ISC Class 12 Mathematics Board Exam Question Paper 2023 with Solutions

ISC Class 12 Mathematics Board Exam Question Paper with Solutions

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

(i) A relation \( R \) on \( \{1, 2, 3\} \) is given by \( R = \{(1, 1), (2, 2), (1, 2), (3, 3), (2, 3)\} \). Then the relation \( R \) is: [1 Mark]
(A) Reflexive
(B) Symmetric
(C) Transitive
(D) Symmetric and transitive

Answer: (A) Reflexive

For a relation to be reflexive, \( (a, a) \in R \) for all \( a \in A \). Here \( (1,1), (2,2), (3,3) \in R \), but it is not symmetric since \( (1,2) \in R \) and \( (2,1) \notin R \).

Teacher's Note:
a) Check each condition of equivalence relations step by step using set elements.
b) Students often confuse non-symmetric relations with anti-symmetric ones; verify directions of ordered pairs carefully.

(ii) If \( A \) is a square matrix of order \( 3 \), then \( |2A| \) is equal to: [1 Mark]
(A) \( 2|A| \)
(B) \( 4|A| \)
(C) \( 8|A| \)
(D) \( 6|A| \)

Answer: (C) \( 8|A| \)

Using the property \( |kA| = k^n|A| \), where \( n \) is the order of the matrix, \( |2A| = 2^3|A| = 8|A| \).

Teacher's Note:
a) Remember the scalar multiplication property of determinants: \( |kA| = k^n |A| \).
b) Avoid the common error of multiplying the matrix determinant directly by the scalar without raising it to the power of the matrix order.

(iii) If the following function is continuous at \( x = 2 \) then the value of \( k \) will be:
\( f(x) = \begin{cases} 2x + 1, & \text{if } x \lt 2 \\ k, & \text{if } x = 2 \\ 3x - 1, & \text{if } x \gt 2 \end{cases} \) [1 Mark]
(A) 2
(B) 3
(C) 5
(D) -1

Answer: (C) 5

For continuity at \( x = 2 \), \( \lim_{x \to 2} f(x) = f(2) \), so \( 2(2) + 1 = 3(2) - 1 = k \), which yields \( k = 5 \).

Teacher's Note:
a) Equate the left-hand limit and right-hand limit to the value of the function at the given point.
b) Students must check both limits to ensure continuity before solving for unknown constants.

(iv) An edge of a variable cube is increasing at the rate of \( 10\text{ cm/sec} \). How fast will the volume of the cube increase if the edge is \( 5\text{ cm} \) long? [1 Mark]
(A) \( 75\text{ cm}^3\text{/sec} \)
(B) \( 750\text{ cm}^3\text{/sec} \)
(C) \( 7500\text{ cm}^3\text{/sec} \)
(D) \( 1250\text{ cm}^3\text{/sec} \)

Answer: (B) \( 750\text{ cm}^3\text{/sec} \)

Volume \( V = x^3 \). Differentiating with respect to time \( t \), \( \frac{dV}{dt} = 3x^2 \frac{dx}{dt} = 3(5)^2(10) = 750\text{ cm}^3\text{/sec} \).

Teacher's Note:
a) Apply chain rule correctly for related rates problems involving volume and surface area.
b) Pay close attention to units and ensure all dimensions are substituted after differentiation.

(v) Let \( f(x) = x^3 \) be a function with domain \( \{0, 1, 2, 3\} \). Then domain of \( f^{-1} \) is: [1 Mark]
(A) \( \{3, 2, 1, 0\} \)
(B) \( \{0, -1, -2, -3\} \)
(C) \( \{0, 1, 8, 27\} \)
(D) \( \{0, -1, -8, -27\} \)

Answer: (C) \( \{0, 1, 8, 27\} \)

The range of \( f(x) \) becomes the domain of \( f^{-1}(x) \). Evaluating \( f(x) \) for each element in the domain gives \( \{0^3, 1^3, 2^3, 3^3\} = \{0, 1, 8, 27\} \).

Teacher's Note:
a) Recall that the domain of an inverse function is identically equal to the range of the original function.
b) Students should explicitly calculate the range elements before identifying the inverse domain.

(vi) For the curve \( y^2 = 2x^3 - 7 \), the slope of the normal at \( (2, 3) \) is: [1 Mark]
(A) 4
(B) \( \frac{1}{4} \)
(C) -4
(D) \( -\frac{1}{4} \)

Answer: (D) \( -\frac{1}{4} \)

Differentiating implicitly, \( 2y \frac{dy}{dx} = 6x^2 \), so \( \frac{dy}{dx} = \frac{3x^2}{y} \). At \( (2, 3) \), slope of tangent is \( 4 \), hence slope of normal is \( -\frac{1}{4} \).

Teacher's Note:
a) The slope of the normal is the negative reciprocal of the slope of the tangent at the given point.
b) Verify coordinates by substituting them into the original curve equation before computing derivatives.

(vii) Evaluate: \( \int \frac{x}{x^2 + 1} dx \) [1 Mark]
(A) \( 2\log(x^2 + 1) + c \)
(B) \( \frac{1}{2}\log(x^2 + 1) + c \)
(C) \( e^{x^2 + 1} + c \)
(D) \( \log x + \frac{x^2}{2} + c \)

Answer: (B) \( \frac{1}{2}\log(x^2 + 1) + c \)

Substitute \( t = x^2 + 1 \), then \( dt = 2x dx \), making the integral \( \frac{1}{2} \int \frac{1}{t} dt = \frac{1}{2}\log|x^2 + 1| + c \).

Teacher's Note:
a) Use substitution method when the numerator is a multiple of the derivative of the denominator.
b) Do not forget to substitute back the original variable and include the constant of integration \( c \).

(viii) The derivative of \( \log x \) with respect to \( \frac{1}{x} \) is: [1 Mark]
(A) \( \frac{1}{x} \)
(B) \( -\frac{1}{x^3} \)
(C) \( -\frac{1}{x} \)
(D) \( -x \)

Answer: (D) \(-x\)

Let \( y = \log x \) and \( t = \frac{1}{x} \). Then \( \frac{dy}{dx} = \frac{1}{x} \) and \( \frac{dt}{dx} = -\frac{1}{x^2} \). Thus, \( \frac{dy}{dt} = \frac{dy/dx}{dt/dx} = \frac{1/x}{-1/x^2} = -x \).

Teacher's Note:
a) Parameter differentiation requires finding the derivative of one function with respect to another using chain rule formulation.
b) Watch out for sign errors during reciprocal differentiation.

(ix) The interval in which the function \( f(x) = 5 + 36x - 3x^2 \) increases will be: [1 Mark]
(A) \( (-\infty, 6) \)
(B) \( (6, \infty) \)
(C) \( (-6, 6) \)
(D) \( (0, -6) \)

Answer: (A) \( (-\infty, 6) \)

Here \( f'(x) = 36 - 6x = -6(x - 6) \). For the function to increase, \( f'(x) \gt 0 \), which gives \( x \lt 6 \), i.e., \( (-\infty, 6) \).

Teacher's Note:
a) A function is increasing where its first derivative is strictly greater than zero.
b) Note the official marking scheme key discrepancy where option (A) is the correct interval \( (-\infty, 6) \) while the OCR text mentions option (B) incorrectly.

(x) Evaluate \( \int_{-1}^{1} x^{17} \cos^4 x \, dx \) [1 Mark]
(A) \( \infty \)
(B) 1
(C) -1
(D) 0

Answer: (D) 0

The integrand \( f(x) = x^{17} \cos^4 x \) is an odd function since \( f(-x) = -f(x) \). Therefore, the definite integral over a symmetric interval \( [-1, 1] \) is zero.

Teacher's Note:
a) Always check the symmetry of the integrand over symmetric limits \( [-a, a] \) before integrating.
b) Definite integrals of odd functions over symmetric bounds evaluate immediately to zero.

(xi) Solve the differential equation: \( \frac{dy}{dx} = \csc y \) [1 Mark]

Answer:
Separating variables gives \( \sin y \, dy = dx \). Integrating both sides yields \( -\cos y = x + C \).

Teacher's Note:
a) Use variable separable form by bringing all terms of \( y \) to one side and \( x \) to the other.
b) Include the constant of integration \( C \) at the end of integration.

(xii) For what value of \( k \) is the matrix \( \begin{pmatrix} 0 & k \\ -6 & 0 \end{pmatrix} \) a skew symmetric matrix? [1 Mark]

Answer:
For a skew-symmetric matrix, \( A = -A^T \). Thus, \( \begin{pmatrix} 0 & k \\ -6 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 6 \\ -k & 0 \end{pmatrix} \), which gives \( k = 6 \).

Teacher's Note:
a) A square matrix is skew-symmetric if its transpose equals its negative matrix.
b) Corresponding elements must satisfy \( a_{ij} = -a_{ji} \).

(xiii) Evaluate \( \int_{0}^{1} |2x + 1| dx \) [1 Mark]

Answer:
Since \( 2x + 1 \gt 0 \) for all \( x \in (0, 1) \), \( \int_{0}^{1} (2x + 1) dx = [x^2 + x]_0^1 = 2 \).

Teacher's Note:
a) Remove the modulus sign by checking the sign of the expression within the interval of integration.
b) Apply standard power rule for polynomial integration.

(xiv) Evaluate \( \int \frac{1 + \cos x}{\sin^2 x} dx \) [1 Mark]

Answer:
\( \int (\csc^2 x + \csc x \cot x) dx = -\cot x - \csc x + C \).

Teacher's Note:
a) Split the fraction into two standard trigonometric integral forms.
b) Recall standard integral formulas for cosecant and cotangent combinations.

(xv) A bag contains 19 tickets, numbered from 1 to 19. Two tickets are drawn randomly in succession with replacement. Find the probability that both the tickets drawn are even numbers. [1 Mark]

Answer:
There are 9 even numbers from 1 to 19. The required probability is \( \frac{9}{19} \times \frac{9}{19} = \frac{81}{361} \).

Teacher's Note:
a) With replacement, the probability of independent events remains constant for each draw.
b) Count the number of favorable outcomes carefully from the given finite set.

2. (i) If \( f(x) = [4 - (x - 7)^3]^{\frac{1}{5}} \) is a real invertible function, then find \( f^{-1}(x) \) [4 Marks]

Answer:
Let \( y = [4 - (x - 7)^3]^{\frac{1}{5}} \).
\( y^5 = 4 - (x - 7)^3 \)
\( (x - 7)^3 = 4 - y^5 \)
\( x - 7 = (4 - y^5)^{\frac{1}{3}} \)
\( x = 7 + (4 - y^5)^{\frac{1}{3}} \)
Replacing \( x \) by \( y \) and \( y \) by \( x \), we get \( f^{-1}(x) = 7 + (4 - x^5)^{\frac{1}{3}} \).

Teacher's Note:
a) To find the inverse, express \( x \) in terms of \( y \) by solving the equation \( y = f(x) \).
b) Interchange the variables \( x \) and \( y \) at the final step to represent the inverse function.

OR

(ii) Let \( A = \mathbb{R} - \{2\} \) and \( B = \mathbb{R} - \{1\} \). If \( f: A \to B \) is a function defined by \( f(x) = \frac{x - 1}{x - 2} \) then show that \( f \) is a one-one and an onto function. [4 Marks]

Answer:
For one-one: Let \( f(x_1) = f(x_2) \implies \frac{x_1 - 1}{x_1 - 2} = \frac{x_2 - 1}{x_2 - 2} \).
\( (x_1 - 1)(x_2 - 2) = (x_2 - 1)(x_1 - 2) \)
\( x_1x_2 - 2x_1 - x_2 + 2 = x_1x_2 - x_1 - 2x_2 + 2 \implies x_1 = x_2 \). Hence, \( f \) is one-one.
For onto: Let \( y \in B \). Then \( y = \frac{x - 1}{x - 2} \implies xy - 2y = x - 1 \implies x(y - 1) = 2y - 1 \implies x = \frac{2y - 1}{y - 1} \). Since \( y \neq 1 \), \( x \in A \), proving \( f \) is onto.

Teacher's Note:
a) Prove injectivity by showing \( f(a) = f(b) \implies a = b \).
b) Prove surjectivity by expressing pre-image \( x \) in terms of \( y \) and showing it belongs to the domain for all \( y \) in co-domain.

3. Evaluate the following determinant without expanding:
\( \begin{vmatrix} 5 & 5 & 5 \\ a & b & c \\ b + c & c + a & a + b \end{vmatrix} \) [3 Marks]

Answer:
Applying \( R_2 \to R_2 + R_3 \):
\( \Delta = \begin{vmatrix} 5 & 5 & 5 \\ a + b + c & a + b + c & a + b + c \\ b + c & c + a & a + b \end{vmatrix} \)
Taking \( 5(a + b + c) \) common from \( R_1 \) and \( R_2 \):
\( \Delta = 5(a + b + c) \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ b + c & c + a & a + b \end{vmatrix} \)
Since \( R_1 \) and \( R_2 \) are identical, \( \Delta = 0 \).

Teacher's Note:
a) Use elementary row or column operations to create identical rows or columns to evaluate determinants efficiently.
b) A determinant with two identical rows or columns always evaluates to zero.

4. The probability of the event \( A \) occurring is \( \frac{1}{3} \) and of the event \( B \) occurring is \( \frac{1}{2} \). If \( A \) and \( B \) are independent events, then find the probability of neither \( A \) nor \( B \) occurring. [3 Marks]

Answer:
\( P(A) = \frac{1}{3}, P(B) = \frac{1}{2} \).
\( P(A \cap B) = P(A) \cdot P(B) = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6} \).
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{3} + \frac{1}{2} - \frac{1}{6} = \frac{2}{3} \).
Probability of neither \( A \) nor \( B \): \( P(A' \cap B') = P(A \cup B)' = 1 - P(A \cup B) = 1 - \frac{2}{3} = \frac{1}{3} \).

Teacher's Note:
a) Use De Morgan's Law: neither \( A \) nor \( B \) is equivalent to the complement of the union of \( A \) and \( B \).
b) For independent events, the probability of intersection is the product of their individual probabilities.

5. Solve for \( x \): \( 5\tan^{-1} x + 3\cot^{-1} x = 2\pi \) [4 Marks]

Answer:
\( 2\tan^{-1} x + 3(\tan^{-1} x + \cot^{-1} x) = 2\pi \)
Since \( \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \):
\( 2\tan^{-1} x + 3\left(\frac{\pi}{2}\right) = 2\pi \)
\( 2\tan^{-1} x = 2\pi - \frac{3\pi}{2} = \frac{\pi}{2} \)
\( \tan^{-1} x = \frac{\pi}{4} \implies x = \tan\left(\frac{\pi}{4}\right) = 1 \).

Teacher's Note:
a) Utilize standard inverse trigonometric identities like \( \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \).
b) Isolate the target inverse trigonometric function before solving for \( x \).

6. (i) Evaluate: \( \int \cos^{-1}(\sin x) dx \) [3 Marks]

Answer:
We know that \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \implies \cos^{-1}(\sin x) = \frac{\pi}{2} - \sin^{-1}(\sin x) = \frac{\pi}{2} - x \).
Therefore, \( \int \left(\frac{\pi}{2} - x\right) dx = \frac{\pi x}{2} - \frac{x^2}{2} + C = \frac{\pi}{2}(\pi - x) + C \) (absorbing constants).

Teacher's Note:
a) Simplify composite trigonometric expressions using standard identities before integration.
b) Integrate term by term once simplified into polynomial form.

(ii) If \( \int x^5 \cos(x^6) dx = k \sin(x^6) + c \), find the value of \( k \) [2 Marks]

Answer:
Let \( t = x^6 \), then \( dt = 6x^5 dx \implies x^5 dx = \frac{1}{6} dt \).
\( \int \cos(t) \frac{1}{6} dt = \frac{1}{6} \sin(t) + c = \frac{1}{6} \sin(x^6) + c \).
Comparing with \( k \sin(x^6) + c \), we get \( k = \frac{1}{6} \).

Teacher's Note:
a) Apply substitution method by letting the argument of trigonometric function equal to a new variable.
b) Equate coefficients after integration to find unknown constants.

7. If \( \tan^{-1}\left(\frac{x - 1}{x + 1}\right) + \tan^{-1}\left(\frac{2x - 1}{2x + 1}\right) = \tan^{-1}\left(\frac{23}{36}\right) \) then prove that \( 24x^2 - 23x - 12 = 0 \) [4 Marks]

Answer:
Using \( \tan^{-1} A + \tan^{-1} B = \tan^{-1}\left(\frac{A + B}{1 - AB}\right) \):
\( \tan^{-1}\left(\frac{\frac{x - 1}{x + 1} + \frac{2x - 1}{2x + 1}}{1 - \frac{x - 1}{x + 1} \cdot \frac{2x - 1}{2x + 1}}\right) = \tan^{-1}\left(\frac{23}{36}\right) \)
Simplifying the fraction inside:
\( \frac{(x - 1)(2x + 1) + (2x - 1)(x + 1)}{(x + 1)(2x + 1) - (x - 1)(2x - 1)} = \frac{23}{36} \)
\( \frac{(2x^2 - x - 1) + (2x^2 + x - 1)}{(2x^2 + 3x + 1) - (2x^2 - 3x + 1)} = \frac{23}{36} \)
\( \frac{4x^2 - 2}{6x} = \frac{23}{36} \implies \frac{2x^2 - 1}{3x} = \frac{23}{36} \)
\( 36(2x^2 - 1) = 3(23)x \implies 12(2x^2 - 1) = 23x \implies 24x^2 - 23x - 12 = 0 \) (Proved).

Teacher's Note:
a) Apply the sum formula for inverse tangent functions carefully, keeping track of denominators.
b) Cross-multiply and simplify algebraic expressions to arrive at the required quadratic equation.

8. If \( y = e^{ax} \cos bx \), then prove that \( \frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + (a^2 + b^2)y = 0 \) [4 Marks]

Answer:
Given \( y = e^{ax} \cos bx \) ... (i)
Differentiating w.r.t. \( x \):
\( \frac{dy}{dx} = ae^{ax} \cos bx - be^{ax} \sin bx = e^{ax}(a \cos bx - b \sin bx) \) ... (ii)
Differentiating again w.r.t. \( x \):
\( \frac{d^2y}{dx^2} = ae^{ax}(a \cos bx - b \sin bx) + e^{ax}(-ab \sin bx - b^2 \cos bx) \)
\( \frac{d^2y}{dx^2} = e^{ax}[(a^2 - b^2)\cos bx - 2ab \sin bx] \) ... (iii)
Substituting into LHS: \( \frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + (a^2 + b^2)y \)
\( = e^{ax}[(a^2 - b^2)\cos bx - 2ab \sin bx] - 2a[e^{ax}(a \cos bx - b \sin bx)] + (a^2 + b^2)e^{ax}\cos bx \)
\( = e^{ax}[(a^2 - b^2 - 2a^2 + a^2 + b^2)\cos bx + (-2ab + 2ab)\sin bx] = 0 \) (Proved).

Teacher's Note:
a) Use product rule twice for higher-order derivatives of exponential and trigonometric products.
b) Substitute first and second derivatives back into the differential equation to verify identity.

9. (i) In a company, 15% of the employees are graduates and 85% of the employees are non-graduates. As per the annual report of the company, 80% of the graduate employees and 10% of the non-graduate employees are in Administrative positions. Find the probability that an employee selected at random from those working in administrative positions will be a graduate. [5 Marks]

Answer:
Let \( G \) and \( G' \) denote graduate and non-graduate employees, and \( A \) denote administrative position.
\( P(G) = \frac{15}{100}, P(G') = \frac{85}{100} \)
\( P(A/G) = \frac{80}{100}, P(A/G') = \frac{10}{100} \)
Using Bayes' theorem:
\( P(G/A) = \frac{P(G) \cdot P(A/G)}{P(G) \cdot P(A/G) + P(G') \cdot P(A/G')} \)
\( P(G/A) = \frac{\frac{15}{100} \times \frac{80}{100}}{\frac{15}{100} \times \frac{80}{100} + \frac{85}{100} \times \frac{10}{100}} = \frac{1200}{1200 + 850} = \frac{1200}{2050} = \frac{24}{41} \).

Teacher's Note:
a) Identify conditional probabilities and marginal probabilities clearly from the problem statement.
b) Apply Bayes' theorem formula accurately to compute posterior probability.

OR

(ii) A problem in Mathematics is given to three students A, B and C. Their chances of solving the problem are \( \frac{1}{2} \), \( \frac{1}{3} \) and \( \frac{1}{4} \) respectively. Find the probability that:
(a) exactly two students will solve the problem.
(b) at least two of them will solve the problem. [5 Marks]

Answer:
\( P(A) = \frac{1}{2}, P(B) = \frac{1}{3}, P(C) = \frac{1}{4} \)
\( P(A') = \frac{1}{2}, P(B') = \frac{2}{3}, P(C') = \frac{3}{4} \)
(a) Exactly two students solve the problem:
\( P(E) = P(A \cap B \cap C') + P(A \cap B' \cap C) + P(A' \cap B \cap C) \)
\( P(E) = \left(\frac{1}{2} \times \frac{1}{3} \times \frac{3}{4}\right) + \left(\frac{1}{2} \times \frac{2}{3} \times \frac{1}{4}\right) + \left(\frac{1}{2} \times \frac{1}{3} \times \frac{1}{4}\right) = \frac{3 + 2 + 1}{24} = \frac{6}{24} = \frac{1}{4} \).
(b) At least two of them will solve the problem:
\( P(\text{at least 2}) = P(\text{exactly 2}) + P(\text{all 3}) = \frac{1}{4} + \left(\frac{1}{2} \times \frac{1}{3} \times \frac{1}{4}\right) = \frac{1}{4} + \frac{1}{24} = \frac{7}{24} \).

Teacher's Note:
a) Break down compound probability statements into mutually exclusive intersections of independent events.
b) 'At least two' includes both the case where exactly two solve and where all three solve.

10. (i) Solve the differential equation: \( (1 + y^2) dx = (\tan^{-1} y - x) dy \) [5 Marks]

Answer:
Rewrite as \( \frac{dx}{dy} + \frac{1}{1 + y^2} x = \frac{\tan^{-1} y}{1 + y^2} \).
Integrating Factor (\( I.F. \)) = \( e^{\int \frac{1}{1 + y^2} dy} = e^{\tan^{-1} y} \).
General solution: \( x \cdot (I.F.) = \int Q \cdot (I.F.) dy + C \)
\( x e^{\tan^{-1} y} = \int \frac{\tan^{-1} y}{1 + y^2} e^{\tan^{-1} y} dy + C \).
Put \( t = \tan^{-1} y \implies dt = \frac{1}{1 + y^2} dy \), so integral becomes \( \int t e^t dt = t e^t - e^t + C \).
Thus, \( x e^{\tan^{-1} y} = (\tan^{-1} y - 1)e^{\tan^{-1} y} + C \implies x = \tan^{-1} y - 1 + C e^{-\tan^{-1} y} \).

Teacher's Note:
a) Recognize linear differential equations of the form \( \frac{dx}{dy} + Px = Q \) where \( P \) and \( Q \) are functions of \( y \).
b) Use integration by parts carefully after substitution to solve the resulting integral.

OR

(ii) Solve the differential equation: \( (x^2 - y^2) dx + 2xy dy = 0 \) [5 Marks]

Answer:
\( \frac{dy}{dx} = \frac{y^2 - x^2}{2xy} \), which is homogeneous.
Put \( y = vx \implies \frac{dy}{dx} = v + x \frac{dv}{dx} \).
\( v + x \frac{dv}{dx} = \frac{v^2x^2 - x^2}{2v x^2} = \frac{v^2 - 1}{2v} \)
\( x \frac{dv}{dx} = \frac{v^2 - 1}{2v} - v = \frac{v^2 - 1 - 2v^2}{2v} = \frac{-1 - v^2}{2v} \)
\( \frac{2v}{1 + v^2} dv = -\frac{dx}{x} \).
Integrating both sides: \( \ln(1 + v^2) = -\ln x + \ln k \implies \ln[x(1 + v^2)] = \ln k \implies x\left(1 + \frac{y^2}{x^2}\right) = k \implies x^2 + y^2 = kx \).

Teacher's Note:
a) Verify homogeneity of degree zero before substituting \( y = vx \).
b) Separate variables in terms of \( v \) and \( x \) and integrate using standard logarithmic forms.

11. Use matrix method to solve the following system of equations:
\( \frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4 \)
\( \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1 \)
\( \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2 \) [6 Marks]

Answer:
Let \( p = \frac{1}{x}, q = \frac{1}{y}, r = \frac{1}{z} \). Matrix equation \( AX = B \) where:
\( A = \begin{pmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{pmatrix}, X = \begin{pmatrix} p \\ q \\ r \end{pmatrix}, B = \begin{pmatrix} 4 \\ 1 \\ 2 \end{pmatrix} \).
\( |A| = 2(120 - 45) - 3(-80 - 30) + 10(36 + 36) = 150 + 330 + 720 = 1200 \).
Adjoint matrix cofactor transpose gives \( A^{-1} = \frac{1}{1200} \begin{pmatrix} 75 & 150 & 75 \\ 110 & -100 & 0 \\ 72 & 0 & -24 \end{pmatrix} \).
\( X = A^{-1}B = \frac{1}{1200} \begin{pmatrix} 75(4) + 150(1) + 75(2) \\ 110(4) - 100(1) + 0(2) \\ 72(4) + 0(1) - 24(2) \end{pmatrix} = \frac{1}{1200} \begin{pmatrix} 600 \\ 400 \\ 240 \end{pmatrix} = \begin{pmatrix} 1/2 \\ 1/3 \\ 1/5 \end{pmatrix} \).
Thus \( p = \frac{1}{2}, q = \frac{1}{3}, r = \frac{1}{5} \implies x = 2, y = 3, z = 5 \).

Teacher's Note:
a) Substitute reciprocals with new variables to convert non-linear systems into standard matrix equations.
b) Double-check determinant calculation and cofactor signs before finding the inverse matrix.

12. (i) Prove that the semi-vertical angle of the right circular cone of given volume and least curved surface is \( \cot^{-1}\sqrt{2} \) [6 Marks]
[Figure: Right circular cone with base radius r, height h, slant height l, and semi-vertical angle theta]

Answer:
Volume \( V = \frac{1}{3}\pi r^2 h \implies h = \frac{3V}{\pi r^2} \).
Curved surface area \( S = \pi r l = \pi r \sqrt{h^2 + r^2} = \pi r \sqrt{\frac{9V^2}{\pi^2 r^4} + r^2} = \frac{\pi}{r} \sqrt{\frac{9V^2 + \pi^2 r^6}{\pi^2}} = \frac{1}{r}\sqrt{9V^2 + \pi^2 r^6} \).
Squaring for convenience: \( S^2 = \frac{9V^2 + \pi^2 r^6}{r^2} = 9V^2 r^{-2} + \pi^2 r^4 \).
Differentiating w.r.t. \( r \): \( 2S \frac{dS}{dr} = -18V^2 r^{-3} + 4\pi^2 r^3 \).
Setting \( \frac{dS}{dr} = 0 \implies 4\pi^2 r^3 = \frac{18V^2}{r^3} \implies 2\pi^2 r^6 = 9V^2 \implies V^2 = \frac{2\pi^2 r^6}{9} \implies V = \frac{\pi r^3 \sqrt{2}}{3} \).
Substitute \( V \) back into height: \( h = \frac{3(\frac{\pi r^3 \sqrt{2}}{3})}{\pi r^2} = r\sqrt{2} \implies \frac{h}{r} = \sqrt{2} \).
Since \( \cot \theta = \frac{h}{r} = \sqrt{2} \), the semi-vertical angle \( \theta = \cot^{-1}\sqrt{2} \) (Minima is verified by second derivative test).

Teacher's Note:
a) Squaring the objective function \( S \) simplifies differentiation significantly in optimization problems.
b) Establish the geometric relation \( \cot \theta = \frac{h}{r} \) at the critical point to complete the proof.

OR

(ii) A running track of \( 440\text{ m} \) is to be laid out enclosing a football field. The football field is in the shape of a rectangle with a semi-circle at each end. If the area of the rectangular portion is to be maximum, then find the length of its sides. Also calculate the area of the football field. [6 Marks]
[Figure: Football field shaped as a rectangle of length x and breadth y with semi-circles of diameter y at both ends]

Answer:
Let length and breadth of rectangle be \( x \) and \( y \). Perimeter of track \( P = 2x + \pi y = 440 \implies y = \frac{440 - 2x}{\pi} \).
Area of rectangle \( A = x y = x \left(\frac{440 - 2x}{\pi}\right) = \frac{440x - 2x^2}{\pi} \).
For maximum area, \( \frac{dA}{dx} = \frac{1}{\pi}(440 - 4x) = 0 \implies 440 - 4x = 0 \implies x = 110\text{ m} \).
Breadth \( y = \frac{440 - 2(110)}{\frac{22}{7}} = \frac{220}{\frac{22}{7}} = 70\text{ m} \).
Length of rectangular field is \( 110\text{ m} \) and breadth is \( 70\text{ m} \).
Total area of football field = Area of rectangle + Area of two semi-circles = \( xy + \pi \left(\frac{y}{2}\right)^2 = (110 \times 70) + \frac{22}{7}(35)^2 = 7700 + 3850 = 11550\text{ sq. meters} \).

Teacher's Note:
a) Express one dimension in terms of the other using the given perimeter constraint.
b) Combine rectangular and circular areas accurately to find the total area of the compound field.

13. (i) Evaluate: \( \int \frac{3e^{2x} - 2e^x}{e^{2x} + 2e^x - 8} dx \) [5 Marks]

Answer:
Substitute \( u = e^x \implies du = e^x dx \). The integral becomes \( \int \frac{3u - 2}{u^2 + 2u - 8} du = \int \frac{3u - 2}{(u - 2)(u + 4)} du \).
Using partial fractions: \( \frac{3u - 2}{(u - 2)(u + 4)} = \frac{A}{u + 4} + \frac{B}{u - 2} \implies 3u - 2 = A(u - 2) + B(u + 4) \).
Equating coefficients gives \( A = \frac{7}{3} \) and \( B = \frac{2}{3} \).
Integral = \( \int \left(\frac{7}{3(u + 4)} + \frac{2}{3(u - 2)}\right) du = \frac{7}{3}\ln|u + 4| + \frac{2}{3}\ln|u - 2| + C \).
Substituting back \( u = e^x \): \( \frac{7}{3}\ln|e^x + 4| + \frac{2}{3}\ln|e^x - 2| + C \).

Teacher's Note:
a) Use exponential substitution to convert rational functions of exponentials into standard algebraic rational forms.
b) Decompose using partial fractions before integrating logarithmic terms.

OR

(ii) Evaluate: \( \int \frac{2}{(1 - x)(1 + x^2)} dx \) [5 Marks]

Answer:
Using partial fractions: \( \frac{2}{(1 - x)(1 + x^2)} = \frac{A}{1 - x} + \frac{Bx + C}{1 + x^2} \).
\( 2 = A(1 + x^2) + (Bx + C)(1 - x) \).
Comparing coefficients gives \( A = 1, B = 1, C = 1 \).
Integral = \( \int \left(\frac{1}{1 - x} + \frac{x + 1}{1 + x^2}\right) dx = -\int \frac{1}{x - 1} dx + \frac{1}{2}\int \frac{2x}{1 + x^2} dx + \int \frac{1}{1 + x^2} dx \).
= \( -\ln|x - 1| + \frac{1}{2}\ln|1 + x^2| + \tan^{-1} x + C \).

Teacher's Note:
a) Linear factors generate simple logarithmic terms while irreducible quadratic factors generate logarithmic and inverse tangent terms.
b) Pay close attention to negative signs arising from linear transformations like \( 1 - x \).

14. A box contains 30 fruits, out of which 10 are rotten. Two fruits are selected at random one by one without replacement from the box. Find the probability distribution of the number of unspoiled fruits. Also find the mean of the probability distribution. [6 Marks]

Answer:
Total fruits = 30, Rotten = 10, Unspoiled = 20.
Let \( X \) be the random variable representing the number of unspoiled fruits, taking values \( 0, 1, 2 \).
\( P(X = 0) = \frac{10}{30} \times \frac{9}{29} = \frac{90}{870} = \frac{9}{87} \)
\( P(X = 1) = \left(\frac{20}{30} \times \frac{10}{29}\right) + \left(\frac{10}{30} \times \frac{20}{29}\right) = \frac{200 + 200}{870} = \frac{400}{870} = \frac{40}{87} \)
\( P(X = 2) = \frac{20}{30} \times \frac{19}{29} = \frac{380}{870} = \frac{38}{87} \)

Mean \( E(X) = \sum x_i P(x_i) = 0\left(\frac{9}{87}\right) + 1\left(\frac{40}{87}\right) + 2\left(\frac{38}{87}\right) = \frac{40 + 76}{87} = \frac{116}{87} \).

Teacher's Note:
a) Calculate probabilities for each possible value of the random variable without replacement using combinations or successive conditional probabilities.
b) Verify that the sum of all probabilities equals 1 before calculating the mean.

15. (i) If \( |\vec{a}| = 3 \), \( |\vec{b}| = \frac{\sqrt{2}}{3} \) and \( \vec{a} \times \vec{b} \) is a unit vector, then the angle between \( \vec{a} \) and \( \vec{b} \) will be: [1 Mark]
(A) \( \frac{\pi}{6} \)
(B) \( \frac{\pi}{4} \)
(C) \( \frac{\pi}{3} \)
(D) \( \frac{\pi}{2} \)

Answer: (B) \( \frac{\pi}{4} \)

Given \( |\vec{a} \times \vec{b}| = 1 \implies |\vec{a}||\vec{b}|\sin\theta = 1 \implies 3 \times \frac{\sqrt{2}}{3} \sin\theta = 1 \implies \sqrt{2}\sin\theta = 1 \implies \sin\theta = \frac{1}{\sqrt{2}} \implies \theta = \frac{\pi}{4} \).

Teacher's Note:
a) Magnitude of cross product is given by \( |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta \).
b) Solve for standard acute angle values matching trigonometric ratios.

(ii) The distance of the point \( 2\hat{i} + \hat{j} - \hat{k} \) from the plane \( \vec{r} \cdot (\hat{i} - 2\hat{j} + 4\hat{k}) = 9 \) will be: [1 Mark]
(a) 13
(b) \( \frac{13}{\sqrt{21}} \)
(c) 21
(d) \( \frac{21}{\sqrt{13}} \)

Answer: (b) \( \frac{13}{\sqrt{21}} \)

Using point-plane distance formula \( \frac{|\vec{a} \cdot \vec{n} - d|}{|\vec{n}|} \):
Distance = \( \frac{|(2\hat{i} + \hat{j} - \hat{k}) \cdot (\hat{i} - 2\hat{j} + 4\hat{k}) - 9|}{\sqrt{1^2 + (-2)^2 + 4^2}} = \frac{|2 - 2 - 4 - 9|}{\sqrt{21}} = \frac{|-13|}{\sqrt{21}} = \frac{13}{\sqrt{21}} \).

Teacher's Note:
a) Substitute vector coordinates into the Cartesian/vector distance formula for points and planes.
b) Take absolute value of the numerator to ensure distance is positive.

(iii) Find the area of the parallelogram whose diagonals are \( \hat{i} - 3\hat{j} + \hat{k} \) and \( \hat{i} + \hat{j} + \hat{k} \) [1 Mark]

Answer:
Area of parallelogram given diagonals \( \vec{d_1} \) and \( \vec{d_2} \) is \( \frac{1}{2} |\vec{d_1} \times \vec{d_2}| \).
\( \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 1 \\ 1 & 1 & 1 \end{vmatrix} = \hat{i}(-3 - 1) - \hat{j}(1 - 1) + \hat{k}(1 - (-3)) = -4\hat{i} + 0\hat{j} + 4\hat{k} \).
Magnitude \( |-4\hat{i} + 4\hat{k}| = \sqrt{(-4)^2 + 4^2} = \sqrt{32} = 4\sqrt{2} \).
Area = \( \frac{1}{2}(4\sqrt{2}) = 2\sqrt{2}\text{ sq. units} \).

Teacher's Note:
a) Remember the formula for area of a parallelogram when diagonals are given: \( \frac{1}{2}|\vec{d_1} \times \vec{d_2}| \).
b) Compute the cross product correctly using determinant expansion.

(iv) Write the equation of the plane passing through the point \( (2, 4, 6) \) and making equal intercepts on the coordinate axes. [1 Mark]

Answer:
Equation of plane with intercepts \( a, b, c \) is \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \). Since intercepts are equal (\( a = b = c \)):
\( \frac{x}{a} + \frac{y}{a} + \frac{z}{a} = 1 \implies x + y + z = a \).
Since it passes through \( (2, 4, 6) \): \( 2 + 4 + 6 = a \implies a = 12 \).
Required equation: \( x + y + z = 12 \).

Teacher's Note:
a) Use intercept form of plane equation when conditions on intercepts are specified.
b) Substitute the given point coordinates to find the unknown intercept parameter.

(v) If the two vectors \( 3\hat{i} + \alpha\hat{j} + \hat{k} \) and \( 2\hat{i} - \hat{j} + 8\hat{k} \) are perpendicular to each other, then find the value of \( \alpha \). [1 Mark]

Answer:
Since vectors are perpendicular, their dot product is zero.
\( (3\hat{i} + \alpha\hat{j} + \hat{k}) \cdot (2\hat{i} - \hat{j} + 8\hat{k}) = 0 \implies (3)(2) + (\alpha)(-1) + (1)(8) = 0 \)
\( 6 - \alpha + 8 = 0 \implies 14 - \alpha = 0 \implies \alpha = 14 \).

Teacher's Note:
a) Two vectors are orthogonal if and only if their scalar dot product equals zero.
b) Multiply corresponding components and sum them up to solve for unknowns.

16. (i) If \( A(1, 2, -3) \) and \( B(-1, -2, 1) \) are the end points of a vector \( \vec{AB} \) then find the unit vector in the direction of \( \vec{AB} \). [3 Marks]

Answer:
Vector \( \vec{AB} = (-1 - 1)\hat{i} + (-2 - 2)\hat{j} + (1 - (-3))\hat{k} = -2\hat{i} - 4\hat{j} + 4\hat{k} \).
Magnitude \( |\vec{AB}| = \sqrt{(-2)^2 + (-4)^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6 \).
Unit vector \( \hat{AB} = \frac{\vec{AB}}{|\vec{AB}|} = \frac{-2\hat{i} - 4\hat{j} + 4\hat{k}}{6} = -\frac{1}{3}\hat{i} - \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k} \).

Teacher's Note:
a) Position vector of vector joining two points is terminal point minus initial point coordinates.
b) Divide the vector by its magnitude to find the corresponding unit vector.

(ii) If \( \hat{a} \) is unit vector and \( (2\vec{x} - 3\hat{a}) \cdot (2\vec{x} + 3\hat{a}) = 91 \), find the value of \( |\vec{x}| \) [3 Marks]

Answer:
Expanding the dot product: \( 4|\vec{x}|^2 + 6\vec{x}\cdot\hat{a} - 6\hat{a}\cdot\vec{x} - 9|\hat{a}|^2 = 91 \).
\( 4|\vec{x}|^2 - 9|\hat{a}|^2 = 91 \).
Since \( \hat{a} \) is a unit vector, \( |\hat{a}| = 1 \):
\( 4|\vec{x}|^2 - 9(1) = 91 \implies 4|\vec{x}|^2 = 100 \implies |\vec{x}|^2 = 25 \implies |\vec{x}| = 5 \).

Teacher's Note:
a) Use algebraic expansion properties for dot products similar to scalar products.
b) Substitute unit vector magnitude property \( |\hat{a}| = 1 \) to simplify equations.

17. (i) Find the equation of the plane passing through the point \( (1, 1, -1) \) and perpendicular to the planes \( x + 2y + 3z = 7 \) and \( 2x - 3y + 4z = 0 \) [5 Marks]

Answer:
Equation of plane passing through \( (1, 1, -1) \) is \( A(x - 1) + B(y - 1) + C(z + 1) = 0 \) ... (1)
Since it is perpendicular to given planes, normal vectors are orthogonal:
\( A + 2B + 3C = 0 \) and \( 2A - 3B + 4C = 0 \).
By cross multiplication rule for direction ratios \( (A, B, C) \):
\( \frac{A}{(2)(4) - (3)(-3)} = \frac{B}{(3)(2) - (1)(4)} = \frac{C}{(1)(-3) - (2)(2)} \)
\( \frac{A}{8 + 9} = \frac{B}{6 - 4} = \frac{C}{-3 - 4} \implies \frac{A}{17} = \frac{B}{2} = \frac{C}{-7} = \lambda \).
Thus \( A = 17\lambda, B = 2\lambda, C = -7\lambda \).
Substituting in (1): \( 17(x - 1) + 2(y - 1) - 7(z + 1) = 0 \implies 17x - 17 + 2y - 2 - 7z - 7 = 0 \implies 17x + 2y - 7z = 26 \).

Teacher's Note:
a) Condition of perpendicularity between planes translates to dot product of normal vectors being zero.
b) Use cross-multiplication method to find direction ratios from two simultaneous equations.

OR

(ii) A line passes through the point \( (2, -1, 3) \) and is perpendicular to the lines \( \vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(2\hat{i} - 2\hat{j} + \hat{k}) \) and \( \vec{r} = (2\hat{i} - \hat{j} - 3\hat{k}) + \mu(\hat{i} + 2\hat{j} + 2\hat{k}) \). Obtain its equation. [5 Marks]

Answer:
Let direction ratios of required line be \( (a_1, a_2, a_3) \).
Since the line is perpendicular to both given lines, its direction ratios satisfy:
\( 2a_1 - 2a_2 + a_3 = 0 \) ... (1)
\( a_1 + 2a_2 + 2a_3 = 0 \) ... (2)
Solving (1) and (2) by cross-multiplication:
\( \frac{a_1}{(-2)(2) - (1)(2)} = \frac{a_2}{(1)(1) - (2)(2)} = \frac{a_3}{(2)(2) - (-2)(1)} \)
\( \frac{a_1}{-4 - 2} = \frac{a_2}{1 - 4} = \frac{a_3}{4 + 2} \implies \frac{a_1}{-6} = \frac{a_2}{-3} = \frac{a_3}{6} \implies \frac{a_1}{2} = \frac{a_2}{1} = \frac{a_3}{-2} \).
Thus, direction ratios are \( (2, 1, -2) \).
Equation of line passing through \( (2, -1, 3) \) in Cartesian form: \( \frac{x - 2}{2} = \frac{y + 1}{1} = \frac{z - 3}{-2} \).

Teacher's Note:
a) Direction ratios of a line perpendicular to two other lines are proportional to the cross product of their direction vectors.
b) Express the final line equation in standard symmetric Cartesian or vector form.

18. Find the area of the region bounded by the curve \( x^2 = 4y \) and the line \( x = 4y - 2 \) [5 Marks]
[Figure: Parabola \( x^2 = 4y \) and line \( x = 4y - 2 \) intersecting at points \( A(-1, 1/4) \) and \( B(2, 1) \) with shaded region OBAO]

Answer:
Solving equations \( x^2 = 4y \) and \( x = 4y - 2 \implies 4y = x + 2 \):
\( x^2 = x + 2 \implies x^2 - x - 2 = 0 \implies (x - 2)(x + 1) = 0 \implies x = -1, 2 \).
Intersection points are \( x = -1 \) and \( x = 2 \).
Required Area = \( \int_{-1}^{2} \left(\frac{x + 2}{4} - \frac{x^2}{4}\right) dx = \frac{1}{4} \int_{-1}^{2} (2 + x - x^2) dx \)
= \( \frac{1}{4} \left[2x + \frac{x^2}{2} - \frac{x^3}{3}\right]_{-1}^{2} \)
= \( \frac{1}{4} \left[\left(4 + 2 - \frac{8}{3}\right) - \left(-2 + \frac{1}{2} + \frac{1}{3}\right)\right] \)
= \( \frac{1}{4} \left[\frac{10}{3} - \left(-\frac{7}{6}\right)\right] = \frac{1}{4} \left[\frac{20 + 7}{6}\right] = \frac{27}{24} = \frac{9}{8}\text{ sq. units} \).

Teacher's Note:
a) Determine limits of integration by finding the abscissas of intersection points of the curve and line.
b) Area between curves is given by integrating upper function minus lower function between intersection limits.

19. (i) If the demand function is given by \( p = 1500 - 2x - x^2 \) then find the marginal revenue when \( x = 10 \) [1 Mark]
(a) 1160
(b) 1600
(c) 1100
(d) 1200

Answer: (a) 1160

Total Revenue \( R = p \cdot x = (1500 - 2x - x^2)x = 1500x - 2x^2 - x^3 \).
Marginal Revenue \( MR = \frac{dR}{dx} = 1500 - 4x - 3x^2 \).
At \( x = 10 \), \( MR = 1500 - 4(10) - 3(10)^2 = 1500 - 40 - 300 = 1160 \).

Teacher's Note:
a) Total revenue is product of price and quantity demanded (\( R = p \cdot x \)).
b) Marginal revenue is the first derivative of total revenue with respect to quantity \( x \).

(ii) If the two regression coefficients are 0.8 and 0.2, then the value of coefficient of correlation \( r \) will be: [1 Mark]
(a) \( \pm 0.4 \)
(b) \( \pm 0.16 \)
(c) 0.4
(d) 0.16

Answer: (c) 0.4

Correlation coefficient \( r = \pm \sqrt{b_{xy} \cdot b_{yx}} = \pm \sqrt{0.8 \times 0.2} = \pm \sqrt{0.16} = \pm 0.4 \). Since both regression coefficients are positive, \( r = 0.4 \).

Teacher's Note:
a) Correlation coefficient is the geometric mean of the two regression coefficients.
b) The sign of \( r \) must match the sign of both regression coefficients.

(iii) Out of the two regression lines \( x + 2y - 5 = 0 \) and \( 2x + 3y = 8 \), find the line of regression of \( y \) on \( x \). [1 Mark]

Answer:
Let \( x + 2y - 5 = 0 \) be regression of \( x \) on \( y \) (\( x = -2y + 5 \implies b_{xy} = -2 \)) and \( 2x + 3y = 8 \) be regression of \( y \) on \( x \) (\( y = -\frac{2}{3}x + \frac{8}{3} \implies b_{yx} = -\frac{2}{3} \)).
Product \( r^2 = b_{xy} \cdot b_{yx} = (-2)\left(-\frac{2}{3}\right) = \frac{4}{3} \gt 1 \), which is impossible since \( -1 \leq r \leq 1 \).
Thus, our assumption is wrong. The regression line of \( y \) on \( x \) is actually \( x + 2y - 5 = 0 \) and \( y \) on \( x \) is \( 2x + 3y = 8 \).

Teacher's Note:
a) Test validity of assumed regression lines by checking if the product of regression coefficients lies between 0 and 1.
b) Reassign regression equations properly if product exceeds unity.

(iv) The cost function \( C(x) = 3x^2 - 6x + 5 \). Find the average cost when \( x = 2 \) [1 Mark]

Answer:
Total Cost \( C(2) = 3(2)^2 - 6(2) + 5 = 12 - 12 + 5 = 5 \).
Average Cost \( AC = \frac{C(x)}{x} = \frac{5}{2} = 2.5 \).

Teacher's Note:
a) Average cost is calculated by dividing total cost by output units \( x \).
b) Evaluate total cost function at given quantity before dividing.

(v) The fixed cost of a product is Rs. 30,000 and its variable cost per unit is Rs. 800. If the demand function is \( p(x) = 4500 - 100x \), find the break-even values. [1 Mark]

Answer:
Total Cost \( TC = 30000 + 800x \).
Total Revenue \( TR = p(x) \cdot x = 4500x - 100x^2 \).
At break-even point, \( TR = TC \):
\( 4500x - 100x^2 = 30000 + 800x \implies 100x^2 - 3700x + 30000 = 0 \implies x^2 - 37x + 300 = 0 \).
\( (x - 25)(x - 12) = 0 \implies x = 25, 12 \).

Teacher's Note:
a) Break-even occurs when total revenue equals total cost.
b) Solve resulting quadratic equation to find all possible break-even output quantities.

20. (i) The total cost function for \( x \) units is given by \( C(x) = \sqrt{6x + 5} + 2500 \). Show that the marginal cost decreases as the output \( x \) increases. [3 Marks]

Answer:
Marginal Cost \( MC = \frac{d}{dx}C(x) = \frac{d}{dx}((6x + 5)^{\frac{1}{2}} + 2500) = \frac{1}{2}(6x + 5)^{-\frac{1}{2}} \cdot 6 = \frac{3}{\sqrt{6x + 5}} \).
Since \( MC \propto \frac{1}{\sqrt{6x + 5}} \), as output \( x \) increases, the denominator increases, causing marginal cost to decrease.

Teacher's Note:
a) Marginal cost is the derivative of total cost function with respect to output \( x \).
b) Analyze the derivative expression with respect to \( x \) to show monotonic decrease.

(ii) The average revenue function is given by \( AR = 25 - \frac{x}{4} \). Find total revenue function and marginal revenue function. [2 Marks]

Answer:
Total Revenue \( TR = AR \cdot x = \left(25 - \frac{x}{4}\right)x = 25x - \frac{x^2}{4} \).
Marginal Revenue \( MR = \frac{d}{dx}(TR) = 25 - \frac{2x}{4} = 25 - \frac{x}{2} \).

Teacher's Note:
a) Total revenue is the product of average revenue and quantity \( x \).
b) Marginal revenue is obtained by differentiating total revenue with respect to quantity.

21. Solve the following Linear Programming Problem graphically:
Maximize \( Z = 5x + 2y \) subject to:
\( x - 2y \leq 2 \)
\( 3x + 2y \leq 12 \)
\( -3x + 2y \leq 3 \)
\( x \geq 0, y \geq 0 \) [6 Marks]
[Figure: LPP feasible region bounded by vertices A(0, 1.5), B(3.5, 0.75), C(2, 0), D(1.5, 3.75), O(0, 0)]

Answer:
Converting inequations to equations and plotting boundary lines:
1. \( x - 2y = 2 \)
2. \( 3x + 2y = 12 \)
3. \( -3x + 2y = 3 \)
Corner points of the feasible region and objective function values \( Z = 5x + 2y \):

The maximum value of \( Z \) is 19 at the point \( (3.5, 0.75) \).

Teacher's Note:
a) Shade the correct half-plane for each constraint inequality and identify the enclosed bounded feasible region.
b) Evaluate objective function at every corner point to determine optimal solution.

22. (i) The following table shows the Mean, the Standard Deviation, and the coefficient of correlation of two variables \( x \) and \( y \):

Calculate:
(a) The regression coefficient \( b_{xy} \) and \( b_{yx} \)
(b) The probable value of \( y \) when \( x = 20 \) [5 Marks]

Answer:
Given: \( \bar{x} = 8, \sigma_x = 12, \bar{y} = 6, \sigma_y = 4, r = 0.6 \).
(a) Regression coefficients:
\( b_{yx} = r \left(\frac{\sigma_y}{\sigma_x}\right) = 0.6 \left(\frac{4}{12}\right) = 0.6 \times \frac{1}{3} = 0.2 \).
\( b_{xy} = r \left(\frac{\sigma_x}{\sigma_y}\right) = 0.6 \left(\frac{12}{4}\right) = 0.6 \times 3 = 1.8 \).
(b) Regression equation of \( y \) on \( x \):
\( y - \bar{y} = b_{yx}(x - \bar{x}) \implies y - 6 = 0.2(x - 8) \implies y = 0.2x - 1.6 + 6 \implies y = 0.2x + 4.4 \).
When \( x = 20 \): \( y = 0.2(20) + 4.4 = 4 + 4.4 = 8.4 \).

Teacher's Note:
a) Use formulas involving correlation coefficient and standard deviations to compute regression coefficients.
b) Construct regression line of \( y \) on \( x \) to estimate values of \( y \) for given \( x \).

OR

(ii) An analyst analysed 102 trips of a travel company. He studied the relation between travel expenses (\( y \)) and the duration (\( x \)) of these trips. He found that the relation between \( x \) and \( y \) was linear. Given the following data, find the regression equation of \( y \) on \( x \):
\( \sum x = 510, \sum y = 7140, \sum x^2 = 4150, \sum y^2 = 740200, \sum xy = 54900 \) [5 Marks]

Answer:
Given \( n = 102, \sum x = 510, \sum y = 7140, \sum x^2 = 4150, \sum y^2 = 740200, \sum xy = 54900 \).
Mean \( \bar{x} = \frac{\sum x}{n} = \frac{510}{102} = 5 \).
Mean \( \bar{y} = \frac{\sum y}{n} = \frac{7140}{102} = 70 \).
Regression coefficient \( b_{yx} = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2} \):
\( b_{yx} = \frac{102(54900) - (510)(7140)}{102(4150) - (510)^2} = \frac{5599800 - 3641400}{423300 - 260100} = \frac{1958400}{163200} = 12 \).
Regression equation of \( y \) on \( x \):
\( y - \bar{y} = b_{yx}(x - \bar{x}) \implies y - 70 = 12(x - 5) \implies y - 70 = 12x - 60 \implies y = 12x + 10 \).

Teacher's Note:
a) Apply least squares regression coefficient formula directly using summary sums.
b) Verify mean values and substitute into point-slope regression line equation.