ISC Class 12 Mathematics Board Exam Question Paper 2015 with Solutions

ISC Class 12 Mathematics Board Exam Question Paper with Solutions

SECTION A

Question 1

(i) Find the value of \(k\) if \(M = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}\) and \(M^2 - kM - I_2 = 0\) [3 Marks]

Answer:
Given \(M = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}\).
\(M^2 = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 1+4 & 2+6 \\ 2+6 & 4+9 \end{bmatrix} = \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix}\).
Now, \(M^2 - kM - I_2 = 0\)
\(\begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix} - k\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
\(\begin{bmatrix} 5 - k - 1 & 8 - 2k \\ 8 - 2k & 13 - 3k - 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
Comparing the corresponding elements:
\(4 - k = 0 \implies k = 4\).
Also, \(8 - 2k = 0 \implies 2k = 8 \implies k = 4\).
Therefore, \(k = 4\).

Teacher's Note:
a) Always compute matrix multiplication carefully row by column to avoid arithmetic errors in intermediate elements.
b) Verify the value of \(k\) by checking multiple corresponding elements of the resulting matrix equation.

(ii) Find the equation of an ellipse whose latus rectum is \(8\) and eccentricity is \(\frac{1}{3}\). [3 Marks]

Answer:
Let the standard equation of the ellipse be \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) (where \(a \gt b\)).
Given eccentricity \(e = \frac{1}{3}\) and length of latus rectum \(\frac{2b^2}{a} = 8 \implies b^2 = 4a\).
We know the relation between \(a\), \(b\), and \(e\) for an ellipse is:
\(b^2 = a^2(1 - e^2)\)
Substitute \(b^2 = 4a\) and \(e = \frac{1}{3}\):
\(4a = a^2\left(1 - \left(\frac{1}{3}\right)^2\right)\)
\(4a = a^2\left(1 - \frac{1}{9}\right)\)
\(4a = \frac{8}{9}a^2\)
Since \(a \neq 0\), dividing both sides by \(a\):
\(4 = \frac{8}{9}a \implies a = 4 \times \frac{9}{8} = \frac{9}{2}\).
Now, find \(b^2\):
\(b^2 = 4a = 4\left(\frac{9}{2}\right) = 18\).
Also find \(a^2\):
\(a^2 = \left(\frac{9}{2}\right)^2 = \frac{81}{4}\).
Thus, the equation of the ellipse is:
\(\frac{x^2}{\frac{81}{4}} + \frac{y^2}{18} = 1 \implies \frac{4x^2}{81} + \frac{y^2}{18} = 1\).

Teacher's Note:
a) Remember the standard formula for the length of the latus rectum, which is \(\frac{2b^2}{a}\) for standard horizontal ellipses.
b) Ensure proper substitution of \(e^2\) into the eccentricity relation \(b^2 = a^2(1 - e^2)\).

(iii) Solve: \(\cos^{-1}\left(\sin\cos^{-1}x\right) = \frac{\pi}{6}\). [3 Marks]

Answer:
Given \(\cos^{-1}\left(\sin\cos^{-1}x\right) = \frac{\pi}{6}\).
Taking cosine on both sides:
\(\sin(\cos^{-1}x) = \cos\left(\frac{\pi}{6}\right)\)
\(\sin(\cos^{-1}x) = \frac{\sqrt{3}}{2}\).
Let \(\cos^{-1}x = \theta\), so \(\cos\theta = x\).
Then \(\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - x^2}\).
Therefore, \(\sqrt{1 - x^2} = \frac{\sqrt{3}}{2}\).
Squaring both sides:
\(1 - x^2 = \frac{3}{4}\)
\(x^2 = 1 - \frac{3}{4} = \frac{1}{4}\)
\(x = \pm \frac{1}{2}\).

Teacher's Note:
a) Convert inverse trigonometric expressions into standard right-triangle or identity-based relationships for easier simplification.
b) Do not forget both positive and negative roots when squaring equations.

(iv) Using L'Hospital's rule, evaluate: \(\lim_{x \to 0} \frac{x - \sin x}{x^2 \sin x}\). [3 Marks]

Answer:
Given limit: \(\lim_{x \to 0} \frac{x - \sin x}{x^2 \sin x}\).
As \(x \to 0\), the numerator \(0 - 0 = 0\) and denominator \(0 \times 0 = 0\), giving the indeterminate form \(\frac{0}{0}\).
Applying L'Hospital's rule by differentiating numerator and denominator with respect to \(x\):
\(= \lim_{x \to 0} \frac{1 - \cos x}{x^2 \cos x + 2x \sin x}\).
Again, substituting \(x = 0\) yields \(\frac{1 - 1}{0 + 0} = \frac{0}{0}\) (indeterminate form).
Applying L'Hospital's rule a second time:
\(= \lim_{x \to 0} \frac{\sin x}{[x^2(-\sin x) + \cos x (2x)] + [2x \cos x + 2 \sin x]}\)
\(= \lim_{x \to 0} \frac{\sin x}{-x^2 \sin x + 2x \cos x + 2x \cos x + 2 \sin x}\)
\(= \lim_{x \to 0} \frac{\sin x}{-x^2 \sin x + 4x \cos x + 2 \sin x}\).
Substituting \(x = 0\) still yields \(\frac{0}{0}\). Alternatively, divide numerator and denominator by \(\sin x\):
\(= \lim_{x \to 0} \frac{1}{-x^2 + 4x \frac{\cos x}{\sin x} + 2}\)
Since \(\lim_{x \to 0} \frac{x}{\sin x} = 1\), rewriting the denominator term \(4x \cot x = \frac{4x}{\sin x}\cos x \to 4(1)(1) = 4\).
Thus, limit = \(\frac{1}{0 + 4 + 2} = \frac{1}{6}\).
Alternatively, using standard expansions:
\(\lim_{x \to 0} \frac{x - (x - \frac{x^3}{6} + \dots)}{x^2 (x - \frac{x^3}{6} + \dots)} = \lim_{x \to 0} \frac{\frac{x^3}{6}}{x^3} = \frac{1}{6}\).

Teacher's Note:
a) L'Hospital's rule can be applied successively as long as the indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) persists.
b) Standard series expansions often provide a quicker check for limits involving trigonometric functions.

(v) Evaluate: \(\int \frac{2y^2}{y^2 + 4} \, dy\). [3 Marks]

Answer:
Given integral: \(\int \frac{2y^2}{y^2 + 4} \, dy\).
Rewrite the integrand by adding and subtracting \(4\) in the numerator:
\(= \int \frac{2(y^2 + 4) - 8}{y^2 + 4} \, dy\)
\(= \int \left(2 - \frac{8}{y^2 + 4}\right) dy\)
\(= \int 2 \, dy - 8 \int \frac{1}{y^2 + 2^2} \, dy\)
\(= 2y - 8 \cdot \frac{1}{2} \tan^{-1}\left(\frac{y}{2}\right) + C\)
\(= 2y - 4 \tan^{-1}\left(\frac{y}{2}\right) + C\).

Teacher's Note:
a) For rational functions where the degree of the numerator equals or exceeds the denominator, perform long division or algebraic adjustment first.
b) Remember the standard integral formula \(\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C\).

(vi) Evaluate: \(\int_{0}^{3} f(x) \, dx\), where \(f(x) = \begin{cases} \cos 2x, & 0 \le x \le \frac{\pi}{2} \\ 3, & \frac{\pi}{2} \le x \le 3 \end{cases}\). [3 Marks]

Answer:
Using the property of definite integrals for piecewise defined functions:
\(\int_{0}^{3} f(x) \, dx = \int_{0}^{\frac{\pi}{2}} f(x) \, dx + \int_{\frac{\pi}{2}}^{3} f(x) \, dx\)
\(= \int_{0}^{\frac{\pi}{2}} \cos 2x \, dx + \int_{\frac{\pi}{2}}^{3} 3 \, dx\)
\(= \left[\frac{\sin 2x}{2}\right]_{0}^{\frac{\pi}{2}} + [3x]_{\frac{\pi}{2}}^{3}\)
\(= \left(\frac{\sin\pi}{2} - \frac{\sin 0}{2}\right) + \left(3(3) - 3\left(\frac{\pi}{2}\right)\right)\)
\(= (0 - 0) + \left(9 - \frac{3\pi}{2}\right)\)
\(= 9 - \frac{3\pi}{2}\).

Teacher's Note:
a) Split the interval of integration at the point where the function's definition changes.
b) Evaluate each definite integral separately before combining the results.

(vii) The two lines of regression are \(4x + 2y - 3 = 0\) and \(3x + 6y + 5 = 0\). Find the correlation co-efficient between \(x\) and \(y\). [3 Marks]

Answer:
Let the first line \(4x + 2y - 3 = 0\) be the regression line of \(y\) on \(x\):
\(2y = -4x + 3 \implies y = -2x + \frac{3}{2}\).
Here, the regression coefficient of \(y\) on \(x\) is \(b_{yx} = -2$.
Let the second line \(3x + 6y + 5 = 0\) be the regression line of \(x\) on \(y\):
\(3x = -6y - 5 \implies x = -2y - \frac{5}{3} \implies x = -2y - \frac{5}{3}\).
Here, the regression coefficient of \(x\) on \(y\) is \(b_{xy} = -2\).
We know that the correlation coefficient \(r\) is given by:
\(r = \pm \sqrt{b_{yx} \cdot b_{xy}}\)
\(r = \pm \sqrt{(-2) \times (-2)} = \pm \sqrt{4} = \pm 2\).
Since the absolute value of the correlation coefficient cannot exceed \(1\) (\(|r| \le 1\)), our assumption about which line is which must be checked. If \(b_{yx} = -0.5\) and \(b_{xy} = -0.5\)...
Let us re-examine the equations:
Assume \(4x + 2y - 3 = 0\) is \(y\) on \(x\): \(b_{yx} = -2$.
Assume \(3x + 6y + 5 = 0\) is \(x\) on \(y\): \(3x = -6y - 5 \implies x = -2y - \frac{5}{3}\), so \(b_{xy} = -2\).
Since \(b_{yx} \cdot b_{xy} = (-2)(-2) = 4 > 1\), this assignment is incorrect. Let us swap them:
Let \(3x + 6y + 5 = 0\) be \(y\) on \(x\): \(6y = -3x - 5 \implies y = -0.5x - \frac{5}{6}\), so \(b_{yx} = -0.5\).
Let \(4x + 2y - 3 = 0\) be \(x\) on \(y\): \(4x = -2y + 3 \implies x = -0.5y + \frac{3}{4}\), so \(b_{xy} = -0.5\).
Then \(r = -\sqrt{b_{yx} \cdot b_{xy}} = -\sqrt{(-0.5)(-0.5)} = -\sqrt{0.25} = -0.5\).

Teacher's Note:
a) The product of regression coefficients \(b_{yx} \cdot b_{xy} = r^2\), and since \(0 \le r^2 \le 1\), the product must not exceed \(1\).
b) The sign of the correlation coefficient \(r\) must match the sign of both regression coefficients.

(viii) A card is drawn from a well shuffled pack of playing cards. What is the probability that it is either a spade or an ace or both? [3 Marks]

Answer:
Total number of cards in a well shuffled pack = \(52\).
Let \(S\) be the event that the card drawn is a spade. Number of spades \(n(S) = 13\).
Let \(A\) be the event that the card drawn is an ace. Number of aces \(n(A) = 4\).
The intersection \(S \cap A\) represents the card being both a spade and an ace (the Ace of Spades). Number of such cards \(n(S \cap A) = 1\).
We need to find the probability of either a spade or an ace, which is \(P(S \cup A)\):
\(P(S \cup A) = P(S) + P(A) - P(S \cap A)\)
\(P(S \cup A) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52}\)
\(P(S \cup A) = \frac{13 + 4 - 1}{52} = \frac{16}{52} = \frac{4}{13}\).

Teacher's Note:
a) Use the addition theorem of probability: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).
b) Account for overlapping elements (like the Ace of Spades) to avoid double counting.

(ix) If \(1\), \(\omega\) and \(\omega^2\) are the cube roots of unity, prove that \(\frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2} = \omega^2\). [3 Marks]

Answer:
Consider the given expression: \(\frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2}\).
Multiply the numerator by \(\omega^3\) (since \(\omega^3 = 1\)):
Numerator \(= (a + b\omega + c\omega^2)\omega^3 = a\omega^3 + b\omega^4 + c\omega^5\).
Since \(\omega^3 = 1\), we have \(\omega^4 = \omega\) and \(\omega^5 = \omega^2\).
So numerator \(= a\omega^3 + b\omega + c\omega^2 = \omega(a\omega^2 + b + c\omega)\) - wait, let us multiply numerator by \(\omega^2\):
Let us multiply numerator and denominator appropriately or use properties of cube roots of unity.
Alternative method:
Consider the denominator: \(c + a\omega + b\omega^2\).
Multiply the numerator by \(\omega^2\):
\(\omega^2 \left(\frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2}\right) = \frac{a\omega^2 + b\omega^3 + c\omega^4}{c + a\omega + b\omega^2} = \frac{a\omega^2 + b(1) + c\omega}{c + a\omega + b\omega^2} = \frac{b + c\omega + a\omega^2}{c + a\omega + b\omega^2}\) - this does not directly match.
Let us multiply the numerator by \(\omega\):
\(\frac{a\omega + b\omega^2 + c\omega^3}{c + a\omega + b\omega^2} = \frac{c + a\omega + b\omega^2}{c + a\omega + b\omega^2} = 1\), so the original expression multiplied by \(\omega\) is \(1\), which means the original expression equals \(\frac{1}{\omega} = \omega^2\).
Detailed steps:
Consider the numerator multiplied by \(\omega\):
\(\omega(a + b\omega + c\omega^2) = a\omega + b\omega^2 + c\omega^3 = c + a\omega + b\omega^2\) (since \(\omega^3 = 1\)).
Thus, \(\omega \times (\text{Numerator}) = \text{Denominator}\).
Therefore, \(\frac{\text{Numerator}}{\text{Denominator}} = \frac{1}{\omega} = \omega^2\). Hence proved.

Teacher's Note:
a) Use the property \(\omega^3 = 1\) to adjust powers of \(\omega\) in algebraic manipulations.
b) Multiplying the numerator or denominator by a power of \(\omega\) often reveals the exact match needed for cancellation.

(x) Solve the differential equation: \(\sin^{-1}\left(\frac{dy}{dx}\right) = x + y\). [3 Marks]

Answer:
Given differential equation: \(\sin^{-1}\left(\frac{dy}{dx}\right) = x + y\).
Taking sine on both sides:
\(\frac{dy}{dx} = \sin(x + y)\).
Let \(v = x + y\). Differentiating with respect to \(x\):
\(\frac{dv}{dx} = 1 + \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{dv}{dx} - 1\).
Substitute into the differential equation:
\(\frac{dv}{dx} - 1 = \sin v\)
\(\frac{dv}{dx} = 1 + \sin v\)
\(\frac{1}{1 + \sin v} \, dv = dx\).
Integrating both sides:
\(\int \frac{1}{1 + \sin v} \, dv = \int 1 \, dx\)
Multiply numerator and denominator by \(1 - \sin v\):
\(\int \frac{1 - \sin v}{1 - \sin^2 v} \, dv = \int \frac{1 - \sin v}{\cos^2 v} \, dv = \int (\sec^2 v - \sec v \tan v) \, dv = x + C\)
\(\tan v - \sec v = x + C\)
Substitute back \(v = x + y\):
\(\tan(x + y) - \sec(x + y) = x + C\).

Teacher's Note:
a) Use substitution for expressions of the form \(ax + by\) inside trigonometric or transcendental functions.
b) Remember standard trigonometric integrals like \(\int \sec^2 v \, dv = \tan v\) and \(\int \sec v \tan v \, dv = \sec v\).

Question 2

(a) Using properties of determinants, prove that: \(\begin{vmatrix} 1 + a^2 - b^2 & 2ab & -2b \\ 2ab & 1 - a^2 + b^2 & 2a \\ 2b & -2a & 1 - a^2 - b^2 \end{vmatrix} = (1 + a^2 + b^2)^3\). [5 Marks]

Answer:
Let \(\Delta = \begin{vmatrix} 1 + a^2 - b^2 & 2ab & -2b \\ 2ab & 1 - a^2 + b^2 & 2a \\ 2b & -2a & 1 - a^2 - b^2 \end{vmatrix}\).
Apply the row operation \(R_1 \to R_1 + bR_3\) and \(R_2 \to R_2 - aR_3\):
For \(R_1\):
Element 1: \(1 + a^2 - b^2 + b(2b) = 1 + a^2 + b^2\)
Element 2: \(2ab + b(-2a) = 0\)
Element 3: \(-2b + b(1 - a^2 - b^2) = -2b + b - ab^2 - b^3 = -b(1 + a^2 + b^2)\) - let us use standard determinant properties.
Alternatively, apply \(C_1 \to C_1 - bC_3\) and \(C_2 \to C_2 + aC_3\):
For \(C_1\):
\(1 + a^2 - b^2 - b(2b) = 1 + a^2 - 3b^2\) (not simplifying nicely).
Let us use the standard property transformation: add \(b \times (\text{Row } 3)\) to Row 1 and subtract \(a \times (\text{Row } 3)\) from Row 2:
Actually, consider \(R_1 \to R_1 - bR_3\)...
Let us write the standard proof steps:
1. Apply \(C_1 \to C_1 - bC_3\) and \(C_2 \to C_2 + aC_3\), or more directly:
Add \(-b\) times column 3 to column 1, and \(a\) times column 3 to column 2.
Let us use row operations: \(R_1 \to R_1 + bR_3\) and \(R_2 \to R_2 - aR_3\).
\(R_1 \to (1 + a^2 + b^2, 0, -b(1 + a^2 + b^2))\).
\(R_2 \to (0, 1 + a^2 + b^2, a(1 + a^2 + b^2))\).
\(R_3 \to (2b, -2a, 1 - a^2 - b^2)\).
Taking out \((1 + a^2 + b^2)\) common from \(R_1\) and \((1 + a^2 + b^2)\) common from \(R_2\):
\(\Delta = (1 + a^2 + b^2)^2 \begin{vmatrix} 1 & 0 & -b \\ 0 & 1 & a \\ 2b & -2a & 1 - a^2 - b^2 \end{vmatrix}\).
Expanding along \(R_1\):
\(= (1 + a^2 + b^2)^2 \left[ 1 \cdot ((1 - a^2 - b^2) - (-2a)(a)) - 0 + (-b)(0 - 2b) \right]\)
\(= (1 + a^2 + b^2)^2 \left[ 1 \cdot (1 - a^2 - b^2 + 2a^2) + 2b^2 \right]\)
\(= (1 + a^2 + b^2)^2 \left[ 1 + a^2 + b^2 \right]\)
\(= (1 + a^2 + b^2)^3\). Hence proved.

Teacher's Note:
a) Strategic row or column operations can simplify complex determinants into triangular or easily expandable forms.
b) Factoring out common terms early simplifies subsequent expansions.

(b) Given two matrices \(A = \begin{bmatrix} 1 & -2 & 3 \\ 1 & 4 & 1 \\ 1 & -3 & 2 \end{bmatrix}\) and \(B = \begin{bmatrix} 11 & -5 & -14 \\ -1 & -1 & 2 \\ -7 & 1 & 6 \end{bmatrix}\), find \(AB\) and use this result to solve the following system of equations: \(x - 2y + 3z = 6\), \(x + 4y + z = 12\), \(x - 3y + 2z = 1\). [5 Marks]

Answer:
1. Find the product \(AB\):
\(AB = \begin{bmatrix} 1 & -2 & 3 \\ 1 & 4 & 1 \\ 1 & -3 & 2 \end{bmatrix} \begin{bmatrix} 11 & -5 & -14 \\ -1 & -1 & 2 \\ -7 & 1 & 6 \end{bmatrix}\)
Row 1: \([1(11) + (-2)(-1) + 3(-7), 1(-5) + (-2)(-1) + 3(1), 1(-14) + (-2)(2) + 3(6)] = [11 + 2 - 21, -5 + 2 + 3, -14 - 4 + 18] = [-8, 0, 0]\)
Row 2: \([1(11) + 4(-1) + 1(-7), 1(-5) + 4(-1) + 1(1), 1(-14) + 4(2) + 1(6)] = [11 - 4 - 7, -5 - 4 + 1, -14 + 8 + 6] = [0, -8, 0]\)
Row 3: \([1(11) + (-3)(-1) + 2(-7), 1(-5) + (-3)(-1) + 2(1), 1(-14) + (-3)(2) + 2(6)] = [11 + 3 - 14, -5 + 3 + 2, -14 - 6 + 12] = [0, 0, -8]\)
So, \(AB = \begin{bmatrix} -8 & 0 & 0 \\ 0 & -8 & 0 \\ 0 & 0 & -8 \end{bmatrix} = -8I_3\).
Therefore, \(A\left(-\frac{1}{8}B\right) = I_3\), which means \(A^{-1} = -\frac{1}{8}B\).
2. Express the given system of equations in matrix form \(AX = C\):
\(\begin{bmatrix} 1 & -2 & 3 \\ 1 & 4 & 1 \\ 1 & -3 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 6 \\ 12 \\ 1 \end{bmatrix}\).
The solution is given by \(X = A^{-1}C\):
\(X = -\frac{1}{8} B \begin{bmatrix} 6 \\ 12 \\ 1 \end{bmatrix}\)
\(X = -\frac{1}{8} \begin{bmatrix} 11 & -5 & -14 \\ -1 & -1 & 2 \\ -7 & 1 & 6 \end{bmatrix} \begin{bmatrix} 6 \\ 12 \\ 1 \end{bmatrix}\)
Multiply matrix \(B\) by column vector \(\begin{bmatrix} 6 \\ 12 \\ 1 \end{bmatrix}\):
Row 1: \(11(6) + (-5)(12) + (-14)(1) = 66 - 60 - 14 = -8\)
Row 2: \(-1(6) + (-1)(12) + 2(1) = -6 - 12 + 2 = -16\)
Row 3: \(-7(6) + 1(12) + 6(1) = -42 + 12 + 6 = -24\)
Thus, \(X = -\frac{1}{8} \begin{bmatrix} -8 \\ -16 \\ -24 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\).
Therefore, \(x = 1\), \(y = 2\), \(z = 3\).

Teacher's Note:
a) When matrix product \(AB\) results in a scalar multiple of the identity matrix, the inverse can be directly determined.
b) Always double check matrix multiplication and scalar multiplication steps during substitution.

Question 3

(a) Solve the equation for \(x\): \(\sin^{-1}\frac{1}{5} + \sin^{-1}\frac{12}{x} = \frac{\pi}{2}\), \(x \neq 0\). [5 Marks]

Answer:
Given equation: \(\sin^{-1}\frac{1}{5} + \sin^{-1}\frac{12}{x} = \frac{\pi}{2}\).
Rearrange terms:
\(\sin^{-1}\frac{12}{x} = \frac{\pi}{2} - \sin^{-1}\frac{1}{5}\).
Since \(\frac{\pi}{2} - \sin^{-1}\theta = \cos^{-1}\theta\):
\(\sin^{-1}\frac{12}{x} = \cos^{-1}\frac{1}{5}\).
Convert \(\cos^{-1}\frac{1}{5}\) to sine inverse:
Let \(\cos^{-1}\frac{1}{5} = \alpha \implies \cos\alpha = \frac{1}{5}\).
Then \(\sin\alpha = \sqrt{1 - \left(\frac{1}{5}\right)^2} = \sqrt{1 - \frac{1}{25}} = \sqrt{\frac{24}{25}} = \frac{2\sqrt{6}}{5}\).
So \(\alpha = \sin^{-1}\left(\frac{2\sqrt{6}}{5}\right)\).
Thus, \(\sin^{-1}\frac{12}{x} = \sin^{-1}\left(\frac{2\sqrt{6}}{5}\right)\).
Equating arguments:
\(\frac{12}{x} = \frac{2\sqrt{6}}{5}\)
\(2\sqrt{6}x = 60 \implies x = \frac{60}{2\sqrt{6}} = \frac{30}{\sqrt{6}} = 5\sqrt{6}\).

Teacher's Note:
a) Use standard identities relating inverse trigonometric functions, such as \(\sin^{-1}\theta + \cos^{-1}\theta = \frac{\pi}{2}\).
b) Convert between sine and cosine inverses using right-triangle relations to equate arguments directly.

(b) A, B and C represent switches in 'on' position and A', B' and C' represent them in 'off' position. Construct a switching circuit representing the polynomial \(ABC + AB'C + A'B'C\). Using Boolean Algebra, prove that the given polynomial can be simplified to \(C(A + B')\). Construct an equivalent switching circuit. [5 Marks]

Answer:
1. Boolean expression: \(E = ABC + AB'C + A'B'C\).
Simplification using Boolean Algebra:
\(E = AC(B + B') + A'B'C\) (since \(ABC + AB'C = AC(B + B')\))
\(E = AC(1) + A'B'C\) (since \(B + B' = 1\))
\(E = AC + A'B'C\)
Factoring out \(C\):
\(E = C(A + A'B')\)
Using the distributive law \((A + A'B') = (A + A')(A + B') = 1 \cdot (A + B') = A + B'\):
\(E = C(A + B')\).
2. Switching Circuits Description:
- For the original polynomial \(ABC + AB'C + A'B'C\): Three parallel paths, where the first path has switches \(A\), \(B\), \(C\) in series; the second path has switches \(A\), \(B'\), \(C\) in series; and the third path has switches \(A'\), \(B'\), \(C\) in series.
- For the simplified polynomial \(C(A + B')\): Switch \(C\) in series with a parallel combination of switch \(A\) and switch \(B'\).

Teacher's Note:
a) Apply standard Boolean laws like complementarity (\(B + B' = 1\)) and absorption/distributive laws to simplify expressions.
b) Clearly describe or sketch the series and parallel combinations for switching circuits.

Question 4

(a) Verify Lagrange's Mean Value Theorem for the following function: \(f(x) = 2\sin x + \sin 2x\) on \([0, \pi]\). [5 Marks]

Answer:
Given function \(f(x) = 2\sin x + \sin 2x\) on the interval \([0, \pi]\).
1. Continuity: \(f(x)$ is continuous on \([0, \pi]\) being a sum of continuous trigonometric functions.
2. Differentiability: \(f(x)\) is differentiable on \((0, \pi)\) with derivative:
\(f'(x) = 2\cos x + 2\cos 2x\).
3. According to Lagrange's Mean Value Theorem (LMVT), there exists at least one real number \(c \in (0, \pi)\) such that:
\(f'(c) = \frac{f(\pi) - f(0)}{\pi - 0}\).
Calculate \(f(\pi)\) and \(f(0)\):
\(f(\pi) = 2\sin(\pi) + \sin(2\pi) = 2(0) + 0 = 0\).
\(f(0) = 2\sin(0) + \sin(0) = 0\).
Thus, \(\frac{f(\pi) - f(0)}{\pi - 0} = \frac{0 - 0}{\pi} = 0\).
Now, set \(f'(c) = 0\):
\(2\cos c + 2\cos 2c = 0\)
\(\cos c + \cos 2c = 0\)
Using the sum-to-product formula \(\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)\) or multiple angle formula \(\cos 2c = 2\cos^2 c - 1\):
\(\cos c + 2\cos^2 c - 1 = 0\)
\(2\cos^2 c + \cos c - 1 = 0\)
Factorizing the quadratic in \(\cos c\):
\((2\cos c - 1)(\cos c + 1) = 0\)
This gives:.
\(\cos c = \frac{1}{2} \implies c = \frac{\pi}{3} \in (0, \pi)\).
Or \(\cos c = -1 \implies c = \pi\) (not in \((0, \pi)\)).
Thus, \(c = \frac{\pi}{3}\), which lies in \((0, \pi)\). Lagrange's Mean Value Theorem is verified.

Teacher's Note:
a) State the conditions for LMVT: continuity on \([a, b]\) and differentiability on \((a, b)\).
b) Ensure the value of \(c$ obtained strictly lies within the open interval \((a, b)\).

(b) Find the equation of the hyperbola whose foci are \((0, \pm\sqrt{10})\) and passing through the point \((2, 3)\). [5 Marks]

Answer:
Since the foci are on the \(y\)-axis at \((0, \pm\sqrt{10})\), the standard equation of the hyperbola is vertical:
\(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\).
Here, the foci are \((0, \pm ae)\), so \(ae = \sqrt{10} \implies a^2 e^2 = 10\).
For a hyperbola, \(b^2 = a^2(e^2 - 1) = a^2 e^2 - a^2 = 10 - a^2\).
The hyperbola passes through the point \((2, 3)\), so substitute \(x = 2\) and \(y = 3\) into the equation:
\(\frac{3^2}{a^2} - \frac{2^2}{b^2} = 1 \implies \frac{9}{a^2} - \frac{4}{10 - a^2} = 1\).
Let \(a^2 = k\). Then:
\(\frac{9}{k} - \frac{4}{10 - k} = 1\)
Multiplying by \(k(10 - k)\):
\(9(10 - k) - 4k = k(10 - k)\)
\(90 - 9k - 4k = 10k - k^2\)
\(90 - 13k = 10k - k^2\)
\(k^2 - 23k + 90 = 0\)
Factorizing the quadratic equation:
\((k - 18)(k - 5) = 0\)
So \(k = 18\) or \(k = 5\).
If \(k = a^2 = 18\), then \(b^2 = 10 - 18 = -8\), which is impossible for real \(b\).
Thus, \(k = a^2 = 5\), which gives \(b^2 = 10 - 5 = 5\).
Therefore, the equation of the hyperbola is:
\(\frac{y^2}{5} - \frac{x^2}{5} = 1 \implies y^2 - x^2 = 5\).

Teacher's Note:
a) Identify the orientation of the conic section (vertical vs horizontal) from the coordinates of the foci.
b) Reject inadmissible values of parameters (like negative \(b^2\)) when solving quadratic relations.

Question 5

(a) If \(y = e^{m\cos^{-1}x}\), prove that: \((1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = m^2y\). [5 Marks]

Answer:
Given \(y = e^{m\cos^{-1}x}\).
Differentiating both sides with respect to \(x\):
\(\frac{dy}{dx} = e^{m\cos^{-1}x} \cdot \frac{d}{dx}(m\cos^{-1}x) = y \cdot \left(-\frac{m}{\sqrt{1 - x^2}}\right)\).
Multiply both sides by \(\sqrt{1 - x^2}\):
\(\sqrt{1 - x^2} \frac{dy}{dx} = -my\).
Squaring both sides:
\((1 - x^2)\left(\frac{dy}{dx}\right)^2 = m^2 y^2\).
Differentiating both sides again with respect to \(x\):
\((1 - x^2) \cdot 2\left(\frac{dy}{dx}\right)\left(\frac{d^2y}{dx^2}\right) + \left(\frac{dy}{dx}\right)^2 \cdot (-2x) = m^2 \cdot 2y \left(\frac{dy}{dx}\right)\).
Dividing through by \(2\frac{dy}{dx}\) (assuming \(\frac{dy}{dx} \neq 0\)):
\((1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = m^2 y\). Hence proved.

Teacher's Note:
a) Squaring the first derivative after clearing the square root denominator simplifies higher-order derivative proofs significantly.
b) Apply the chain rule carefully when differentiating exponential and inverse trigonometric functions.

(b) Show that the rectangle of maximum perimeter which can be inscribed in a circle of radius \(10\text{ cm}\) is a square of side \(10\sqrt{2}\text{ cm}\). [5 Marks]

Answer:
Let the rectangle have length \(x$ and breadth \(y\). The diagonal of the rectangle inscribed in a circle of radius \(R = 10\text{ cm}\) is equal to the diameter of the circle, so \(\sqrt{x^2 + y^2} = 2(10) = 20\).
Thus, \(x^2 + y^2 = 400 \implies y = \sqrt{400 - x^2}\).
The perimeter \(P\) of the rectangle is given by:
\(P = 2(x + y) = 2\left(x + \sqrt{400 - x^2}\right)\).
To maximize \(P$, differentiate with respect to \(x\):
\(\frac{dP}{dx} = 2\left(1 + \frac{1}{2\sqrt{400 - x^2}} \cdot (-2x)\right) = 2\left(1 - \frac{x}{\sqrt{400 - x^2}}\right)\).
Set \(\frac{dP}{dx} = 0\):
\(1 - \frac{x}{\sqrt{400 - x^2}} = 0 \implies x = \sqrt{400 - x^2}\)
Squaring both sides:
\(x^2 = 400 - x^2 \implies 2x^2 = 400 \implies x^2 = 200 \implies x = 10\sqrt{2}\text{ cm}\).
Then \(y = \sqrt{400 - 200} = \sqrt{200} = 10\sqrt{2}\text{ cm}\).
Since \(x = y = 10\sqrt{2}\text{ cm}\), the rectangle is a square.
Checking the second derivative for maximum:\br />\(\frac{d^2P}{dx^2} = 2 \cdot \frac{d}{dx}\left(1 - \frac{x}{\sqrt{400 - x^2}}\right) \lt 0\) at \(x = 10\sqrt{2}\), confirming maximum perimeter.

Teacher's Note:
a) Express one variable in terms of the other using geometric constraints (like the diagonal of a rectangle inscribed in a circle equaling the diameter).
b) Use the first and second derivative tests to establish local maxima for optimization problems.

Question 6

(a) Evaluate: \(\int \frac{\sec x}{1 + \csc x} \, dx\). [5 Marks]

Answer:
Given integral: \(\int \frac{\sec x}{1 + \csc x} \, dx\).
Express in terms of sine and cosine:
\(\sec x = \frac{1}{\cos x}\) and \(\csc x = \frac{1}{\sin x}\).
So the integrand becomes:
\(\frac{\frac{1}{\cos x}}{1 + \frac{1}{\sin x}} = \frac{\frac{1}{\cos x}}{\frac{\sin x + 1}{\sin x}} = \frac{\sin x}{\cos x (\sin x + 1)} = \frac{\tan x}{\sin x + 1}\).
Alternatively, let us rewrite: \(\int \frac{1}{\cos x (1 + \frac{1}{\sin x})} \, dx = \int \frac{\sin x}{\cos x(\sin x + 1)} \, dx = \int \frac{\tan x}{\sin x + 1} \, dx\).
Let us multiply numerator and denominator by \((1 - \sin x)\):
\(\int \frac{\sin x (1 - \sin x)}{\cos x (1 + \sin x)(1 - \sin x)} \, dx = \int \frac{\sin x - \sin^2 x}{\cos x \cos^2 x} \, dx = \int \frac{\sin x - \sin^2 x}{\cos^3 x} \, dx\)
\(= \int \left(\frac{\sin x}{\cos^3 x} - \frac{\sin^2 x}{\cos^3 x}\right) dx = \int (\tan x \sec^2 x - \tan^2 x \sec x) \, dx\)
\(= \int \tan x \sec^2 x \, dx - \int (\sec^2 x - 1)\sec x \, dx\)
\(= \frac{\sec^2 x}{2} - \int \sec^3 x \, dx + \int \sec x \, dx\).
Alternatively, using substitution: put \(t = \sin x\), or convert to sine/cosine:
\(\int \frac{\sin x}{\cos^2 x(1 + \sin x)} \, dx\). Let \(u = \sin x\), then \(du = \cos x \, dx$, so \(\cos^2 x = 1 - u^2\):
\(\int \frac{u}{(1 - u^2)(1 + u)} \, du = \int \frac{u}{(1-u)(1+u)^2} \, du\).
Using partial fractions:
\(\frac{u}{(1-u)(1+u)^2} = \frac{A}{1-u} + \frac{B}{1+u} + \frac{C}{(1+u)^2}\)
\(u = A(1+u)^2 + B(1-u)(1+u) + C(1-u)\).
At \(u = 1\): \(1 = A(4) \implies A = \frac{1}{4}\).
At \(u = -1\): \(-1 = C(2) \implies C = -\frac{1}{2}\).
Comparing coefficients of \(u^2\): \(0 = A - B \implies B = A = \frac{1}{4}\).
Thus, the integral is:
\(\int \left(\frac{1/4}{1-u} + \frac{1/4}{1+u} - \frac{1/2}{(1+u)^2}\right) du\)
\(= -\frac{1}{4}\ln|1-u| + \frac{1}{4}\ln|1+u| + \frac{1}{2(1+u)} + C\)
\(= \frac{1}{4}\ln\left|\frac{1+u}{1-u}\right| + \frac{1}{2(1+u)} + C\).
Substitute back \(u = \sin x\):
\(= \frac{1}{4}\ln\left|\frac{1+\sin x}{1-\sin x}\right| + \frac{1}{2(1+\sin x)} + C\).

Teacher's Note:
a) Converting trigonometric functions into sine and cosine simplifies rational trigonometric integrands.
b) Partial fraction decomposition is a powerful technique for integrating rational algebraic functions in terms of substituted variables.

(b) Find the smaller area enclosed by the circle \(x^2 + y^2 = 2\) and the line \(x + y = 2\). [5 Marks]

Answer:
Given circle: \(x^2 + y^2 = 2$ (centre \((0, 0)\), radius \(r = \sqrt{2}\)).
Given line: \(x + y = 2 \implies y = 2 - x\).
Find the points of intersection of the line and the circle:
\(x^2 + (2 - x)^2 = 2\)
\(x^2 + 4 - 4x + x^2 = 2\)
\(2x^2 - 4x + 2 = 0 \implies x^2 - 2x + 1 = 0 \implies (x - 1)^2 = 0 \implies x = 1\).
When \(x = 1$, \(y = 2 - 1 = 1\).
The line touches the circle at the single point \((1, 1)\).
Wait, let us check the distance from the centre \((0, 0)\) to the line \(x + y - 2 = 0\):
\(d = \frac{|0 + 0 - 2|}{\sqrt{1^2 + 1^2}} = \frac{2}{\sqrt{2}} = \sqrt{2}\).
Since the radius of the circle is \(\sqrt{2}\) and the perpendicular distance from the centre to the line is also \(\sqrt{2}\), the line is a tangent to the circle at \((1, 1)\).
Therefore, the line does not intersect the circle in two distinct points; it only touches it at \((1, 1)\). There is no enclosed area between the circle and the line, or the area is \(0\).
*(Note: If the line was \(x + y = \sqrt{2}\) or similar, an area would be enclosed. As stated, the line \(x + y = 2\) is outside the circle since its distance from origin is \(\sqrt{2}\) and radius is \(\sqrt{2}\), hence they touch at one point.)*

Teacher's Note:
a) Always check the perpendicular distance from the circle's centre to the line before setting up definite integrals for area.
b) When a line is tangent to a circle, the enclosed area between them is zero.

Question 7

(a) Given that the observations are: \((9, -4), (10, -3), (11, -1), (12, 0), (13, 1), (14, 3), (15, 5), (16, 8)\). Find the two lines of regression and estimate the value of \(y\) when \(x = 13.5\). [5 Marks]

Answer:
Number of observations \(n = 8\).
Let us calculate summary statistics:
\(x_i$: \(9, 10, 11, 12, 13, 14, 15, 16 \implies \sum x = 100\), \(\bar{x} = \frac{100}{8} = 12.5\).
\(y_i$: \(-4, -3, -1, 0, 1, 3, 5, 8 \implies \sum y = 9\), \(\bar{y} = \frac{9}{8} = 1.125\).
Let \(u = x - 12.5\) and \(v = y - 1.125\) or compute directly:
\(\sum x^2 = 9^2 + 10^2 + 11^2 + 12^2 + 13^2 + 14^2 + 15^2 + 16^2 = 81 + 100 + 121 + 144 + 169 + 196 + 225 + 256 = 1292\).
\(\sum y^2 = (-4)^2 + (-3)^2 + (-1)^2 + 0^2 + 1^2 + 3^2 + 5^2 + 8^2 = 16 + 9 + 1 + 0 + 1 + 9 + 25 + 64 = 125\).
\(\sum xy = 9(-4) + 10(-3) + 11(-1) + 12(0) + 13(1) + 14(3) + 15(5) + 16(8)\)
\(= -36 - 30 - 11 + 0 + 13 + 42 + 75 + 128 = 181\).
Calculations for covariances and variances:
\(S_{xx} = \sum x^2 - n\bar{x}^2 = 1292 - 8(12.5)^2 = 1292 - 8(156.25) = 1292 - 1250 = 42\).
\(S_{yy} = \sum y^2 - n\bar{y}^2 = 125 - 8(1.125)^2 = 125 - 8(1.265625) = 125 - 10.125 = 114.875\).
\(S_{xy} = \sum xy - n\bar{x}\bar{y} = 181 - 8(12.5)(1.125) = 181 - 112.5 = 68.5\).
1. Regression line of \(y$ on \(x\):
\(y - \bar{y} = b_{yx} (x - \bar{x})\), where \(b_{yx} = \frac{S_{xy}}{S_{xx}} = \frac{68.5}{42} = 1.631\).
\(y - 1.125 = 1.631(x - 12.5) \implies y = 1.631x - 20.387 + 1.125 \implies y = 1.631x - 19.262\).
2. Regression line of \(x$ on \(y\):
\(x - \bar{x} = b_{xy} (y - \bar{y})\), where \(b_{xy} = \frac{S_{xy}}{S_{yy}} = \frac{68.5}{114.875} = 0.596\).
\(x - 12.5 = 0.596(y - 1.125) \implies x = 0.596y + 11.83\).
3. Estimate \(y$ when \(x = 13.5\):
Using the regression line of \(y$ on \(x\):
\(y = 1.631(13.5) - 19.262 = 22.0185 - 19.262 = 2.7565\).

Teacher's Note:
a) Carefully compute summary statistics (\(\sum x$, \(\sum y$, \(\sum x^2$, \(\sum y^2$, \(\sum xy\)) to avoid errors in regression coefficients.
b) Use the regression line of \(y$ on \(x\) to estimate \(y$ for a given value of \(x\).

(b) In a contest the competitors are awarded marks out of \(20\) by two judges. The scores of the \(10\) competitors are given below. Calculate Spearman's rank correlation. [5 Marks]

Answer:
Let us rank the scores given by Judge A (\(R_1\)) and Judge B (\(R_2\)) from highest to lowest (rank 1 for highest score):
Scores for Judge A: 2, 11, 11, 18, 6, 5, 8, 16, 13, 15.
Sorted descending: 18 (1), 16 (2), 15 (3), 13 (4), 11 (5), 11 (6), 8 (7), 6 (8), 5 (9), 2 (10).
Since 11 appears twice (ranks 5 and 6), average rank is \(\frac{5+6}{2} = 5.5\).
Ranks for Judge A (\(R_1\)):
A: 10
B: 5.5
C: 5.5
D: 1
E: 8
F: 9
G: 7
H: 2
I: 4
J: 3
Scores for Judge B: 6, 11, 16, 9, 14, 20, 4, 3, 13, 17.
Sorted descending: 20 (1), 17 (2), 16 (3), 14 (4), 13 (5), 11 (6), 9 (7), 6 (8), 4 (9), 3 (10).
Ranks for Judge B (\(R_2\)):
A: 8
B: 6
C: 3
D: 7
E: 4
F: 1
G: 9
H: 10
I: 5
J: 2
Calculate differences \(d = R_1 - R_2\) and \(d^2\):
- A: \(10 - 8 = 2 \implies d^2 = 4\)
- B: \(5.5 - 6 = -0.5 \implies d^2 = 0.25\)
- C: \(5.5 - 3 = 2.5 \implies d^2 = 6.25\)
- D: \(1 - 7 = -6 \implies d^2 = 36\)
- E: \(8 - 4 = 4 \implies d^2 = 16\)
- F: \(9 - 1 = 8 \implies d^2 = 64\)
- G: \(7 - 9 = -2 \implies d^2 = 4\)
- H: \(2 - 10 = -8 \implies d^2 = 64\)
- I: \(4 - 5 = -1 \implies d^2 = 1\)
- J: \(3 - 2 = 1 \implies d^2 = 1\)
Sum of squared differences \(\sum d^2 = 4 + 0.25 + 6.25 + 36 + 16 + 64 + 4 + 64 + 1 + 1 = 197.5\).
Since there are tied ranks (11 appears twice in Judge A), apply correction factor \(CF = \frac{\sum (m^3 - m)}{12}\):
For \(m = 2\): \(\frac{2^3 - 2}{12} = \frac{6}{12} = 0.5\).
Spearman's rank correlation coefficient \(R = 1 - \frac{6\left(\sum d^2 + CF\right)}{n(n^2 - 1)}\):
\(R = 1 - \frac{6(197.5 + 0.5)}{10(100 - 1)} = 1 - \frac{6(198)}{10(99)} = 1 - \frac{1188}{990} = 1 - 1.2 = -0.2\).

Teacher's Note:
a) Assign average ranks when there are ties in the data scores.
b) Include the correction factor in Spearman's rank formula when ties occur.

Question 8

(a) An urn contains \(2\) white and \(2\) black balls. A ball is drawn at random. If it is white, it is not replaced into the urn. Otherwise, it is replaced with another ball of the same colour. The process is repeated. Find the probability that the third ball drawn is black. [5 Marks]

Answer:
Initial state of the urn: \(2\) white (\(W\)) and \(2\) black (\(B\)). Total \(= 4\).
We want the third ball drawn to be black (\(B_3\)).
Let us consider all possible sequences of draws for the first two balls that lead to a black ball on the third draw:
1. Draw 1: White (\(W_1\)), Draw 2: White (\(W_2\)), Draw 3: Black (\(B_3\)):
- Initial: \(2W, 2B$ (Total 4). Prob of \(W_1 = \frac{2}{4} = \frac{1}{2}\).
- Since white is not replaced, remaining: \(1W, 2B$ (Total 3). Prob of \(W_2\) given \(W_1 = \frac{1}{3}\).
- Remaining after two whites: \(0W, 2B$ (Total 2). Prob of \(B_3\) given \(W_1W_2 = \frac{2}{2} = 1\).
Probability of this path \(= \frac{1}{2} \times \frac{1}{3} \times 1 = \frac{1}{6}\).
2. Draw 1: White (\(W_1\)), Draw 2: Black (\(B_2\)), Draw 3: Black (\(B_3\)):
- Initial: \(2W, 2B$ (Total 4). Prob of \(W_1 = \frac{2}{4} = \frac{1}{2}\).
- Remaining after \(W_1\): \(1W, 2B$ (Total 3). Prob of \(B_2 = \frac{2}{3}\).
- Since black is replaced with another black, urn becomes: \(1W, 3B$ (Total 4). Prob of \(B_3 = \frac{3}{4}\).
Probability of this path \(= \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} = \frac{6}{24} = \frac{1}{4}\).
3. Draw 1: Black (\(B_1\)), Draw 2: White (\(W_2\)), Draw 3: Black (\(B_3\)):
- Initial: \(2W, 2B$ (Total 4). Prob of \(B_1 = \frac{2}{4} = \frac{1}{2}\).
- Since black is replaced with another black, urn becomes: \(2W, 3B$ (Total 5). Prob of \(W_2 = \frac{2}{5}\).
- Since white is not replaced, remaining: \(1W, 3B$ (Total 4). Prob of \(B_3 = \frac{3}{4}\).
Probability of this path \(= \frac{1}{2} \times \frac{2}{5} \times \frac{3}{4} = \frac{6}{40} = \frac{3}{20}\).
4. Draw 1: Black (\(B_1\)), Draw 2: Black (\(B_2\)), Draw 3: Black (\(B_3\)):
- Initial: \(2W, 2B$ (Total 4). Prob of \(B_1 = \frac{2}{4} = \frac{1}{2}\).
- Urn becomes \(2W, 3B$ (Total 5). Prob of \(B_2 = \frac{3}{5}\).
- Urn becomes \(2W, 4B$ (Total 6). Prob of \(B_3 = \frac{4}{6} = \frac{2}{3}\).
Probability of this path \(= \frac{1}{2} \times \frac{3}{5} \times \frac{2}{3} = \frac{6}{30} = \frac{1}{5}\).
Total probability \(= \frac{1}{6} + \frac{1}{4} + \frac{3}{20} + \frac{1}{5}\):
LCM of \(6, 4, 20, 5 = 60\).
\(= \frac{10}{60} + \frac{15}{60} + \frac{9}{60} + \frac{12}{60} = \frac{46}{60} = \frac{23}{30}\).

Teacher's Note:
a) Track the composition of the urn meticulously after each draw based on the replacement rules.
b) Sum the probabilities of all mutually exclusive paths that result in the desired final outcome.

(b) Three persons A, B and C shoot to hit a target. If A hits the target four times in five trials, B hits it three times in four trials and C hits it two times in three trials, find the probability that: [5 Marks]
(i) Exactly two persons hit the target.
(ii) At least two persons hit the target.
(iii) None hit the target.

Answer:
Let \(P(A) = \frac{4}{5}\), so probability of A missing \(P(A') = 1 - \frac{4}{5} = \frac{1}{5}\).
Let \(P(B) = \frac{3}{4}\), so probability of B missing \(P(B') = 1 - \frac{3}{4} = \frac{1}{4}\).
Let \(P(C) = \frac{2}{3}\), so probability of C missing \(P(C') = 1 - \frac{2}{3} = \frac{1}{3}\).
(i) Exactly two persons hit the target:
This can happen in three mutually exclusive ways: A and B hit, C misses; A and C hit, B misses; or B and C hit, A misses.
\(P(\text{A and B hit, C misses}) = P(A) \cdot P(B) \cdot P(C') = \frac{4}{5} \times \frac{3}{4} \times \frac{1}{3} = \frac{12}{60} = \frac{1}{5}\).
\(P(\text{A and C hit, B misses}) = P(A) \cdot P(B') \cdot P(C) = \frac{4}{5} \times \frac{1}{4} \times \frac{2}{3} = \frac{8}{60} = \frac{2}{15}\).
\(P(\text{B and C hit, A misses}) = P(A') \cdot P(B) \cdot P(C) = \frac{1}{5} \times \frac{3}{4} \times \frac{2}{3} = \frac{6}{60} = \frac{1}{10}\).
Total probability \(= \frac{1}{5} + \frac{2}{15} + \frac{1}{10} = \frac{6 + 4 + 3}{30} = \frac{13}{30}\).
(ii) At least two persons hit the target:
This means either exactly two persons hit or all three persons hit.
\(P(\text{All three hit}) = P(A) \cdot P(B) \cdot P(C) = \frac{4}{5} \times \frac{3}{4} \times \frac{2}{3} = \frac{24}{60} = \frac{2}{5}\).
Total probability \(= P(\text{Exactly two}) + P(\text{All three}) = \frac{13}{30} + \frac{2}{5} = \frac{13 + 12}{30} = \frac{25}{30} = \frac{5}{6}\).
(iii) None hit the target:
\(P(\text{None hit}) = P(A') \cdot P(B') \cdot P(C') = \frac{1}{5} \times \frac{1}{4} \times \frac{1}{3} = \frac{1}{60}\).

Teacher's Note:
a) Calculate individual probabilities of success and failure for each independent event.
b) Group mutually exclusive scenarios correctly for compound probability questions.

Question 9

(a) If \(z = x + iy\), \(w = \frac{2 - iz}{2z - i}\) and \(|w| = 1\), find the locus of \(z\) and illustrate it in the Argand Plane. [5 Marks]

Answer:
Given \(|w| = 1 \implies \left|\frac{2 - iz}{2z - i}\right| = 1 \implies |2 - iz| = |2z - i|\).
Substitute \(z = x + iy\):
\(|2 - i(x + iy)| = |2(x + iy) - i|\)
\(|2 - ix + y| = |2x + i(2y - 1)|\)
\(\mid(2 + y) - ix\mid = \mid2x + i(2y - 1)\mid\)
Taking the modulus squared on both sides:
\((2 + y)^2 + (-x)^2 = (2x)^2 + (2y - 1)^2\)
\(4 + 4y + y^2 + x^2 = 4x^2 + 4y^2 - 4y + 1\)
Rearranging terms:
\(3x^2 + 3y^2 - 8y - 3 = 0\)
Dividing by \(3\):
\(x^2 + y^2 - \frac{8}{3}y - 1 = 0\).
This represents a circle in the Argand plane.
Completing the square for \(y\):
\(x^2 + \left(y - \frac{4}{3}\right)^2 = 1 + \left(\frac{4}{3}\right)^2 = 1 + \frac{16}{9} = \frac{25}{9} = \left(\frac{5}{3}\right)^2\).
Thus, the locus of \(z\) is a circle with centre \(\left(0, \frac{4}{3}\right)\) and radius \(\frac{5}{3}\).

Teacher's Note:
a) Use the property \(\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}\) to simplify complex modulus equations.
b) The resulting equation of second degree in \(x$ and \(y\) represents a circle in the Argand plane.

(b) Solve the differential equation: \(e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right) + \left(1 + e^{\frac{x}{y}}\right)\frac{dx}{dy} = 0\) when \(x = 0\), \(y = 1\). [5 Marks]

Answer:
Given differential equation: \(e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right) + \left(1 + e^{\frac{x}{y}}\right)\frac{dx}{dy} = 0\).
Rearranging:
\(\frac{dx}{dy} = -\frac{e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right)}{1 + e^{\frac{x}{y}}}\).
This is a homogeneous differential equation in \(\frac{x}{y}\). Let \(x = vy\), so \(\frac{dx}{dy} = v + y\frac{dv}{dy}\).
Substitute into the differential equation:
\(v + y\frac{dv}{dy} = -\frac{e^v(1 - v)}{1 + e^v} = \frac{e^v(v - 1)}{1 + e^v}\)
\(y\frac{dv}{dy} = \frac{ve^v - e^v}{1 + e^v} - v = \frac{ve^v - e^v - v - ve^v}{1 + e^v} = \frac{-e^v - v}{1 + e^v} = -\frac{v + e^v}{1 + e^v}\)
Separating variables:
\(\frac{1 + e^v}{v + e^v} \, dv = -\frac{1}{y} \, dy\).
Integrating both sides:
\(\int \frac{1 + e^v}{v + e^v} \, dv = -\int \frac{1}{y} \, dy\)
\(\ln|v + e^v| = -\ln|y| + \ln C = \ln\left(\frac{C}{y}\right)\)
\(v + e^v = \frac{C}{y}\).
Substitute back \(v = \frac{x}{y}\):
\(\frac{x}{y} + e^{\frac{x}{y}} = \frac{C}{y} \implies x + y e^{\frac{x}{y}} = C\).
Using the given condition \(x = 0$, \(y = 1\):
\(0 + 1 \cdot e^{\frac{0}{1}} = C \implies 0 + 1 \cdot 1 = C \implies C = 1\).
Therefore, the particular solution is:
\(x + y e^{\frac{x}{y}} = 1\).

Teacher's Note:
a) Recognize homogeneous differential equations in \(\frac{x}{y}\) and use the substitution \(x = vy\).
b) Apply initial conditions carefully to determine the exact value of the constant of integration \(C\).

SECTION B

Question 10

(a) Using vectors, prove that angle in a semicircle is a right angle. [5 Marks]

Answer:
Let \(O\) be the centre of a circle with radius \(R\). Let \(AB$ be the diameter of the circle.
Let \(C\) be any point on the circumference of the semicircle.
Choose \(O\) as the origin. Then position vector of \(A\) is \(-\vec{r}\), position vector of \(B\) is \(\vec{r}\), where \(|\vec{r}| = R\).
The position vector of point \(C\) is \(\vec{c}\) such that \(|\vec{c}| = R\).
We need to prove that \(\angle ACB = 90^{\circ}\), which means vector \(\vec{CA}\) is perpendicular to vector \(\vec{CB}\), i.e., \(\vec{CA} \cdot \vec{CB} = 0\).
Vector \(\vec{CA} = \vec{A} - \vec{C} = -\vec{r} - \vec{c}\).
Vector \(\vec{CB} = \vec{B} - \vec{C} = \vec{r} - \vec{c}\).
Now calculate the dot product \(\vec{CA} \cdot \vec{CB}\):
\(\vec{CA} \cdot \vec{CB} = (-\vec{r} - \vec{c}) \cdot (\vec{r} - \vec{c}) = -(\vec{r} + \vec{c}) \cdot (\vec{r} - \vec{c})\)
\(= -(\vec{r} \cdot \vec{r} - \vec{c} \cdot \vec{c}) = -(|\vec{r}|^2 - |\vec{c}|^2)\).
Since \(A$, \(B$, and \(C\) lie on the circle with centre \(O\), \(|\vec{r}| = |\vec{c}| = R\).
Therefore, \(|\vec{r}|^2 - |\vec{c}|^2 = R^2 - R^2 = 0\).
Thus, \(\vec{CA} \cdot \vec{CB} = 0\), which proves that \(\vec{CA}\) is perpendicular to \(\vec{CB}\), so \(\angle ACB = 90^{\circ}\).

Teacher's Note:
a) Set up position vectors relative to the centre of the circle to simplify vector dot products.
b) Equality of radii from the centre to any point on the circle is key to proving orthogonality.

(b) Find the volume of a parallelopiped whose edges are represented by the vectors: \(\vec{a} = 2\hat{i} - 3\hat{j} - 4\hat{k}\), \(\vec{b} = \hat{i} + 2\hat{j} - \hat{k}\), and \(\vec{c} = 3\hat{i} + \hat{j} + 2\hat{k}\). [5 Marks]

Answer:
The volume of a parallelopiped formed by vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) is given by the scalar triple product \(|\vec{a} \cdot (\vec{b} \times \vec{c})|\), which is evaluated as the determinant of the components:
\(\text{Volume} = \begin{vmatrix} 2 & -3 & -4 \\ 1 & 2 & -1 \\ 3 & 1 & 2 \end{vmatrix}\).
Expanding the determinant along the first row:
\(= 2[2(2) - (-1)(1)] - (-3)[1(2) - (-1)(3)] + (-4)[1(1) - 2(3)]\)
\(= 2[4 + 1] + 3[2 + 3] - 4[1 - 6]\)
\(= 2[5] + 3[5] - 4[-5]\)
\(= 10 + 15 + 20 = 45\).\ Therefore, the volume of the parallelopiped is \(45\) cubic units.

Teacher's Note:
a) The volume of a parallelopiped is the absolute value of the scalar triple product of its coterminous edge vectors.
b) Compute the determinant carefully, paying close attention to signs of individual cofactor terms.

Question 11

(a) Find the equation of the plane passing through the intersection of the planes: \(x + y + z + 1 = 0\) and \(2x - 3y + 5z - 2 = 0\) and the point \((-1, 2, 1)\). [5 Marks]

Answer:
The equation of any plane passing through the line of intersection of the two given planes is:
\((x + y + z + 1) + \lambda(2x - 3y + 5z - 2) = 0\).
Since this plane passes through the point \((-1, 2, 1)\), substitute \(x = -1\), \(y = 2\), \(z = 1\):
\((-1 + 2 + 1 + 1) + \lambda(2(-1) - 3(2) + 5(1) - 2) = 0\)
\((3) + \lambda(-2 - 6 + 5 - 2) = 0\)
\(3 + \lambda(-5) = 0 \implies 5\lambda = 3 \implies \lambda = \frac{3}{5}\).
Substitute \(\lambda = \frac{3}{5}\) back into the plane equation:
\((x + y + z + 1) + \frac{3}{5}(2x - 3y + 5z - 2) = 0\)
Multiply through by \(5\):
\(5(x + y + z + 1) + 3(2x - 3y + 5z - 2) = 0\)
\(5x + 5y + 5z + 5 + 6x - 9y + 15z - 6 = 0\)
\(11x - 4y + 20z - 1 = 0\).

Teacher's Note:
a) Use the family of planes formula \(P_1 + \lambda P_2 = 0\) for planes passing through the intersection of two given planes.
b) Substitute the given point coordinates to determine the parameter \(\lambda\).

(b) Find the shortest distance between the lines \(\vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k})\) and \(\vec{r} = 2\hat{i} + 4\hat{j} + 5\hat{k} + \mu(4\hat{i} + 6\hat{j} + 8\hat{k})\). [5 Marks]

Answer:
Let the two lines be given by:
\(\vec{r}_1 = \vec{a}_1 + \lambda \vec{b}_1\) where \(\vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}\) and \(\vec{b}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}\).
\(\vec{r}_2 = \vec{a}_2 + \mu \vec{b}_2\) where \(\vec{a}_2 = 2\hat{i} + 4\hat{j} + 5\hat{k}\) and \(\vec{b}_2 = 4\hat{i} + 6\hat{j} + 8\hat{k}\).
Notice that \(\vec{b}_2 = 2(2\hat{i} + 3\hat{j} + 4\hat{k}) = 2\vec{b}_1\).
Since the direction vectors \(\vec{b}_1\) and \(\vec{b}_2$ are proportional, the two lines are parallel.
The shortest distance \(d\) between two parallel lines is given by the formula:
\(d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}\), where \(\vec{b} = \vec{b}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}\).
Calculate \(\vec{a}_2 - \vec{a}_1\):
\(\vec{a}_2 - \vec{a}_1 = (2 - 1)\hat{i} + (4 - 2)\hat{j} + (5 - 3)\hat{k} = \hat{i} + 2\hat{j} + 2\hat{k}\).
Calculate the cross product \((\vec{a}_2 - \vec{a}_1) \times \vec{b}\):
\((\vec{a}_2 - \vec{a}_1) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 3 & 4 \end{vmatrix}\)
\(= \hat{i}(2(4) - 2(3)) - \hat{j}(1(4) - 2(2)) + \hat{k}(1(3) - 2(2))\)
\(= \hat{i}(8 - 6) - \hat{j}(4 - 4) + \hat{k}(3 - 4)\)
\(= 2\hat{i} + 0\hat{j} - \hat{k} = 2\hat{i} - \hat{k}\).
Magnitude of this cross product:
\(|(\vec{a}_2 - \vec{a}_1) \times \vec{b}| = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}\).
Magnitude of \(\vec{b}\):
\(|\vec{b}| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}\).
Therefore, the shortest distance \(d\) is:
\(d = \frac{\sqrt{5}}{\sqrt{29}} = \sqrt{\frac{5}{29}}\) units.

Teacher's Note:
a) Always check if direction vectors are parallel before applying general skew-line distance formulas.
b) For parallel lines, use the cross product formula involving the difference of position vectors and the shared direction vector.

Question 12

(a) Box I contains two white and three black balls. Box II contains four white and one black balls and box III contains three white and four black balls. A dice having three red, two yellow and one green face, is thrown to select the box. If red face turns up, we pick up box I, if a yellow face turns up we pick up box II, otherwise, we pick up box III. Then, we draw a ball from the selected box. If the ball drawn is white, what is the probability that the dice had turned up with a red face? [5 Marks]

Answer:
Let \(E_1\) be the event of selecting Box I, \(E_2\) Box II, and \(E_3\) Box III.
The dice has 6 faces: 3 red, 2 yellow, 1 green.
- Probability of selecting Box I (Red face): \(P(E_1) = \frac{3}{6} = \frac{1}{2}\).
- Probability of selecting Box II (Yellow face): \(P(E_2) = \frac{2}{6} = \frac{1}{3}\).
- Probability of selecting Box III (Green face): \(P(E_3) = \frac{1}{6}\).
Let \(W\) be the event that the ball drawn is white.
- Probability of drawing a white ball from Box I (\(2W, 3B\)): \(P(W|E_1) = \frac{2}{5}\).
- Probability of drawing a white ball from Box II (\(4W, 1B\)): \(P(W|E_2) = \frac{4}{5}\).
- Probability of drawing a white ball from Box III (\(3W, 4B\)): \(P(W|E_3) = \frac{3}{7}\).
We need to find the conditional probability \(P(E_1|W)\) using Bayes' Theorem:
\(P(E_1|W) = \frac{P(E_1)P(W|E_1)}{P(E_1)P(W|E_1) + P(E_2)P(W|E_2) + P(E_3)P(W|E_3)}\).
Calculate numerator: \(P(E_1)P(W|E_1) = \frac{1}{2} \times \frac{2}{5} = \frac{1}{5} = \frac{21}{105}\).
Calculate denominator terms:
- Term 1: \(\frac{1}{2} \times \frac{2}{5} = \frac{1}{5}\)
- Term 2: \(\frac{1}{3} \times \frac{4}{5} = \frac{4}{15}\)
- Term 3: \(\frac{1}{6} \times \frac{3}{7} = \frac{3}{42} = \frac{1}{14}\)
Sum of denominator: \(\frac{1}{5} + \frac{4}{15} + \frac{1}{14}\).
LCM of \(5, 15, 14 = 210\).
\(= \frac{42}{210} + \frac{56}{210} + \frac{15}{210} = \frac{113}{210}\).
Numerator = \(\frac{1}{5} = \frac{42}{210}\).
Therefore, \(P(E_1|W) = \frac{42/210}{113/210} = \frac{42}{113}\).

Teacher's Note:
a) Apply Bayes' Theorem when finding the probability of a specific prior cause given an observed outcome.
b) Compute probabilities of selection based on the frequency of face types on the dice.

(b) Five dice are thrown simultaneously. If the occurrence of an odd number in a single dice is considered a success, find the probability of maximum three successes. [5 Marks]

Answer:
Number of trials \(n = 5\).
On a single die, odd numbers are \(1, 3, 5\), so number of successful outcomes = \(3\).
Probability of success in a single trial \(p = \frac{3}{6} = \frac{1}{2}\).
Probability of failure \(q = 1 - \frac{1}{2} = \frac{1}{2}\).
We need to find the probability of maximum three successes, i.e., \(P(X \le 3)\), where \(X$ follows a binomial distribution \(B(5, \frac{1}{2})\):
\(P(X \le 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)\).
Alternatively, use the complement rule: \(P(X \le 3) = 1 - [P(X = 4) + P(X = 5)]\).
Binomial probability formula: \(P(X = r) = {^nC_r} p^r q^{n-r} = {^5C_r} \left(\frac{1}{2}\right)^r \left(\frac{1}{2}\right)^{5-r} = {^5C_r} \left(\frac{1}{2}\right)^5 = \frac{^5C_r}{32}\).
Calculate probabilities for \(r = 4\) and \(r = 5\):
- \(P(X = 4) = \frac{^5C_4}{32} = \frac{5}{32}\)
- \(P(X = 5) = \frac{^5C_5}{32} = \frac{1}{32}\)
Sum of \(P(X = 4)\) and \(P(X = 5)\) = \(\frac{5}{32} + \frac{1}{32} = \frac{6}{32} = \frac{3}{16}\).
Therefore, \(P(X \le 3) = 1 - \frac{6}{32} = \frac{26}{32} = \frac{13}{16}\).

Teacher's Note:
a) Use the binomial probability distribution formula for independent trial successes.
b) The complement rule simplifies calculations when finding probabilities like "at most" or "maximum" values.

SECTION C

Question 13

(a) Mr. Nirav borrowed Rs. \(50,000\) from the bank for \(5\) years. The rate of interest is \(9\%\) per annum compounded monthly. Find the payment he makes monthly if he pays back at the beginning of each month. [5 Marks]

Answer:
Principal amount \(P = \text{Rs. } 50,000\).
Time \(n = 5\text{ years} = 60\text{ months}\).
Annual interest rate \(r = 9\% = 0.09\).
Monthly interest rate \(i = \frac{0.09}{12} = 0.0075\).
For payments made at the beginning of each month (Annuity Due), the present value \(P\) formula is:
\(P = R \cdot \frac{1 - (1 + i)^{-n}}{i} \cdot (1 + i)\), where \(R\) is the monthly payment.
Substitute the values:
\(50000 = R \cdot \frac{1 - (1 + 0.0075)^{-60}}{0.0075} \cdot (1 + 0.0075)\)
Given \((1.0075)^{-60} \approx 0.6387\):
\(\frac{1 - 0.6387}{0.0075} = \frac{0.3613}{0.0075} = 48.1733\)
Multiply by \((1.0075)\):
\(48.1733 \times 1.0075 = 48.5346\)
So \(50000 = R \times 48.5346\)
\(R = \frac{50000}{48.5346} \approx \text{Rs. } 1,030.19\).

Teacher's Note:
a) Use the present value formula for an annuity due when payments are made at the beginning of each period.
b) Adjust the annual interest rate and total periods to monthly compounding terms correctly.

(b) A dietician wishes to mix two kinds of food X and Y in such a way that the mixture contains at least \(10\) units of vitamin A, \(12\) units of vitamin B and \(8\) units of vitamin C. The vitamin contents of one kg food is given below: [5 Marks]

One kg of food X costs Rs. \(24\) and one kg of food Y costs Rs. \(36\). Using Linear Programming, find the least cost of the total mixture which will contain the required vitamins.

Answer:
Let \(x\) kg of food X and \(y\) kg of food Y be mixed.
Objective function: Minimize Cost \(Z = 24x + 36y\).
Subject to constraints:
1. Vitamin A: \(1x + 2y \ge 10\)
2. Vitamin B: \(2x + 2y \ge 12 \implies x + y \ge 6\)
3. Vitamin C: \(3x + 1y \ge 8\)
4. Non-negativity: \(x \ge 0\), \(y \ge 0\).
Find the corner points of the feasible region defined by the constraints:
- Intersection of \(x + 2y = 10\) and \(x + y = 6\):
Subtracting equations: \(y = 4\). Then \(x = 2\). Point \((2, 4)\).
- Intersection of \(x + y = 6\) and \(3x + y = 8\):
Subtracting equations: \(2x = 2 \implies x = 1\). Then \(y = 5\). Point \((1, 5)\).
- Boundary intercepts:
For \(x + 2y = 10\): when \(y = 0\), \(x = 10\). Point \((10, 0)\).
For \(3x + y = 8\): when \(x = 0\), \(y = 8\). Point \((0, 8)\).
Evaluate the objective function \(Z = 24x + 36y\) at feasible corner points:
- At \((10, 0)\): \(Z = 24(10) + 36(0) = \text{Rs. } 240\)
- At \((2, 4)\): \(Z = 24(2) + 36(4) = 48 + 144 = \text{Rs. } 192\)
- At \((1, 5)\): \(Z = 24(1) + 36(5) = 24 + 180 = \text{Rs. } 204\)
- At \((0, 8)\): \(Z = 24(0) + 36(8) = \text{Rs. } 288\)
The minimum cost is Rs. \(192\) at \(x = 2\text{ kg}\) and \(y = 4\text{ kg}\).

Teacher's Note:
a) Formulate LP problems clearly by defining decision variables, objective function, and constraints.
b) Test all corner points of the feasible region to determine the optimal minimum or maximum value.

Question 14

(a) A bill for Rs. \(7,650\) was drawn on \(8^{\text{th}}\) March, 2013, at \(7\) months. It was discounted on \(18^{\text{th}}\) May, 2013 and the holder of the bill received Rs. \(7,497\). What isellipse... rate of interest charged by the bank? [5 Marks]

Answer:
Face value (FV) = Rs. \(7,650\).
Date of drawing = \(8^{\text{th}}\) March, 2013.
Term of the bill = \(7\) months.
Legal due date = \(8^{\text{th}}\) October, 2013 (adding 3 grace days: \(11^{\text{th}}\) October, 2013).
Date of discounting = \(18^{\text{th}}\) May, 2013.
Calculate the unexpired period (discount period) from \(18^{\text{th}}\) May, 2013 to \(11^{\text{th}}\) October, 2013:
- May remaining days: \(31 - 18 = 13\) days
- June: \(30\) days
- July: \(31\) days
- August: \(31\) days
- September: \(30\) days
- October: \(11\) days
Total unexpired days \(t = 13 + 30 + 31 + 31 + 30 + 11 = 146\text{ days}\) (or \(\frac{146}{365} = \frac{2}{5}\text{ year}\)).
Amount received by holder = Rs. \(7,497\).
Bank Discount (BD) = Face Value - Cash Value = \(7650 - 7497 = \text{Rs. } 153\).
We know \(BD = \frac{FV \cdot r \cdot t}{365}\):
\(153 = \frac{7650 \cdot r \cdot 146}{365}\)
\(153 = 7650 \cdot r \cdot \frac{2}{5} = 3060 \cdot r\)
\(r = \frac{153}{3060} = \frac{1}{20} = 0.05 = 5\%\).\ Therefore, the rate of interest charged by the bank is \(5\%\) per annum.

Teacher's Note:
a) Add three days of grace to the nominal due date to determine the legal due date.
b) Count the exact number of unexpired days from the date of discounting to the legal due date.

(b) The average cost function, AC for a commodity is given by \(AC = x + 5 + \frac{36}{x}\), in terms of output \(x\). Find: [5 Marks]
(i) The total cost, C and marginal cost, MC as a function of \(x\).
(ii) The outputs for which AC increases.

Answer:
Given average cost \(AC = x + 5 + \frac{36}{x}\).
(i) Total Cost \(C\) and Marginal Cost \(MC\):
Total cost \(C = AC \times x = \left(x + 5 + \frac{36}{x}\right)x = x^2 + 5x + 36\).
Marginal cost \(MC = \frac{dC}{dx} = \frac{d}{dx}(x^2 + 5x + 36) = 2x + 5\).
(ii) Outputs for which AC increases:
For AC to increase, its derivative with respect to \(x\) must be greater than zero (\(\frac{d(AC)}{dx} \gt 0\)):
\(\frac{d(AC)}{dx} = \frac{d}{dx}\left(x + 5 + \frac{36}{x}\right) = 1 - \frac{36}{x^2}\).
Set \(\frac{d(AC)}{dx} \gt 0\):
\(1 - \frac{36}{x^2} \gt 0 \implies \frac{x^2 - 36}{x^2} \gt 0\).
Since \(x^2 \gt 0\) for all \(x \neq 0\), this implies \(x^2 - 36 \gt 0\)
\((x - 6)(x + 6) \gt 0\).
Since output \(x \gt 0\), the AC increases when \(x \gt 6\).

Teacher's Note:
a) Total cost is obtained by multiplying average cost by output \(x\), and marginal cost is the first derivative of total cost.
b) A function is increasing where its first derivative is positive.

Question 15

(a) Calculate the index number for the year 2014, with 2010 as the base year by the weighted aggregate method from the following data: [5 Marks]

Answer:
Let \(p_0\) be the price in 2010 (base year), \(p_1\) be the price in 2014 (current year), and \(w\) be the weights.
Using the weighted aggregate method (Laspeyres/weighted index formula):
\(I_{01} = \frac{\sum p_1 w}{\sum p_0 w} \times 100\).
Let us construct the calculation table:
- Commodity A: \(p_0 = 2\), \(p_1 = 4\), \(w = 8 \implies p_0 w = 2 \times 8 = 16\), \(p_1 w = 4 \times 8 = 32\).
- Commodity B: \(p_0 = 5\), \(p_1 = 6\), \(w = 10 \implies p_0 w = 5 \times 10 = 50\), \(p_1 w = 6 \times 10 = 60\).
- Commodity C: \(p_0 = 4\), \(p_1 = 5\), \(w = 14 \implies p_0 w = 4 \times 14 = 56\), \(p_1 w = 5 \times 14 = 70\).
- Commodity D: \(p_0 = 2\), \(p_1 = 2\), \(w = 19 \implies p_0 w = 2 \times 19 = 38\), \(p_1 w = 2 \times 19 = 38\).
Summation totals:
\(\sum p_0 w = 16 + 50 + 56 + 38 = 160\).
\(\sum p_1 w = 32 + 60 + 70 + 38 = 200\).
Calculate the index number:
\(I_{01} = \frac{200}{160} \times 100 = \frac{5}{4} \times 100 = 125\).

Teacher's Note:
a) Apply the weighted aggregate index formula \(\frac{\sum p_1 w}{\sum p_0 w} \times 100\).
b) Double check product sums for base and current price-weight columns.

(b) The quarterly profits of a small scale industry (in thousands of rupees) is as follows: [5 Marks]

Calculate four quarterly moving averages. Display these and the original figures graphically on the same graph sheet.

Answer:
To calculate four-quarter moving averages:
1. Compute 4-period moving totals.
2. Center the 4-period moving totals by taking 2-period moving averages of the totals.
Let us list the data sequentially by quarters (\(Y_{t}\)):
2012 Q1: 39
2012 Q2: 47
2012 Q3: 20
2012 Q4: 56
2013 Q1: 68
2013 Q2: 59
2013 Q3: 66
2013 Q4: 72
2014 Q1: 88
2014 Q2: 60
2014 Q3: 60
2014 Q4: 67
- 4-quarter moving totals:
- \(39 + 47 + 20 + 56 = 162\) (centered between Q2 and Q3 of 2012)
- \(47 + 20 + 56 + 68 = 191\)
- \(20 + 56 + 68 + 59 = 203\)
- \(56 + 68 + 59 + 66 = 249\)
- \(68 + 59 + 66 + 72 = 265\)
- \(59 + 66 + 72 + 88 = 285\)
- \(66 + 72 + 88 + 60 = 286\)
- \(72 + 88 + 60 + 60 = 280\)
- \(88 + 60 + 60 + 67 = 275\)
- 2-period centered moving totals (sum of consecutive 4-quarter moving totals):
- \(162 + 191 = 353\)
- \(191 + 203 = 394\)
- \(203 + 249 = 452\)
- \(249 + 265 = 514\)
- \(265 + 285 = 550\)
- \(285 + 286 = 571\)
- \(286 + 280 = 566\)
- \(280 + 275 = 555\)
- Divided by 8 (Centered 4-quarter moving averages):
- \(353 / 8 = 44.13\) (centered at 2012 Q3)
- \(394 / 8 = 49.25\) (centered at 2012 Q4)
- \(452 / 8 = 56.50\) (centered at 2013 Q1)
- \(514 / 8 = 64.25\) (centered at 2013 Q2)
- \(550 / 8 = 68.75\) (centered at 2013 Q3)
- \(571 / 8 = 71.38\) (centered at 2013 Q4)
- \(566 / 8 = 70.75\) (centered at 2014 Q1)
- \(555 / 8 = 69.38\) (centered at 2014 Q2)
Graphical representation: Plot original quarterly data and the centered moving averages on the same graph sheet to illustrate trend and seasonal fluctuations.

Teacher's Note:
a) For an even period moving average (like 4 quarters), centering requires taking the 2-period moving average of the 4-period totals and dividing by 8.
b) Centered moving averages smooth out seasonal variations to reveal the underlying secular trend.