) --- (Lower x-intercept among the two lines)

3. \( B(30, 20) \) --- (Intersection point)

4. \( C(0, 50) \) --- (Lower y-intercept among the two lines)

Step 4: Evaluate the objective function \( Z = 10500x + 9000y \)

At \( O(0, 0) \): \( Z = 0 \).

At \( A(40, 0) \): \( Z = 10500(40) + 0 = 4,20,000 \).

At \( B(30, 20) \): \( Z = 10500(30) + 9000(20) = 3,15,000 + 1,80,000 = 4,95,000 \).

At \( C(0, 50) \): \( Z = 0 + 9000(50) = 4,50,000 \).

Step 5: Find the maximum value

Comparing all values, the maximum value is \( 4,95,000 \).

It occurs at the point \( (30, 20) \).

Quick Tip: For \( \leq \) constraints with positive coefficients, the feasible region is always 'towards' the origin.
Always verify the intersection point by plugging it back into both constraint equations.

Concept:

Let \( p \) be the probability of hitting the target and \( q \) be the probability of missing.

Since hitting and missing are complementary events, \( p + q = 1 \).

Bernoulli trials are used for repeated independent experiments.


Step 1: Calculate individual probabilities \( p \) and \( q \)

Given that probability of hitting is three times the probability of missing:
\( p = 3q \)

Since \( p + q = 1 \):
\( 3q + q = 1 \implies 4q = 1 \implies q = \frac{1}{4} \)

Then \( p = 3 \times \frac{1}{4} = \frac{3}{4} \).

Step 2: Find the probability that the target is hit

The target is hit if at least one of the two shots hits.
\( P(target is hit) = 1 - P(both shots miss) \)
\( P(target is hit) = 1 - (q \times q) \)
\( P(target is hit) = 1 - \left(\frac{1}{4}\right)^2 = 1 - \frac{1}{16} = \frac{15}{16} \).

Step 3: Find the probability that atleast one shot misses

At least one shot misses is the complement of both shots hitting.
\( P(atleast one miss) = 1 - P(both shots hit) \)
\( P(atleast one miss) = 1 - (p \times p) \)
\( P(atleast one miss) = 1 - \left(\frac{3}{4}\right)^2 = 1 - \frac{9}{16} = \frac{7}{16} \). Quick Tip: "At least one" problems are almost always solved faster using the complement: \( 1 - P(none) \).
Always ensure that the sum of elementary probabilities \( p \) and \( q \) equals 1.

Concept:



The sample space \( S \) for 3 people (M, F, S) consists of \( 3! = 6 \) arrangements.

Conditional probability \( P(E/F) = \frac{P(E \cap F)}{P(F)} = \frac{n(E \cap F)}{n(F)} \).


Step 1: Identify the sample space and Event F


Sample space \( S = \{MFS, MSF, FMS, FSM, SMF, SFM\} \).


Event F (Father in the middle):

\( F = \{MFS, SFM\} \)


So, \( n(F) = 2 \).


Step 2: Identify Event E and intersection \( E \cap F \)


Event E (Son on one end):

\( E = \{MFS, FMS, SMF, SFM\} \)


The intersection \( E \cap F \) (Father in middle AND Son on an end):


In \( MFS \), Father is middle, Son is on an end.


In \( SFM \), Father is middle, Son is on an end.


So, \( E \cap F = \{MFS, SFM\} \) and \( n(E \cap F) = 2 \).


Step 3: Calculate the conditional probability

\( P(E/F) = \frac{n(E \cap F)}{n(F)} = \frac{2}{2} = 1 \). Quick Tip: If every arrangement in event F also satisfies event E, the conditional probability \( P(E/F) \) is always 1.
List out permutations for small sets to avoid logical errors.

Concept:

Matrix multiplication \( QP \).

If \( QP = kI \), then \( P^{-1} = \frac{1}{k}Q \).

A system of equations \( AX = B \) is solved as \( X = A^{-1}B \).


Step 1: Calculate the product \( QP \)
\[ QP = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix} \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} \]

Performing matrix multiplication:
\( R_1: (2\cdot1+2\cdot2-4\cdot0), (2\cdot-1+2\cdot3-4\cdot1), (2\cdot0+2\cdot4-4\cdot2) = [6, 0, 0] \)
\( R_2: (-4\cdot1+2\cdot2-4\cdot0), (-4\cdot-1+2\cdot3-4\cdot1), (-4\cdot0+2\cdot4-4\cdot2) = [0, 6, 0] \)
\( R_3: (2\cdot1-1\cdot2+5\cdot0), (2\cdot-1-1\cdot3+5\cdot1), (2\cdot0-1\cdot4+5\cdot2) = [0, 0, 6] \)

So, \( QP = 6I \).

Step 2: Express the system in matrix form \( AX = B \)

The equations are:
\( \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix} \)

This is of the form \( PX = B \), where \( P \) is the matrix given in the question.

Step 3: Solve for X

From \( QP = 6I \), we have \( P^{-1} = \frac{1}{6}Q \).
\( X = P^{-1}B = \frac{1}{6}QB \)
\( \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{6} \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix} \begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix} \)
\( x = \frac{1}{6}(2\cdot3 + 2\cdot17 - 4\cdot7) = \frac{1}{6}(6 + 34 - 28) = \frac{12}{6} = 2 \).
\( y = \frac{1}{6}(-4\cdot3 + 2\cdot17 - 4\cdot7) = \frac{1}{6}(-12 + 34 - 28) = \frac{-6}{6} = -1 \).
\( z = \frac{1}{6}(2\cdot3 - 1\cdot17 + 5\cdot7) = \frac{1}{6}(6 - 17 + 35) = \frac{24}{6} = 4 \).

Solution: \( x = 2, y = -1, z = 4 \). Quick Tip: The "hence" keyword implies you must use the matrix product result to find the inverse, rather than calculating adj(A) manually.
Verify the solution by plugging \( (2, -1, 4) \) back into the original equations.

Concept:

Properties of determinants.

Factoring out terms from rows or columns.


Step 1: Factor out x, y, z from columns

Since \( x, y, z \neq 0 \), factor out \( x \) from \( C_1 \), \( y \) from \( C_2 \), and \( z \) from \( C_3 \):
\( \Delta = xyz \begin{vmatrix} \frac{1}{x}+1 & \frac{1}{y} & \frac{1}{z} \\ \frac{1}{x} & \frac{1}{y}+1 & \frac{1}{z} \\ \frac{1}{x} & \frac{1}{y} & \frac{1}{z}+1 \end{vmatrix} \)

Step 2: Apply row transformation

Perform \( R_1 \to R_1 + R_2 + R_3 \):
\( \Delta = xyz \begin{vmatrix} 1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} & 1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} & 1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \\ \frac{1}{x} & \frac{1}{y}+1 & \frac{1}{z} \\ \frac{1}{x} & \frac{1}{y} & \frac{1}{z}+1 \end{vmatrix} \)

Factoring out \( (1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}) \) from \( R_1 \):
\( \Delta = xyz(1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}) \begin{vmatrix} 1 & 1 & 1 \\ \frac{1}{x} & \frac{1}{y}+1 & \frac{1}{z} \\ \frac{1}{x} & \frac{1}{y} & \frac{1}{z}+1 \end{vmatrix} \)

Step 3: Simplify the remaining determinant

Perform \( C_2 \to C_2 - C_1 \) and \( C_3 \to C_3 - C_1 \):
\( \Delta = xyz(1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}) \begin{vmatrix} 1 & 0 & 0 \\ \frac{1}{x} & 1 & 0 \\ \frac{1}{x} & 0 & 1 \end{vmatrix} \)

The determinant of the identity-like matrix is 1.
\( \Delta = xyz(1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}) = xyz + yz + xz + xy \).

Step 4: Prove the given condition

If \( \Delta = 0 \):
\( xyz(1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}) = 0 \)

Since \( x, y, z \neq 0 \), the product \( xyz \neq 0 \).
\( 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \)
\( x^{-1} + y^{-1} + z^{-1} = -1 \). (Proved) Quick Tip: Factoring out variables to create common terms in rows/columns is a very common trick for symmetric-looking determinants.
Always look for transformations that result in a row or column of 1s.

Concept:

A function is increasing where \( f'(x) > 0 \) and decreasing where \( f'(x) < 0 \).

Derivative of \( \cot^{-1} u \) is \( \frac{-1}{1+u^2} \frac{du}{dx} \).


Step 1: Find the derivative \( f'(x) \)

Let \( u = \sin x + \cos x \). Then \( f(x) = \cot^{-1} u \).
\( f'(x) = \frac{-1}{1 + (\sin x + \cos x)^2} \times (\cos x - \sin x) \)
\( f'(x) = \frac{\sin x - \cos x}{1 + (\sin x + \cos x)^2} \).

Step 2: Identify the sign of the derivative

The denominator \( 1 + (\sin x + \cos x)^2 \) is always positive for all \( x \).

The sign of \( f'(x) \) depends solely on the numerator \( \sin x - \cos x \).

Set \( f'(x) = 0 \):
\( \sin x - \cos x = 0 \implies \tan x = 1 \).

In the interval \( (0, \pi) \), \( x = \frac{\pi}{4} \).

Step 3: Determine intervals

1. For \( x \in (0, \pi/4) \): \( \cos x > \sin x \implies \sin x - \cos x < 0 \).

Thus, \( f'(x) < 0 \), so \( f(x) \) is decreasing.

2. For \( x \in (\pi/4, \pi) \):

- In \( (\pi/4, \pi/2) \), \( \sin x > \cos x \).

- In \( (\pi/2, \pi) \), \( \sin x > 0 \) and \( \cos x < 0 \), so \( \sin x - \cos x > 0 \).

In both cases, \( f'(x) > 0 \), so \( f(x) \) is increasing. Quick Tip: Use test values like \( x = \pi/6 \) and \( x = \pi/2 \) to check signs in sub-intervals quickly.
Ignore denominators that are always positive (like \( 1 + u^2 \)) when checking for inequalities.

Concept:



Let the sides of the rectangle be \( r \) and \( h \).

Volume of cylinder \( V = \pi r^2 h \).

Perimeter of rectangle \( 2(r + h) = 36 \).


Step 1: Express Volume as a single-variable function


Given \( 2(r + h) = 36 \implies r + h = 18 \implies h = 18 - r \).


Volume \( V(r) = \pi r^2 (18 - r) = \pi (18r^2 - r^3) \).


Step 2: Find critical points

\( V'(r) = \pi (36r - 3r^2) \).


Set \( V'(r) = 0 \):

\( 3\pi r (12 - r) = 0 \implies r = 12 \) (since \( r=0 \) is not possible).


Step 3: Verify maximum using second derivative test

\( V''(r) = \pi (36 - 6r) \).


At \( r = 12 \): \( V''(12) = \pi (36 - 72) = -36\pi < 0 \).


Since \( V''(r) \) is negative, the volume is maximum at \( r = 12 \).


Step 4: Determine dimensions


When \( r = 12 cm \):

\( h = 18 - 12 = 6 cm \).


The dimensions of the rectangle are 12 cm and 6 cm. Quick Tip: When a rectangle is revolved around side \( h \), that side becomes the height and the other side \( r \) becomes the radius.
In optimization, the variable with the higher power in the formula (here \( r \)) usually takes a larger share of the total sum.

Concept:

Domain of \( \cos^{-1} u \) is \( -1 \leq u \leq 1 \).

Range of \( \cos^{-1} x \) (principal) is \( [0, \pi] \).


Step 1: Find the domain

For \( \cos^{-1} (x^2 - 1) \) to be defined:
\( -1 \leq x^2 - 1 \leq 1 \)

Add 1 to all sides:
\( 0 \leq x^2 \leq 2 \)

Taking the square root:
\( -\sqrt{2} \leq x \leq \sqrt{2} \).

Domain: \( [-\sqrt{2}, \sqrt{2}] \).

Step 2: Solve for \( g(x) = \pi/3 \)
\( \cos^{-1} (x^2 - 1) = \frac{\pi}{3} \)
\( x^2 - 1 = \cos(\pi/3) = \frac{1}{2} \)
\( x^2 = 1 + \frac{1}{2} = \frac{3}{2} \)
\( x = \pm \sqrt{\frac{3}{2}} \).

Step 3: Write the range other than principal branch

The principal branch range is \( [0, \pi] \).

Any other interval of the form \( [k\pi, (k+1)\pi] \) for \( k \in Z, k \neq 0 \) is a valid branch.

Examples: \( [\pi, 2\pi] \) or \( [-\pi, 0] \). Quick Tip: When solving \( x^2 \leq a \), the solution is always the interval \( [-\sqrt{a}, \sqrt{a}] \).
The cosine function repeats its range every \( \pi \) units in different "chunks" for its inverse to exist.

Concept:

Equation of line through two points \( (x_1,y_1,z_1) \) and \( (x_2,y_2,z_2) \).

Intersection of lines: equating general points.

Perpendicularity: direction ratios are given by the cross product of the two lines' directions.


Step 1: Find the equation of line AB

Direction ratios of AB: \( (5-1, 8-2, 11-3) = (4, 6, 8) \to (2, 3, 4) \).

Equation of Line 1: \( \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \mu(2\hat{i} + 3\hat{j} + 4\hat{k}) \).

Step 2: Find the point of intersection

General point on Line 1: \( (1+2\mu, 2+3\mu, 3+4\mu) \).

General point on Line 2: \( (4+5\lambda, 1+2\lambda, \lambda) \).

Equating the coordinates:

1) \( 1 + 2\mu = 4 + 5\lambda \implies 2\mu - 5\lambda = 3 \)

2) \( 2 + 3\mu = 1 + 2\lambda \implies 3\mu - 2\lambda = -1 \)

3) \( 3 + 4\mu = \lambda \implies \lambda - 4\mu = 3 \).

Substitute (3) into (1): \( 2\mu - 5(3 + 4\mu) = 3 \implies 2\mu - 15 - 20\mu = 3 \implies -18\mu = 18 \implies \mu = -1 \).

Then \( \lambda = 3 + 4(-1) = -1 \).

Point of intersection: \( (1+2(-1), 2+3(-1), 3+4(-1)) = (-1, -1, -1) \).

Step 3: Find the perpendicular direction

Directions of the two lines: \( \vec{d_1} = (2, 3, 4) \) and \( \vec{d_2} = (5, 2, 1) \).

Direction of perpendicular line \( \vec{n} = \vec{d_1} \times \vec{d_2} \):
\( \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & 3 & 4
5 & 2 & 1 \end{vmatrix} = \hat{i}(3-8) - \hat{j}(2-20) + \hat{k}(4-15) = -5\hat{i} + 18\hat{j} - 11\hat{k} \).

Step 4: Write the required equation

The line passes through \( (-1, -1, -1) \) with DRs \( (-5, 18, -11) \).

Equation: \( \frac{x+1}{-5} = \frac{y+1}{18} = \frac{z+1}{-11} \). Quick Tip: Always check the intersection point in the third coordinate equation to ensure the lines actually intersect.
The cross product of two vectors gives a vector perpendicular to both, which is perfect for finding the direction of a common perpendicular.

Concept:



Probability of an event is the ratio of the number of favorable outcomes to the total number of outcomes.
\( P(E) = \frac{n(E)}{n(S)} \).


Step 1: Identify the total number of smokers and the number of female smokers


The total number of smokers in the colony is given as \( 50 \).


The number of male smokers is given as \( 30 \).


The number of female smokers is \( 50 - 30 = 20 \).

Step 2: Calculate the probability of selecting a female


Let \( F \) be the event that the selected person is a female.

\[ P(F) = \frac{Number of females}{Total number of smokers} \]
\[ P(F) = \frac{20}{50} \]

\[ P(F) = \frac{2}{5} = 0.4 \] Quick Tip: Always identify the total sample space clearly before calculating probabilities. Reducing fractions to decimals can make comparisons easier in multi-part questions.

Concept:

The probability of a complement event \( P(E') = 1 - P(E) \).
Conditional probability \( P(L'|M) \) represents the probability of not having lung problems given the person is male.


Step 1: Determine the probability of a male smoker having lung problems

From the study, \( 170 \) out of \( 1000 \) males develop complications.

Let \( M \) be the event that the person is male and \( L \) be the event that the person has lung problems.
\[ P(L|M) = \frac{170}{1000} = 0.17 \]

Step 2: Calculate the probability of not having lung problems

The probability that he will not suffer from lung problems is the complement:
\[ P(L'|M) = 1 - P(L|M) \]
\[ P(L'|M) = 1 - 0.17 \]
\[ P(L'|M) = 0.83 \] Quick Tip: Subtraction from 1 is the fastest way to find the probability of "not" happening. Ensure you use the rate provided for the specific gender mentioned in the condition.

Concept:

Bayes' Theorem is used to find reverse conditional probabilities.
\( P(F|L) = \frac{P(F) \cdot P(L|F)}{P(M) \cdot P(L|M) + P(F) \cdot P(L|F)} \).


Step 1: Define the known probabilities

Let \( M \): Male smoker, \( F \): Female smoker, \( L \): Person has lung complications.
\( P(M) = \frac{30}{50} = 0.6 \)
\( P(F) = \frac{20}{50} = 0.4 \)
\( P(L|M) = \frac{170}{1000} = 0.17 \)
\( P(L|F) = \frac{120}{1000} = 0.12 \)

Step 2: Apply Bayes' Theorem

We need to find \( P(F|L) \):
\[ P(F|L) = \frac{0.4 \times 0.12}{(0.6 \times 0.17) + (0.4 \times 0.12)} \]
\[ P(F|L) = \frac{0.048}{0.102 + 0.048} \]
\[ P(F|L) = \frac{0.048}{0.150} \]

Step 3: Simplify the fraction
\[ P(F|L) = \frac{48}{150} \]

Dividing by 6:
\[ P(F|L) = \frac{8}{25} = 0.32 \] Quick Tip: The denominator in Bayes' Theorem represents the "Total Probability" of the condition occurring. Double-check your decimal multiplications (e.g., \( 0.4 \times 0.12 \)) to avoid calculation errors.

Concept:

Bayes' Theorem for the complement event \( L' \).
Total probability of not having lung problems must be calculated first.


Step 1: Calculate complement conditional probabilities
\( P(L'|M) = 1 - 0.17 = 0.83 \)
\( P(L'|F) = 1 - 0.12 = 0.88 \)

Step 2: Apply Bayes' Theorem for \( P(M|L') \)
\[ P(M|L') = \frac{P(M) \cdot P(L'|M)}{P(M) \cdot P(L'|M) + P(F) \cdot P(L'|F)} \]
\[ P(M|L') = \frac{0.6 \times 0.83}{(0.6 \times 0.83) + (0.4 \times 0.88)} \]
\[ P(M|L') = \frac{0.498}{0.498 + 0.352} \]
\[ P(M|L') = \frac{0.498}{0.850} \]

Step 3: Simplify the final probability
\[ P(M|L') = \frac{498}{850} = \frac{249}{425} \approx 0.585 \] Quick Tip: When dealing with "not having problems," ensure you use the complement rates (0.83 and 0.88). Bayes' Theorem works exactly the same way for "success" or "failure" conditions.

Concept:

An ellipse equation \( Ax^2 + By^2 = C \) can be rearranged to express one variable in terms of another.
Standard form is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).


Step 1: Isolate the \( y^2 \) term

Starting with the equation:
\[ 9x^2 + 16y^2 = 144 \]

Subtract \( 9x^2 \) from both sides:
\[ 16y^2 = 144 - 9x^2 \]

Step 2: Solve for \( y \)

Divide by 16:
\[ y^2 = \frac{144 - 9x^2}{16} \]
\[ y^2 = \frac{9(16 - x^2)}{16} \]

Taking the square root:
\[ y = \frac{3}{4} \sqrt{16 - x^2} \] (considering the upper half for the function). Quick Tip: Factoring out constants before dividing helps in simplifying square roots. In area problems, we usually take the positive root to represent the curve in the first and second quadrants.

Concept:

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

We need to evaluate \( \int \frac{3}{4} \sqrt{16 - x^2} \, dx \).

Identify \( a^2 = 16 \), so \( a = 4 \).

Step 2: Apply the standard integration formula
\[ \int \frac{3}{4} \sqrt{4^2 - x^2} \, dx = \frac{3}{4} \left[ \frac{x}{2} \sqrt{16 - x^2} + \frac{16}{2} \sin^{-1} \left( \frac{x}{4} \right) \right] \]
\[ = \frac{3}{4} \left[ \frac{x}{2} \sqrt{16 - x^2} + 8 \sin^{-1} \left( \frac{x}{4} \right) \right] \]
\[ = \frac{3x}{8} \sqrt{16 - x^2} + 6 \sin^{-1} \left( \frac{x}{4} \right) + C \] Quick Tip: Memorize the \( \sqrt{a^2 - x^2} \) integral formula; it is the most common formula in area-under-curve problems. Distribute the outside constant carefully to each term in the bracket.

Concept:

Area of an ellipse is \( 4 \times \) area in the first quadrant.
Total area = \( 4 \int_{0}^{a} y \, dx \).


Step 1: Determine the limits of integration

From the equation \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \), the semi-major axis is \( a = 4 \).

The limits for the first quadrant are \( x = 0 \) to \( x = 4 \).

Step 2: Calculate the definite integral

Area = \( 4 \int_{0}^{4} \frac{3}{4} \sqrt{16 - x^2} \, dx = 3 \int_{0}^{4} \sqrt{16 - x^2} \, dx \).

Using the integrated result from (ii):
\[ Area = 3 \left[ \frac{x}{2} \sqrt{16 - x^2} + 8 \sin^{-1} \left( \frac{x}{4} \right) \right]_{0}^{4} \]
\[ Area = 3 \left[ (0 + 8 \sin^{-1}(1)) - (0 + 8 \sin^{-1}(0)) \right] \]
\[ Area = 3 \left[ 8 \cdot \frac{\pi}{2} - 0 \right] \]
\[ Area = 12\pi sq m \] Quick Tip: For an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), the area is simply \( \pi ab \).
Calculation: \( \pi \times 4 \times 3 = 12\pi \). Use this to verify your integration result.

Concept:

The outer boundary is formed by adding the width to the inner semi-axes.
Area of triangle bounded by a line and axes is \( \int y \, dx \).


Step 1: Find the outer coordinates

Inner semi-axes: \( a = 4, b = 3 \).

Width of track = \( 3 m \).

Outer semi-axes: \( A = 4 + 3 = 7, B = 3 + 3 = 6 \).

Point \( P \) (on \( x \)-axis): \( (7, 0) \).

Point \( Q \) (on \( y \)-axis): \( (0, 6) \).

Origin \( O \): \( (0, 0) \).

Step 2: Find the equation of line PQ and integrate

Equation of line in intercept form: \( \frac{x}{7} + \frac{y}{6} = 1 \implies y = 6 \left( 1 - \frac{x}{7} \right) \).

Area \( \Delta POQ = \int_{0}^{7} 6 \left( 1 - \frac{x}{7} \right) \, dx \).
\[ Area = 6 \left[ x - \frac{x^2}{14} \right]_{0}^{7} \]
\[ Area = 6 \left[ 7 - \frac{49}{14} \right] = 6 [ 7 - 3.5 ] = 6 \times 3.5 = 21 sq m \] Quick Tip: Always visualize the width correctly; it is added to both dimensions. Area of a right-angled triangle is \( \frac{1}{2} \times base \times height = \frac{1}{2} \cdot 7 \cdot 6 = 21 \).

Concept:

Power Rule of differentiation: \( \frac{d}{dx}(x^n) = nx^{n-1} \).


Step 1: Identify the function in the given interval

For the interval \( 0 < x < 3 \), the function is defined as:
\[ f(x) = x^4 - 4x^2 + 4 \]

Step 2: Differentiate term by term
\[ f'(x) = \frac{d}{dx}(x^4) - \frac{d}{dx}(4x^2) + \frac{d}{dx}(4) \]
\[ f'(x) = 4x^{4-1} - 4(2x^{2-1}) + 0 \]
\[ f'(x) = 4x^3 - 8x \] Quick Tip: The derivative of a constant is always zero. Ensure you are using the correct piece of the piecewise function for the given domain.

Concept:



For a piecewise function, determine the derivative of the specific piece containing the point.


Step 1: Identify the relevant piece of the function


Since \( x = 4 \) satisfies the condition \( x \geq 3 \), we use the second piece:

\[ f(x) = x^2 + 40 \]

Step 2: Differentiate and substitute


Derivative for \( x > 3 \):

\[ f'(x) = \frac{d}{dx}(x^2 + 40) = 2x \]


Substitute \( x = 4 \):

\[ f'(4) = 2(4) = 8 \] Quick Tip: Always check the domain conditions \( x < 3 \) or \( x \geq 3 \) before differentiating. Pointwise derivatives are calculated from the general derivative of that interval.

Concept:



A function is continuous at \( x = c \) if LHL = RHL = \( f(c) \).


Step 1: Calculate Left Hand Limit (LHL)

\[ LHL = \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (x^4 - 4x^2 + 4) \]

\[ LHL = 3^4 - 4(3^2) + 4 = 81 - 36 + 4 = 49 \]


Step 2: Calculate Right Hand Limit (RHL) and \( f(3) \)

\[ RHL = \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (x^2 + 40) \]

\[ RHL = 3^2 + 40 = 9 + 40 = 49 \]


Also, \( f(3) = 3^2 + 40 = 49 \).


Step 3: Conclude the test


Since \( LHL = RHL = f(3) = 49 \), the function \( f(x) \) is continuous at \( x = 3 \). Quick Tip: Continuity ensures there are no jumps or holes in the track design. If any one limit differs, the function is immediately classified as discontinuous.

Concept:

A function is differentiable at \( x = c \) if LHD = RHD.


Step 1: Calculate Left Hand Derivative (LHD)

For \( x < 3 \), \( f'(x) = 4x^3 - 8x \).
\[ LHD = \lim_{x \to 3^-} (4x^3 - 8x) \]
\[ LHD = 4(3^3) - 8(3) = 4(27) - 24 = 108 - 24 = 84 \]

Step 2: Calculate Right Hand Derivative (RHD)

For \( x > 3 \), \( f'(x) = 2x \).
\[ RHD = \lim_{x \to 3^+} (2x) \]
\[ RHD = 2(3) = 6 \]

Step 3: Conclude the test

Since \( LHD = 84 \) and \( RHD = 6 \), and \( LHD \neq RHD \), the function \( f(x) \) is not differentiable at \( x = 3 \). Quick Tip: A function can be continuous but not differentiable (a "sharp corner"). Always calculate the limits of the derived pieces from both sides to check differentiability.