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

Concept:

The principal value branch of the inverse cosine function, \( \cos^{-1}x \), is defined for \( x \in [-1, 1] \).
The range of the principal value branch for \( \cos^{-1}x \) is \( [0, \pi] \).
This implies that for any valid input \( x \), the output value lies between \( 0 \) and \( \pi \) inclusive.


Step 1: Determine the range of the standard inverse function

By definition, the range of the principal value of \( \cos^{-1}x \) is: \[ 0 \le \cos^{-1}x \le \pi \]

Step 2: Apply the scalar multiplier to the inequality

The given equation is \( y = 2 \cos^{-1}x \).
To find the range of \( y \), we multiply every term in the inequality established in Step 1 by the positive constant \( 2 \): \[ 2 \times 0 \le 2 \times \cos^{-1}x \le 2 \times \pi \] \[ 0 \le 2 \cos^{-1}x \le 2\pi \]

Step 3: Substitute the variable and identify the final range

Replacing the middle term with \( y \), we get: \[ 0 \le y \le 2\pi \]
Comparing this result with the given options, it matches option (C). Quick Tip: For range problems involving \( k \cdot f^{-1}(x) \), find the standard range of \( f^{-1}(x) \) first and then multiply the bounds by the constant \( k \).
Recall: \( \sin^{-1}x \in [-\pi/2, \pi/2] \), \( \cos^{-1}x \in [0, \pi] \), and \( \tan^{-1}x \in (-\pi/2, \pi/2) \).

Concept:

A matrix is classified based on its dimensions (rows \( \times \) columns).
A row matrix is a specific type of matrix that contains only one row.
In the standard notation \( m \times n \), \( m \) represents the number of rows and \( n \) represents the number of columns.
For a row matrix, the condition \( m = 1 \) must strictly hold true.


Step 1: Define the mathematical requirement for a row matrix

For any matrix of order \( m \times n \) to be a row matrix, the number of rows (\( m \)) must be exactly 1.
The number of columns (\( n \)) can be any natural number.

Step 2: Analyze each provided option


Option (B): \( 1 \times 2 \). Here, \( m = 1 \). This has 1 row and 2 columns. It is a row matrix.
Option (C): \( 1 \times 1 \). Here, \( m = 1 \). This has 1 row and 1 column. It is both a row matrix and a column matrix.
Option (D): \( 1 \times n \). Here, \( m = 1 \). This represents the general form of any row matrix with \( n \) elements.
Option (A): \( 2 \times 1 \). Here, \( m = 2 \). Since there are two rows, this does not satisfy the definition of a row matrix. It is actually a column matrix.


Step 3: Conclusion

Since option (A) has more than one row, it cannot be a row matrix. Quick Tip: Row matrix \( \implies \) Rows = 1 (Order is \( 1 \times n \)).
Column matrix \( \implies \) Columns = 1 (Order is \( m \times 1 \)).
A \( 1 \times 1 \) matrix is the only matrix that is simultaneously a row matrix and a column matrix.

Concept:

The transpose of a matrix (denoted by \( A' \)) is obtained by swapping rows and columns.
Key properties of transposes:

Sum Rule: \( (A + B)' = A' + B' \)
Difference Rule: \( (A - B)' = A' - B' \)
Reversal Law for Multiplication: \( (AB)' = B'A' \)
Scalar Rule: \( (kA)' = kA' \)



Step 1: Evaluate statement (i)

Property (i) states \( (A + B)' = A' + B' \).
This is a standard property of matrix addition and transposition. Addition is commutative and linear under transposition.
Conclusion: (i) is TRUE.

Step 2: Evaluate statement (ii)

Property (ii) states \( (A - B)' = B' - A' \).
The correct property is \( (A - B)' = A' - B' \).
Since matrix subtraction is not commutative (\( A' - B' \neq B' - A' \)), statement (ii) is false.
Conclusion: (ii) is FALSE.

Step 3: Evaluate statement (iii)

Property (iii) states \( (AB)' = A'B' \).
According to the reversal law of transposes, the order of multiplication must be reversed: \( (AB)' = B'A' \).
Since matrix multiplication is generally non-commutative, \( B'A' \neq A'B' \).
Conclusion: (iii) is FALSE.

Step 4: Evaluate statement (iv)

Property (iv) states \( (kAB)' = kB'A' \).
Using the scalar rule: \( (k \cdot [AB])' = k \cdot (AB)' \).
Now, applying the reversal law to \( (AB)' \): \( k \cdot (AB)' = k(B'A') = kB'A' \).
Conclusion: (iv) is TRUE.

Step 5: Final Selection

Since only (i) and (iv) are true, the correct option is (D). Quick Tip: Always remember the "Reversal Law": When you take the transpose or inverse of a product, the individual matrices flip their positions: \( (AB)' = B'A' \).
Linear operations like addition and subtraction do NOT flip the order.

Concept:

The determinant of a diagonal matrix is the product of the elements on its principal diagonal.
For a general matrix, the determinant can be computed by expansion along any row or column.
Determinants also follow property rules: swapping two rows changes the sign of the determinant.


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

The matrix in \( \Delta_1 \) is a diagonal matrix. \[ \Delta_1 = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{vmatrix} \]
Using the property of diagonal determinants: \[ \Delta_1 = 1 \times 2 \times 3 = 6 \]

Step 2: Calculate the value of \( \Delta_2 \)
\[ \Delta_2 = \begin{vmatrix} 0 & 2 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 6 \end{vmatrix} \]
Expand along the first row (R1): \[ \Delta_2 = 0 \cdot minor - 2 \cdot \begin{vmatrix} 1 & 0 \\ 0 & 6 \end{vmatrix} + 0 \cdot minor \] \[ \Delta_2 = -2(1 \times 6 - 0 \times 0) \] \[ \Delta_2 = -2(6) = -12 \]

Step 3: Find the relationship between the two values

We have \( \Delta_1 = 6 \) and \( \Delta_2 = -12 \).
To express \( \Delta_2 \) in terms of \( \Delta_1 \): \[ \Delta_2 = -12 = -2 \times (6) \] \[ \Delta_2 = -2\Delta_1 \]
This matches option (B). Quick Tip: You can also see that swapping \( R_1 \) and \( R_2 \) in \( \Delta_2 \) gives \[ \begin{vmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 6 \end{vmatrix}. \] Swapping rows changes the sign of the determinant, so \[ \Delta_2 = - \begin{vmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 6 \end{vmatrix}. \].
Taking factor 2 out from the last row/column also works, but direct calculation is fastest here.

Concept:

For a \( 2 \times 2 \) determinant \( \begin{vmatrix} a & b \\ c & d \end{vmatrix} \), the value is \( ad - bc \).
Trigonometric double angle identity: \( 2 \sin \theta \cos \theta = \sin 2\theta \).
Principal values for trigonometric equations like \( \sin \theta = 1 \).


Step 1: Expand the determinant expression

Given determinant: \[ \begin{vmatrix} \cos x & \sin x \\ -\cos x & \sin x \end{vmatrix} = 1 \]
Applying the determinant formula: \[ (\cos x)(\sin x) - (\sin x)(-\cos x) = 1 \] \[ \sin x \cos x + \sin x \cos x = 1 \] \[ 2 \sin x \cos x = 1 \]

Step 2: Simplify using trigonometric identities

The expression on the left-hand side is the standard double angle formula for sine: \[ 2 \sin x \cos x = \sin 2x \]
Substituting this back into the equation: \[ \sin 2x = 1 \]

Step 3: Solve for \( x \)

For a sine function to be equal to \( 1 \), the angle must be \( \pi/2 \) (in the first rotation): \[ 2x = \frac{\pi}{2} \]
Dividing by 2: \[ x = \frac{\pi}{4} \]

Step 4: Check options

The calculated value \( x = \pi/4 \) corresponds to option (B). Quick Tip: Always simplify determinants with trig functions using identities like \( \sin^2 x + \cos^2 x = 1 \) or double angle formulas.
If you forget the identity, you can quickly test the options: for \( x = \pi/4 \), \( \sin x = \cos x = 1/\sqrt{2} \). The det becomes \( (1/\sqrt{2})(1/\sqrt{2}) - (1/\sqrt{2})(-1/\sqrt{2}) = 1/2 + 1/2 = 1 \).

Concept:

A square matrix \( M \) is skew-symmetric if its transpose is equal to its negative, i.e., \( M' = -M \).
Properties of transpose: \( (A \pm B)' = A' \pm B' \), \( (AB)' = B'A' \), and \( (kA)' = kA' \).
Given: \( A' = -A \) and \( B' = -B \).


Step 1: Test option (A): \( AB \)

Take the transpose of \( AB \): \[ (AB)' = B'A' \]
Since \( A \) and \( B \) are skew-symmetric: \[ (AB)' = (-B)(-A) = BA \]
For \( AB \) to be skew-symmetric, \( (AB)' \) must be equal to \( -AB \). Since \( BA \) is not necessarily \( -AB \), this is not always skew-symmetric.

Step 2: Test option (B): \( AB + BA \)

Take the transpose: \[ (AB + BA)' = (AB)' + (BA)' \] \[ = B'A' + A'B' \] \[ = (-B)(-A) + (-A)(-B) = BA + AB \]
Since \( (AB + BA)' = AB + BA \), this matrix is symmetric, not skew-symmetric.

Step 3: Test option (C): \( (A + B)^2 \)

Let \( C = A + B \). Then \( C' = (A + B)' = A' + B' = -A - B = -(A + B) = -C \).
So \( A + B \) is skew-symmetric.
Now take the transpose of \( C^2 \): \[ (C^2)' = (C \cdot C)' = C'C' = (-C)(-C) = C^2 \]
Since its transpose equals itself, \( (A + B)^2 \) is symmetric.

Step 4: Test option (D): \( A - B \)

Take the transpose: \[ (A - B)' = A' - B' \]
Substitute \( A' = -A \) and \( B' = -B \): \[ (A - B)' = (-A) - (-B) = -A + B \]
Factor out the negative sign: \[ (A - B)' = -(A - B) \]
Since the transpose of \( (A - B) \) results in the negative of itself, \( A - B \) is skew-symmetric. Quick Tip: Any linear combination of skew-symmetric matrices (like \( A \pm B \)) is always skew-symmetric.
The square of a skew-symmetric matrix is always symmetric.

Concept:

To find the absolute minimum (least value) of a continuous function on a closed interval \( [a, b] \), we check the values at the critical points and the endpoints.
Critical points occur where the first derivative \( f'(x) = 0 \).


Step 1: Find the first derivative of the function

Given \( f(x) = x^3 - 12x \).
Differentiating with respect to \( x \): \[ f'(x) = 3x^2 - 12 \]

Step 2: Identify critical points within the given interval

Set \( f'(x) = 0 \): \[ 3x^2 - 12 = 0 \] \[ 3x^2 = 12 \implies x^2 = 4 \] \[ x = 2, -2 \]
The interval is \( [0, 3] \). Therefore, only \( x = 2 \) is a valid critical point to consider. \( x = -2 \) is excluded.

Step 3: Evaluate the function at critical points and endpoints

We need to calculate \( f(x) \) at \( x = 0 \) (start), \( x = 2 \) (critical point), and \( x = 3 \) (end):

At \( x = 0 \): \( f(0) = (0)^3 - 12(0) = 0 \)
At \( x = 2 \): \( f(2) = (2)^3 - 12(2) = 8 - 24 = -16 \)
At \( x = 3 \): \( f(3) = (3)^3 - 12(3) = 27 - 36 = -9 \)


Step 4: Identify the least value

Comparing the values: \( 0, -16, -9 \).
The least value is \( -16 \). Quick Tip: Always check if the critical point lies within the specified domain. If it's outside, ignore it.
For "least value" on a closed interval, checking endpoints is mandatory as the minimum can occur there.

Concept:

Use the method of substitution for integration.
If the numerator is a multiple of the derivative of the denominator, the integral takes the form \( \int \frac{f'(x)}{f(x)} dx = \log |f(x)| + C \).


Step 1: Apply substitution to the integral

Let the denominator be \( u \): \[ u = b^2 + c^2x^2 \]
Differentiate both sides with respect to \( x \): \[ \frac{du}{dx} = 2c^2x \implies du = 2c^2x \, dx \]
Rearranging for \( x \, dx \): \[ x \, dx = \frac{du}{2c^2} \]

Step 2: Rewrite the integral in terms of \( u \)

Substitute the values into the original integral: \[ I = \int \frac{3a \cdot x}{b^2 + c^2x^2} dx = \int \frac{3a}{u} \cdot \frac{du}{2c^2} \]
Take the constants outside: \[ I = \frac{3a}{2c^2} \int \frac{1}{u} du \]

Step 3: Perform the integration

The integral of \( \frac{1}{u} \) is \( \log|u| \): \[ I = \frac{3a}{2c^2} \log |u| + K \]
Substitute back \( u = b^2 + c^2x^2 \): \[ I = \frac{3a}{2c^2} \log |b^2 + c^2x^2| + K \]

Step 4: Compare with the given expression

The given form is \( A \log |b^2 + c^2x^2| + K \).
By comparison: \[ A = \frac{3a}{2c^2} \]
This matches option (D). Quick Tip: Whenever you see \( x^2 \) in the denominator and \( x \) in the numerator, the integral will almost always involve a log of the denominator.
Quick check: Differentiate the result. \( \frac{d}{dx}[A \log(b^2+c^2x^2)] = A \cdot \frac{2c^2x}{b^2+c^2x^2} \). Setting this equal to \( \frac{3ax}{b^2+c^2x^2} \) gives \( 2Ac^2 = 3a \implies A = \frac{3a}{2c^2} \).

Concept:

Definite integral property: \( \int_{-a}^{a} f(x) dx = 0 \) if \( f(x) \) is an odd function.
A function is odd if \( f(-x) = -f(x) \).


Step 1: Define the integrand function

Let \( f(x) = \frac{x^3}{x^2 + 2|x| + 1} \).

Step 2: Check if the function is even or odd

Replace \( x \) with \( -x \): \[ f(-x) = \frac{(-x)^3}{(-x)^2 + 2|-x| + 1} \]
Using properties \( (-x)^3 = -x^3 \), \( (-x)^2 = x^2 \), and \( |-x| = |x| \): \[ f(-x) = \frac{-x^3}{x^2 + 2|x| + 1} \] \[ f(-x) = - \left[ \frac{x^3}{x^2 + 2|x| + 1} \right] \] \[ f(-x) = -f(x) \]

Step 3: Conclusion based on properties

Since \( f(x) \) satisfies the condition for an odd function and the limits of integration are symmetric about the origin (\( -1 \) to \( 1 \)): \[ \int_{-1}^{1} \frac{x^3}{x^2 + 2|x| + 1} dx = 0 \]
This matches option (A). Quick Tip: Before trying to integrate complex rational or modular functions over symmetric limits \( [-a, a] \), always check for odd/even property first.
An odd power of \( x \) in the numerator often makes the whole fraction odd if the denominator involves only even powers or absolute values.

Concept:

Area bounded by a curve \( y = f(x) \) and the x-axis is given by \( \int_a^b |y| dx \).
The function \( y = x|x| \) is defined as \( y = x^2 \) for \( x \ge 0 \) and \( y = -x^2 \) for \( x < 0 \).


Step 1: Define the function over the interval

For the interval \( [-1, 1] \): \[ y = \begin{cases} -x^2, & if -1 \le x < 0
x^2, & if 0 \le x \le 1 \end{cases} \]

Step 2: Set up the area integral

Area \( A = \int_{-1}^{1} |y| dx = \int_{-1}^{1} |x|x|| dx \).
Since \( |x|x|| = |x^2| = x^2 \): \[ A = \int_{-1}^{1} x^2 dx \]

Step 3: Calculate the integral
\[ A = \left[ \frac{x^3}{3} \right]_{-1}^{1} \] \[ A = \left( \frac{1^3}{3} \right) - \left( \frac{(-1)^3}{3} \right) \] \[ A = \frac{1}{3} - \left( -\frac{1}{3} \right) \] \[ A = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} sq. units \]

Step 4: Final Check

Alternatively, due to symmetry, Area \( A = 2 \times \int_{0}^{1} x^2 dx = 2 \times \left[ \frac{1}{3} \right] = \frac{2}{3} \).
This matches option (C). Quick Tip: "Area bounded" is always a non-negative quantity. Do not confuse it with a simple definite integral which might result in 0 for odd functions.
The graph of \( y = x|x| \) is in the 1st and 3rd quadrants. To find area, we take the absolute value of the function.

Concept:

A first-order linear differential equation in the form \( \frac{dx}{dy} + p(y)x = q(y) \) has an integrating factor.
The Integrating Factor (I.F.) is defined as \( e^{\int p(y) dy} \).
This factor is used to transform the left side of the equation into a total derivative.


Step 1: Convert the given equation into standard linear form

The given equation is: \[ R\frac{dx}{dy} + Px = Q \]
To isolate the derivative term \( \frac{dx}{dy} \), we divide the entire equation by \( R \) (since \( P, Q, R \) are functions of \( y \)): \[ \frac{dx}{dy} + \frac{P}{R}x = \frac{Q}{R} \]

Step 2: Identify the coefficient function \( p(y) \)

Comparing this to the standard form \( \frac{dx}{dy} + p(y)x = q(y) \): \[ p(y) = \frac{P}{R} \]

Step 3: Calculate the Integrating Factor

Using the formula for I.F.: \[ I.F. = e^{\int p(y) dy} \]
Substituting \( p(y) = \frac{P}{R} \): \[ I.F. = e^{\int \frac{P}{R} dy} \]
This matches option (C). Quick Tip: Always ensure the coefficient of the highest derivative term is 1 before identifying \( p(y) \) or \( p(x) \).
Pay close attention to whether the independent variable is \( x \) or \( y \); if it's \( dx/dy \), integrate with respect to \( y \).

Concept:

Order: The order of a differential equation is the order of the highest-order derivative appearing in it.
Degree: The degree of a differential equation is the power of the highest-order derivative, provided the equation is a polynomial in its derivatives.
Use the chain rule for differentiation: \( \frac{d}{dx}[f(y)] = f'(y) \frac{dy}{dx} \).


Step 1: Perform the differentiation on the left side

The given equation is \( \frac{d}{dx}(e^y) = 0 \).
Applying the chain rule: \[ e^y \cdot \frac{dy}{dx} = 0 \]

Step 2: Simplify the differential equation

Since \( e^y \) is never zero for any real value of \( y \), we can divide both sides by \( e^y \): \[ \frac{dy}{dx} = 0 \]

Step 3: Determine order and degree


Order: The highest derivative present is \( \frac{dy}{dx} \), which is a first-order derivative. Thus, Order = 1.
Degree: The highest derivative \( \frac{dy}{dx} \) is raised to the power of 1. Since the equation is a polynomial in its derivatives, Degree = 1.

The result is (1, 1), which matches option (B). Quick Tip: Expand or simplify all "outer" derivatives before determining order and degree.
Remember: Degree is defined only if the equation can be expressed as a polynomial in the derivatives (e.g., no \( \sin(y') \) or \( e^{y'} \)).

Concept:



Two non-zero vectors \( \vec{a} \) and \( \vec{b} \) are perpendicular (orthogonal) if and only if their dot product is zero: \( \vec{a} \cdot \vec{b} = 0 \).
The dot product of \( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \) and \( \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \) is \( a_1b_1 + a_2b_2 + a_3b_3 \).


Step 1: Set up the dot product equation


Let \( \vec{a} = 1\hat{i} + 2\hat{j} + 3\hat{k} \) and \( \vec{b} = 2\hat{i} - p\hat{j} + 1\hat{k} \).
Since they are perpendicular: \[ \vec{a} \cdot \vec{b} = 0 \] \[ (1)(2) + (2)(-p) + (3)(1) = 0 \]


Step 2: Solve for \( p \)


Expand the arithmetic: \[ 2 - 2p + 3 = 0 \]
Combine constant terms: \[ 5 - 2p = 0 \]
Isolate \( p \): \[ 2p = 5 \implies p = \frac{5}{2} \]
This matches option (C). Quick Tip: Perpendicular \( \implies \) Dot Product = 0.
Parallel \( \implies \) Cross Product = 0 or coefficients are proportional.
Dot product involves multiplying corresponding components and summing them into a scalar.

Concept:

Three points \( A, B, C \) are collinear if the vectors formed by them, say \( \vec{AB} \) and \( \vec{BC} \), are parallel.
Two vectors are parallel if their components are proportional: \( \frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} \).


Step 1: Determine the vectors \( \vec{AB} \) and \( \vec{BC} \)

Let the points be \( A(-1, -1, 2) \), \( B(2, m, 5) \), and \( C(3, 11, 6) \). \[ \vec{AB} = (2 - (-1))\hat{i} + (m - (-1))\hat{j} + (5 - 2)\hat{k} = 3\hat{i} + (m + 1)\hat{j} + 3\hat{k} \] \[ \vec{BC} = (3 - 2)\hat{i} + (11 - m)\hat{j} + (6 - 5)\hat{k} = 1\hat{i} + (11 - m)\hat{j} + 1\hat{k} \]

Step 2: Apply the collinearity condition

Since \( A, B, C \) are collinear, \( \vec{AB} \parallel \vec{BC} \). Therefore: \[ \frac{3}{1} = \frac{m+1}{11-m} = \frac{3}{1} \]

Step 3: Solve the resulting equation

From the proportion: \[ 3 = \frac{m + 1}{11 - m} \]
Multiply both sides by \( (11 - m) \): \[ 3(11 - m) = m + 1 \] \[ 33 - 3m = m + 1 \] \[ 32 = 4m \] \[ m = 8 \]
This matches option (A). Quick Tip: To test collinearity, the simplest method is checking the proportionality of direction ratios of vectors connecting the points.
Alternatively, for three points to be collinear, the area of the triangle formed by them must be zero.

Concept:

Dot Product formula: \( \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos \theta \).
Cross Product magnitude formula: \( |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| \sin \theta \).
Lagrange's Identity: \( |\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 \).


Step 1: Apply Lagrange's Identity

Using the identity relating cross and dot products: \[ (12)^2 + (\vec{a} \cdot \vec{b})^2 = (8)^2 \cdot (3)^2 \]

Step 2: Substitute values and simplify
\[ 144 + (\vec{a} \cdot \vec{b})^2 = 64 \cdot 9 \] \[ 144 + (\vec{a} \cdot \vec{b})^2 = 576 \]

Step 3: Solve for the dot product magnitude
\[ (\vec{a} \cdot \vec{b})^2 = 576 - 144 \] \[ (\vec{a} \cdot \vec{b})^2 = 432 \]
Taking the square root to find \( |\vec{a} \cdot \vec{b}| \): \[ |\vec{a} \cdot \vec{b}| = \sqrt{432} \]

Step 4: Simplify the radical

Factor out square numbers from 432: \[ \sqrt{432} = \sqrt{144 \times 3} \] \[ = 12\sqrt{3} \]
This matches option (C). Quick Tip: Lagrange's Identity \( (|\vec{a}|^2|\vec{b}|^2 = |\vec{a} \cdot \vec{b}|^2 + |\vec{a} \times \vec{b}|^2) \) is the vector version of the trigonometric identity \( \sin^2\theta + \cos^2\theta = 1 \).
It is extremely useful when the angle \( \theta \) is unknown.

Concept:

The line \( \frac{x}{1} = \frac{y}{0} = \frac{z}{0} \) represents the x-axis because its direction ratios are \( (1, 0, 0) \) and it passes through the origin \( (0, 0, 0) \).
The perpendicular distance of a point \( P(x_1, y_1, z_1) \) from the x-axis is given by the formula \( \sqrt{y_1^2 + z_1^2} \).
This is because the foot of the perpendicular from \( P \) to the x-axis is always \( Q(x_1, 0, 0) \).


Step 1: Identify the coordinates of the point and the nature of the line

The given point is \( P(2, 5, 7) \).
The given line equation is \( \frac{x-0}{1} = \frac{y-0}{0} = \frac{z-0}{0} \).
This line passes through the origin \( O(0, 0, 0) \) and has direction ratios \( (1, 0, 0) \).
A line with direction ratios proportional to \( (1, 0, 0) \) is the x-axis.

Step 2: Find the foot of the perpendicular

The foot of the perpendicular from a point \( (x, y, z) \) to the x-axis is found by keeping the x-coordinate same and making the other coordinates zero.
Foot of perpendicular \( Q = (2, 0, 0) \).

Step 3: Calculate the distance using the distance formula

The length of the perpendicular is the distance between \( P(2, 5, 7) \) and \( Q(2, 0, 0) \): \[ d = \sqrt{(2 - 2)^2 + (5 - 0)^2 + (7 - 0)^2} \] \[ d = \sqrt{0^2 + 5^2 + 7^2} \] \[ d = \sqrt{0 + 25 + 49} \] \[ d = \sqrt{74} \]
Comparing with options, (C) is the correct choice. Quick Tip: Distance of \( (x, y, z) \) from x-axis is \( \sqrt{y^2 + z^2} \).
Distance of \( (x, y, z) \) from y-axis is \( \sqrt{x^2 + z^2} \).
Distance of \( (x, y, z) \) from z-axis is \( \sqrt{x^2 + y^2} \).

Concept:

According to the Corner Point Theorem, the optimal (maximum or minimum) value of an objective function in a Linear Programming Problem occurs at the corner points (vertices) of the feasible region.
We evaluate the objective function \( Z = ax + by \) at each vertex and identify the highest and lowest values.


Step 1: Identify the corner points of the feasible region

From the provided graph, the vertices of the shaded feasible region are:

\( A(0, 2) \)
\( B(3, 4) \)
\( C(6, 0) \)


Step 2: Evaluate the objective function \( Z = 5x + 7y \) at each vertex


At \( A(0, 2) \): \( Z = 5(0) + 7(2) = 0 + 14 = 14 \)
At \( B(3, 4) \): \( Z = 5(3) + 7(4) = 15 + 28 = 43 \)
At \( C(6, 0) \): \( Z = 5(6) + 7(0) = 30 + 0 = 30 \)


Step 3: Determine the difference between the maximum and minimum values

From the evaluations:
Maximum value of \( Z = 43 \) (at point \( B \))
Minimum value of \( Z = 14 \) (at point \( A \))
Calculation: \[ Difference = Z_{max} - Z_{min} \] \[ Difference = 43 - 14 = 29 \]
This matches option (B). Quick Tip: Always list all vertices clearly before calculating \( Z \) values.
Check the scale of the axes in the graph carefully to ensure you read the coordinate values correctly.

Concept:

Linear Programming Problems (LPP) involve "linear" relationships.
The objective function \( Z \) represents the quantity to be maximized or minimized (e.g., profit or cost).
By definition, in LPP, the objective function must be a linear function of the decision variables.


Step 1: Define the mathematical form of the objective function

In a standard linear programming problem with two variables \( x \) and \( y \), the objective function is represented as: \[ Z = ax + by \]
where \( a \) and \( b \) are constants.

Step 2: Determine the degree of the expression

The degree of a polynomial term is the sum of the exponents of its variables.

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

Since the highest power of any variable in the entire expression is 1, the degree of the function is 1.

Step 3: Conclusion

Linear functions always have a degree of 1. Therefore, the degree of an objective function in LPP is always 1.
This corresponds to option (B). Quick Tip: The word "Linear" in Linear Programming directly implies that all expressions (objective function and constraints) are of degree 1.
Degree 0 would mean a constant function, and degree 2 would be a Quadratic Programming problem.

Concept:



Conditional Probability: \( P(A|B) \) is the probability of event \( A \) occurring given that event \( B \) has already occurred.
The correct formula is \( P(A|B) = \frac{P(A \cap B)}{P(B)} \).


Step 1: Verify Assertion (A)


Let the sample space be \( S = \{1, 2, 3, 4, 5, 6\} \).
Let \( B \) be the event that the number appearing is odd: \( B = \{1, 3, 5\} \). Thus, \( n(B) = 3 \).


Let \( A \) be the event that the number appearing is prime: \( A = \{2, 3, 5\} \).


The intersection \( A \cap B \) represents numbers that are both odd and prime: \( A \cap B = \{3, 5\} \). Thus, \( n(A \cap B) = 2 \).


Using the conditional probability formula:

\[ P(A|B) = \frac{n(A \cap B)}{n(B)} = \frac{2}{3} \]
Assertion (A) is TRUE.


Step 2: Verify Reason (R)


The Reason states: \( P(A|B) = \frac{P(A \cup B)}{P(B)} \).


However, the standard definition of conditional probability uses the intersection (\( \cap \)), not the union (\( \cup \)).


The correct formula is \( P(A|B) = \frac{P(A \cap B)}{P(B)} \).
Reason (R) is FALSE.


Step 3: Final determination


Since the assertion is true but the reason provided is a mathematically incorrect formula, option (C) is correct. Quick Tip: In Assertion-Reason questions, check the truth of both statements independently first.
Remember the mnemonic: "Intersection in the numerator, given condition in the denominator."

Concept:

Two lines are perpendicular if the dot product of their direction vectors is zero.
For lines in Cartesian form \( \frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1} \) and \( \frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2} \), the condition is \( a_1a_2 + b_1b_2 + c_1c_2 = 0 \).


Step 1: Verify Reason (R)

Reason (R) states that lines are perpendicular if the dot product of their direction vectors \( \vec{b}_1 \) and \( \vec{b}_2 \) is zero.
This is a fundamental property of vectors and line geometry.
Reason (R) is TRUE.

Step 2: Find the direction vectors for the lines in Assertion (A)

Line 1: \( x = py + q \implies y = \frac{x-q}{p} \); \( z = ry + s \implies y = \frac{z-s}{r} \).

Combining these: \( \frac{x-q}{p} = \frac{y-0}{1} = \frac{z-s}{r} \). Direction vector \( \vec{b}_1 = (p, 1, r) \).

Line 2: \( x = p'y + q' \implies y = \frac{x-q'}{p'} \); \( z = r'y + s' \implies y = \frac{z-s'}{r'} \).

Combining these: \( \frac{x-q'}{p'} = \frac{y-0}{1} = \frac{z-s'}{r'} \). Direction vector \( \vec{b}_2 = (p', 1, r') \).


Step 3: Evaluate the perpendicularity condition for Assertion (A)

Applying \( \vec{b}_1 \cdot \vec{b}_2 = 0 \): \[ (p)(p') + (1)(1) + (r)(r') = 0 \] \[ pp' + 1 + rr' = 0 \] \[ pp' + rr' = -1 \]
The Assertion states the condition is \( pp' + rr' = 1 \). This is mathematically incorrect.
Assertion (A) is FALSE.

Step 4: Final Determination

Since Assertion is false and Reason is true, the correct option is (D). Quick Tip: To find direction ratios from equations like \( x = py + q \), always solve for the common variable (in this case, \( y \)) to put the equation in symmetric form.
Perpendicular condition is always "Sum of products = 0", never 1.

Concept:

A function \( f(x) \) is continuous at \( x = a \) if the Left-Hand Limit (LHL), Right-Hand Limit (RHL), and the function value at that point \( f(a) \) are all equal.
LHL at \( x = 3 \): \( \lim_{x \to 3^-} f(x) \)
RHL at \( x = 3 \): \( \lim_{x \to 3^+} f(x) \)
Definition of Absolute Value: \( |x-3| = -(x-3) \) for \( x < 3 \).


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

For \( x < 3 \), \( f(x) = \frac{|x-3|}{2(x-3)} \).
As \( x \to 3^- \), \( x - 3 < 0 \), so \( |x-3| = -(x-3) \). \[ LHL = \lim_{x \to 3^-} \frac{-(x-3)}{2(x-3)} \] \[ LHL = \lim_{x \to 3^-} \left(-\frac{1}{2}\right) = -0.5 \]

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

For \( x \ge 3 \), \( f(x) = \frac{x-6}{6} \). \[ RHL = \lim_{x \to 3^+} \frac{x-6}{6} \] \[ RHL = \frac{3-6}{6} = \frac{-3}{6} = -0.5 \]

Step 3: Find the function value at \( x = 3 \)

Using the definition for \( x \ge 3 \): \[ f(3) = \frac{3-6}{6} = -0.5 \]

Step 4: Compare LHL, RHL, and \( f(3) \)

Since \( LHL = RHL = f(3) = -0.5 \), all three values match.
Therefore, the function \( f(x) \) is continuous at \( x = 3 \). Quick Tip: Whenever a function involves modulus signs, always redefine the function by removing the absolute value bars for the specific intervals first.
For continuity, check: \( \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a) \).

Concept:

Implicit Differentiation: Differentiate both sides of the equation with respect to \( x \), treating \( y \) as a function of \( x \).
Product Rule: \( \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx} \).
Chain Rule: \( \frac{d}{dx}(y^n) = ny^{n-1} \frac{dy}{dx} \).


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

The given equation is \( \sqrt{3}(x^2 + y^2) = 4xy \).
Differentiating both sides: \[ \sqrt{3}\left(2x + 2y \frac{dy}{dx}\right) = 4\left(x \frac{dy}{dx} + y \cdot 1\right) \]

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

Expand the brackets: \[ 2\sqrt{3}x + 2\sqrt{3}y \frac{dy}{dx} = 4x \frac{dy}{dx} + 4y \]
Collect terms involving \( \frac{dy}{dx} \) on one side: \[ 2\sqrt{3}y \frac{dy}{dx} - 4x \frac{dy}{dx} = 4y - 2\sqrt{3}x \] \[ \frac{dy}{dx} (2\sqrt{3}y - 4x) = 4y - 2\sqrt{3}x \] \[ \frac{dy}{dx} = \frac{4y - 2\sqrt{3}x}{2\sqrt{3}y - 4x} \]
Simplifying by dividing numerator and denominator by 2: \[ \frac{dy}{dx} = \frac{2y - \sqrt{3}x}{\sqrt{3}y - 2x} \]

Step 3: Substitute the point \( \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \)

Put \( x = \frac{1}{2} \) and \( y = \frac{\sqrt{3}}{2} \): \[ \frac{dy}{dx} = \frac{2\left(\frac{\sqrt{3}}{2}\right) - \sqrt{3}\left(\frac{1}{2}\right)}{\sqrt{3}\left(\frac{\sqrt{3}}{2}\right) - 2\left(\frac{1}{2}\right)} \] \[ \frac{dy}{dx} = \frac{\sqrt{3} - \frac{\sqrt{3}}{2}}{\frac{3}{2} - 1} \] \[ \frac{dy}{dx} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \] Quick Tip: In implicit differentiation, always remember to attach a \( \frac{dy}{dx} \) factor whenever you differentiate a term containing \( y \).
Simplifying the derivative expression before substitution can often prevent calculation errors.

Concept:

Volume of a cone: \( V = \frac{1}{3}\pi r^2 h \).
Relation between radius \( r \) and height \( h \) using semi-vertical angle \( \alpha \): \( \tan \alpha = \frac{r}{h} \).
Rates of change: relate \( \frac{dV}{dt} \) to \( \frac{dh}{dt} \) using the derivative.


Step 1: Express radius in terms of height

Given semi-vertical angle \( \alpha = \frac{\pi}{6} \). \[ \tan\left(\frac{\pi}{6}\right) = \frac{r}{h} \implies \frac{1}{\sqrt{3}} = \frac{r}{h} \] \[ r = \frac{h}{\sqrt{3}} \]

Step 2: Express volume as a function of height only

Substitute \( r = \frac{h}{\sqrt{3}} \) into the volume formula: \[ V = \frac{1}{3}\pi \left(\frac{h}{\sqrt{3}}\right)^2 h \] \[ V = \frac{1}{3}\pi \frac{h^2}{3} h = \frac{\pi h^3}{9} \]

Step 3: Differentiate with respect to time \( t \)

Using the chain rule: \[ \frac{dV}{dt} = \frac{d}{dt}\left(\frac{\pi h^3}{9}\right) \] \[ \frac{dV}{dt} = \frac{\pi}{9} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi h^2}{3} \frac{dh}{dt} \]

Step 4: Substitute given values and solve for \( \frac{dh}{dt} \)

Given: \( \frac{dV}{dt} = -1 mm^3/min \) (negative because volume is decreasing) and \( h = 10 mm \). \[ -1 = \frac{\pi (10)^2}{3} \frac{dh}{dt} \] \[ -1 = \frac{100\pi}{3} \frac{dh}{dt} \] \[ \frac{dh}{dt} = -\frac{3}{100\pi} mm/min \]
The rate at which the level is dropping is \( \frac{3}{100\pi} mm/min \). Quick Tip: Always use geometry (like similar triangles or trigonometry) to eliminate one variable so that the volume is a function of only the variable whose rate is being studied.
Be careful with signs: "decreasing" implies a negative rate of change for the quantity itself.

Concept:

Displacement Vector: \( \vec{QP} = Position Vector of P - Position Vector of Q \).
Unit Vector: \( \hat{n} = \frac{\vec{a}}{|\vec{a}|} \), which points in the direction of \( \vec{a} \) and has magnitude 1.
Required Vector: A vector of magnitude \( \lambda \) in the direction of \( \vec{a} \) is \( \lambda \cdot \hat{n} \).


Step 1: Find the vector \( \vec{QP} \)

Given \( P(1, 3, 2) \) and \( Q(-1, 0, 8) \). \[ \vec{QP} = (1 - (-1))\hat{i} + (3 - 0)\hat{j} + (2 - 8)\hat{k} \] \[ \vec{QP} = 2\hat{i} + 3\hat{j} - 6\hat{k} \]

Step 2: Calculate the magnitude of \( \vec{QP} \)
\[ |\vec{QP}| = \sqrt{2^2 + 3^2 + (-6)^2} \] \[ |\vec{QP}| = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]

Step 3: Find the unit vector in the direction of \( \vec{QP} \)
\[ \hat{u} = \frac{\vec{QP}}{|\vec{QP}|} = \frac{2\hat{i} + 3\hat{j} - 6\hat{k}}{7} \]

Step 4: Calculate the final vector with magnitude 14
\[ \vec{v} = 14 \cdot \hat{u} \] \[ \vec{v} = 14 \left( \frac{2\hat{i} + 3\hat{j} - 6\hat{k}}{7} \right) \] \[ \vec{v} = 2(2\hat{i} + 3\hat{j} - 6\hat{k}) = 4\hat{i} + 6\hat{j} - 12\hat{k} \] Quick Tip: Direction of \( \vec{QP} \) means "from Q to P", so subtract coordinates of Q from P.
Scaling a unit vector is the standard way to construct a vector of any desired length in a given direction.

Concept:

Parallelogram Law: If \( \vec{a} \) and \( \vec{b} \) are adjacent sides, then the two diagonals are represented by \( \vec{d}_1 = \vec{a} + \vec{b} \) and \( \vec{d}_2 = \vec{a} - \vec{b} \).
Length of a vector \( x\hat{i} + y\hat{j} + z\hat{k} \) is \( \sqrt{x^2 + y^2 + z^2} \).


Step 1: Find the diagonal vectors

First diagonal \( \vec{d}_1 = \vec{a} + \vec{b} \): \[ \vec{d}_1 = (3+1)\hat{i} + (-2+0)\hat{j} + (2+2)\hat{k} = 4\hat{i} - 2\hat{j} + 4\hat{k} \]
Second diagonal \( \vec{d}_2 = \vec{a} - \vec{b} \): \[ \vec{d}_2 = (3-1)\hat{i} + (-2-0)\hat{j} + (2-2)\hat{k} = 2\hat{i} - 2\hat{j} + 0\hat{k} = 2\hat{i} - 2\hat{j} \]

Step 2: Find the length of the first diagonal
\[ |\vec{d}_1| = \sqrt{4^2 + (-2)^2 + 4^2} \] \[ |\vec{d}_1| = \sqrt{16 + 4 + 16} = \sqrt{36} = 6 units \]

Step 3: Find the length of the second diagonal
\[ |\vec{d}_2| = \sqrt{2^2 + (-2)^2 + 0^2} \] \[ |\vec{d}_2| = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} units \] Quick Tip: In a parallelogram, \( \vec{a}+\vec{b} \) is the long diagonal and \( \vec{a}-\vec{b} \) is the short diagonal (or vice versa depending on the angle between the sides).
Always simplify square roots like \( \sqrt{8} \) to \( 2\sqrt{2} \) in final answers.

Concept:

Tangent addition formula: \( \tan\left(\frac{\pi}{4} - \theta\right) = \frac{1 - \tan \theta}{1 + \tan \theta} \).
Divide numerator and denominator by \( \cos \theta \) to convert a \( \cos / \sin \) expression into tangents.
Property of inverse functions: \( \tan^{-1}(\tan \theta) = \theta \) if \( \theta \) is in the principal branch \( (-\pi/2, \pi/2) \).


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

Inside the inverse tangent function: \[ \frac{\frac{\cos 2x}{\cos 2x} - \frac{\sin 2x}{\cos 2x}}{\frac{\cos 2x}{\cos 2x} + \frac{\sin 2x}{\cos 2x}} = \frac{1 - \tan 2x}{1 + \tan 2x} \]

Step 2: Use the tangent identity

Since \( \tan\left(\frac{\pi}{4}\right) = 1 \), the expression is equivalent to: \[ \frac{\tan\left(\frac{\pi}{4}\right) - \tan 2x}{1 + \tan\left(\frac{\pi}{4}\right)\tan 2x} = \tan\left(\frac{\pi}{4} - 2x\right) \]

Step 3: Apply the inverse function property

The original expression becomes: \[ \tan^{-1}\left[ \tan\left(\frac{\pi}{4} - 2x\right) \right] \]
Check the range: Given \( 0 < x < \frac{\pi}{4} \), we have \( 0 < 2x < \frac{\pi}{2} \).
Subtracting from \( \frac{\pi}{4} \): \[ \frac{\pi}{4} - \frac{\pi}{2} < \frac{\pi}{4} - 2x < \frac{\pi}{4} \] \[ -\frac{\pi}{4} < \frac{\pi}{4} - 2x < \frac{\pi}{4} \]
This range lies safely within the principal interval \( (-\pi/2, \pi/2) \).
Therefore: \[ Simplified form = \frac{\pi}{4} - 2x \] Quick Tip: A numerator and denominator both containing \( \sin \) and \( \cos \) of the same angle is a strong signal to divide by \( \cos \) to get \( \tan \).
Always verify that the resulting angle falls within the inverse function's range to justify the final cancellation.

Concept:

Principal values: \( \sin^{-1}(1) = \frac{\pi}{2} \).
Inverse cosine of negative values: \( \cos^{-1}(-x) = \pi - \cos^{-1}(x) \).
Known value: \( \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \).


Step 1: Determine the individual inverse function values

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

Step 2: Substitute values into the tangent argument
\[ Argument = \frac{\pi}{2} - \frac{2\pi}{3} \]
Find a common denominator (6): \[ \frac{3\pi}{6} - \frac{4\pi}{6} = -\frac{\pi}{6} \]

Step 3: Evaluate the tangent function
\[ \tan\left(-\frac{\pi}{6}\right) = -\tan\left(\frac{\pi}{6}\right) \]
Using the standard value \( \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \): \[ Value = -\frac{1}{\sqrt{3}} \] Quick Tip: Always find the numerical angles first when constants are given inside inverse functions.
Recall that \( \tan(-\theta) = -\tan(\theta) \), which helps in handling negative angles easily.

Concept:

Integration by Parts (ILATE rule): \( \int u \, dv = uv - \int v \, du \).
Since the inverse trigonometric function is listed first in ILATE, we set \( u = \tan^{-1}x \).
Standard integral: \( \int \frac{x^2}{1+x^2} dx = \int \frac{(x^2+1)-1}{x^2+1} dx = \int (1 - \frac{1}{x^2+1}) dx \).


Step 1: Apply Integration by Parts

Let \( u = \tan^{-1}x \) and \( dv = x \, dx \).
Then \( du = \frac{1}{1+x^2} dx \) and \( v = \frac{x^2}{2} \).
Using the formula: \[ \int x \tan^{-1}x \, dx = \frac{x^2}{2} \tan^{-1}x - \int \frac{x^2}{2(1+x^2)} dx \]

Step 2: Solve the resulting integral

Consider the integral \( I_r = \frac{1}{2} \int \frac{x^2}{1+x^2} dx \).
Add and subtract 1 in the numerator: \[ I_r = \frac{1}{2} \int \left( \frac{x^2 + 1 - 1}{x^2 + 1} \right) dx \] \[ I_r = \frac{1}{2} \int \left( 1 - \frac{1}{1+x^2} \right) dx \] \[ I_r = \frac{1}{2} [x - \tan^{-1}x] \]

Step 3: Combine and apply limits

The indefinite integral is \( \frac{x^2}{2} \tan^{-1}x - \frac{1}{2}x + \frac{1}{2}\tan^{-1}x \).
Applying limits from 0 to 1: \[ \left[ \frac{x^2+1}{2} \tan^{-1}x - \frac{x}{2} \right]_{0}^{1} \] \[ = \left( \frac{1+1}{2} \tan^{-1}(1) - \frac{1}{2} \right) - \left( \frac{0+1}{2} \tan^{-1}(0) - 0 \right) \] \[ = (1 \cdot \frac{\pi}{4} - \frac{1}{2}) - (0) = \frac{\pi}{4} - \frac{1}{2} \] Quick Tip: In Integration by Parts, choose 'u' based on the ILATE rule (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential).
For integrals like \( \frac{x^2}{1+x^2} \), adding and subtracting 1 is a faster technique than long division.

Concept:

Rationalization: Multiply the numerator and denominator by the numerator's expression to remove the radical from the top.
Standard formula: \( \int \frac{1}{\sqrt{x^2 - a^2}} dx = \log |x + \sqrt{x^2 - a^2}| + C \).
Substitution rule: For \( \int \frac{x}{\sqrt{x^2 - a^2}} dx \), use \( u = x^2 - a^2 \).


Step 1: Rationalize the integrand

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

Step 2: Split the integral into two parts
\[ I = \int \frac{x}{\sqrt{x^2 - 4}} \, dx + \int \frac{2}{\sqrt{x^2 - 4}} \, dx \]

Step 3: Integrate Part 1

For \( \int \frac{x}{\sqrt{x^2 - 4}} dx \), let \( u = x^2 - 4 \), then \( du = 2x \, dx \): \[ \int \frac{1}{2\sqrt{u}} du = \sqrt{u} = \sqrt{x^2 - 4} \]

Step 4: Integrate Part 2

Using the standard formula \( \int \frac{1}{\sqrt{x^2 - a^2}} dx \): \[ 2 \int \frac{1}{\sqrt{x^2 - 2^2}} dx = 2 \log |x + \sqrt{x^2 - 4}| \]

Step 5: Combine the results
\[ I = \sqrt{x^2 - 4} + 2 \log |x + \sqrt{x^2 - 4}| + C \] Quick Tip: Rationalizing the numerator is a standard trick to convert "nested square root" fractions into a sum of simple integrable functions.
Always try to split complex numerators over a common radical denominator.

Concept:

Partial Fractions: Decompose the algebraic fraction into simpler parts.
Substitution for partial fractions: Temporarily replace \( x^2 \) with \( t \) to find the constants, then replace \( t \) with \( x^2 \) before integrating.
Standard Integral: \( \int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \).


Step 1: Decompose using partial fractions

Let \( x^2 = t \). Consider the fraction \( \frac{t}{(t+9)(t+16)} \). \[ \frac{t}{(t+9)(t+16)} = \frac{A}{t+9} + \frac{B}{t+16} \]
Multiplying by the denominator: \[ t = A(t+16) + B(t+9) \]

Step 2: Solve for constants A and B


Put \( t = -9 \): \( -9 = A(-9+16) \implies -9 = 7A \implies A = -9/7 \).
Put \( t = -16 \): \( -16 = B(-16+9) \implies -16 = -7B \implies B = 16/7 \).


Step 3: Rewrite the integral

Replace \( t \) with \( x^2 \): \[ \int \frac{x^2}{(x^2+9)(x^2+16)} dx = \int \left( \frac{-9/7}{x^2+9} + \frac{16/7}{x^2+16} \right) dx \] \[ = -\frac{9}{7} \int \frac{1}{x^2+3^2} dx + \frac{16}{7} \int \frac{1}{x^2+4^2} dx \]

Step 4: Perform integration

Apply the \( \tan^{-1} \) formula: \[ = -\frac{9}{7} \left( \frac{1}{3} \tan^{-1}\left(\frac{x}{3}\right) \right) + \frac{16}{7} \left( \frac{1}{4} \tan^{-1}\left(\frac{x}{4}\right) \right) + C \] \[ = -\frac{3}{7} \tan^{-1}\left(\frac{x}{3}\right) + \frac{4}{7} \tan^{-1}\left(\frac{x}{4}\right) + C \] Quick Tip: When the entire integrand consists of \( x^2 \) terms, substitution \( t=x^2 \) is only for partial fractions. Don't differentiate to find \( dt \) unless you want to use it for the final integration.
Always check your partial fraction constants by adding the terms back together mentally.

Concept:

Trigonometric identity: \( 1 + \cos 2\theta = 2\cos^2 \theta \).
Property of Definite Integrals: For an even function \( f(x) \), \( \int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx \).
The functions \( \sec^2 x \) and \( |x| \) are both even.


Step 1: Evaluate integral \( I_1 \)

Using identity \( 1 + \cos 2x = 2\cos^2 x \): \[ I_1 = \int_{-\pi/4}^{\pi/4} \frac{dx}{2\cos^2 x} = \frac{1}{2} \int_{-\pi/4}^{\pi/4} \sec^2 x \, dx \]
Since \( \sec^2 x \) is even: \[ I_1 = \frac{1}{2} \cdot 2 \int_{0}^{\pi/4} \sec^2 x \, dx = [\tan x]_{0}^{\pi/4} \] \[ I_1 = \tan(\pi/4) - \tan(0) = 1 - 0 = 1 \]

Step 2: Evaluate integral \( I_2 \)

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

Step 3: Show the final relation

We need to prove \( I_1 - 4I_2 = 0 \).
Substitute the calculated values: \[ LHS = 1 - 4(1/4) = 1 - 1 = 0 \]
LHS = RHS. Hence proved. Quick Tip: Recognizing even functions simplifies the limits and prevents errors with negative signs.
Recall that \( |x| = x \) for the positive interval \( [0, 1/2] \), which makes integration straightforward.

Concept:



Homogeneous differential equation: An equation where the total degree of every term is the same.
Substitution: Let \( y = vx \), then \( \frac{dy}{dx} = v + x\frac{dv}{dx} \).
Method of variables separable is then used to solve the equation in terms of \( v \) and \( x \).


Step 1: Check the homogeneity and substitute


Rearranging: \( \frac{dy}{dx} = \frac{x^2 + xy + y^2}{x^2} = 1 + \frac{y}{x} + \left(\frac{y}{x}\right)^2 \).
Substituting \( y = vx \) and \( \frac{dy}{dx} = v + x\frac{dv}{dx} \): \[ v + x\frac{dv}{dx} = 1 + v + v^2 \]

Step 2: Separate the variables


Subtract \( v \) from both sides: \[ x\frac{dv}{dx} = 1 + v^2 \] \[ \frac{dv}{1+v^2} = \frac{dx}{x} \]

Step 3: Integrate both sides

\[ \int \frac{dv}{1+v^2} = \int \frac{dx}{x} \] \[ \tan^{-1}v = \log |x| + C \]

Step 4: Substitute back for y


Since \( v = y/x \): \[ \tan^{-1}\left(\frac{y}{x}\right) = \log |x| + C \]
This is the required general solution. Quick Tip: A quick way to spot homogeneous equations is to see if dividing the whole numerator and denominator by \( x^n \) leaves only terms of \( y/x \).
Don't forget to add the constant of integration \( C \) only once the integration steps are completed.

Concept:

Variables Separable: Rearrange the equation so that all \( y \) terms are with \( dy \) and all \( x \) terms are with \( dx \).
Particular Solution: Use the initial condition (given coordinates) to find the specific value of the constant \( C \).


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

Step 2: Integrate both sides
\[ \int (1 - \frac{2}{y+2}) dy = \int (1 + \frac{2}{x}) dx \] \[ y - 2\log|y+2| = x + 2\log|x| + C \]

Step 3: Apply initial condition \( y(1) = -1 \)

Substitute \( x=1, y=-1 \): \[ -1 - 2\log|-1+2| = 1 + 2\log|1| + C \] \[ -1 - 2\log(1) = 1 + 2(0) + C \] \[ -1 - 0 = 1 + C \implies C = -2 \]

Step 4: Write final particular solution
\[ y - 2\log|y+2| = x + 2\log|x| - 2 \] \[ y - x + 2 = 2\log|x| + 2\log|y+2| \] \[ y - x + 2 = 2\log|x(y+2)| \] Quick Tip: When integrating \( \frac{y}{y+2} \), using "plus-minus adjustment" in the numerator is often easier than long division.
Remember \( \log 1 = 0 \), which simplifies many particular solution problems involving boundaries at \( x=1 \) or \( y=1 \).

Concept:

Convert inequalities into linear equations to find boundary lines.
Plot the feasible region defined by the intersection of all constraints.
Corner Point Method: Evaluate the objective function at each vertex of the feasible region.


Step 1: Find intercepts for boundary lines


Line \( L_1: x + y = 7 \). Intercepts: (7, 0) and (0, 7).
Line \( L_2: 2x - 3y = -6 \). Intercepts: Put \( y=0 \to x=-3 \), intercept is (-3, 0). Put \( x=0 \to y=2 \), intercept is (0, 2).


Step 2: Determine the Feasible Region


For \( x+y \le 7 \), testing (0,0): \( 0 \le 7 \) (True). Shading is towards origin.
For \( 2x-3y \ge -6 \), testing (0,0): \( 0 \ge -6 \) (True). Shading is towards origin.
Since \( x, y \ge 0 \), the region is in the first quadrant.


Step 3: Find intersection points (Vertices)

Solve \( x+y=7 \) and \( 2x-3y=-6 \):
From \( L_1 \), \( y = 7-x \). Substitute in \( L_2 \): \[ 2x - 3(7-x) = -6 \implies 2x - 21 + 3x = -6 \implies 5x = 15 \implies x = 3 \]
Then \( y = 7-3 = 4 \). Intersection point is (3, 4).
The vertices of the feasible region are \( A(0, 2) \), \( B(3, 4) \), \( C(7, 0) \), and \( O(0, 0) \).

Step 4: Evaluate Z at vertices


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


Step 5: Conclusion




Comparing values: 0, -30, -21, 91.
The minimum value of \( Z \) is \( -30 \) at point (0, 2). Quick Tip: Always test the origin (0,0) in inequalities to quickly determine which side of the line to shade.
If you're asked to minimize and have negative values, the "most negative" number is your answer.

Concept:



Law of Total Probability: If \( E_1, E_2, \dots, E_n \) is a partition of sample space \( S \), then for any event \( A \), \( P(A) = \sum P(E_i)P(A|E_i) \).
Here, the choice of bag depends on the outcome of the die throw.


Step 1: Define the events and their probabilities


Let \( E_1 \) be the event that the number on the die is less than 3, i.e., \( \{1, 2\} \).
\( P(E_1) = \frac{2}{6} = \frac{1}{3} \)


Let \( E_2 \) be the event that the number on the die is 3 or more, i.e., \( \{3, 4, 5, 6\} \).
\( P(E_2) = \frac{4}{6} = \frac{2}{3} \)


Let \( R \) be the event of drawing a red ball.

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


Bag I has 3 red and 4 white balls (Total = 7).

\( P(R|E_1) = \frac{3}{7} \)

Bag II has 8 red and 6 white balls (Total = 14).

\( P(R|E_2) = \frac{8}{14} = \frac{4}{7} \)

Step 3: Apply the Law of Total Probability

\[ P(R) = P(E_1)P(R|E_1) + P(E_2)P(R|E_2) \] \[ P(R) = \left( \frac{1}{3} \right) \left( \frac{3}{7} \right) + \left( \frac{2}{3} \right) \left( \frac{4}{7} \right) \] \[ P(R) = \frac{1}{7} + \frac{8}{21} = \frac{3 + 8}{21} = \frac{11}{21} \] Quick Tip: Always identify the initial experiment (die throw) as the basis for partitioning the sample space into mutually exclusive events.
Check if the conditional probabilities (drawing from bag) are calculated using the correct total number of balls in that specific bag.

Concept:

\( P(atleast one) = P(X \cup Y) \).
\( P(exactly one) = P(X \cup Y) - P(X \cap Y) \).
Complementary probability: \( P(X') = 1 - P(X) \).


Step 1: Express given information in terms of probabilities

We are given: \[ P(X \cup Y) = a \] \[ P(exactly one) = b \implies P(X \cup Y) - P(X \cap Y) = b \]
Substitute the value of \( P(X \cup Y) \): \[ a - P(X \cap Y) = b \implies P(X \cap Y) = a - b \]

Step 2: Use the addition theorem for two events
\[ P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) \] \[ a = P(X) + P(Y) - (a - b) \] \[ P(X) + P(Y) = a + a - b = 2a - b \]

Step 3: Calculate the required expression \( P(X') + P(Y') \)
\[ P(X') + P(Y') = (1 - P(X)) + (1 - P(Y)) \] \[ P(X') + P(Y') = 2 - (P(X) + P(Y)) \]
Substituting the value from Step 2: \[ P(X') + P(Y') = 2 - (2a - b) = 2 - 2a + b \]
Hence proved. Quick Tip: Remember the Venn diagram interpretation: "Exactly one" is the total union minus the common intersection.
Complementary logic \( P(X') = 1 - P(X) \) is essential when a proof involves primes/complements.

Concept:

An equivalence relation must be Reflexive, Symmetric, and Transitive.
Reflexive: \( (x, x) \in R \) for all \( x \in Z \).
Symmetric: \( (x, y) \in R \implies (y, x) \in R \).
Transitive: \( (x, y) \in R and (y, z) \in R \implies (x, z) \in R \).


Step 1: Check for Reflexivity

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

Since \( 0 \) is divisible by any prime number \( p \) (\( 0 = 0 \times p \)), we have \( (x, x) \in R \).

Thus, \( R \) is reflexive.

Step 2: Check for Symmetry

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

We know that \( |x - y| = |y - x| \).

So, \( |y - x| \) is also divisible by \( p \), which implies \( (y, x) \in R \).

Thus, \( R \) is symmetric.

Step 3: Check for Transitivity

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

Then \( |x - y| = k_1 p \) and \( |y - z| = k_2 p \), where \( k_1, k_2 \in \{0, 1, 2, \dots \} \).

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

Adding these equations: \[ (x - y) + (y - z) = (\pm k_1 \pm k_2)p \] \[ x - z = ( \pm k_1 \pm k_2)p \]
Taking absolute values: \[ |x - z| = |(\pm k_1 \pm k_2)|p \]
Since \( |x - z| \) is a multiple of \( p \), \( (x, z) \in R \).

Thus, \( R \) is transitive.

Step 4: Conclusion

Since the relation \( R \) is reflexive, symmetric, and transitive, it is an equivalence relation. Quick Tip: When checking transitivity for divisibility relations, it is often easier to write the condition as \( x - y = kp \) rather than using the absolute value directly.
Recall that \( 0 \) is a multiple of every integer except itself.

Concept:

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


Step 1: Prove that the function is one-one

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

Step 2: Prove that the function is onto

Let \( y \in R - \{3/5\} \). We need to find \( x \) such that \( f(x) = y \). \[ y = \frac{3x + 2}{5x - 3} \] \[ y(5x - 3) = 3x + 2 \] \[ 5xy - 3y = 3x + 2 \] \[ 5xy - 3x = 3y + 2 \implies x(5y - 3) = 3y + 2 \] \[ x = \frac{3y + 2}{5y - 3} \]
Since \( y \neq 3/5 \), the denominator \( 5y - 3 \neq 0 \), so \( x \) is a real number.
Also, check if \( x = 3/5 \): \[ \frac{3y + 2}{5y - 3} = \frac{3}{5} \implies 15y + 10 = 15y - 9 \implies 10 = -9 (not possible) \]
So, \( x \in Domain \). Thus, \( f \) is onto. Quick Tip: To prove onto, solve for \( x \) in terms of \( y \) and demonstrate that this \( x \) always belongs to the domain for any valid \( y \).
Simplifying cross-multiplication carefully is key for one-one proofs of rational functions.

Concept:

Matrix Inverse: \( A^{-1} = \frac{1}{|A|} adj(A) \).
Solving \( AX = B \implies X = A^{-1}B \).
Note the system of equations needs to be mapped to the matrix \( A \).


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

Step 2: Find the adjoint of A

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

Step 3: Solve the system of equations

The system is: \[ \begin{bmatrix} 0 & -2 & 1 \\ 2 & -1 & -1 \\ 1 & -2 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 7 \\ 8 \\ 10 \end{bmatrix} \]
The coefficient matrix is \( A^T \). So \( (A^T)X = B \implies X = (A^T)^{-1}B = (A^{-1})^T B \). \[ X = \begin{bmatrix} 2 & 2 & -3 \\ 1 & 1 & -2 \\ 3 & 2 & -4 \end{bmatrix} \begin{bmatrix} 7 \\ 8 \\ 10 \end{bmatrix} = \begin{bmatrix} 14 + 16 - 30 \\ 7 + 8 - 20 \\ 21 + 16 - 40 \end{bmatrix} = \begin{bmatrix} 0 \\ -5 \\ -3 \end{bmatrix} \]
Solution: \( x = 0, y = -5, z = -3 \). Quick Tip: Always check if the coefficient matrix of the system is \( A \) or \( A^T \). If it's \( A^T \), use \( (A^{-1})^T \) to solve.
Verify \( A \cdot A^{-1} = I \) if time permits to ensure calculations are correct.

Concept:

A matrix is singular if its determinant is zero.
Expand the determinant and solve the resulting trigonometric equation.


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

Step 2: Solve the trigonometric equation

Using \( 1 - \cos 2x = 2\sin^2 x \) and \( \sin 3x = 3\sin x - 4\sin^3 x \): \[ 2(2\sin^2 x) - (3\sin x - 4\sin^3 x) = 0 \] \[ 4\sin^2 x - 3\sin x + 4\sin^3 x = 0 \] \[ \sin x (4\sin^2 x + 4\sin x - 3) = 0 \]

Step 3: Find the values of x in the given interval

Case 1: \( \sin x = 0 \implies x = 0 \in [0, \pi/2] \).

Case 2: \( 4\sin^2 x + 4\sin x - 3 = 0 \).

Using quadratic formula: \( \sin x = \frac{-4 \pm \sqrt{16 + 48}}{8} = \frac{-4 \pm 8}{8} \).
\( \sin x = 1/2 \) or \( \sin x = -3/2 \) (Reject).
\( \sin x = 1/2 \implies x = \pi/6 \in [0, \pi/2] \).

The values are \( x = 0, \pi/6 \). Quick Tip: Singular matrix problems are essentially determinant equations. Factorize the final equation to find all possible roots.
Double-check roots against the specified interval \( [0, \pi/2] \).

Concept:

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


Step 1: Find the first derivative \( \frac{dy}{dx} \)
\[ \frac{dx}{dt} = -\sin t, \quad \frac{dy}{dt} = -m\sin mt \] \[ \frac{dy}{dx} = \frac{-m\sin mt}{-\sin t} = \frac{m\sin mt}{\sin t} \]

Step 2: Find the second derivative \( \frac{d^2y}{dx^2} \)
\[ \frac{d^2y}{dx^2} = \frac{d}{dt} \left( \frac{m\sin mt}{\sin t} \right) \cdot \frac{dt}{dx} \]
Using quotient rule for the t-derivative: \[ \frac{d^2y}{dx^2} = \frac{m^2\cos mt \sin t - m\sin mt \cos t}{\sin^2 t} \cdot \left( \frac{1}{-\sin t} \right) \] \[ \frac{d^2y}{dx^2} = \frac{m\sin mt \cos t - m^2\cos mt \sin t}{\sin^3 t} \]

Step 3: Substitute into the target equation

LHS \( = (1 - \cos^2 t) \left( \frac{m\sin mt \cos t - m^2\cos mt \sin t}{\sin^3 t} \right) - \cos t \left( \frac{m\sin mt}{\sin t} \right) + m^2\cos mt \) \[ = \sin^2 t \left( \frac{m\sin mt \cos t - m^2\cos mt \sin t}{\sin^3 t} \right) - \frac{m\cos t \sin mt}{\sin t} + m^2\cos mt \] \[ = \frac{m\sin mt \cos t - m^2\cos mt \sin t}{\sin t} - \frac{m\sin mt \cos t}{\sin t} + m^2\cos mt \] \[ = -m^2\cos mt + m^2\cos mt = 0 \]
Hence proved. Quick Tip: For parametric differentiation, don't forget the \( \frac{dt}{dx} \) factor when calculating the second derivative.
Simplifying \( \sin^2 t / \sin^3 t \) early helps avoid cumbersome expressions.

Concept:

Lines are parallel if their direction ratios are proportional.
Intersecting lines: Solve the parametric equations of both lines to find common parameters.


Step 1: Check for parallelism

Direction ratios of Line 1 (\( L_1 \)): \( (a_1, b_1, c_1) = (2, 3, 4) \).

Direction ratios of Line 2 (\( L_2 \)): \( (a_2, b_2, c_2) = (5, 2, 1) \).

Since \( \frac{2}{5} \neq \frac{3}{2} \), the lines are NOT parallel.

Step 2: Set up parametric equations and solve

General point on \( L_1 \): \( P(2\lambda + 1, 3\lambda + 2, 4\lambda + 3) \).

General point on \( L_2 \): \( Q(5\mu + 4, 2\mu + 1, \mu) \).

For intersection, \( P = Q \): \[ 2\lambda + 1 = 5\mu + 4 \implies 2\lambda - 5\mu = 3 \dots(1) \] \[ 3\lambda + 2 = 2\mu + 1 \implies 3\lambda - 2\mu = -1 \dots(2) \] \[ 4\lambda + 3 = \mu \dots(3) \]

Step 3: Solve the system of equations

Substitute \( \mu = 4\lambda + 3 \) from (3) into (1): \[ 2\lambda - 5(4\lambda + 3) = 3 \implies 2\lambda - 20\lambda - 15 = 3 \] \[ -18\lambda = 18 \implies \lambda = -1 \]
Substitute \( \lambda = -1 \) into (3): \( \mu = 4(-1) + 3 = -1 \).
Check values in (2): \( 3(-1) - 2(-1) = -3 + 2 = -1 \). (Satisfied).

Step 4: Find the point of intersection

Using \( \lambda = -1 \) in parametric form of \( L_1 \): \[ x = 2(-1) + 1 = -1, \quad y = 3(-1) + 2 = -1, \quad z = 4(-1) + 3 = -1 \]
The point of intersection is \( (-1, -1, -1) \). Quick Tip: Always check the third equation with the results of the first two to confirm the lines actually intersect rather than being skew.
If direction ratios are not proportional, the lines can either be skew or intersecting.

Concept:



A linear relationship exists between the price increase and the loss of subscribers.
Rate of change: 10 subscribers lost per \( Rs 1 \) increase.


Step 1: Identify the rate of decrease


For every \( Rs 1 \) increase, the number of subscribers lost is 10.

Step 2: Calculate total decrease for an increase of \( x \)


Since the increase is \( Rs x \), the total number of people leaving is: \[ Subscribers discontinuing = 10 \times x = 10x \] Quick Tip: Always define the variables clearly. Here, \( x \) represents the unit increase in price, and the total loss is simply the unit loss multiplied by \( x \).

Concept:



Revenue = (New Subscription Fee) \( \times \) (Number of remaining Subscribers).
New Fee = Original Fee + Increase.
Remaining Subscribers = Original Subscribers - Discontinuing Subscribers.


Step 1: Find the new fee and subscriber count


Original fee = \( Rs 300 \). New fee = \( 300 + x \).


Original subscribers = 5000. Discontinuing = \( 10x \).


Remaining subscribers = \( 5000 - 10x \).

Step 2: Formulate the Revenue function

\[ R(x) = (300 + x)(5000 - 10x) \]
Expanding the product: \[ R(x) = 300(5000) - 300(10x) + x(5000) - x(10x) \] \[ R(x) = 1,500,000 - 3000x + 5000x - 10x^2 \] \[ R(x) = 1,500,000 + 2000x - 10x^2 \] Quick Tip: Revenue is always "Price \( \times \) Quantity". If price goes up and quantity goes down, find the algebraic expression for both before multiplying.

Concept:



For a function to have a maximum value, its first derivative must be zero (\( R'(x) = 0 \)).
The second derivative test must satisfy \( R''(x) < 0 \) for a local maximum.


Step 1: Find the first derivative of the revenue function


From the previous part, \( R(x) = -10x^2 + 2000x + 1,500,000 \). \[ R'(x) = -20x + 2000 \]

Step 2: Find the critical point


Set \( R'(x) = 0 \): \[ -20x + 2000 = 0 \] \[ 20x = 2000 \implies x = 100 \]

Step 3: Verify maximum using second derivative

\[ R''(x) = -20 \]
Since \( R''(100) = -20 < 0 \), the revenue is maximized when the price increase \( x \) is \( Rs 100 \). Quick Tip: The point where the derivative of a quadratic function (downward parabola) is zero is always its global maximum.

Concept:



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


Step 1: Identify the sign of the derivative


From the previous part, \( R'(x) = 2000 - 20x = 20(100 - x) \).

Step 2: Determine the increasing interval


For \( R(x) \) to be increasing: \[ 20(100 - x) > 0 \implies 100 - x > 0 \implies x < 100 \]
Given the domain starts from 0, the interval is \( (0, 100) \).

Step 3: Determine the decreasing interval


For \( R(x) \) to be decreasing: \[ 20(100 - x) < 0 \implies 100 - x < 0 \implies x > 100 \]
Given the upper limit is 5000, the interval is \( (100, 5000) \). Quick Tip: The critical point (\( x = 100 \)) splits the entire domain into two intervals. Test a sample value from each to check for increasing/decreasing behavior.

Concept:

Possible outcomes are defined by the prize structure.
Not winning any prize is also a possibility with a non-zero probability.


Step 1: List the prize amounts

The prize structure offers:

\( Rs 3,00,000 \) (First prize)
\( Rs 2,00,000 \) (Second prize)
\( Rs 50,000 \) (Third prize)


Step 2: Consider the "no win" scenario

Since only 6 tickets out of 1,00,000 win prizes, the amount Rohan can win if his ticket is not drawn is \( Rs 0 \).

Step 3: Conclusion

The possible amounts are \( \{Rs 3,00,000, Rs 2,00,000, Rs 50,000, Rs 0\} \). Quick Tip: In probability, don't forget the case where no event occurs. Winning \( Rs 0 \) is a valid and most frequent outcome.

Concept:



Winning "atleast \( Rs 2,00,000 \)" means winning either the first prize or the second prize.
Probability = (Number of favorable tickets) / (Total tickets).


Step 1: Count the favorable tickets


Number of 1st prize tickets = 1.


Number of 2nd prize tickets = 2.


Total favorable tickets = \( 1 + 2 = 3 \).


Step 2: Calculate the probability


Total tickets = 1,00,000. \[ P(atleast 2,00,000) = \frac{3}{1,00,000} \] Quick Tip: "At least" indicates we should sum the counts of all categories that meet or exceed the specified value.

Concept:



Probability of not winning = 1 - (Probability of winning any prize).
Total winning tickets = sum of all tickets assigned to prizes.


Step 1: Find total winning tickets


Winning tickets = 1 (1st) + 2 (2nd) + 3 (3rd) = 6.


Step 2: Find non-winning tickets


Non-winning tickets = \( 1,00,000 - 6 = 99,994 \).


Step 3: Calculate the probability

\[ P(no win) = \frac{99,994}{1,00,000} \]
Alternatively: \( 1 - \frac{6}{1,00,000} = \frac{99,994}{1,00,000} \). Quick Tip: For "none" cases, it is often simpler to calculate the probability of the complement (at least one win) and subtract from 1.

Concept:



Let \( A \) be winning the first jackpot and \( B \) be winning the second.
Winning "exactly one" means (Wins A and Fails B) OR (Fails A and Wins B).
\( P(exactly one) = P(A)P(B') + P(A')P(B) \).


Step 1: Identify probabilities for both jackpots


Jackpot 1 (from previous parts): \( P(A) = \frac{6}{1,00,000} \), \( P(A') = \frac{99,994}{1,00,000} \).


Jackpot 2: \( P(B) = \frac{1}{1,00,000} \), \( P(B') = \frac{99,999}{1,00,000} \).


Step 2: Apply the formula for exactly one occurrence

\[ P = \left(\frac{6}{10^5}\right)\left(\frac{99,999}{10^5}\right) + \left(\frac{99,994}{10^5}\right)\left(\frac{1}{10^5}\right) \] \[ P = \frac{599,994 + 99,994}{10^{10}} = \frac{699,988}{10^{10}} \] Quick Tip: When events are independent, the probability of "exactly one" is calculated by summing the mutually exclusive cases of winning one while losing the other.

Concept:



Equations of form \( x^2 + y^2 = r^2 \) represent circles centered at origin \( (0,0) \) with radius \( r \).
\( C_1 \) has radius 8; \( C_2 \) has radius 2.


Step 1: Identify geometric parameters


For \( C_1 \): \( r_1 = \sqrt{64} = 8 \).


For \( C_2 \): \( r_2 = \sqrt{4} = 2 \).

Step 2: Describe the diagram




The diagram consists of two concentric circles. The outer circle has radius 8, and the inner circle has radius 2. Quick Tip: Concentric circles share the same center point. Always label the radii and the axes to provide a complete geometric representation.

Concept:



Rearrange the circle equation \( x^2 + y^2 = r^2 \) to solve for \( y \).
Use the positive root for the upper semi-circle.


Step 1: Rearrange for \( C_1 \)

\[ x^2 + y^2 = 64 \implies y^2 = 64 - x^2 \implies y = \pm \sqrt{64 - x^2} \]


Step 2: Rearrange for \( C_2 \)

\[ x^2 + y^2 = 4 \implies y^2 = 4 - x^2 \implies y = \pm \sqrt{4 - x^2} \] Quick Tip: A circle is not a function unless we restrict it to either the upper (\( + \)) or lower (\( - \)) part. In integration, we usually use the positive root for the first quadrant calculation.

Concept:
Concept:

Area of circle = \( 4 \times \int_0^r \sqrt{r^2 - x^2} \, dx \).
Area of roundabout = Area of \( C_1 \) - Area of \( C_2 \).
Formula: \( \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} \).


Step 1: Calculate area of \( C_1 \) using integration
\[ A_1 = 4 \int_{0}^{8} \sqrt{8^2 - x^2} \, dx = 4 \left[ \frac{x}{2}\sqrt{64 - x^2} + \frac{64}{2}\sin^{-1}\frac{x}{8} \right]_{0}^{8} \] \[ A_1 = 4 [ 0 + 32 \sin^{-1}(1) - 0 ] = 4 \times 32 \times \frac{\pi}{2} = 64\pi \]

Step 2: Calculate area of \( C_2 \) using integration
\[ A_2 = 4 \int_{0}^{2} \sqrt{2^2 - x^2} \, dx = 4 \left[ \frac{x}{2}\sqrt{4 - x^2} + \frac{4}{2}\sin^{-1}\frac{x}{2} \right]_{0}^{2} \] \[ A_2 = 4 [ 0 + 2 \sin^{-1}(1) - 0 ] = 4 \times 2 \times \frac{\pi}{2} = 4\pi \]

Step 3: Find the final area
\[ Area of roundabout = A_1 - A_2 = 64\pi - 4\pi = 60\pi sq. units \] Quick Tip: Use the symmetry of the circle. Calculating area in the first quadrant and multiplying by 4 simplifies the evaluation of limits.

Concept:



The pond area corresponds to the interior of the smaller circle \( C_2 \).
Standard circle area integration from Step 2 of previous part.


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


Radius = 2. Area is \( 4 \times First Quadrant Area \). \[ Area = 4 \int_{0}^{2} \sqrt{4 - x^2} \, dx \]


Step 2: Evaluate the integral

\[ = 4 \left[ \frac{x}{2}\sqrt{4 - x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right) \right]_{0}^{2} \] \[ = 4 \left[ \left( 0 + 2\sin^{-1}(1) \right) - (0 + 0) \right] \] \[ = 4 \times 2 \times \frac{\pi}{2} = 4\pi sq. units \] Quick Tip: Recall the definite integral result for a full circle: \( \pi r^2 \). Use this to verify your integration steps. Here \( \pi(2)^2 = 4\pi \).