ISC Class 12 Mathematics Board Exam Question Paper 2025 with Solutions

ISC Class 12 Mathematics Board Exam Question Paper with Solutions

SECTION - A (65 MARKS)

Question 1

In subparts (i) to (xi) choose the correct options and in subparts (xii) to (xv), answer the questions as instructed.

(i) If \( A = \begin{bmatrix} 0 & a \\ 0 & 0 \end{bmatrix} \), then \( A^{16} \) is: [1 Mark]
(A) Unit matrix
(B) Null matrix
(C) Diagonal matrix
(D) Skew matrix

Answer: (B) Null matrix

\( A^2 = A \times A = \begin{bmatrix} 0 & a \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & a \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \). Since \( A^2 \) is the null matrix, any higher power \( A^{16} \) is also the null matrix.

Teacher's Note:
a) For nilpotent matrices of index 2, any power greater than or equal to 2 is a null matrix.
b) Do not confuse a null matrix with a diagonal or unit matrix.

(ii) Which of the following is a homogenous differential equation? [1 Mark]
(A) \((4x^2 + 6y + 5) dy - (3y^2 + 2x + 4) dx = 0\)
(B) \((xy) dx - (x^3 + y^3) dy = 0\)
(C) \((x^3 + 2y^2) dx + 2xy dy = 0\)
(D) \(y^2 dx + (x^2 - xy - y^2) dy = 0\)

Answer: (D) \(y^2 dx + (x^2 - xy - y^2) dy = 0\)

Every term in each expression of option (D) is of degree 2, making it homogeneous.

Teacher's Note:
a) A differential equation is homogeneous if every term has the same total degree in variables \( x \) and \( y \).
b) Check the degree of constant terms and mixed terms carefully.

(iii) Consider the graph of the function \( f(x) \) shown below:
[Figure: A curve on a Cartesian plane passing through origin, with local minimum at \( x = 1/2 \) and local maximum at \( x = 1 \), becoming constant for \( x \gt 1 \). Coordinates marked on x-axis are \( 1/2, 1, 2 \), and y-axis has label \( 1 \).]
Statement 1: The function \( f(x) \) is increasing in \( \left(\frac{1}{2}, 2\right) \).
Statement 2: The function \( f(x) \) is strictly increasing in \( \left(\frac{1}{2}, 1\right) \).
Which of the following is correct with respect to the above statements? [1 Mark]
(A) Statement 1 is true and Statement 2 is false.
(B) Statement 2 is true and Statement 1 is first.
(C) Both the statements are true.
(D) Both the statements are false.

Answer: (C) Both the statements are true.

The function increases from \( x = 1/2 \) to \( x = 1 \) strictly, and remains constant from \( x = 1 \) to \( x = 2 \), fulfilling the definition of an increasing function over the larger interval.

Teacher's Note:
a) A function is increasing if \( f'(x) \ge 0 \) and strictly increasing if \( f'(x) \gt 0 \).
b) Flat regions where derivative is zero still satisfy the condition for a non-strictly increasing function.

(iv) \( \int \frac{x^4 - 1}{x^2 + 1} dx \) is equal to: [1 Mark]
(A) \( \frac{2}{3} \)
(B) \( \frac{1}{3} \)
(C) \( \frac{-2}{3} \)
(D) \( 0 \)

Answer: (C) \( \frac{-2}{3}\) (Note: The option values in question appear from definite integral limits 0 to 1 implicitly evaluated in options, or standard printing convention; based on step evaluation \( \frac{x^3}{3} - x \) evaluated from 0 to 1 gives \( -2/3 \)).

Factorising numerator as \( (x^2 - 1)(x^2 + 1) \) and canceling \( x^2 + 1 \) leaves \( \int (x^2 - 1) dx \).

Teacher's Note:
a) Simplify algebraic rational functions by factorisation before integration.
b) Watch out for definite integral limits if bounds are implied.

(v) Assertion: Consider the two events \( A \) and \( B \) such that \( n(A) = n(B) \) and \( P\left(\frac{A}{B}\right) = P\left(\frac{B}{A}\right) \).
Reason: The events \( A \) and \( B \) are mutually exclusive. [1 Mark]
(A) Both Assertion and Reason are true and Reason is the correct explanation for Assertion.
(B) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.
(C) Assertion is true and Reason is false.
(D) Assertion is false and Reason is true.

Answer: (C) Assertion is true and Reason is false.

Mutual exclusivity is not necessary for the conditional probabilities to be equal when \( n(A) = n(B) \).

Teacher's Note:
a) Mutually exclusive events have \( P(A \cap B) = 0 \).
b) Always test conditional probability formulas thoroughly.

(vi) The existence of unique solution of the system of equations \( x + y = \lambda \) and \( 5x + ky = 2 \) depends on: [1 Mark]
(A) \( \lambda \) only
(B) \( \frac{2}{k} = 1 \)
(C) both \( k \) and \( \lambda \)
(D) \( k \) only

Answer: (D) \( k \) only

Unique solution requires determinant \( \Delta \neq 0 \), hence \( 1(k) - 1(5) \neq 0 \implies k \neq 5 \), which depends solely on \( k \).

Teacher's Note:
a) Consistency and uniqueness of linear systems are determined by the coefficient matrix determinant.
b) Parameter \( \lambda \) only affects the specific values of solution, not uniqueness.

(vii) A cylindrical popcorn tub of radius \( 10\text{ cm} \) is being filled with popcorn at the rate of \( 314\text{ cm}^3 \) per minute. The level of the popcorns in the tub is increasing at the rate of: [1 Mark]
(A) \( 1\text{ cm/minute} \)
(B) \( 0.1\text{ cm/minute} \)
(C) \( 1.1\text{ cm/minute} \)
(D) \( 0.5\text{ cm/minute} \)

Answer: (A) \( 1\text{ cm/minute} \)

\( V = \pi r^2 h \implies \frac{dV}{dt} = \pi r^2 \frac{dh}{dt} \implies 314 = (3.14)(10)^2 \frac{dh}{dt} \implies \frac{dh}{dt} = 1 \).

Teacher's Note:
a) Relate rates of change using standard volume formulas for cylinders.
b) Substitute numerical values carefully keeping units consistent.

(viii) If \( f(x) = \begin{cases} x + 2, & x \lt 0 \\ -x^2 - 2, & 0 \le x \lt 1 \\ x, & x \ge 1 \end{cases} \), then the number of point(s) of discontinuity of \( f(x) \), is/are: [1 Mark]
(A) \( 1 \)
(B) \( 3 \)
(C) \( 2 \)
(D) \( 0 \)

Answer: (C) \( 2 \)

Checking continuity at \( x = 0 \) and \( x = 1 \) reveals left-hand and right-hand limits do not match at both points.

Teacher's Note:
a) Evaluate LHL, RHL, and function value at boundary points.
b) Points of definition change are prime candidates for discontinuity.

(ix) Assertion: If Set A has \( m \) elements, Set B has \( n \) elements and \( n \lt m \), then the number of one-one function(s) from \( A \to B \) is zero.
Reason: A function \( f: A \to B \) is defined only if all elements in Set A have an image in Set B. [1 Mark]
(A) Both Assertion and Reason are true and Reason is the correct explanation for Assertion.
(B) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.
(C) Assertion is true and Reason is false.
(D) Assertion is false and Reason is true.

Answer: (B) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.

The reason states a general property of functions rather than explaining why injective mapping fails when domain size exceeds codomain size.

Teacher's Note:
a) Injective functions require domain cardinality less than or equal to codomain cardinality.
b) Distinguish between general function definition and injectivity constraints.

(x) Let \( X \) be a discrete random variable. The probability distribution of \( X \) is given below:

Answer: (B) \( 4 \)

\( E(X) = \sum X P(X) = 30\left(\frac{1}{5}\right) + 10\left(\frac{3}{10}\right) + (-10)\left(\frac{1}{2}\right) = 6 + 3 - 5 = 4 \).

Teacher's Note:
a) Expected value is the weighted average of random variable values by their probabilities.
b) Verify that the sum of probabilities equals \( 1 \) before computing.

(xi) Statement 1: If \( A \) is an invertible matrix, then \( (A^2)^{-1} = (A^{-1})^2 \).
Statement 2: If \( A \) is an invertible matrix, then \( |A^{-1}| = |A|^{-1} \). [1 Mark]
(A) Statement 1 is true and Statement 2 is false.
(B) Statement 2 is true and Statement 1 is false.
(C) Both the statements are true.
(D) Both the statements are false.

Answer: (C) Both the statements are true.

Inverse properties of matrices apply identically to powers and determinants.

Teacher's Note:
a) \( (A^n)^{-1} = (A^{-1})^n \) for any invertible matrix \( A \).
b) Determinant of an inverse is the reciprocal of the determinant.

(xii) Write the smallest equivalence relation from the set \( A \) to \( A \), where \( A = \{1, 2, 3\} \). [1 Mark]

Answer:
The smallest equivalence relation is the identity relation: \( R = \{(1, 1), (2, 2), (3, 3)\} \).

Teacher's Note:
a) An equivalence relation must be reflexive, symmetric, and transitive.
b) The identity relation is always the minimum relation satisfying all three conditions.

(xiii) For what value of \( x \), is \( A = \begin{bmatrix} 0 & -1 & x \\ -1 & 0 & -3 \\ x & -3 & 0 \end{bmatrix} \) a skew symmetric matrix? [1 Mark]

Answer:
\( x = 2 \).

For skew symmetric matrix \( A^T = -A \), comparing elements gives \( x = 2 \).

Teacher's Note:
a) A skew-symmetric matrix has zeroes on its principal diagonal and \( a_{ij} = -a_{ji} \).
b) Check corresponding off-diagonal elements carefully.

(xiv) Three critics review a book. Odds in favour of the book are \( 5:2 \), \( 4:3 \) and \( 3:4 \), respectively for the three critics. Find the probability that all critics are in favour of the book. [1 Mark]

Answer:
\( \frac{60}{343} \).

Probabilities are \( \frac{5}{7} \), \( \frac{4}{7} \), and \( \frac{3}{7} \). Product equals \( \frac{60}{343} \).

Teacher's Note:
a) Convert odds in favour \( a:b \) to probability \( \frac{a}{a+b} \).
b) Multiply independent probabilities for joint occurrence.

(xv) Evaluate: \( \int \frac{5}{\sqrt{2x + 7}} dx \) [1 Mark]

Answer:
\( 5\sqrt{2x + 7} + C \).

Substitute \( u = 2x + 7 \), \( du = 2 dx \), resulting in \( \frac{5}{2} \int u^{-1/2} du = 5\sqrt{u} + C \).

Teacher's Note:
a) Use linear substitution for integrals of composite power functions.
b) Do not forget the constant of integration \( C \).

Question 2 [2 Marks]
Find the point on the curve \( y = 2x^2 - 6x - 4 \) at which the tangent is parallel to the x-axis.

Answer:
\( \left(\frac{3}{2}, -\frac{17}{2}\right) \).

Derivative \( \frac{dy}{dx} = 4x - 6 = 0 \implies x = \frac{3}{2} \). Substituting \( x \) into the curve gives \( y = -\frac{17}{2} \).

Teacher's Note:
a) Tangent parallel to the x-axis implies slope \( \frac{dy}{dx} = 0 \).
b) Substitute the critical x-value back into the original equation to find the y-coordinate.

Question 3 [2 Marks]
Find the value of \( \tan^{-1} x - \cot^{-1} x \), if \( (\tan^{-1} x)^2 - (\cot^{-1} x)^2 = \frac{5\pi}{8} \).

Answer:
\( \frac{5}{4} \pi \).

Using \( \tan^{-1}x + \cot^{-1}x = \frac{\pi}{2} \) and factoring the difference of squares gives \( \tan^{-1}x - \cot^{-1}x = \frac{5}{4} \).

Teacher's Note:
a) Use standard inverse trigonometric identity sum properties.
b) Apply algebraic factorisation \( a^2 - b^2 = (a-b)(a+b) \).

Question 4 [2 Marks]
(i) If \( x^y = e^{x-y} \), prove that \( \frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2} \)

Answer:
Taking logarithm on both sides gives \( y \log x = x - y \implies y(1 + \log x) = x \implies y = \frac{x}{1 + \log x} \). Differentiating with respect to \( x \) using quotient rule yields \( \frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2} \).

Teacher's Note:
a) Take logarithms to simplify expressions involving variables in exponents.
b) Apply the quotient rule carefully during differentiation.

OR

(ii) If \( f(x) = \log(1 + x) + \frac{1}{1 + x} \), show that \( f(x) \) attains its minimum value at \( x = 0 \). [2 Marks]

Answer:
First derivative \( f'(x) = \frac{1}{1+x} - \frac{1}{(1+x)^2} = \frac{x}{(1+x)^2} \). Setting \( f'(x) = 0 \) gives \( x = 0 \). Second derivative test confirms \( f''(0) = 1 \gt 0 \), hence minimum at \( x = 0 \).

Teacher's Note:
a) Find critical points by equating the first derivative to zero.
b) Use the second derivative test to determine local extrema.

Question 5 [2 Marks]
Three shopkeepers Gaurav, Rizwan and Jacob use carry bags made of polythene, handmade paper and newspaper. The number of polythene bags, handmade bags and newspaper bags used by Gaurav, Rizwan and Jacob are \( (20, 30, 40) \), \( (30, 40, 20) \) and \( (40, 20, 30) \). One polythene bag costs Rs. \( 1 \), one handmade bag is for Rs. \( 5 \) and one newspaper bag costs Rs. \( 2 \). Gaurav, Rizwan and Jacob spend Rs. \( A \), Rs. \( B \) and Rs. \( C \), respectively on these carry bags.
Using the concepts of matrices and determinants, answer the following questions:
(i) Represent the above information in matrix form.
(ii) Find the values of \( A \), \( B \) and \( C \).

Answer:
(i) Matrix representation: Usage matrix \( U = \begin{bmatrix} 20 & 30 & 40 \\ 30 & 40 & 20 \\ 40 & 20 & 30 \end{bmatrix} \), Cost vector \( C = \begin{bmatrix} 1 \\ 5 \\ 2 \end{bmatrix} \).
(ii) Expenditure matrix \( S = U \times C = \begin{bmatrix} 250 \\ 270 \\ 200 \end{bmatrix} \implies A = 250, B = 270, C = 200 \).

Teacher's Note:
a) Formulate matrix multiplication correctly matching rows and columns.
b) Calculate dot products row by column to find individual expenditures.

Question 6 [2 Marks]
(i) Differentiate \( \sin^{-1} \left( \frac{2^{x+1} \cdot 3^x}{1 + (36)^x} \right) \) with respect to \( x \).

Answer:
Rewrite expression as \( \sin^{-1} \left( \frac{2 \cdot 6^x}{1 + (6^x)^2} \right) \). Substitute \( 6^x = \tan \theta \) to get \( 2 \tan^{-1}(6^x) \). Differentiating gives \( \frac{2 \cdot 6^x \log 6}{1 + 36^x} \).

Teacher's Note:
a) Use trigonometric substitution \( \frac{2t}{1+t^2} = \sin(2\theta) \).
b) Apply chain rule for exponential functions.

OR

(ii) Show that \( \tan^{-1} x + \tan^{-1} y = C \) is the general solution of the differential equation \( (1 + x^2) dy + (1 + y^2) dx = 0 \) [2 Marks]

Answer:
Separating variables gives \( \frac{dy}{1+y^2} + \frac{dx}{1+x^2} = 0 \). Integrating both sides yields \( \tan^{-1}y + \tan^{-1}x = C \).

Teacher's Note:
a) Separate variables into respective derivative terms.
b) Standard integral of \( \frac{1}{1+x^2} \) is \( \tan^{-1}x \).

Question 7 [4 Marks]
If \( x + y + z = 0 \) then show that \( \begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ x^3 & y^3 & z^3 \end{vmatrix} = 0 \), using properties of determinant.

Answer:
Applying operations \( C_2 \to C_2 - C_1 \) and \( C_3 \to C_3 - C_1 \) and factoring out \( (x-y) \) and \( (z-x) \), the determinant simplifies to a product containing \( (x+y+z) \), which is \( 0 \).

Teacher's Note:
a) Use elementary column operations to create zeros in determinants.
b) Factor out common terms to simplify evaluation.

Question 8 [4 Marks]
(i) Evaluate: \( \int \frac{\cos x}{3\cos x - 5} dx \)

Answer:
Express numerator as \( \frac{1}{3}(3\cos x - 5) + \frac{5}{3} \) and integrate into standard logarithmic and half-angle substitution forms to get \( \frac{x}{3} - \frac{5}{6}\tan^{-1}\left(\frac{1}{2}\tan\frac{x}{2}\right) + C \).

Teacher's Note:
a) Adjust numerators in trigonometric integrals using linear combinations of denominators.
b) Apply tangent half-angle substitution where necessary.

OR

(ii) Evaluate: \( \int (\log x)^2 dx \) [4 Marks]

Answer:
Using integration by parts twice, \( \int (\log x)^2 dx = x(\log x)^2 - 2x \log x + 2x + C \).

Teacher's Note:
a) Apply integration by parts taking \( (\log x)^2 \) as the first function and \( 1 \) as the second.
b) Repeat integration by parts for the remaining integral term.

Question 9 [4 Marks]
(i) If \( x = \tan \left(\frac{1}{a} \log y\right) \) then show that \( (1 + x^2) \frac{d^2 y}{dx^2} + (2x - a) \frac{dy}{dx} = 0 \)

Answer:
Differentiating successively with respect to \( x \) and eliminating inverse trigonometric functions yields the required second-order differential equation.

Teacher's Note:
a) Differentiate implicitly and use chain rule systematically.
b) Cross-multiply derivative terms to clear fractions before second differentiation.

OR

(ii) The graph of \( f(x) = -x^3 + 27x - 2 \) is given below:
[Figure: Cubic curve with local maximum and local minimum points on a Cartesian plane.]
(a) Find the slope of the above graph.
(b) Find the co-ordinates of turning points, A and B.
(c) Evaluate \( f''(-2), f(0) \) and \( f'(3) \) and arrange them in ascending order. [4 Marks]

Answer:
(a) Slope \( f'(x) = -3x^2 + 27 \).
(b) Turning points at \( (-3, -56) \) and \( (3, 52) \).
(c) \( f''(-2) = 12, f(0) = -2, f'(3) = 0 \); arranged in ascending order: \( f(0) \lt f'(3) \lt f''(-2) \).

Teacher's Note:
a) Turning points occur where the first derivative equals zero.
b) Evaluate derivatives carefully at specified points.

Question 10 [4 Marks]
Pia, Sia and Dia displayed their paintings in an art exhibition. The three artists displayed \( 15, 5 \) and \( 10 \) of their paintings, respectively. A person bought three paintings from the exhibition.
(i) Find the probability that he bought one painting from each of them.
(ii) Find the probability that he bought all the three paintings from the same person.

Answer:
(i) \( \frac{75}{406} \)
(ii) \( \frac{117}{812} \)

Teacher's Note:
a) Use combinations to calculate favorable and total selection ways.
b) Sum mutually exclusive cases for part (ii).

Question 11 [6 Marks]
(i) Prove: \( \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{x dx}{1 + \sin x} = (\sqrt{2} - 1)\pi \)

Answer:
Using definite integral property \( \int_a^b f(x) dx = \int_a^b f(a+b-x) dx \), combining integrals, multiplying numerator and denominator by \( 1 - \sin x \), and integrating yields \( \pi(\sqrt{2}-1) \).

Teacher's Note:
a) King property of definite integrals \( \int_a^b f(x)dx = \int_a^b f(a+b-x)dx \) is extremely helpful here.
b) Rationalize denominators using trigonometric conjugates.

OR

(ii) Evaluate: \( \int \frac{x^2}{(x-1)^2(x^2+1)} dx \) [6 Marks]

Answer:
Using partial fractions: \( \frac{1}{2}\ln|x-1| - \frac{1}{2(x-1)} - \frac{1}{4}\ln(x^2+1) + C \).

Teacher's Note:
a) Express rational functions with repeated linear and irreducible quadratic factors using partial fraction decomposition.
b) Integrate individual terms using standard logarithmic and algebraic rules.

Question 12 [6 Marks]
(i) Solve the differential equation: \( (x + 5y^2) \frac{dy}{dx} = y \) when \( x = 2 \) and \( y = 1 \)

Answer:
Rewriting as linear differential equation in \( x \) with respect to \( y \): \( x = 5y^2 - 3y \).

Teacher's Note:
a) Treat \( x \) as dependent and \( y \) as independent variable if standard separation fails.
b) Use integrating factor method for linear differential equations.

OR

(ii) Find the particular solution of the differential equation: \( (x^2 - 2y^2) dx + 2xydy = 0 \), when \( x = 1 \) and \( y = 1 \) [6 Marks]

Answer:
Using homogeneous substitution \( y = vx \), the particular solution is \( \left(\frac{y}{x}\right)^2 + \ln|x| = 1 \).

Teacher's Note:
a) Substitute \( y = vx \) for homogeneous differential equations.
b) Substitute given initial conditions to find the constant of integration \( C \).

Question 13 [6 Marks]
Observe the two graphs, Graph 1 and Graph 2 given below and answer the questions that follow.
[Figure: Graph 1 shows principal branch of sine curve; Graph 2 shows arcsine curve in restricted domain.]
(i) Which one of the graphs represents \( y = \sin^{-1} x \)?
(ii) Write the domain and range of \( y = \sin^{-1} x \).
(iii) Prove that \( \sin^{-1} \frac{1}{\sqrt{5}} + \sin^{-1} \frac{2}{\sqrt{5}} = \frac{\pi}{2} \).
(iv) Find the value of \( \tan^{-1} \left[ 2 \sin \left( 2 \cos^{-1} \frac{\sqrt{3}}{2} \right) \right] \).

Answer:
(i) Graph 2 represents \( y = \sin^{-1} x \).
(ii) Domain: \( [-1, 1] \), Range: \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).
(iii) LHS evaluated using \( \sin^{-1}A + \sin^{-1}B \) formula equals \( \frac{\pi}{2} \).
(iv) Value equals \( \frac{\pi}{3} \).

Teacher's Note:
a) Memorise standard domain and range restrictions for inverse trigonometric functions.
b) Evaluate composite trigonometric expressions step-by-step from inside out.

Question 14 [6 Marks]
An international conference takes place in a metropolitan city. International leaders, scientists and industrialists participate in it.
The organisers of the conference appoint three agencies namely X, Y and Z for the security of the participants. The track record of the success of X, Y and Z in providing security services is \( 99\%, 98.5\% \) and \( 98\% \), respectively. The organisers assign the responsibility of ensuring the security of \( 1000 \) people to agency X, \( 2000 \) people to agency Y and \( 3000 \) people to agency Z.
At the end of the conference, one participant goes missing from the conference room. What is the probability that the missing participant was placed under the responsibility of the security agency X?

Answer:
\( 0.1 \).

Using Bayes' theorem, \( P(X|M) = \frac{P(M|X)P(X)}{P(M)} = \frac{0.01 \times \frac{1}{6}}{0.016667} = 0.1 \).

Teacher's Note:
a) Apply Bayes' theorem for conditional probability problems involving multiple causes.
b) Calculate total probability in the denominator carefully summing all mutually exclusive branches.

SECTION - B (15 MARKS)

Question 15

In subparts (i) and (ii) choose the correct options and in subparts (iii) and (iv), answer the questions as instructed.

(i) Assertion: \( (\vec{a} + \vec{b})^2 + (\vec{b} - \vec{a})^2 = 2(\vec{a}^2 + \vec{b}^2) \)
Reason: Dot product of any two vectors is commutative. [1 Mark]
(A) Both Assertion and Reason are true and Reason is the correct explanation for Assertion.
(B) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.
(C) Assertion is true and Reason is false.
(D) Assertion is false and Reason is true.

Answer: (B) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.

Vector expansion uses commutativity of dot product, but the identity itself follows from algebraic expansion rules of dot products.

Teacher's Note:
a) Dot product satisfies commutative law \( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \).
b) Vector magnitude squared equals self dot product.

(ii) The angle between the two planes \( x + y + 2z = 9 \) and \( 2x - y + z = 15 \) is: [1 Mark]
(A) \( \frac{\pi}{2} \)
(B) \( \frac{\pi}{3} \)
(C) \( \pi \)
(D) \( \frac{3\pi}{4} \)

Answer: (B) \( \frac{\pi}{3} \)

Using normal vectors \( \vec{n_1} = \langle 1, 1, 2 \rangle \) and \( \vec{n_2} = \langle 2, -1, 1 \rangle \), \( \cos\theta = \frac{3}{\sqrt{6}\sqrt{6}} = \frac{1}{2} \implies \theta = \frac{\pi}{3} \).

Teacher's Note:
a) Angle between two planes is the angle between their normal vectors.
b) Use dot product formula for vectors to find cosine of the angle.

(iii) Show that points \( P(-2, 3, 5) \), \( Q(1, 2, 3) \) and \( R(7, 0, -1) \) are collinear. [1 Mark]

Answer:
Determinant formed by direction ratios or area of triangle formed by three points equals zero, proving collinearity.

Teacher's Note:
a) Three points are collinear if direction ratios of segments between them are proportional.
b) Alternatively, check if \( |\vec{PQ} \times \vec{QR}| = 0 \).

(iv) Two honeybees are flying parallel to each other in the garden to collect the nectar. The path traced by the bees is given \( \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k}) \).
[Figure: Two honeybees flying in parallel paths.]
(a) Write the above-mentioned equation in Cartesian form.
(b) Find the equation of the path traced by the other honeybee passing through the point \( (2, 4, 5) \). [1 Mark]

Answer:
(a) Cartesian form: \( \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} \).
(b) Equation passing through \( (2, 4, 5) \) with same direction ratios: \( \frac{x-2}{2} = \frac{y-4}{3} = \frac{z-5}{4} \).

Teacher's Note:
a) Convert vector line equations to Cartesian form using point coordinates and direction ratios.
b) Parallel lines share identical direction ratios.

Question 16 [2 Marks]
(i) Find the equation of the plane passing through the points \( (2, 2, -1) \), \( (3, 4, 2) \) and \( (7, 0, 6) \).

Answer:
\( 5x + 2y - 3z = 17 \).

Teacher's Note:
a) Use determinant formula for equation of plane passing through three non-collinear points.
b) Substitute points to verify correctness.

OR

(ii) Find the equation of the plane passing through the points \( (2, 3, 1) \), \( (4, -5, 3) \) and parallel to x-axis. [2 Marks]

Answer:
\( y + 4z = 7 \).

Teacher's Note:
a) Parallel to x-axis implies the x-coefficient in normal vector is zero (\( a = 0 \)).
b) Use given passing points to determine remaining plane coefficients.

Question 17 [2 Marks]
Consider the position vectors of A, B and C as \( \vec{OA} = 2\hat{i} - 2\hat{j} + \hat{k} \), \( \vec{OB} = \hat{i} + 2\hat{j} - 2\hat{k} \) and \( \vec{OC} = 2\hat{i} - \hat{j} + 4\hat{k} \).
(i) Calculate \( \vec{AB} \) and \( \vec{BC} \).
(ii) Find the projection of \( \vec{AB} \) on \( \vec{BC} \).
(iii) Find the area of the triangle ABC whose sides are \( \vec{AB} \) and \( \vec{BC} \).

Answer:
(i) \( \vec{AB} = -\hat{i} + 4\hat{j} - 3\hat{k} \), \( \vec{BC} = \hat{i} - 3\hat{j} + 6\hat{k} \).
(ii) Projection = \( \frac{-31}{\sqrt{46}} \).
(iii) Area = \( \frac{1}{2}\sqrt{235}\text{ sq. units} \).

Teacher's Note:
a) Vector subtraction yields side vectors from position vectors.
b) Triangle area is half the magnitude of cross product of adjacent side vectors.

Question 18 [4 Marks]
(i) The equation \( y = 4 - x^2 \) represents a parabola.
(a) Make a rough sketch of the graph of the given function.
(b) Determine the area enclosed between the curve, the x-axis, the lines \( x = 0 \) and \( x = 2 \).
(c) Hence, find the area bounded by the parabola and the x-axis.

Answer:
(a) Downward opening parabola with vertex at \( (0, 4) \).
(b) Area = \( \frac{16}{3}\text{ sq. units} \).
(c) Total area = \( \frac{32}{3}\text{ sq. units} \).

OR

(ii) A farmer has a field bounded by three lines \( x + 2y = 2 \), \( y - x = 1 \), \( 2x + y = 7 \). Using integration, find the area of the region bounded by these lines. [4 Marks]

Answer:
\( 6\text{ sq. units} \).

Teacher's Note:
a) Find intersection points of lines to set up integration limits.
b) Split integrals appropriately along regions to compute enclosed area.

SECTION - C (15 MARKS)

Question 19

In subparts (i) and (ii) choose the correct options and in subpart (iii), answer the questions as instructed.

(i) The total revenue received from the sale of \( x \) units of a product is \( R(x) = 36x + 3x^2 + 5 \). Then, the actual revenue for selling the \( 10^{\text{th}} \) item will be: [1 Mark]
(A) \( 27 \)
(B) \( 90 \)
(C) \( 93 \)
(D) \( 33 \)

Answer: (B) \( 90 \)

Marginal revenue at \( x = 9 \) or \( R(10) - R(9) \) gives \( 90 \).

Teacher's Note:
a) Actual revenue for the \( n^{\text{th}} \) item is \( R(n) - R(n-1) \).
b) Substitute values carefully into the revenue function.

(ii) Read the following statements and choose the correct option.
(I) The correlation coefficient and the regression coefficients are of the same sign.
(II) The correlation coefficient is the arithmetic mean between the regression coefficients.
(III) The product of two regression coefficients is always equal to \( 1 \).
(IV) Both the regression coefficients cannot be numerically greater than unity. [1 Mark]
(A) Only (IV) is correct.
(B) Only (I) and (II) are correct.
(C) Only (I) and (IV) are correct.
(D) Only (III) and (IV) are correct.

Answer: (C) Only (I) and (IV) are correct.

Correlation coefficient is the geometric mean of regression coefficients, and both cannot exceed unity in magnitude.

Teacher's Note:
a) Properties of regression coefficients dictate \( r = \pm\sqrt{b_{yx}b_{xy}} \).
b) Individual regression coefficients can exceed \( 1 \), but their product cannot exceed \( 1 \).

(iii) Consider the following data:

Answer:
(a) \( \bar{x} = 3, \bar{y} = 4 \).
(b) Differences calculated and tabulated.
(c) \( b_{xy} = -1 \).

Teacher's Note:
a) Compute means \( \bar{x} \) and \( \bar{y} \) first.
b) Use formula \( b_{xy} = \frac{\sum(x-\bar{x})(y-\bar{y})}{\sum(y-\bar{y})^2} \).

Question 20 [4 Marks]
(i) Find the regression line of best fit from the following data.
\( \sum x = 24, \sum y = 44, \sum xy = 306, \sum x^2 = 164, \sum y^2 = 576, n = 4 \)

Answer:
\( y = 2.1x - 1.6 \).

Teacher's Note:
a) Calculate regression coefficient \( b_{yx} \) using summary sums.
b) Use point-slope form with means to formulate the line equation.

OR

(ii) Two lines of regression are given as \( 4x + 3y + 7 = 0 \) and \( 3x + 4y + 8 = 0 \). Identify the line of regression of \( x \) on \( y \). [4 Marks]

Answer:
\( 3x + 4y + 8 = 0 \) represents the regression line of \( x \) on \( y \).

Teacher's Note:
a) Express equations in form \( x = byy + c \) or \( y = bxy + c \).
b) Verify that product of regression coefficients lies between \( 0 \) and \( 1 \).

Question 21 [4 Marks]
(i) A utensil manufacturer produces \( x \) dinner sets per week and sells each set at \( p \), where \( x = \frac{600 - p}{8} \).
(a) Write the revenue function.
(b) Write the profit function.
(c) Calculate the number of dinner sets to be produced and sold per week to ensure maximum profit.

Answer:
(a) Revenue \( R(x) = 600x - 8x^2 \).
(b) Profit \( P(x) = -9x^2 + 522x - 2000 \).
(c) \( x = 29 \) dinner sets per week.

OR

(ii) The Average Cost of producing \( x \) units of commodity is given by: \( AC = \frac{x^2}{200} - \frac{x}{50} - 30 + \frac{5000}{x} \)
(a) Find the Cost function.
(b) Find the Marginal Cost function.
(c) Find the Marginal Average Cost function.
(d) Verify that \( \frac{d}{dx}(AC) = \frac{MC - AC}{x} \) [4 Marks]

Answer:
(a) Cost \( C(x) = \frac{x^3}{200} - \frac{x^2}{50} - 30x + 5000 \).
(b) \( MC = \frac{3x^2}{200} - \frac{x}{25} - 30 \).
(c) \( MAC = \frac{x}{100} - \frac{1}{50} - \frac{5000}{x^2} \).
(d) Verified successfully.

Teacher's Note:
a) Cost is product of Average Cost and quantity \( x \).
b) Marginal cost is derivative of total cost function.

Question 22 [4 Marks]
Two different types of books have to be stacked in the shelf of a library. The first type of book weighs \( 1\text{ kg} \) and has a thickness of \( 6\text{ cm} \). The second type of book weighs \( 1.5\text{ kg} \) and has a thickness of \( 4\text{ cm} \). The shelf is \( 96\text{ cm} \) long and can support a maximum weight of \( 21\text{ kg} \).
How should both the types of books be placed in the shelf to include the maximum number of books? Formulate a Linear Programming Problem and solve it graphically.

Answer:
Maximum number of books is \( 18 \) at \( (12, 6) \) (Books of type I = \( 12 \), Books of type II = \( 6 \)).

Teacher's Note:
a) Formulate constraints based on thickness and weight limits.
b) Evaluate objective function \( Z = x + y \) at all corner points of the feasible region to find maximum.