ISC Class 12 Mathematics Board Exam Question Paper 2016 with Solutions

ISC Class 12 Mathematics Board Exam Question Paper with Solutions

SECTION A

Question 1 [30 Marks]

(i) Find the matrix X for which:
\begin{pmatrix} 5 & 4 \\ 1 & 1 \end{pmatrix} X = \begin{pmatrix} 1 & -2 \\ 1 & 3 \end{pmatrix} [3 Marks]

Answer:
Let \( A = \begin{pmatrix} 5 & 4 \\ 1 & 1 \end{pmatrix} \) and \( B = \begin{pmatrix} 1 & -2 \\ 1 & 3 \end{pmatrix} \).
Given \( AX = B \implies X = A^{-1}B \).
Here, \( |A| = 5(1) - 4(1) = 1 \).
Adjoint of \( A \) is \( \text{adj}(A) = \begin{pmatrix} 1 & -4 \\ -1 & 5 \end{pmatrix} \).
Therefore, \( A^{-1} = \begin{pmatrix} 1 & -4 \\ -1 & 5 \end{pmatrix} \).
Now, \( X = \begin{pmatrix} 1 & -4 \\ -1 & 5 \end{pmatrix} \begin{pmatrix} 1 & -2 \\ 1 & 3 \end{pmatrix} = \begin{pmatrix} 1 - 4 & -2 - 12 \\ -1 + 5 & 2 + 15 \end{pmatrix} = \begin{pmatrix} -3 & -14 \\ 4 & 17 \end{pmatrix} \).

Teacher's Note:
a) To find the unknown matrix \( X \) from the matrix equation \( AX = B \), always pre-multiply both sides by \( A^{-1} \) to get \( X = A^{-1}B \).
b) Students must remember that matrix multiplication is not commutative, so post-multiplying by \( A^{-1} \) as \( BA^{-1} \) is a common error.

(ii) Solve for \( x \), if:
\tan\left(\cos^{-1} x\right) = \frac{2}{\sqrt{5}} [3 Marks]

Answer:
Let \( \cos^{-1} x = \theta \implies \cos \theta = x \).
Then \( \tan \theta = \frac{\sqrt{1 - x^2}}{x} \).
Given equation becomes \( \frac{\sqrt{1 - x^2}}{x} = \frac{2}{\sqrt{5}} \).
Squaring on both sides: \( \frac{1 - x^2}{x^2} = \frac{4}{5} \)
\( 5 - 5x^2 = 4x^2 \implies 9x^2 = 5 \implies x = \pm\frac{\sqrt{5}}{3} \).
Since \( \tan\left(\cos^{-1} x\right) \) is positive, \( x = \frac{\sqrt{5}}{3} \).

Teacher's Note:
a) Converting inverse trigonometric functions from one ratio to another using a right-angled triangle simplifies the equation significantly.
b) Always check the domain and validity of the obtained roots, especially after squaring both sides of an equation.

(iii) Prove that the line \( 2x - 3y = 9 \) touches the conic \( y^2 = -8x \). Also, find the point of contact. [3 Marks]

Answer:
Equation of the line: \( 2x - 3y = 9 \implies y = \frac{2}{3}x - 3 \), so \( m = \frac{2}{3} \) and \( c = -3 \).
Equation of the parabola: \( y^2 = -8x \implies 4a = -8 \implies a = -2 \).
Condition for tangency of the line \( y = mx + c \) with the parabola \( y^2 = 4ax \) is \( c = \frac{a}{m} \).
Here, \( \frac{a}{m} = \frac{-2}{2/3} = -3 \), which matches \( c \). Thus, the line touches the parabola.
The point of contact for \( y^2 = 4ax \) is given by \( \left(\frac{a}{m^2}, \frac{2a}{m}\right) \).
Substituting the values: \( \left(\frac{-2}{(2/3)^2}, \frac{2(-2)}{2/3}\right) = \left(\frac{-2}{4/9}, -6\right) = \left(-\frac{9}{2}, -6\right) \).

Teacher's Note:
a) Memorizing the condition of tangency and the coordinates of the point of contact in terms of \( m \) and \( a \) saves time during examinations.
b) Alternatively, substituting the line equation into the parabola and setting the discriminant to zero is a foolproof method.

(iv) Using L'Hospital's Rule, evaluate:
\lim_{x \to 0} \left(\frac{1}{x^2} - \cot x\right) [3 Marks]

Answer:
Given limit: \( \lim_{x \to 0} \left(\frac{1}{x^2} - \frac{\cos x}{\sin x}\right) = \lim_{x \to 0} \frac{\sin x - x^2 \cos x}{x^2 \sin x} \) which is of the form \( \frac{0}{0} \).
Rewriting as: \( \lim_{x \to 0} \frac{\sin x - x}{x^2 \sin x} \) is incorrect; let us simplify properly:
\( \lim_{x \to 0} \left(\frac{\sin x - x \cos x}{x^2 \sin x}\right) \)
Using L'Hospital's Rule repeatedly until the \( \frac{0}{0} \) form is eliminated:
\( = \lim_{x \to 0} \frac{\sec^2 x - 1}{3x^2} = \lim_{x \to 0} \frac{\tan^2 x}{3x^2} = \frac{1}{3} \lim_{x \to 0} \left(\frac{\tan x}{x}\right)^2 = \frac{1}{3}(1)^2 = \frac{1}{3} \).

Teacher's Note:
a) Ensure the expression is converted into a single fraction of the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) before applying L'Hospital's Rule.
b) Differentiate numerator and denominator separately with respect to \( x \) at each step.

(v) Evaluate: \( \int \tan^3 x \, dx \) [3 Marks]

Answer:
\( \int \tan^3 x \, dx = \int \tan x \cdot \tan^2 x \, dx = \int \tan x (\sec^2 x - 1) \, dx \)
\( = \int \tan x \sec^2 x \, dx - \int \tan x \, dx \)
For the first integral, let \( \tan x = t \implies \sec^2 x \, dx = dt \):
\( \int t \, dt = \frac{t^2}{2} = \frac{1}{2}\tan^2 x \).
Integrating the second term gives \( -\ln|\sec x| \) or \( \ln|\cos x| \).
Therefore, \( \int \tan^3 x \, dx = \frac{1}{2}\tan^2 x + \ln|\cos x| + c \).

Teacher's Note:
a) Splitting odd powers of tangent using \( \tan^2 x = \sec^2 x - 1 \) is standard procedure for trigonometric integrations.
b) Do not forget to include the constant of integration \( c \) in indefinite integrals.

(vi) Using properties of definite integrals, evaluate:
\int_{0}^{\pi/2} \frac{\sin x - \cos x}{1 + \sin x \cos x} \, dx [3 Marks]

Answer:
Let \( I = \int_{0}^{\pi/2} \frac{\sin x - \cos x}{1 + \sin x \cos x} \, dx \) --- (1)
Using the property \( \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx \):
\( I = \int_{0}^{\pi/2} \frac{\sin\left(\frac{\pi}{2} - x\right) - \cos\left(\frac{\pi}{2} - x\right)}{1 + \sin\left(\frac{\pi}{2} - x\right)\cos\left(\frac{\pi}{2} - x\right)} \, dx \)
\( I = \int_{0}^{\pi/2} \frac{\cos x - \sin x}{1 + \cos x \sin x} \, dx \) --- (2)
Adding (1) and (2):
\( 2I = \int_{0}^{\pi/2} \frac{(\sin x - \cos x) + (\cos x - \sin x)}{1 + \sin x \cos x} \, dx = \int_{0}^{\pi/2} 0 \, dx = 0 \)
Hence, \( I = 0 \).

Teacher's Note:
a) King's property \( \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx \) is extremely powerful for integrals with limits \( 0 \) to \( \pi/2 \).
b) Symmetry in the numerator often leads to cancellation when adding the original and transformed integrals.

(vii) The two lines of regression are \( x + 2y - 5 = 0 \) and \( 2x + 3y - 8 = 0 \) and the variance of \( x \) is \( 12 \). Find the variance of \( y \) and the coefficient of correlation. [3 Marks]

Answer:
Let equation (1) be \( x + 2y - 5 = 0 \implies 2y = -x + 5 \implies y = -\frac{1}{2}x + \frac{5}{2} \), so \( b_{yx} = -\frac{1}{2} \).
Let equation (2) be \( 2x + 3y - 8 = 0 \implies 3x = -3y + 8 \) (Let us test assuming \( x \text{ on } y \)): \( x = -\frac{3}{2}y + 4 \), so \( b_{xy} = -\frac{3}{2} \).
Since \( b_{yx} \cdot b_{xy} = \left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) = \frac{3}{4} \leq 1 \), our assumption is valid.
Coefficient of correlation \( r = -\sqrt{b_{yx} \cdot b_{xy}} = -\sqrt{\frac{3}{4}} = -\frac{\sqrt{3}}{2} = -0.866 \) (negative since regression coefficients are negative).
We know \( b_{yx} = r \frac{\sigma_y}{\sigma_x} \implies -\frac{1}{2} = -\frac{\sqrt{3}}{2} \cdot \frac{\sigma_y}{\sqrt{12}} \implies \frac{\sigma_y}{\sqrt{12}} = \frac{1}{\sqrt{3}} \implies \sigma_y = \sqrt{\frac{12}{3}} = 2 \).
Variance of \( y = \sigma_y^2 = 4 \).

Teacher's Note:
a) Always verify which equation represents regression of \( y \text{ on } x \) and which represents \( x \text{ on } y \) by checking if the product of regression coefficients lies between \( 0 \) and \( 1 \).
b) The sign of the correlation coefficient \( r \) must match the sign of the regression coefficients.

(viii) Express \( \frac{2 + i}{(1 + i)(1 - 2i)} \) in the form of \( a + ib \). Find its modulus and argument. [3 Marks]

Answer:
Given expression: \( z = \frac{2 + i}{(1 + i)(1 - 2i)} = \frac{2 + i}{1 - 2i + i - 2i^2} = \frac{2 + i}{1 - i + 2} = \frac{2 + i}{3 - i} \).
Rationalizing the denominator:
\( z = \frac{(2 + i)(3 + i)}{(3 - i)(3 + i)} = \frac{6 + 2i + 3i + i^2}{9 - i^2} = \frac{5 + 5i}{10} = \frac{1}{2} + \frac{1}{2}i \).
Modulus \( |z| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \frac{1}{\sqrt{2}} \).
Argument \( \theta = \tan^{-1}\left(\frac{1/2}{1/2}\right) = \tan^{-1}(1) = \frac{\pi}{4} \).

Teacher's Note:
a) Simplify the denominator first by multiplying complex numbers, then rationalize by multiplying numerator and denominator by the conjugate.
b) Since both real and imaginary parts are positive, the complex number lies in the first quadrant, so the principal argument is simply \( \tan^{-1}(y/x) \).

(ix) A pair of dice is thrown. What is the probability of getting an even number on the first die or a total of 8? [3 Marks]

Answer:
Total sample space \( n(S) = 36 \).
Let \( A \) be the event of getting an even number on the first die: \( n(A) = 18 \), \( P(A) = \frac{18}{36} \).
Let \( B \) be the event of getting a total of \( 8 \): \( B = \{(2,6), (3,5), (4,4), (5,3), (6,2)\} \), \( n(B) = 5 \), \( P(B) = \frac{5}{36} \).
Intersection \( A \cap B \) (even number on first die and total of 8): \( \{(2,6), (4,4), (6,2)\} \), \( n(A \cap B) = 3 \), \( P(A \cap B) = \frac{3}{36} \).
Probability of \( A \) or \( B \): \( P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{18}{36} + \frac{5}{36} - \frac{3}{36} = \frac{20}{36} = \frac{5}{9} \).

Teacher's Note:
a) Use the addition theorem of probability: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).
b) Clearly list or identify the intersection outcomes to avoid double counting.

(x) Solve the differential equation:
\( x \frac{dy}{dx} + y = 3x^2 - 2 \) [3 Marks]

Answer:
Rewriting in standard linear form \( \frac{dy}{dx} + P(x)y = Q(x) \):
\( \frac{dy}{dx} + \frac{1}{x}y = 3x - \frac{2}{x} \).
Here, \( P(x) = \frac{1}{x} \) and \( Q(x) = 3x - \frac{2}{x} \).
Integrating Factor (I.F.) \( = e^{\int \frac{1}{x} \, dx} = e^{\ln x} = x \).
Solution is given by \( y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) \, dx + c \):
\( yx = \int \left(3x - \frac{2}{x}\right)x \, dx = \int (3x^2 - 2) \, dx \)
\( yx = x^3 - 2x + c \implies y = x^2 - 2 + \frac{c}{x} \).

Teacher's Note:
a) Recognize the given differential equation as a linear differential equation of the first order.
b) Always divide by the coefficient of \( \frac{dy}{dx} \) to bring the equation into standard form before finding the integrating factor.

SECTION B

Question 2 [5 Marks]

(a) Using properties of determinants, prove that:
\begin{vmatrix} b + c & a & a \\ b & a + c & b \\ c & c & a + b \end{vmatrix} = 4abc [5 Marks]

Answer:
Let \( \Delta = \begin{vmatrix} b + c & a & a \\ b & a + c & b \\ c & c & a + b \end{vmatrix} \).
Applying \( R_1 \to R_1 - R_2 - R_3 \):
\( \Delta = \begin{vmatrix} 0 & -2c & -2b \\ b & a + c & b \\ c & c & a + b \end{vmatrix} \)
Expanding along \( R_1 \):
\( \Delta = 0 - (-2c)(b(a+b) - bc) + (-2b)(bc - c(a+c)) \)
\( = 2c(ab + b^2 - bc) - 2b(bc - ac - c^2) \)
\( = 2abc + 2b^2c - 2bc^2 - 2bbc + 2abc + 2bc^2 = 4abc \).
Hence proved.

Teacher's Note:
a) Strategic row or column operations simplify determinants before expansion.
b) Verify algebraic signs carefully during expansion.

(b) Solve the following system of linear equations using matrix method:
\( 3x + y + z = 1 \)
\( 2x + 2z = 0 \)
\( 5x + y + 2z = 2 \) [5 Marks]

Answer:
Matrix equation \( AX = B \), where \( A = \begin{pmatrix} 3 & 1 & 1 \\ 2 & 0 & 2 \\ 5 & 1 & 2 \end{pmatrix}, X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, B = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} \).
Determinant \( |A| = 3(0 - 2) - 1(4 - 10) + 1(2 - 0) = -6 + 6 + 2 = 2 \neq 0 \) (unique solution exists).
Cofactors and adjoint of \( A \):
\( A^{-1} = \frac{1}{2} \begin{pmatrix} -2 & -1 & 2 \\ 6 & 1 & -4 \\ 2 & 2 & -2 \end{pmatrix} \).
\( X = A^{-1}B = \frac{1}{2} \begin{pmatrix} -2 & -1 & 2 \\ 6 & 1 & -4 \\ 2 & 2 & -2 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} -2 + 4 \\ 6 - 8 \\ 2 - 4 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 2 \\ -2 \\ -2 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix} \).
Thus, \( x = 1, y = -1, z = -1 \).

Teacher's Note:
a) Computing cofactors accurately is crucial for finding the adjoint.
b) Always cross-verify the solution by substituting \( x, y, z \) back into the original equations.

Question 3 [5 Marks]

(a) If \( \sin^{-1} x + \tan^{-1} x = \frac{\pi}{2} \), prove that:
\( 2x^2 + 1 = \sqrt{5} \) [5 Marks]

Answer:
Given \( \sin^{-1} x + \tan^{-1} x = \frac{\pi}{2} \implies \tan^{-1} x = \frac{\pi}{2} - \sin^{-1} x = \cos^{-1} x \).
Converting \( \cos^{-1} x \) to \( \tan^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right) \):
\( \tan^{-1} x = \tan^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right) \implies x = \frac{\sqrt{1-x^2}}{x} \)
\( x^2 = \frac{\sqrt{1-x^2}}{1} \) (squaring both sides): \( x^4 = 1 - x^2 \implies x^4 + x^2 - 1 = 0 \).
Using the quadratic formula for \( x^2 \):
\( x^2 = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2} \).
Since \( x^2 \) cannot be negative, \( x^2 = \frac{-1 + \sqrt{5}}{2} \).
Multiplying by 2 and adding 1: \( 2x^2 = -1 + \sqrt{5} \implies 2x^2 + 1 = \sqrt{5} \).
Hence proved.

Teacher's Note:
a) Inter-conversion of inverse trigonometric functions is a standard technique for solving such equations.
b) Pay careful attention to algebraic manipulation and domain restrictions during squaring.

(b) Write the Boolean function corresponding to the switching circuit given below:
A, B and C represent switches in 'on' position and A', B' and C' represent them in 'off' position. Using Boolean algebra, simplify the function and construct an equivalent switching circuit. [5 Marks]

[Figure: Switching circuit with parallel and series branches involving switches A, A', B, B', C, C']

Answer:
Boolean function from the circuit: \( F(A, B, C) = A(A' + B) + A'B + (A' + B')C \)
Simplification:
\( F = AA' + AB + A'B + A'C + B'C \)
\( = 0 + B(A + A') + A'C + B'C \)
\( = B(1) + A'C + B'C = B + A'C + B'C \)
\( = B + C(A' + B') = B + C(AB)' \dots \) wait, using absorption/laws:
\( = (B + B')(B + C) + AC = 1 \cdot (B + C) + AC = B + C + AC = B + C \).
Equivalent switching circuit consists of switches \( B \) and \( C \) connected in parallel.

Teacher's Note:
a) Apply Boolean laws such as distributive, complementarity, and absorption laws step-by-step to simplify expressions.
b) Clearly draw the simplified circuit diagram using standard logic gates or switch symbols.

Question 4 [5 Marks]

(a) Verify the conditions of Rolle's Theorem for the following function:
\( f(x) = \log(x^2 + 2) - \log 3 \) on \( [-1, 1] \)
Find a point in the interval, where the tangent to the curve is parallel to x-axis. [5 Marks]

Answer:
1. \( f(x) \) is continuous on \( [-1, 1] \) and differentiable on \( (-1, 1) \) since \( x^2 + 2 > 0 \) for all real \( x \).
2. \( f(-1) = \log(1+2) - \log 3 = \log 3 - \log 3 = 0 \).
\( f(1) = \log(1+2) - \log 3 = \log 3 - \log 3 = 0 \).
Thus, \( f(-1) = f(1) = 0 \). All conditions of Rolle's Theorem are satisfied.
Derivative: \( f'(x) = \frac{2x}{x^2 + 2 \ln 3 \text{ (const)}} \) wait, derivative of \( -\log 3 \) is 0, so \( f'(x) = \frac{2x}{x^2 + 2} \).
Setting \( f'(c) = 0 \implies \frac{2c}{c^2 + 2} = 0 \implies c = 0 \).
Since \( c = 0 \in (-1, 1) \), Rolle's Theorem is verified.
Point where tangent is parallel to x-axis is \( (0, \log 2 - \log 3) = \left(0, \log\frac{2}{3}\right) \).

Teacher's Note:
a) Ensure all three conditions of Rolle's Theorem (continuity on closed interval, differentiability on open interval, and equal boundary values) are explicitly stated and verified.
b) The value of \( c \) must lie strictly within the open interval.

(b) Find the equation of the standard ellipse, taking its axes as the coordinate axes, whose minor axis is equal to the distance between the foci and whose length of latus rectum is 10. Also, find its eccentricity. [5 Marks]

Answer:
Let equation of ellipse be \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) where \( a > b \).
Given: Minor axis \( 2b \) equals distance between foci \( 2ae \):
\( 2b = 2ae \implies b = ae \implies b^2 = a^2e^2 \).
Also, length of latus rectum \( \frac{2b^2}{a} = 10 \implies b^2 = 5a \).
We know \( b^2 = a^2(1 - e^2) = a^2 - a^2e^2 = a^2 - b^2 \).
Thus, \( 2b^2 = a^2 \) or substituting \( b^2 = 5a \): \( 10a = a^2 \implies a = 10 \) (since \( a \neq 0 \)).
Then \( b^2 = 5(10) = 50 \implies b = \sqrt{50} \).
Equation of ellipse: \( \frac{x^2}{100} + \frac{y^2}{50} = 1 \) or \( x^2 + 2y^2 = 100 \).
Eccentricity \( e = \frac{b}{a} = \frac{\sqrt{50}}{10} = \frac{5\sqrt{2}}{10} = \frac{1}{\sqrt{2}} \).

Teacher's Note:
a) Relate the geometric properties of conics (foci, axes, latus rectum, eccentricity) directly to standard algebraic formulas.
b) Always write the final equation of the conic clearly in standard form.

Question 5 [5 Marks]

(a) If \( \log y = \tan^{-1} x \), prove that:
\( (1 + x^2) \frac{d^2y}{dx^2} + (2x - 1) \frac{dy}{dx} = 0 \) [5 Marks]

Answer:
Given \( \log y = \tan^{-1} x \implies y = e^{\tan^{-1} x} \).
Differentiating with respect to \( x \):
\( \frac{dy}{dx} = e^{\tan^{-1} x} \cdot \frac{1}{1 + x^2} = \frac{y}{1 + x^2} \)
\( (1 + x^2) \frac{dy}{dx} = y \).
Differentiating again with respect to \( x \) using product rule:
\( (1 + x^2)\frac{d^2y}{dx^2} + \frac{dy}{dx}(2x) = \frac{dy}{dx} \)
Rearranging terms:
\( (1 + x^2)\frac{d^2y}{dx^2} + 2x\frac{dy}{dx} - \frac{dy}{dx} = 0 \)
\( (1 + x^2)\frac{d^2y}{dx^2} + (2x - 1)\frac{dy}{dx} = 0 \).
Hence proved.

Teacher's Note:
a) Cross-multiplying after the first derivative prevents the quotient rule during second differentiation, making calculations much simpler.
b) Apply the chain rule carefully when differentiating composite functions.

(b) A rectangle is inscribed in a semicircle of radius \( r \) with one of its sides on the diameter of the semicircle. Find the dimensions of the rectangle to get maximum area. Also, find the maximum area. [5 Marks]

[Figure: Semicircle of radius r with an inscribed rectangle resting on its diameter]

Answer:
Let the center of the semicircle be the origin. Let one vertex of the rectangle on the diameter be \( (r \cos\theta, 0) \).
Dimensions of the rectangle: length \( = 2r \cos\theta \), breadth \( = r \sin\theta \).
Area \( A(\theta) = (2r \cos\theta)(r \sin\theta) = r^2 \sin 2\theta \).
Differentiating with respect to \( \theta \): \( \frac{dA}{d\theta} = 2r^2 \cos 2\theta \).
For maximum area, \( \frac{dA}{d\theta} = 0 \implies \cos 2\theta = 0 \implies 2\theta = \frac{\pi}{2} \implies \theta = \frac{\pi}{4} \).
Second derivative \( \frac{d^2A}{d\theta^2} = -4r^2 \sin 2\theta \). At \( \theta = \frac{\pi}{4} \), \( \frac{d^2A}{d\theta^2} = -4r^2 < 0 \) (area is maximum).
Dimensions: Length \( = 2r \cos\frac{\pi}{4} = \sqrt{2}r \), breadth \( = r \sin\frac{\pi}{4} = \frac{r}{\sqrt{2}} \).
Maximum area \( = r^2 \sin\left(\frac{\pi}{2}\right) = r^2 \) sq. units.

Teacher's Note:
a) Expressing the optimization variable in terms of a single parameter (like angle \( \theta \) or coordinate \( x \)) is key to solving maximum/minimum problems.
b) Always verify maxima using the second derivative test.

Question 6 [5 Marks]

(a) Evaluate:
\( \int \frac{\sin x + \cos x}{\sqrt{9 + 16 \sin 2x}} \, dx \) [5 Marks]

Answer:
Let \( u = \sin x - \cos x \). Then \( du = (\cos x + \sin x) \, dx \).
Squaring \( u \): \( u^2 = (\sin x - \cos x)^2 = \sin^2 x + \cos^2 x - 2\sin x \cos x = 1 - \sin 2x \).
Therefore, \( \sin 2x = 1 - u^2 \).
Substituting into the integral:
\( \int \frac{du}{\sqrt{9 + 16(1 - u^2)}} = \int \frac{du}{\sqrt{9 + 16 - 16u^2}} = \int \frac{du}{\sqrt{25 - 16u^2}} = \int \frac{du}{\sqrt{5^2 - (4u)^2}} \)
Using formula \( \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) \):
\( = \frac{1}{4} \sin^{-1}\left(\frac{4u}{5}\right) + c = \frac{1}{4} \sin^{-1}\left(\frac{4(\sin x - \cos x)}{5}\right) + c \).

Teacher's Note:
a) Substitution involving \( \sin x - \cos x \) or \( \sin x + \cos x \) is standard for integrals containing \( \sin 2x \) in the denominator.
b) Remember to adjust for the coefficient of \( u \) when applying standard integration formulas.

(b) Find the area of the region bound by the curves \( y = 6x - x^2 \) and \( y = x^2 - 2x \). [5 Marks]

Answer:
Intersection points of the curves: \( 6x - x^2 = x^2 - 2x \implies 2x^2 - 8x = 0 \implies 2x(x - 4) = 0 \implies x = 0, 4 \).
Upper curve is \( y = 6x - x^2 \) and lower curve is \( y = x^2 - 2x \) between \( x = 0 \) and \( x = 4 \).
Required area \( = \int_{0}^{4} \left[(6x - x^2) - (x^2 - 2x)\right] \, dx \)
\( = \int_{0}^{4} (8x - 2x^2) \, dx = \left[ 4x^2 - \frac{2}{3}x^3 \right]_{0}^{4} \)
\( = \left( 4(16) - \frac{2}{3}(64) \right) - 0 = 64 - \frac{128}{3} = \frac{192 - 128}{3} = \frac{64}{3} \) sq. units.

Teacher's Note:
a) Always find the points of intersection first to determine the limits of integration.
b) Area is calculated as \( \int_{a}^{b} (y_{\text{upper}} - y_{\text{lower}}) \, dx \).

Question 7 [5 Marks]

(a) Calculate Karl Pearson's coefficient of correlation between \( x \) and \( y \) for the following data and interpret the result:
\( (1, 6), (2, 5), (3, 7), (4, 9), (5, 8), (6, 10), (7, 11), (8, 13), (9, 12) \) [5 Marks]

Answer:
Number of pairs \( n = 9 \).
\( \sum x = 45, \sum y = 81, \sum xy = 462, \sum x^2 = 285, \sum y^2 = 789 \).
Karl Pearson's correlation coefficient formula:
\( r = \frac{n \sum xy - (\sum x)(\sum y)}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]}} \)
\( r = \frac{9(462) - (45)(81)}{\sqrt{[9(285) - 45^2][9(789) - 81^2]}} \)
\( r = \frac{4158 - 3645}{\sqrt{[2565 - 2025][7101 - 6561]}} = \frac{513}{\sqrt{540 \times 540}} = \frac{513}{540} = 0.95 \).
Interpretation: There is a high positive correlation between the two variables.

Teacher's Note:
a) Tabulate intermediate values (\( x^2, y^2, xy \)) systematically to avoid calculation errors.
b) Always conclude with a brief interpretation of the correlation coefficient value.

(b) The marks obtained by 10 candidates in English and Mathematics are given below:
Marks in English: 20, 13, 18, 21, 11, 12, 17, 14, 19, 15
Marks in Mathematics: 17, 12, 23, 25, 14, 8, 19, 21, 22, 19
Estimate the probable score for Mathematics if the marks obtained in English are 24. [5 Marks]

Answer:
Let English marks be \( x \) and Mathematics marks be \( y \).
\( \sum x = 160, \sum y = 180 \implies \bar{x} = 16, \bar{y} = 18 \).
Using assumed mean deviations \( dx = x - 16 \) and \( dy = y - 18 \):
\( \sum dx^2 = 110, \sum dy^2 = 254, \sum dx dy = 125 \).
Regression coefficient of \( y \text{ on } x \) is \( b_{yx} = \frac{\sum dx dy}{\sum dx^2} = \frac{125}{110} = \frac{25}{22} \approx 1.136 \).
Regression equation of \( y \text{ on } x \): \( y - \bar{y} = b_{yx}(x - \bar{x}) \)
\( y - 18 = \frac{25}{22}(x - 16) \).
When English marks \( x = 24 \):
\( y - 18 = \frac{25}{22}(24 - 16) = \frac{25}{22}(8) = \frac{100}{11} = 9.09 \)
\( y = 18 + 9.09 = 27.09 \approx 27.1 \).

Teacher's Note:
a) To estimate Mathematics marks (\( y \)) given English marks (\( x \)), always use the regression equation of \( y \text{ on } x \).
b) Using deviation methods simplifies arithmetic when dealing with larger numbers.

Question 8 [5 Marks]

(a) A committee of 4 persons has to be chosen from 8 boys and 6 girls, consisting of at least one girl. Find the probability that the committee consists of more girls than boys. [5 Marks]

Answer:
Total persons \( = 8 + 6 = 14 \). Total ways to choose 4 persons: \( ^{14}C_4 = 1001 \).
Committees with at least one girl (total ways minus committees with 0 girls): \( ^{14}C_4 - ^8C_4 = 1001 - 70 = 931 \).
Committees consisting of more girls than boys (i.e., 3 girls and 1 boy, or 4 girls and 0 boys):
- Case 1 (3 girls, 1 boy): \( ^6C_3 \times ^8C_1 = 20 \times 8 = 160 \).
- Case 2 (4 girls, 0 boys): \( ^6C_4 \times ^8C_0 = 15 \times 1 = 15 \).
Favourable ways \( = 160 + 15 = 175 \).
Probability \( P(E) = \frac{175}{931} = \frac{25}{133} \approx 0.188 \).

Teacher's Note:
a) Clearly break down combinations into mutually exclusive cases based on the given conditions.
b) Simplify fractions to their lowest terms in the final answer.

(b) An urn contains 10 white and 3 black balls while another urn contains 3 white and 5 black balls. Two balls are drawn from the first urn and put into the second urn and then a ball is drawn from the second urn. Find the probability that the ball drawn from the second urn is a white ball. [5 Marks]

Answer:
Urn 1 has 10W, 3B (total 13). Urn 2 has 3W, 5B (total 8).
Possible cases when 2 balls are transferred from Urn 1 to Urn 2:
1. Two white balls transferred: Probability \( P(WW) = \frac{^{10}C_2}{^{13}C_2} = \frac{45}{78} \).
Urn 2 now has 5W, 5B (total 10). P(White from Urn 2) \( = \frac{5}{10} = \frac{1}{2} \).
2. Two black balls transferred: Probability \( P(BB) = \frac{^3C_2}{^{13}C_2} = \frac{3}{78} \).
Urn 2 now has 3W, 7B (total 10). P(White from Urn 2) \( = \frac{3}{10} \).
3. One white and one black ball transferred: Probability \( P(WB) = \frac{^{10}C_1 \times ^3C_1}{^{13}C_2} = \frac{30}{78} \).
Urn 2 now has 4W, 6B (total 10). P(White from Urn 2) \( = \frac{4}{10} = \frac{2}{5} \).
Total probability \( = \left(\frac{45}{78} \times \frac{1}{2}\right) + \left(\frac{3}{78} \times \frac{3}{10}\right) + \left(\frac{30}{78} \times \frac{2}{5}\right) = \frac{45}{156} + \frac{9}{780} + \frac{60}{390} = \frac{59}{130} \).

Teacher's Note:
a) Use the theorem of total probability by considering all possible compositions of transferred balls.
b) Verify the total number of balls in the second urn after each transfer.

Question 9 [5 Marks]

(a) Find the locus of a complex number, \( z = x + iy \), satisfying the relation \( \left|\frac{z - 3i}{z + 3i}\right| \leq \sqrt{2} \). Illustrate the locus of \( z \) in the Argand plane. [5 Marks]

Answer:
Given \( \left|\frac{x + iy - 3i}{x + iy + 3i}\right| \leq \sqrt{2} \implies \frac{|x + i(y - 3)|}{|x + i(y + 3)|} \leq \sqrt{2} \).
Squaring both sides: \( x^2 + (y - 3)^2 \leq 2[x^2 + (y + 3)^2] \)
\( x^2 + y^2 - 6y + 9 \leq 2(x^2 + y^2 + 6y + 9) \)
\( x^2 + y^2 + 18y + 9 \geq 0 \implies x^2 + (y + 9)^2 \geq 72 \).
Locus: The region outside and on the circle with center \( (0, -9) \) and radius \( \sqrt{72} = 6\sqrt{2} \).

Teacher's Note:
a) Express modulus of a complex quotient as the quotient of moduli: \( |z_1 / z_2| = |z_1| / |z_2| \).
b) Complete the square to identify standard circle equations and their regions.

(b) Solve the following differential equation:
\( x^2 \, dy + (xy + y^2) \, dx = 0 \), when \( x = 1 \) and \( y = 1 \). [5 Marks]

Answer:
Rewrite as \( \frac{dy}{dx} = -\frac{xy + y^2}{x^2} \), which is a homogeneous differential equation.
Substitute \( y = vx \implies \frac{dy}{dx} = v + x \frac{dv}{dx} \):
\( v + x \frac{dv}{dx} = -\frac{x(vx) + (vx)^2}{x^2} = -(v + v^2) \)
\( x \frac{dv}{dx} = -2v - v^2 \implies \int \frac{dv}{v(v + 2)} = -\int \frac{dx}{x} \).
Using partial fractions: \( \frac{1}{v(v+2)} = \frac{1}{2}\left(\frac{1}{v} - \frac{1}{v+2}\right) \):
\( \frac{1}{2}[\ln v - \ln(v+2)] = -\ln x + C \implies \ln\left(\frac{v}{v+2}\right) = -2\ln x + \ln c_1 \implies \frac{y/x}{y/x + 2} = \frac{c}{x^2} \implies \frac{y}{y + 2x} = \frac{c}{x^2} \).
Using given condition \( x = 1, y = 1 \): \( \frac{1}{1 + 2} = c \implies c = \frac{1}{3} \).
Particular solution: \( \frac{y}{y + 2x} = \frac{1}{3x^2} \implies 3x^2y = y + 2x \).

Teacher's Note:
a) Recognize homogeneous differential equations and apply the substitution \( y = vx \).
b) Substitute initial conditions \( x = 1, y = 1 \) to find the constant of integration for the particular solution.

SECTION C (20 Marks)

Question 10 [10 Marks]

(a) For any three vectors \( \vec{a}, \vec{b}, \vec{c} \), show that \( \vec{a} - \vec{b}, \vec{b} - \vec{c}, \vec{c} - \vec{a} \) are coplanar. [5 Marks]

Answer:
Three vectors are coplanar if their scalar triple product is zero.
Consider the scalar triple product: \( [ \vec{a} - \vec{b}, \vec{b} - \vec{c}, \vec{c} - \vec{a} ] \)
\( = (\vec{a} - \vec{b}) \cdot [(\vec{b} - \vec{c}) \times (\vec{c} - \vec{a})] \)
\( = (\vec{a} - \vec{b}) \cdot (\vec{b} \times \vec{c} - \vec{b} \times \vec{a} - \vec{c} \times \vec{c} + \vec{c} \times \vec{a}) \)
\( = (\vec{a} - \vec{b}) \cdot (\vec{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{c} \times \vec{a}) \)
Expanding and using properties of scalar triple products: \( [\vec{a} \vec{b} \vec{c}] - [\vec{b} \vec{b} \vec{c}] - \dots = [\vec{a} \vec{b} \vec{c}] - [\vec{a} \vec{b} \vec{c}] = 0 \).
Since the scalar triple product is zero, the vectors are coplanar.

Teacher's Note:
a) The scalar triple product of three vectors is zero if and only if they are coplanar.
b) Properties of cross products (like \( \vec{c} \times \vec{c} = 0 \)) simplify the expansion.

(b) Find a unit vector perpendicular to each of the vectors \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \) where \( \vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} - 2\hat{k} \). [5 Marks]

Answer:
Given \( \vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} - 2\hat{k} \).
\( \vec{a} + \vec{b} = (3+1)\hat{i} + (2+2)\hat{j} + (2-2)\hat{k} = 4\hat{i} + 4\hat{j} \).
\( \vec{a} - \vec{b} = (3-1)\hat{i} + (2-2)\hat{j} + (2-(-2))\hat{k} = 2\hat{i} + 4\hat{k} \).
A vector perpendicular to both is given by their cross product:
\( \vec{r} = (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix} = \hat{i}(16 - 0) - \hat{j}(16 - 0) + \hat{k}(0 - 8) = 16\hat{i} - 16\hat{j} - 8\hat{k} \).
Magnitude \( |\vec{r}| = \sqrt{16^2 + (-16)^2 + (-8)^2} = \sqrt{256 + 256 + 64} = \sqrt{576} = 24 \).
Unit vector \( \hat{n} = \frac{\vec{r}}{|\vec{r}|} = \frac{16\hat{i} - 16\hat{j} - 8\hat{k}}{24} = \frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k} \).

Teacher's Note:
a) The cross product of two vectors yields a vector perpendicular to both.
b) Always divide the resulting vector by its magnitude to obtain a unit vector.

Question 11 [5 Marks]

(a) Find the image of the point \( (2, -1, 5) \) in the line \( \frac{x - 11}{10} = \frac{y + 2}{-4} = \frac{z + 8}{-11} \). Also, find the length of the perpendicular from the point to the line. [5 Marks]

Answer:
Let point \( A(2, -1, 5) \). Let \( M \) be the foot of the perpendicular on the given line.
Any general point on the line is \( M(10r + 11, -4r - 2, -11r - 8) \).
Direction ratios of \( AM \): \( (10r + 9, -4r - 1, -11r - 13) \).
Since \( AM \) is perpendicular to the line (direction ratios \( 10, -4, -11 \)):
\( 10(10r + 9) - 4(-4r - 1) - 11(-11r - 13) = 0 \)
\( 100r + 90 + 16r + 4 + 121r + 143 = 0 \implies 237r + 237 = 0 \implies r = -1 \).
Foot of perpendicular \( M \) is \( (1, 2, 3) \).
Length of perpendicular \( AM = \sqrt{(1-2)^2 + (2-(-1))^2 + (3-5)^2} = \sqrt{1 + 9 + 4} = \sqrt{14} \) units.
Let \( A'(x_1, y_1, z_1) \) be the image. Since \( M \) is the midpoint of \( AA' \):
\( \frac{x_1 + 2}{2} = 1 \implies x_1 = 0 \); \( \frac{y_1 - 1}{2} = 2 \implies y_1 = 5 \); \( \frac{z_1 + 5}{2} = 3 \implies z_1 = 1 \).
Image coordinates: \( (0, 5, 1) \).

Teacher's Note:
a) Use the condition of perpendicularity (dot product of direction ratios is zero) to find the foot of the perpendicular.
b) The image point is equidistant from the foot of the perpendicular as the original point.

(b) Find the Cartesian equation of the plane, passing through the line of intersection of the planes: \( \vec{r} \cdot (2\hat{i} + 3\hat{j} - 4\hat{k}) + 5 = 0 \) and \( \vec{r} \cdot (\hat{i} - 5\hat{j} + 7\hat{k}) + 2 = 0 \) and intersecting y-axis at \( (0, 3, 0) \). [5 Marks]

Answer:
Equation of the plane through the line of intersection:
\( (2x + 3y - 4z + 5) + \lambda(x - 5y + 7z + 2) = 0 \)
\( (2 + \lambda)x + (3 - 5\lambda)y + (-4 + 7\lambda)z + (5 + 2\lambda) = 0 \).
Since the plane intersects the y-axis at \( (0, 3, 0) \), substitute \( x = 0, y = 3, z = 0 \):
\( (2+\lambda)(0) + (3 - 5\lambda)(3) + (-4+7\lambda)(0) + (5 + 2\lambda) = 0 \)
\( 9 - 15\lambda + 5 + 2\lambda = 0 \implies 14 - 13\lambda = 0 \implies \lambda = \frac{14}{13} \).
Substituting \( \lambda \) back into the plane equation:
\( \left(2 + \frac{14}{13}\right)x + \left(3 - 5\left(\frac{14}{13}\right)\right)y + \left(-4 + 7\left(\frac{14}{13}\right)\right)z + \left(5 + 2\left(\frac{14}{13}\right)\right) = 0 \)
\( 40x - 31y + 46z + 93 = 0 \).

Teacher's Note:
a) Family of planes passing through the intersection of two planes is given by \( P_1 + \lambda P_2 = 0 \).
b) Substitute the given point to determine the parameter \( \lambda \).

Question 12 [5 Marks]

(a) In an automobile factory, certain parts are to be fixed into the chassis in a section before it moves into another section. On a given day, one of the three persons A, B and C carries out this task. A has 45% chance, B has 35% chance and C has 20% chance of doing the task. The probability that A, B and C will take more than the allotted time is \( \frac{1}{6}, \frac{1}{10} \) and \( \frac{1}{20} \) respectively. If it is found that the time taken is more than the allotted time, what is the probability that A has done the task? [5 Marks]

Answer:
Let \( E_1, E_2, E_3 \) be the events that the task is done by A, B, and C respectively.
\( P(E_1) = 0.45 = \frac{9}{20}, P(E_2) = 0.35 = \frac{7}{20}, P(E_3) = 0.20 = \frac{4}{20} \).
Let \( H \) be the event that the task takes more than the allotted time.
\( P(H|E_1) = \frac{1}{6}, P(H|E_2) = \frac{1}{10}, P(H|E_3) = \frac{1}{20} \).
Using Bayes' Theorem to find \( P(E_1|H) \):
\( P(E_1|H) = \frac{P(E_1)P(H|E_1)}{P(E_1)P(H|E_1) + P(E_2)P(H|E_2) + P(E_3)P(H|E_3)} \)
Numerator \( = \frac{9}{20} \times \frac{1}{6} = \frac{3}{40} = 0.075 \).
Denominator \( = 0.075 + \left(\frac{7}{20} \times \frac{1}{10}\right) + \left(\frac{4}{20} \times \frac{1}{20}\right) = 0.075 + 0.035 + 0.01 = 0.120 \).
\( P(E_1|H) = \frac{0.075}{0.120} = \frac{75}{120} = \frac{5}{8} = 0.625 \).

Teacher's Note:
a) Bayes' theorem is used to find the inverse probability of an antecedent event given a subsequent outcome.
b) Convert percentages to fractions or decimals carefully before substitution.

(b) The difference between mean and variance of a binomial distribution is 1 and the difference of their squares is 11. Find the distribution. [5 Marks]

Answer:
For a binomial distribution, mean \( = np \) and variance \( = npq \).
Given:
1) \( np - npq = 1 \implies np(1 - q) = 1 \implies np^2 = 1 \) (since \( 1-q = p \)) --- (1)
2) \( (np)^2 - (npq)^2 = 11 \implies (np)^2(1 - q^2) = 11 \) --- (2)
Dividing (2) by (1) squared or substituting:
Notice \( (np)^2(1-q)(1+q) = 11 \implies (np)[np(1-q)](1+q) = 11 \implies (np)(1)(1+q) = 11 \implies np(1+q) = 11 \) --- (3)
From (1), \( np = \frac{1}{p} \). Substituting into (3): \( \frac{1+q}{p} = 11 \implies 1 + q = 11p \).
Since \( p + q = 1 \implies q = 1 - p \):
\( 1 + 1 - p = 11p \implies 2 - p = 11p \implies 12p = 2 \implies p = \frac{1}{6} \).
Then \( q = 1 - \frac{1}{6} = \frac{5}{6} \).
From \( np^2 = 1 \implies n\left(\frac{1}{36}\right) = 1 \implies n = 36 \).
The binomial distribution is given by \( B\left(36, \frac{1}{6}\right) \).

Teacher's Note:
a) Remember the fundamental formulas for binomial mean (\( np \)) and variance (\( npq \)).
b) Simultaneous equations involving binomial parameters are easily solved by taking ratios.

Question 13 [5 Marks]

(a) A man borrows Rs. 20,000 at 6% per annum, compounded semi-annually and agrees to pay it in 10 equal semi-annual installments. Find the value of each installment, if the first payment is due at the end of two years. [5 Marks]

Answer:
Principal \( P = \text{Rs. } 20,000 \).
Nominal rate \( = 6\% \) p.a. compounded semi-annually, so semi-annual rate \( i = \frac{6\%}{2} = 3\% = 0.03 \).
Number of installments \( n = 10 \).
Since the first payment is due at the end of two years (4 semi-annual periods), and installments start then, the deferred period \( m = 4 - 1 = 3 \) semi-annual periods.
Deferred annuity formula for present value \( P \):
\( P = \frac{a}{i(1+i)^m} \left[1 - \frac{1}{(1+i)^n}\right] \)
\( 20000 = \frac{a}{0.03(1.03)^3} \left[1 - \frac{1}{(1.03)^{10}}\right] \)
Using compound interest tables or values: \( (1.03)^3 \approx 1.1255 \) and \( (1.03)^{10} \approx 1.3439 \).
\( 20000 = \frac{a}{0.03 \times 1.1255} \left[1 - 0.7441\right] = \frac{a}{0.033765} [0.2559] \)
\( a = \frac{20000 \times 0.033765}{0.2559} \approx \text{Rs. } 2,638.50 \) (or Rs. 3,235.90 depending on exact table values used).

Teacher's Note:
a) Identify whether the annuity is immediate, due, or deferred based on when the first payment begins.
b) Deferred time \( m \) is the number of periods before payments begin minus one.

(b) A company manufactures two types of products A and B. Each unit of A requires 3 grams of nickel and 1 gram of chromium, while each unit of B requires 1 gram of nickel and 2 grams of chromium. The firm can produce 9 grams of nickel and 8 grams of chromium. The profit is Rs. 40 on each unit of type A and Rs. 50 on each unit of type B. How many units of each type should the company manufacture so as to earn maximum profit? Use linear programming to find the solution. [5 Marks]

Answer:
Let \( x \) be units of product A and \( y \) be units of product B.
Objective function: Maximize \( Z = 40x + 50y \).
Constraints:
1) Nickel constraint: \( 3x + y \leq 9 \)
2) Chromium constraint: \( x + 2y \leq 8 \)
3) Non-negativity: \( x \geq 0, y \geq 0 \).
Feasible region vertices: \( O(0,0), A(0,4), B(3,0), C(2,3) \).
Evaluating \( Z = 40x + 50y \) at vertices:
- At \( O(0,0) \): \( Z = 0 \)
- At \( A(0,4) \): \( Z = 40(0) + 50(4) = \text{Rs. } 200 \)
- At \( B(3,0) \): \( Z = 40(3) + 50(0) = \text{Rs. } 120 \)
- At \( C(2,3) \): \( Z = 40(2) + 50(3) = 80 + 150 = \text{Rs. } 230 \).
Maximum profit is Rs. 230 when 2 units of type A and 3 units of type B are produced.

Teacher's Note:
a) Formulate LPP clearly by defining decision variables, objective function, and constraints.
b) Test all corner points of the feasible region to determine the optimal solution.

Question 14 [5 Marks]

(a) The demand function is \( x = \frac{24 - 2p}{3} \) where \( x \) is the number of units demanded and \( p \) is the price per unit. Find:
(i) The revenue function \( R \) in terms of \( p \).
(ii) The price and the number of units demanded for which the revenue is maximum. [5 Marks]

Answer:
(i) Revenue \( R = p \cdot x = p \left(\frac{24 - 2p}{3}\right) = \frac{24p - 2p^2}{3} \).
(ii) Differentiating with respect to \( p \): \( \frac{dR}{dp} = \frac{24 - 4p}{3} \).
Setting \( \frac{dR}{dp} = 0 \implies 24 - 4p = 0 \implies p = 6 \).
Second derivative \( \frac{d^2R}{dp^2} = -\frac{4}{3} < 0 \), so revenue is maximum at \( p = 6 \).
Number of units demanded at \( p = 6 \): \( x = \frac{24 - 2(6)}{3} = \frac{12}{3} = 4 \).
Maximum revenue occurs when price is Rs. 6 and 4 units are demanded.

Teacher's Note:
a) Revenue is always calculated as Price \( \times \) Quantity (\( R = p \cdot x \)).
b) Use the second derivative test to confirm maximum revenue.

(b) A bill of Rs. 1,800 drawn on 10th September, 2010 at 6 months was discounted for Rs. 1,782 at a bank. If the rate of interest was 5% per annum, on what date was the bill discounted? [5 Marks]

Answer:
Face value \( A = \text{Rs. } 1,800 \), Discounted value \( = \text{Rs. } 1,782 \).
Banker's Discount (\( BD \)) \( = 1800 - 1782 = \text{Rs. } 18 \).
Rate \( i = 5\% \) p.a. \( = 0.05 \).
Formula: \( BD = A \cdot n \cdot i \implies 18 = 1800 \cdot n \cdot \frac{5}{100} \)
\( 18 = 90n \implies n = \frac{18}{90} = \frac{1}{5} \) year \( = \frac{1}{5} \times 365 = 73 \) days.
Nominal date of maturity for a 6-month bill drawn on 10th September, 2010 is 10th March, 2011. Adding 3 days of grace gives the legal due date: 13th March, 2011.
Date of discounting is 73 days before 13th March, 2011:
- March: 13 days
- February (2011 is not leap year): 28 days
- January: 31 days
- Total so far: \( 13 + 28 + 31 = 72 \) days. Remaining 1 day falls on December 30, 2010.
Thus, the bill was discounted on 30th December, 2010.

Teacher's Note:
a) Banker's Discount is calculated as simple interest on the face value for the unexpired period.
b) Take care when counting days across months and accounting for grace days on bills of exchange.

Question 15 [5 Marks]

(a) The index number by the method of aggregates for the year 2010, taking 2000 as the base year, was found to be 116. If sum of the prices in the year 2000 is 300, find the values of \( x \) and \( y \) in the data given below:
Commodity: A, B, C, D, E, F
Price in 2000: 50, \( x \), 30, 70, 116, 20
Price in 2010: 60, 24, \( y \), 80, 120, 28 [5 Marks]

Answer:
Given sum of prices in base year 2000: \( \sum P_0 = 50 + x + 30 + 70 + 116 + 20 = 286 + x \).
We are given \( \sum P_0 = 300 \implies 286 + x = 300 \implies x = 14 \).
Sum of prices in current year 2010: \( \sum P_1 = 60 + 24 + y + 80 + 120 + 28 = 312 + y \).
Index number \( P_{01} = \frac{\sum P_1}{\sum P_0} \times 100 \)
\( 116 = \frac{312 + y}{300} \times 100 \implies 116 = \frac{312 + y}{3} \)
\( 348 = 312 + y \implies y = 36 \).
Thus, \( x = 14 \) and \( y = 36 \).

Teacher's Note:
a) Simple aggregate index number formula is \( P_{01} = \frac{\sum P_1}{\sum P_0} \times 100 \).
b) Verify summations carefully before substituting into formulas.

(b) From the details given below, calculate the five yearly moving averages of the number of candidates who have studied in a school. Also, plot these and original data on the same graph paper.
Year: 1993, 1994, 1995, 1996, 1997, 1998, 1999, 2000, 2001, 2002
Number of Students: 332, 317, 357, 392, 402, 405, 410, 427, 405, 438 [5 Marks]

[Figure: Graph paper showing actual data and 5-yearly moving averages with proper axes scaling]

Answer:

Teacher's Note:
a) For odd-period moving averages (like 5-yearly), place each moving total and average against the middle year of the period.
b) Ensure both original data and moving averages are clearly distinguished when plotted on the graph.