ISC Class 12 Mathematics Board Exam Question Paper 2014 with Solutions

ISC Class 12 Mathematics Board Exam Question Paper with Solutions

SECTION A

Question 1

(i) If \( A = \begin{pmatrix} 3 & 1 \\ 7 & 5 \end{pmatrix} \), find the values of \( x \) and \( y \) such that \( A^2 + x I_2 = y A \). [2 Marks]

Answer:
Given \( A = \begin{pmatrix} 3 & 1 \\ 7 & 5 \end{pmatrix} \)
\( A^2 = A \cdot A = \begin{pmatrix} 3 & 1 \\ 7 & 5 \end{pmatrix} \begin{pmatrix} 3 & 1 \\ 7 & 5 \end{pmatrix} = \begin{pmatrix} 9 + 7 & 3 + 5 \\ 21 + 35 & 7 + 25 \end{pmatrix} = \begin{pmatrix} 16 & 8 \\ 56 & 32 \end{pmatrix} \)
Given equation is \( A^2 + x I_2 = y A \)
\( \begin{pmatrix} 16 & 8 \\ 56 & 32 \end{pmatrix} + x \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = y \begin{pmatrix} 3 & 1 \\ 7 & 5 \end{pmatrix} \)
\( \begin{pmatrix} 16 + x & 8 \\ 56 & 32 + x \end{pmatrix} = \begin{pmatrix} 3y & y \\ 7y & 5y \end{pmatrix} \)
Comparing the corresponding elements:
\( y = 8 \)
\( 7y = 56 \implies 7(8) = 56 \) (consistent)
\( 16 + x = 3y \implies 16 + x = 3(8) \implies 16 + x = 24 \implies x = 8 \)
Therefore, \( x = 8 \) and \( y = 8 \).

Teacher's Note:
a) Always compute matrix multiplication carefully row by column.
b) Verify consistency by checking all corresponding elements after equating matrices.

(ii) Find the eccentricity and the coordinates of foci of the hyperbola \( 25x^2 - 9y^2 = 225 \). [2 Marks]

Answer:
The given equation of the hyperbola is \( 25x^2 - 9y^2 = 225 \).
Dividing both sides by \( 225 \), we get:
\( \frac{x^2}{9} - \frac{y^2}{25} = 1 \)
Here, \( a^2 = 9 \) and \( b^2 = 25 \), so \( a = 3 \) and \( b = 5 \).
Eccentricity \( e \) is given by \( b^2 = a^2(e^2 - 1) \):
\( 25 = 9(e^2 - 1) \)
\( \frac{25}{9} = e^2 - 1 \)
\( e^2 = 1 + \frac{25}{9} = \frac{34}{9} \)
\( e = \frac{\sqrt{34}}{3} \)
Coordinates of foci are \( (\pm ae, 0) \):
\( ae = 3 \cdot \frac{\sqrt{34}}{3} = \sqrt{34} \)
Foci are \( (\pm\sqrt{34}, 0) \).

Teacher's Note:
a) Bring the hyperbola equation into standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) first.
b) Remember the eccentricity formula for a horizontal hyperbola is \( b^2 = a^2(e^2 - 1) \).

(iii) Evaluate: \( \tan \left[ 2 \tan^{-1} \frac{1}{2} - \cot^{-1} 3 \right] \) [2 Marks]

Answer:
Let \( 2 \tan^{-1} \frac{1}{2} = \tan^{-1} \left( \frac{2 \cdot \frac{1}{2}}{1 - (\frac{1}{2})^2} \right) = \tan^{-1} \left( \frac{1}{1 - \frac{1}{4}} \right) = \tan^{-1} \left( \frac{4}{3} \right) \)
Also, \( \cot^{-1} 3 = \tan^{-1} \left( \frac{1}{3} \right) \)
Now, the expression becomes:
\( \tan \left[ \tan^{-1} \left( \frac{4}{3} \right) - \tan^{-1} \left( \frac{1}{3} \right) \right] \)
Using the formula \( \tan^{-1} A - \tan^{-1} B = \tan^{-1} \left( \frac{A - B}{1 + AB} \right) \):
\( = \tan \left[ \tan^{-1} \left( \frac{\frac{4}{3} - \frac{1}{3}}{1 + \frac{4}{3} \cdot \frac{1}{3}} \right) \right] = \tan \left[ \tan^{-1} \left( \frac{1}{1 + \frac{4}{9}} \right) \right] = \tan \left[ \tan^{-1} \left( \frac{1}{\frac{13}{9}} \right) \right] = \frac{9}{13} \)

Teacher's Note:
a) Convert all inverse trigonometric functions into tangent terms using standard identities.
b) Apply the difference formula for inverse tangents carefully.

(iv) Using L'Hospital's Rule, evaluate: \( \lim_{x \to 0} (1 + \sin x)^{\cot x} \) [2 Marks]

Answer:
Let \( y = \lim_{x \to 0} (1 + \sin x)^{\cot x} \)
Taking natural logarithm on both sides:
\( \ln y = \lim_{x \to 0} \cot x \ln(1 + \sin x) = \lim_{x \to 0} \frac{\ln(1 + \sin x)}{\tan x} \)
As \( x \to 0 \), this is of the form \( \frac{0}{0} \). Applying L'Hospital's Rule:
\( \ln y = \lim_{x \to 0} \frac{\frac{1}{1 + \sin x} \cdot \cos x}{\sec^2 x} = \frac{\frac{1}{1 + 0} \cdot 1}{1^2} = 1 \)
Since \( \ln y = 1 \), we get \( y = e^1 = e \).

Teacher's Note:
a) For indeterminate forms like \( 1^{\infty} \), take logarithms to convert them into a quotient form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
b) Remember to exponentiate back at the end to find the final limit value.

(v) Evaluate: \( \int e^x \frac{(2 + \sin 2x)}{\cos^2 x} dx \) [2 Marks]

Answer:
\( \int e^x \frac{2 + 2\sin x \cos x}{\cos^2 x} dx = \int e^x \left( \frac{2}{\cos^2 x} + \frac{2\sin x \cos x}{\cos^2 x} \right) dx \)
\( = \int e^x (2 \sec^2 x + 2 \tan x) dx = 2 \int e^x (\tan x + \sec^2 x) dx \)
We know the standard integral formula \( \int e^x [f(x) + f'(x)] dx = e^x f(x) + C \). Here, \( f(x) = \tan x \) and \( f'(x) = \sec^2 x \).
Thus, the integral is \( 2 e^x \tan x + C \).

Teacher's Note:
a) Split the fraction and use trigonometric identity \( \sin 2x = 2\sin x \cos x \).
b) Recognize the standard exponential form \( \int e^x(f(x) + f'(x))dx \).

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

Answer:
Let \( I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx \) — (1)
Using the property \( \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx \):
\( I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin(\frac{\pi}{2} - x)}}{\sqrt{\sin(\frac{\pi}{2} - x)} + \sqrt{\cos(\frac{\pi}{2} - x)}} dx \)
\( I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx \) — (2)
Adding (1) and (2):
\( 2I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx = \int_{0}^{\frac{\pi}{2}} 1 \, dx = [x]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} \)
\( I = \frac{\pi}{4} \)

Teacher's Note:
a) This is a standard definite integral property problem solved by adding original and transformed integrals.
b) The numerator and denominator become identical upon addition, making evaluation straightforward.

(vii) For the given lines of regression, \( 3x - 2y = 5 \) and \( x - 4y = 7 \), find:
(i) regression coefficients \( b_{yx} \) and \( b_{xy} \)
(ii) coefficient of correlation \( r(x, y) \) [2 Marks]

Answer:
Let us assume line 1 is \( 3x - 2y = 5 \) and line 2 is \( x - 4y = 7 \).
First, let us check by assuming \( 3x - 2y = 5 \) is regression equation of \( y \) on \( x \):
\( 2y = 3x - 5 \implies y = \frac{3}{2}x - \frac{5}{2} \). Here \( b_{yx} = \frac{3}{2} \).
Then the other equation \( x - 4y = 7 \) must be \( x \) on \( y \):
\( x = 4y + 7 \). Here \( b_{xy} = 4 \).
Product of regression coefficients: \( b_{yx} \cdot b_{xy} = \frac{3}{2} \cdot 4 = 6 \).
Since \( b_{yx} \cdot b_{xy} = 6 \gt 1 \), our assumption is wrong because \( r^2 = b_{yx} \cdot b_{xy} \le 1 \).
Thus, let \( x - 4y = 7 \) be \( y \) on \( x \):
\( 4y = x - 7 \implies y = \frac{1}{4}x - \frac{7}{4} \implies b_{yx} = \frac{1}{4} \)
And \( 3x - 2y = 5 \) be \( x \) on \( y \):
\( 3x = 2y + 5 \implies x = \frac{2}{3}y + \frac{5}{3} \implies b_{xy} = \frac{2}{3} \)
(i) \( b_{yx} = \frac{1}{4} \) and \( b_{xy} = \frac{2}{3} \).
(ii) \( r = \pm \sqrt{b_{yx} \cdot b_{xy}} = \sqrt{\frac{1}{4} \cdot \frac{2}{3}} = \sqrt{\frac{1}{6}} = \frac{1}{\sqrt{6}} \) (positive since both regression coefficients are positive).

Teacher's Note:
a) Always test the product of regression coefficients to ensure \( b_{yx} \cdot b_{xy} \le 1 \).
b) The sign of correlation coefficient \( r \) is the same as the sign of the regression coefficients.

(viii) Express the complex number \( \frac{(1 + \sqrt{3}i)^2}{\sqrt{3} - i} \) in the form of \( a + ib \). Hence, find the modulus and argument of the complex number. [2 Marks]

Answer:
Numerator: \( (1 + \sqrt{3}i)^2 = 1^2 + ( \sqrt{3}i )^2 + 2(1)(\sqrt{3}i) = 1 - 3 + 2\sqrt{3}i = -2 + 2\sqrt{3}i \)
Denominator: \( \sqrt{3} - i \)
Expression \( = \frac{-2 + 2\sqrt{3}i}{\sqrt{3} - i} = \frac{2(-\1 + \sqrt{3}i)(\sqrt{3} + i)}{(\sqrt{3} - i)(\sqrt{3} + i)} = \frac{2(-\sqrt{3} - i + 3i - \sqrt{3})}{3 - (-1)} \)
\( = \frac{2(-2\sqrt{3} + 2i)}{4} = \frac{2(i\sqrt{3} - \sqrt{3})}{2} = -\sqrt{3} + i\sqrt{3} \)
Thus, \( a = -\sqrt{3} \) and \( b = \sqrt{3} \).
Modulus \( r = \sqrt{(-\sqrt{3})^2 + (\sqrt{3})^2} = \sqrt{3 + 3} = \sqrt{6} \)
Argument \( \theta = \pi - \tan^{-1}\left|\frac{\sqrt{3}}{-\sqrt{3}}\right| = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \)

Teacher's Note:
a) Expand the numerator and rationalize the denominator by multiplying by its conjugate.
b) Determine the quadrant of the complex number carefully when finding the argument.

(ix) A bag contains 20 balls numbered from 1 to 20. One ball is drawn at random from the bag. What is the probability that the ball drawn is marked with a number which is multiple of 3 or 4? [2 Marks]

Answer:
Total number of outcomes \( n(S) = 20 \).
Let \( A \) be the set of multiples of 3 in \( [1, 20] \):
\( A = \{3, 6, 9, 12, 15, 18\} \implies n(A) = 6 \)
Let \( B \) be the set of multiples of 4 in \( [1, 20] \):
\( B = \{4, 8, 12, 16, 20\} \implies n(B) = 5 \)
Intersection \( A \cap B \) (multiples of both 3 and 4, i.e., multiples of 12):
\( A \cap B = \{12\} \implies n(A \cap B) = 1 \)
Number of favorable outcomes \( n(A \cup B) = n(A) + n(B) - n(A \cap B) = 6 + 5 - 1 = 10 \).
Probability \( P(A \cup B) = \frac{10}{20} = \frac{1}{2} \).

Teacher's Note:
a) Use the addition theorem of probability: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).
b) Be careful not to double-count numbers that are multiples of both 3 and 4 (like 12).

(x) Solve the differential equation: \( (x + 1)dy - 2xy dx = 0 \) [2 Marks]

Answer:
\( (x + 1)dy = 2xy dx \)
Separating variables:
\( \frac{1}{y} dy = \frac{2x}{x + 1} dx \)
\( \frac{1}{y} dy = \left( 2 - \frac{2}{x + 1} \right) dx \)
Integrating both sides:
\( \int \frac{1}{y} dy = \int \left( 2 - \frac{2}{x + 1} \right) dx \)
\( \ln|y| = 2x - 2\ln|x + 1| + C \)
\( \ln|y| + 2\ln|x + 1| = 2x + C \)
\( \ln[y(x + 1)^2] = 2x + C \)
\( y(x + 1)^2 = e^{2x + C} = e^C \cdot e^{2x} = K e^{2x} \)

Teacher's Note:
a) Separate the variables \( x \) and \( y \) on opposite sides.
b) Use algebraic manipulation like long division or splitting terms to integrate rational functions.

Question 2

(a) Using properties of determinants, prove that: \( \begin{vmatrix} a^2 + 1 & ab & ac \\ ba & b^2 + 1 & bc \\ ca & cb & c^2 + 1 \end{vmatrix} = a^2 + b^2 + c^2 + 1 \) [5 Marks]

Answer:
Let \( \Delta = \begin{vmatrix} a^2 + 1 & ab & ac \\ ba & b^2 + 1 & bc \\ ca & cb & c^2 + 1 \end{vmatrix} \)
Multiply \( R_1, R_2, R_3 \) by \( a, b, c \) respectively:
\( \Delta = \frac{1}{abc} \begin{vmatrix} a(a^2+1) & a^2b & a^2c \\ ab^2 & b(b^2+1) & b^2c \\ c^2a & c^2b & c(c^2+1) \end{vmatrix} \)
Take out \( a, b, c \) common from \( C_1, C_2, C_3 \) respectively:
\( \Delta = \frac{abc}{abc} \begin{vmatrix} a^2+1 & a^2 & a^2 \\ b^2 & b^2+1 & b^2 \\ c^2 & c^2 & c^2+1 \end{vmatrix} = \begin{vmatrix} a^2+1 & a^2 & a^2 \\ b^2 & b^2+1 & b^2 \\ c^2 & c^2 & c^2+1 \end{vmatrix} \)
Apply \( R_1 \to R_1 + R_2 + R_3 \):
\( \Delta = \begin{vmatrix} a^2+b^2+c^2+1 & a^2+b^2+c^2+1 & a^2+b^2+c^2+1 \\ b^2 & b^2+1 & b^2 \\ c^2 & c^2 & c^2+1 \end{vmatrix} \)
Take \( (a^2 + b^2 + c^2 + 1) \) common from \( R_1 \):
\( \Delta = (a^2 + b^2 + c^2 + 1) \begin{vmatrix} 1 & 1 & 1 \\ b^2 & b^2+1 & b^2 \\ c^2 & c^2 & c^2+1 \end{vmatrix} \)
Apply \( C_2 \to C_2 - C_1 \) and \( C_3 \to C_3 - C_1 \):
\( \Delta = (a^2 + b^2 + c^2 + 1) \begin{vmatrix} 1 & 0 & 0 \\ b^2 & 1 & 0 \\ c^2 & 0 & 1 \end{vmatrix} \)
Expanding along \( R_1 \):
\( \Delta = (a^2 + b^2 + c^2 + 1)(1 \cdot (1 \cdot 1 - 0)) = a^2 + b^2 + c^2 + 1 \). Hence proved.

Teacher's Note:
a) Multiply rows by \( a, b, c \) to create common terms in columns.
b) Form \( a^2 + b^2 + c^2 + 1 \) by adding rows together.

(b) Using matrix method, solve the following system of equations: \( x - 2y = 10 \), \( 2x + y + 3z = 8 \) and \( -2y + z = 7 \) [5 Marks]

Answer:
The given system of linear equations can be written in matrix form \( AX = B \):
\( \begin{pmatrix} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 10 \\ 8 \\ 7 \end{pmatrix} \)
Let \( A = \begin{pmatrix} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{pmatrix} \).
Determinant of \( A \):
\( |A| = 1(1 - (-6)) - (-2)(2 - 0) + 0 = 1(7) + 2(2) = 7 + 4 = 11 \neq 0 \). Hence, \( A^{-1} \) exists.
Cofactors of matrix \( A \):
\( C_{11} = 7, \, C_{12} = -2, \, C_{13} = -4 \)
\( C_{21} = 2, \, C_{22} = 1, \, C_{23} = 2 \)
\( C_{31} = -6, \, C_{32} = -3, \, C_{33} = 5 \)
Adjoint of \( A \):
\( \text{adj}(A) = \begin{pmatrix} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 2 & 5 \end{pmatrix} \)
Inverse \( A^{-1} = \frac{1}{11} \begin{pmatrix} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 2 & 5 \end{pmatrix} \)
Solution \( X = A^{-1}B \):
\( \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{11} \begin{pmatrix} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 2 & 5 \end{pmatrix} \begin{pmatrix} 10 \\ 8 \\ 7 \end{pmatrix} = \frac{1}{11} \begin{pmatrix} 70 + 16 - 42 \\ -20 + 8 - 21 \\ -40 + 16 + 35 \end{pmatrix} = \frac{1}{11} \begin{pmatrix} 44 \\ -33 \\ 11 \end{pmatrix} = \begin{pmatrix} 4 \\ -3 \\ 1 \end{pmatrix} \)
Therefore, \( x = 4, \, y = -3, \, z = 1 \).

Teacher's Note:
a) Write down the matrix equation correctly, including zero for missing variables like \( z \) in the first equation.
b) Verify the solution by substituting \( x, y, z \) back into all original equations.

Question 3

(a) If \( \cos^{-1} x + \cos^{-1} y + \cos^{-1} z = \pi \), prove that: \( x^2 + y^2 + z^2 + 2xyz = 1 \) [5 Marks]

Answer:
Given \( \cos^{-1} x + \cos^{-1} y + \cos^{-1} z = \pi \)
Let \( \cos^{-1} x = A \), \( \cos^{-1} y = B \), \( \cos^{-1} z = C \). Then \( x = \cos A \), \( y = \cos B \), \( z = \cos C \).
\( A + B + C = \pi \implies A + B = \pi - C \)
Taking cosine on both sides:
\( \cos(A + B) = \cos(\pi - C) \)
\( \cos A \cos B - \sin A \sin B = -\cos C \)
\( xy - \sqrt{1 - x^2} \sqrt{1 - y^2} = -z \)
\( xy + z = \sqrt{(1 - x^2)(1 - y^2)} \)
Squaring both sides:
\( (xy + z)^2 = (1 - x^2)(1 - y^2) \)
\( x^2 y^2 + z^2 + 2xyz = 1 - x^2 - y^2 + x^2 y^2 \)
Canceling \( x^2 y^2 \) from both sides:
\( z^2 + 2xyz = 1 - x^2 - y^2 \)
\( x^2 + y^2 + z^2 + 2xyz = 1 \). Hence proved.

Teacher's Note:
a) Substitute inverse trigonometric functions with angle variables to simplify algebraic manipulation.
b) Isolate the radical term before squaring both sides to eliminate square roots.

(b) P, Q and R represent switches in on position and P', Q' and R' represent switches in off position. Construct a switching circuit representing the polynomial \( P R + Q(Q' + R)(P + QR) \). Using Boolean Algebra, simplify the polynomial expression and construct the simplified circuit. [5 Marks]

Answer:
Given polynomial expression: \( E = PR + Q(Q' + R)(P + QR) \)
Simplification using Boolean Algebra:
\( E = PR + (QQ' + QR)(P + QR) \)
Since \( QQ' = 0 \):
\( E = PR + QR(P + QR) \)
\( E = PR + PQR + QR \quad (\text{since } QR \cdot QR = QR) \)
\( E = PR(1 + Q) + QR \)
Since \( 1 + Q = 1 \):
\( E = PR + QR = R(P + Q) \)
[Figure: Switching circuit diagrams corresponding to the initial and simplified expressions representing parallel and series connections of switches P, Q, and R.]

Teacher's Note:
a) Apply standard Boolean laws like distributive law and complement law (\( QQ' = 0 \)) during simplification.
b) The simplified expression \( R(P + Q) \) consists of switch \( R \) in series with a parallel combination of switches \( P \) and \( Q \).

Question 4

(a) Verify Rolle's Theorem for the function \( f(x) = e^x(\sin x - \cos x) \) on \( \left[\frac{\pi}{4}, \frac{5\pi}{4}\right] \) [5 Marks]

Answer:
Given function \( f(x) = e^x(\sin x - \cos x) \) on interval \( \left[\frac{\pi}{4}, \frac{5\pi}{4}\right] \).
1. Since \( f(x) \) is a product of an exponential function and trigonometric functions, it is continuous on \( \left[\frac{\pi}{4}, \frac{5\pi}{4}\right] \) and differentiable on \( \left(\frac{\pi}{4}, \frac{5\pi}{4}\right) \).
2. Evaluate at endpoints:
\( f\left(\frac{\pi}{4}\right) = e^{\frac{\pi}{4}}\left(\sin\frac{\pi}{4} - \cos\frac{\pi}{4}\right) = e^{\frac{\pi}{4}}\left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) = 0 \)
\( f\left(\frac{5\pi}{4}\right) = e^{\frac{5\pi}{4}}\left(\sin\frac{5\pi}{4} - \cos\frac{5\pi}{4}\right) = e^{\frac{5\pi}{4}}\left(-\frac{1}{\sqrt{2}} - \left(-\frac{1}{\sqrt{2}}\right)\right) = 0 \)
Thus, \( f\left(\frac{\pi}{4}\right) = f\left(\frac{5\pi}{4}\right) = 0 \). All conditions of Rolle's Theorem are satisfied.
3. Now, find \( f'(x) \):
\( f'(x) = e^x(\sin x - \cos x) + e^x(\cos x + \sin x) = e^x(2\sin x) = 2e^x \sin x \)
According to Rolle's Theorem, there exists at least one \( c \in \left(\frac{\pi}{4}, \frac{5\pi}{4}\right) \) such that \( f'(c) = 0 \):
\( 2e^c \sin c = 0 \implies \sin c = 0 \) (since \( e^c \neq 0 \))
Within \( \left(\frac{\pi}{4}, \frac{5\pi}{4}\right) \), \( \sin c = 0 \) gives \( c = \pi \).
Since \( \pi \in \left(\frac{\pi}{4}, \frac{5\pi}{4}\right) \), Rolle's Theorem is verified.

Teacher's Note:
a) Check continuity and differentiability explicitly before applying Rolle's Theorem.
b) Simplify derivative using product rule and trigonometric cancellations.

(b) Find the equation of the parabola with latus rectum joining points \( (4, 6) \) and \( (4, -2) \) [5 Marks]

Answer:
The endpoints of the latus rectum are given as \( (4, 6) \) and \( (4, -2) \).
The length of the latus rectum is the distance between these two points:
\( \text{Length} = |6 - (-2)| = 8 \)
Since the x-coordinate of both endpoints is \( 4 \) and the length is \( 8 \), the axis of the parabola is parallel to the x-axis, and it opens to the left or right.
The midpoint of the latus rectum gives the focus of the parabola:
\( \text{Focus } S = \left(\frac{4+4}{2}, \frac{6+(-2)}{2}\right) = (4, 2) \)
The length of latus rectum \( 4a = 8 \implies a = 2 \).
Since the x-coordinate of the focus is greater than the x-coordinate of the vertex, the parabola opens to the left (standard form: \( (y - k)^2 = -4a(x - h) \)) or right.
Let us check orientation: Latus rectum endpoints are at \( x = 4 \), and the length is opening towards negative x-axis because the directrix would be to the right. Vertex \( V = (4 + 2, 2) = (6, 2) \) if opening left, or focus is to the left of vertex.
Let's use standard form for horizontal parabola with vertex \( (h, k) \) and focus \( (h+a, k) \) or \( (h-a, k) \):
Focus \( S(4, 2) \), length of latus rectum \( 4a = 8 \implies a = 2 \).
Since the x-coordinate of endpoints of latus rectum is 4 and they lie on \( x = h - a \):
\( h - a = 4 \implies h - 2 = 4 \implies h = 6 \).
The y-coordinate of vertex is the same as focus, so \( k = 2 \). Vertex \( V = (6, 2) \).
Since focus is at \( (4, 2) \) and vertex is at \( (6, 2) \), the parabola opens to the left.
Equation: \( (y - 2)^2 = -8(x - 6) \)
\( y^2 - 4y + 4 = -8x + 48 \)
\( y^2 + 8x - 4y - 44 = 0 \).

Teacher's Note:
a) The distance between the endpoints of the latus rectum gives \( 4a \).
b) Determine the vertex and direction of opening using the coordinates of the focus and latus rectum.

Question 5

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

Answer:
Given \( y = \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \implies y \sqrt{1 - x^2} = x \sin^{-1} x \)
Differentiating both sides with respect to \( x \):
\( y \cdot \frac{-2x}{2\sqrt{1 - x^2}} + \sqrt{1 - x^2} \frac{dy}{dx} = x \cdot \frac{1}{\sqrt{1 - x^2}} + \sin^{-1} x \cdot 1 \)
Multiply the entire equation by \( \sqrt{1 - x^2} \):
\( -xy + (1 - x^2)\frac{dy}{dx} = x + \sqrt{1 - x^2} \sin^{-1} x \)
From the original equation, \( \sin^{-1} x = \frac{y \sqrt{1 - x^2}}{x} \). Substitute this in the expression:\
\( \sqrt{1 - x^2} \sin^{-1} x = \sqrt{1 - x^2} \left( \frac{y \sqrt{1 - x^2}}{x} \right) = \frac{y(1 - x^2)}{x} \)
Substituting back:
\( -xy + (1 - x^2)\frac{dy}{dx} = x + \frac{y(1 - x^2)}{x} = \frac{x}{x} + \frac{y}{x} - yx \) (Wait, let's simplify carefully)
\( (1 - x^2)\frac{dy}{dx} - xy = x + \frac{y}{x}(1 - x^2) = x + \frac{y}{x} - yx \)
Adding \( xy \) to both sides:
\( (1 - x^2)\frac{dy}{dx} = x + \frac{y}{x} \). Hence proved.

Teacher's Note:
a) Cross-multiply the denominator to simplify differentiation using the product rule.
b) Substitute original terms back cleverly to match the required target expression.

(b) A wire of length \( 50 \) m is cut into two pieces. One piece of the wire is bent in the shape of a square and the other in the shape of a circle. What should be the length of each piece so that the combined area of the two is minimum? [5 Marks]

Answer:
Let the length of the wire cut for the square be \( x \) m. Then the length of the wire for the circle is \( 50 - x \) m.
Side of the square \( s = \frac{x}{4} \). Area of the square \( A_1 = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16} \).
Circumference of the circle \( 2\pi r = 50 - x \implies r = \frac{50 - x}{2\pi} \). Area of the circle \( A_2 = \pi r^2 = \pi \left(\frac{50 - x}{2\pi}\right)^2 = \frac{(50 - x)^2}{4\pi} \).
Total combined area \( A = A_1 + A_2 = \frac{x^2}{16} + \frac{(50 - x)^2}{4\pi} \)
Differentiating with respect to \( x \):
\( \frac{dA}{dx} = \frac{2x}{16} + \frac{2(50 - x)(-1)}{4\pi} = \frac{x}{8} - \frac{50 - x}{2\pi} \)
For critical points, set \( \frac{dA}{dx} = 0 \):
\( \frac{x}{8} = \frac{50 - x}{2\pi} \implies \frac{x}{4} = \frac{50 - x}{\pi} \implies \pi x = 200 - 4x \implies x(\pi + 4) = 200 \implies x = \frac{200}{\pi + 4} \)
Second derivative test: \( \frac{d^2A}{dx^2} = \frac{1}{8} - \left(-\frac{1}{2\pi}\right) = \frac{1}{8} + \frac{1}{2\pi} \gt 0 \).
Since \( \frac{d^2A}{dx^2} \gt 0 \), the area is minimum when \( x = \frac{200}{\pi + 4} \) m.
Length of wire for square \( = \frac{200}{\pi + 4} \) m, and length of wire for circle \( = 50 - \frac{200}{\pi + 4} = \frac{50\pi}{\pi + 4} \) m.

Teacher's Note:
a) Set up the area function correctly in terms of a single variable \( x \).
b) Use the second derivative test to confirm that the critical point yields a minimum area.

Question 6

(a) Evaluate: \( \int \frac{x + \sin x}{1 + \cos x} dx \) [5 Marks]

Answer:
\( \int \frac{x + \sin x}{1 + \cos x} dx = \int \frac{x + 2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}} dx \)
\( = \int \left( \frac{x}{2\cos^2\frac{x}{2}} + \frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}} \right) dx \)
\( = \int \left( \frac{1}{2} x \sec^2\frac{x}{2} + \tan\frac{x}{2} \right) dx \)
Let us integrate by parts for \( \int \frac{1}{2} x \sec^2\frac{x}{2} dx \):
\( = \frac{1}{2} x \left( 2 \tan\frac{x}{2} \right) - \int 1 \cdot \left( 2 \tan\frac{x}{2} \right) \cdot \frac{1}{2} dx \)
\( = x \tan\frac{x}{2} - \int \tan\frac{x}{2} dx \)
Adding the remaining term \( \int \tan\frac{x}{2} dx \) from the expansion:
\( \int \frac{x + \sin x}{1 + \cos x} dx = x \tan\frac{x}{2} - \int \tan\frac{x}{2} dx + \int \tan\frac{x}{2} dx = x \tan\frac{x}{2} + C \).

Teacher's Note:
a) Use half-angle formulas \( \sin x = 2\sin\frac{x}{2}\cos\frac{x}{2} \) and \( 1 + \cos x = 2\cos^2\frac{x}{2} \).
b) The integration by parts cancels out the tangent term neatly.

(b) Sketch the graphs of the curves \( y^2 = x \) and \( y^2 = 4 - 3x \) and find the area enclosed between them. [5 Marks]

Answer:
To find the points of intersection of \( y^2 = x \) and \( y^2 = 4 - 3x \):
\( x = 4 - 3x \implies 4x = 4 \implies x = 1 \)
When \( x = 1 \), \( y^2 = 1 \implies y = \pm 1 \).
Intersection points are \( (1, 1) \) and \( (1, -1) \).
[Figure: Symmetric parabolas \( y^2 = x \) opening to the right and \( y^2 = 4 - 3x \) opening to the left intersecting at \( (1, 1) \) and \( (1, -1) \).]
The required area is symmetric about the x-axis, so we can calculate twice the area from \( y = 0 \) to \( y = 1 \), or integrate with respect to \( y \):
From \( y^2 = x \), we have \( x_1 = y^2 \). From \( y^2 = 4 - 3x \), we have \( 3x = 4 - y^2 \implies x_2 = \frac{4 - y^2}{3} \).
\( \text{Area} = \int_{-1}^{1} (x_2 - x_1) dy = 2 \int_{0}^{1} \left( \frac{4 - y^2}{3} - y^2 \right) dy \)
\( = 2 \int_{0}^{1} \left( \frac{4}{3} - \frac{4}{3}y^2 \right) dy = \frac{8}{3} \int_{0}^{1} (1 - y^2) dy \)
\( = \frac{8}{3} \left[ y - \frac{y^3}{3} \right]_{0}^{1} = \frac{8}{3} \left( 1 - \frac{1}{3} \right) = \frac{8}{3} \left(\frac{2}{3}\right) = \frac{16}{9} \text{ sq. units} \).

Teacher's Note:
a) Integrating with respect to \( y \) is much easier here since both curves are expressed as functions of \( y \).
b) Use symmetry across the x-axis to simplify the definite integral limits.

Question 7

(a) A psychologist selected a random sample of 22 students. He grouped them in 11 pairs so that the students in each pair have nearly equal scores in an intelligence test. In each pair, one student was taught by method A and the other by method B and examined after the course. The marks obtained by them after the course are as follows:

Calculate Spearman's Rank correlation. [5 Marks]

Answer:

Number of pairs \( n = 11 \).
Note duplicate ranks: In A, 30 appears twice (ranks 1,2 avg 1.5 - wait, ranks 1,2 avg 1.5; let's rank descending properly: 30(1,2), 29(3), 28(4), 27(5), 24(6), 20(7), 19(8,9), 14(10), 11(11)). Let's use standard formula with correction factor if needed, or standard \( r_s = 1 - \frac{6 \sum d^2}{n(n^2 - 1)} \).
Using \( \sum d^2 = 269.5 \):
\( r_s = 1 - \frac{6 \times 269.5}{11(11^2 - 1)} = 1 - \frac{1617}{11 \times 120} = 1 - \frac{1617}{1320} = 1 - 1.225 = -0.225 \).

Teacher's Note:
a) Assign ranks carefully in descending order of marks, taking averages for tied values.
b) Apply Spearman's rank correlation formula correctly with \( n = 11 \).

(b) The coefficient of correlation between the values denoted by X and Y is 0.5. The mean of X is 3 and that of Y is 5. Their standard deviations are 5 and 4 respectively. Find:
(i) the two lines of regression
(ii) the expected value of Y, when X is given 14
(iii) the expected value of X, when Y is given 9. [5 Marks]

Answer:
Given: \( \bar{x} = 3, \, \bar{y} = 5, \, \sigma_x = 5, \, \sigma_y = 4, \, r = 0.5 \).
(i) Regression coefficient of \( y \) on \( x \):
\( b_{yx} = r \cdot \frac{\sigma_y}{\sigma_x} = 0.5 \cdot \frac{4}{5} = 0.5 \cdot 0.8 = 0.4 \)
Regression equation of \( y \) on \( x \):
\( y - \bar{y} = b_{yx}(x - \bar{x}) \implies y - 5 = 0.4(x - 3) \implies y = 0.4x - 1.2 + 5 \implies y = 0.4x + 3.8 \)
Regression coefficient of \( x \) on \( y \):
\( b_{xy} = r \cdot \frac{\sigma_x}{\sigma_y} = 0.5 \cdot \frac{5}{4} = 0.5 \cdot 1.25 = 0.625 \)
Regression equation of \( x \) on \( y \):
\( x - \bar{x} = b_{xy}(y - \bar{y}) \implies x - 3 = 0.625(y - 5) \implies x = 0.625y - 3.125 + 3 \implies x = 0.625y - 0.125 \)
(ii) Expected value of \( Y \) when \( X = 14 \):
\( y = 0.4(14) + 3.8 = 5.6 + 3.8 = 9.4 \)
(iii) Expected value of \( X \) when \( Y = 9 \):
\( x = 0.625(9) - 0.125 = 5.625 - 0.125 = 5.5 \).

Teacher's Note:
a) Use formulas \( b_{yx} = r(\sigma_y / \sigma_x) \) and \( b_{xy} = r(\sigma_x / \sigma_y) \) to find regression coefficients.
b) Substitute given values directly into the derived regression lines to find expected values.

Question 8

(a) In a college, 70% students pass in Physics, 75% pass in Mathematics and 10% students fail in both. One student is chosen at random. What is the probability that:
(i) He passes in Physics and Mathematics.
(ii) He passes in Mathematics given that he passes in Physics.
(iii) He passes in Physics given that he passes in Mathematics. [5 Marks]

Answer:
Let \( P \) be the event of passing in Physics and \( M \) be the event of passing in Mathematics.
Given: \( P(P) = 0.70 \), \( P(M) = 0.75 \), and \( P(P' \cap M') = 0.10 \).
Probability of passing in at least one subject: \( P(P \cup M) = 1 - P(P' \cap M') = 1 - 0.10 = 0.90 \).
We know that \( P(P \cup M) = P(P) + P(M) - P(P \cap M) \):
\( 0.90 = 0.70 + 0.75 - P(P \cap M) \)
\( 0.90 = 1.45 - P(P \cap M) \implies P(P \cap M) = 1.45 - 0.90 = 0.55 \).
(i) Probability that he passes in both Physics and Mathematics: \( P(P \cap M) = 0.55 \).
(ii) Probability that he passes in Mathematics given that he passes in Physics:
\( P(M | P) = \frac{P(P \cap M)}{P(P)} = \frac{0.55}{0.70} = \frac{55}{70} = \frac{11}{14} \approx 0.7857 \)
(iii) Probability that he passes in Physics given that he passes in Mathematics:
\( P(P | M) = \frac{P(P \cap M)}{P(M)} = \frac{0.55}{0.75} = \frac{55}{75} = \frac{11}{15} \approx 0.7333 \).

Teacher's Note:
a) Use De Morgan's Law or the union formula to find the intersection probability from the probability of failing both.
b) Apply conditional probability formula \( P(A|B) = P(A \cap B) / P(B) \) accurately.

(b) A bag contains 5 white and 4 black balls and another bag contains 7 white and 9 black balls. A ball is drawn from the first bag and two balls drawn from the second bag. What is the probability of drawing one white and two black balls? [5 Marks]

Answer:
Bag 1 contains 5 white and 4 black balls (Total 9).
Bag 2 contains 7 white and 9 black balls (Total 16).
We need to draw 1 white and 2 black balls in total. This can happen in two mutually exclusive cases:
Case 1: White ball drawn from Bag 1 (probability \( \frac{5}{9} \)) AND 2 black balls drawn from Bag 2 (probability \( \frac{\binom{9}{2}}{\binom{16}{2}} \)).
\( P(\text{Case 1}) = \frac{5}{9} \times \frac{\frac{9 \times 8}{2}}{\frac{16 \times 15}{2}} = \frac{5}{9} \times \frac{36}{120} = \frac{5}{9} \times \frac{3}{10} = \frac{15}{90} = \frac{1}{6} \)
Case 2: Black ball drawn from Bag 1 (probability \( \frac{4}{9} \)) AND 1 white and 1 black ball drawn from Bag 2 (probability \( \frac{\binom{7}{1} \times \binom{9}{1}}{\binom{16}{2}} \)).
\( P(\text{Case 2}) = \frac{4}{9} \times \frac{7 \times 9}{120} = \frac{4}{9} \times \frac{63}{120} = \frac{252}{1080} = \frac{7}{30} \)
Total Probability \( = P(\text{Case 1}) + P(\text{Case 2}) = \frac{1}{6} + \frac{7}{30} = \frac{5 + 7}{30} = \frac{12}{30} = \frac{2}{5} \).

Teacher's Note:
a) Identify all possible mutually exclusive combinations of draws from both bags.
b) Use combinations \( \binom{n}{r} \) to calculate probabilities for selections from the second bag.

Question 9

(a) Using De Moivre's theorem, find the least positive integer \( n \) such that \( \left(\frac{2i}{1 + i}\right)^n \) is a positive integer. [5 Marks]

Answer:
Consider the complex number \( z = \frac{2i}{1 + i} \).
Rationalize the denominator:
\( z = \frac{2i(1 - i)}{(1 + i)(1 - i)} = \frac{2i - 2i^2}{1 - i^2} = \frac{2 + 2i}{2} = 1 + i \)
Convert \( 1 + i \) to polar form:
Modulus \( r = \sqrt{1^2 + 1^2} = \sqrt{2} \)
Argument \( \theta = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4} \)
So, \( 1 + i = \sqrt{2} \left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right) \)
Therefore, \( \left(\frac{2i}{1 + i}\right)^n = (\sqrt{2})^n \left(\cos\frac{n\pi}{4} + i\sin\frac{n\pi}{4}\right) \)
For this to be a positive integer:
1. The imaginary part must be zero: \( \sin\frac{n\pi}{4} = 0 \implies \frac{n\pi}{4} = k\pi \implies n = 4k \) for integer \( k \).
2. The real part must be positive: \( (\sqrt{2})^n \cos(k\pi) \gt 0 \implies (-1)^k \gt 0 \implies k \) must be an even integer (or positive even integer).
For the least positive integer \( n \), let \( k = 2 \), which gives \( n = 4(2) = 8 \).
(Check for \( n = 4 \): \( (\sqrt{2})^4 \cos(\pi) = 4(-1) = -4 \) (negative integer). For \( n = 8 \): \( (\sqrt{2})^8 \cos(2\pi) = 16(1) = 16 \) (positive integer)).
Thus, the least positive integer \( n \) is \( 8 \).

Teacher's Note:
a) Simplify the base expression first before applying De Moivre's theorem.
b) Check both conditions for the result to be a positive integer: imaginary part equals zero and real part is strictly positive.

(b) Solve the following differential equation: \( (3xy + y^2)dx + (x^2 + xy)dy = 0 \) [5 Marks]

Answer:
Given equation: \( (3xy + y^2)dx + (x^2 + xy)dy = 0 \)
\( \frac{dy}{dx} = -\frac{3xy + y^2}{x^2 + xy} \)
This is a homogeneous differential equation. Put \( y = vx \), so \( \frac{dy}{dx} = v + x\frac{dv}{dx} \):
\( v + x\frac{dv}{dx} = -\frac{3x(vx) + (vx)^2}{x^2 + x(vx)} = -\frac{3vx^2 + v^2x^2}{x^2 + v x^2} = -\frac{3v + v^2}{1 + v} \)
\( x\frac{dv}{dx} = -\frac{3v + v^2}{1 + v} - v = \frac{-3v - v^2 - v - v^2}{1 + v} = \frac{-4v - 2v^2}{1 + v} = -\frac{2v(2 + v)}{1 + v} \)
Separating variables:
\( \frac{1 + v}{v(2 + v)} dv = -2 \frac{dx}{x} \)
Using partial fractions on LHS: \( \frac{1 + v}{v(2 + v)} = \frac{1/2}{v} + \frac{1/2}{2 + v} \)
\( \frac{1}{2} \left[ \frac{1}{v} + \frac{1}{2 + v} \right] dv = -2 \frac{dx}{x} \)
\( \left( \frac{1}{v} + \frac{1}{2 + v} \right) dv = -4 \frac{dx}{x} \)
Integrating both sides:
\( \ln|v| + \ln|2 + v| = -4\ln|x| + C \)
\( \ln|v(2 + v)| = \ln|x^{-4}| + \ln K \implies v(2 + v) = \frac{K}{x^4} \)
Substitute \( v = \frac{y}{x} \):
\( \frac{y}{x} \left(2 + \frac{y}{x}\right) = \frac{K}{x^4} \implies \frac{y(2x + y)}{x^2} = \frac{K}{x^4} \)
\( y(2x + y)x^2 = K \implies x^2 y(2x + y) = K \).

Teacher's Note:
a) Recognize the differential equation as homogeneous and substitute \( y = vx \).
b) Use partial fraction decomposition to integrate the resulting rational function in \( v \).

SECTION B

Question 10

(a) In a triangle ABC, using vectors, prove that \( c^2 = a^2 + b^2 - 2ab \cos C \) [5 Marks]

Answer:
Let \( \vec{BC} = \vec{a} \), \( \vec{CA} = \vec{b} \), and \( \vec{AB} = \vec{c} \) be the side vectors of triangle ABC.
By triangle law of addition:
\( \vec{a} + \vec{b} + \vec{c} = \vec{0} \implies \vec{a} + \vec{b} = -\vec{c} \)
Taking the dot product of \( (-\vec{c}) \) with itself:
\( c^2 = (-\vec{c}) \cdot (-\vec{c}) = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) \)
\( c^2 = \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + 2(\vec{a} \cdot \vec{b}) \)
\( c^2 = a^2 + b^2 + 2|\vec{a}||\vec{b}|\cos(\pi - C) \)
Since the angle between vectors \( \vec{a} \) and \( \vec{b} \) taken in order is \( \pi - C \), and \( \cos(\pi - C) = -\cos C \):
\( c^2 = a^2 + b^2 + 2ab(-\cos C) \)
\( c^2 = a^2 + b^2 - 2ab \cos C \). Hence proved.

Teacher's Note:
a) Use vector addition in a closed triangle: \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \).
b) Pay careful attention to the angle between vectors when expanding the dot product.

(b) Prove that: \( \vec{a} \cdot (\vec{b} + \vec{c}) \times (\vec{a} + 2\vec{b} + 3\vec{c}) = [\vec{a} \vec{b} \vec{c}] \) [5 Marks]

Answer:
Consider the expression \( \vec{a} \cdot [(\vec{b} + \vec{c}) \times (\vec{a} + 2\vec{b} + 3\vec{c})] \).
First, expand the cross product:kon
\( (\vec{b} + \vec{c}) \times (\vec{a} + 2\vec{b} + 3\vec{c}) = (\vec{b} \times \vec{a}) + 2(\vec{b} \times \vec{b}) + 3(\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a}) + 2(\vec{c} \times \vec{b}) + 3(\vec{c} \times \vec{c}) \)
Since \( \vec{b} \times \vec{b} = \vec{0} \) and \( \vec{c} \times \vec{c} = \vec{0} \):
\( = (\vec{b} \times \vec{a}) + 3(\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a}) - 2(\vec{b} \times \vec{c}) \quad (\text{since } \vec{c} \times \vec{b} = -\vec{b} \times \vec{c}) \)
\( = (\vec{b} \times \vec{a}) + (\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a}) \)
Now, take the dot product with \( \vec{a} \):
\( \vec{a} \cdot [(\vec{b} \times \vec{a}) + (\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a})] \)
\( = \vec{a} \cdot (\vec{b} \times \vec{a}) + \vec{a} \cdot (\vec{b} \times \vec{c}) + \vec{a} \cdot (\vec{c} \times \vec{a}) \)
Since scalar triple product with repeating vectors is zero, \( \vec{a} \cdot (\vec{b} \times \vec{a}) = 0 \) and \( \vec{a} \cdot (\vec{c} \times \vec{a}) = 0 \):
\( = 0 + [\vec{a} \vec{b} \vec{c}] + 0 = [\vec{a} \vec{b} \vec{c}] \). Hence proved.

Teacher's Note:
a) Use properties of cross products such as \( \vec{a} \times \vec{a} = \vec{0} \) and \( \vec{b} \times \vec{c} = -\vec{c} \times \vec{b} \).
b) Scalar triple products with any repeated vector evaluate to zero.

Question 11

(a) Find the equation of a line passing through the points P (-1, 3, 2) and Q (-4, 2, -2). Also, if the point R (5, 5, \( \lambda \)) is collinear with the points P and Q, then find the value of \( \lambda \). [5 Marks]

Answer:
Equation of line passing through \( P(-1, 3, 2) \) and \( Q(-4, 2, -2) \):
Direction ratios are \( a = -4 - (-1) = -3 \), \( b = 2 - 3 = -1 \), \( c = -2 - 2 = -4 \).
Line equation in Cartesian form:
\( \frac{x + 1}{-3} = \frac{y - 3}{-1} = \frac{z - 2}{-4} \) (or with direction ratios \( 3, 1, 4 \): \( \frac{x + 1}{3} = \frac{y - 3}{1} = \frac{z - 2}{4} \))
Since point \( R(5, 5, \lambda) \) lies on this line, it must satisfy the equation:
\( \frac{5 + 1}{3} = \frac{5 - 3}{1} = \frac{\lambda - 2}{4} \)
\( \frac{6}{3} = \frac{2}{1} = \frac{\lambda - 2}{4} \)
\( 2 = 2 = \frac{\lambda - 2}{4} \)
\( \frac{\lambda - 2}{4} = 2 \implies \lambda - 2 = 8 \implies \lambda = 10 \).

Teacher's Note:
a) Find direction ratios by taking differences of coordinates between two points.
b) Substitute the coordinates of the collinear point into the line equation to solve for unknown parameters.

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

Answer:
Equation of any plane passing through \( (2, -3, 1) \) is:
\( A(x - 2) + B(y + 3) + C(z - 1) = 0 \) — (1)
Since it passes through \( (-1, 1, -7) \):
\( A(-1 - 2) + B(1 + 3) + C(-7 - 1) = 0 \)
\( -3A + 4B - 8C = 0 \implies 3A - 4B + 8C = 0 \) — (2)
Since the plane is perpendicular to the plane \( x - 2y + 5z + 1 = 0 \), the normal vectors are perpendicular:
\( A(1) + B(-2) + C(5) = 0 \implies A - 2B + 5C = 0 \) — (3)
Solving (2) and (3) using cross multiplication rule for \( A, B, C \):
\( \frac{A}{(-4)(5) - (-2)(8)} = \frac{B}{(8)(1) - (3)(5)} = \frac{C}{(3)(-2) - (-4)(1)} \)
\( \frac{A}{-20 + 16} = \frac{B}{8 - 15} = \frac{C}{-6 + 4} \)
\( \frac{A}{-4} = \frac{B}{-7} = \frac{C}{-2} \implies A : B : C = 4 : 7 : 2 \)
Substitute \( A = 4, B = 7, C = 2 \) into equation (1):
\( 4(x - 2) + 7(y + 3) + 2(z - 1) = 0 \)
\( 4x - 8 + 7y + 21 + 2z - 2 = 0 \)
\( 4x + 7y + 2z + 11 = 0 \).

Teacher's Note:
a) Use the general point equation for a plane and apply point substitution and perpendicularity conditions.
b) Solve the resulting system of linear equations for direction ratios \( A, B, C \).

Question 12

(a) In a bolt factory, three machines A, B and C manufacture 25%, 35% and 40% of the total production respectively. Of their respective outputs, 5%, 4% and 2% are defective. A bolt is drawn at random from the total production and it is found to be defective. Find the probability that it was manufactured by machine C. [5 Marks]

Answer:
Let \( E_1, E_2, E_3 \) be the events that the bolt is manufactured by machines A, B and C respectively.
\( P(E_1) = 0.25, \, P(E_2) = 0.35, \, P(E_3) = 0.40 \)
Let \( D \) be the event that the bolt is defective.
Given probabilities of defect given the machine:
\( P(D|E_1) = 0.05, \, P(D|E_2) = 0.04, \, P(D|E_3) = 0.02 \)
We need to find \( P(E_3|D) \) using Bayes' Theorem:
\( P(E_3|D) = \frac{P(E_3)P(D|E_3)}{P(E_1)P(D|E_1) + P(E_2)P(D|E_2) + P(E_3)P(D|E_3)} \)
Denominator calculation:
\( P(D) = (0.25 \times 0.05) + (0.35 \times 0.04) + (0.40 \times 0.02) \)
\( P(D) = 0.0125 + 0.0140 + 0.0080 = 0.0345 \)
Numerator calculation:
\( P(E_3)P(D|E_3) = 0.40 \times 0.02 = 0.0080 \)
Applying Bayes' Theorem:
\( P(E_3|D) = \frac{0.0080}{0.0345} = \frac{80}{345} = \frac{16}{69} \approx 0.2319 \).

Teacher's Note:
a) Identify this as a standard Bayes' Theorem problem with mutually exclusive and exhaustive machine hypotheses.
b) Compute total probability in the denominator by summing products of individual machine probabilities and defect rates.

(b) On dialling certain telephone numbers, a telephone number out of five is busy. Five telephone numbers are randomly selected and dialled. Find the probability that at least three of them will be busy. [5 Marks]

Answer:
This is a binomial distribution problem with \( n = 5 \).
Probability of success (number is busy) \( p = \frac{1}{5} = 0.2 \).
Probability of failure \( q = 1 - \frac{1}{5} = \frac{4}{5} = 0.8 \).
Probability of \( r \) successes is given by \( P(X = r) = \binom{n}{r} p^r q^{n-r} \).
We need to find the probability that at least three of them will be busy, i.e., \( P(X \ge 3) = P(X = 3) + P(X = 4) + P(X = 5) \):
\( P(X = 3) = \binom{5}{3} (0.2)^3 (0.8)^2 = 10 \times 0.008 \times 0.64 = 0.0512 \)
\( P(X = 4) = \binom{5}{4} (0.2)^4 (0.8)^1 = 5 \times 0.0016 \times 0.8 = 0.0064 \)
\( P(X = 5) = \binom{5}{5} (0.2)^5 (0.8)^0 = 1 \times 0.00032 \times 1 = 0.00032 \)
Total Probability \( P(X \ge 3) = 0.0512 + 0.0064 + 0.00032 = 0.05792 \).

Teacher's Note:
a) Recognize binomial distribution parameters \( n = 5, p = 1/5, q = 4/5 \).
b) Compute probabilities for \( r = 3, 4, 5 \) individually and sum them up.

SECTION C

Question 13

(a) A person borrows Rs. 68,962 on the condition that he repay the money with compound interest at 5% per annum in 4 equal annual instalments, the first one being payable at the end of the first year. Find the value of each instalment. [5 Marks]

Answer:
Let the value of each annual instalment be \( R \).
Principal amount \( P = \text{Rs. } 68,962 \), rate of interest \( i = 5\% = 0.05 \), number of instalments \( n = 4 \).
Using the present value formula for instalments with compound interest:
\( P = \frac{R}{1 + i} + \frac{R}{(1 + i)^2} + \frac{R}{(1 + i)^3} + \frac{R}{(1 + i)^4} \)
\( 68962 = R \left[ \frac{1}{1.05} + \frac{1}{(1.05)^2} + \frac{1}{(1.05)^3} + \frac{1}{(1.05)^4} \right] \)
Let \( v = \frac{1}{1.05} \approx 0.95238 \):
\( v^1 = 0.95238 \)
\( v^2 = 0.90703 \)
\( v^3 = 0.86384 \)
\( v^4 = 0.82270 \)
Sum of discount factors \( = 0.95238 + 0.90703 + 0.86384 + 0.82270 = 3.54595 \)
\( 68962 = R \times 3.54595 \)
\( R = \frac{68962}{3.54595} \approx \text{Rs. } 19,451.05 \).

Teacher's Note:
a) Use the present value of annuity formula to equate the borrowed sum to the sum of discounted instalment values.
b) Compute powers of discount factor \( (1 + i)^{-t} \) carefully to avoid rounding errors.

(b) A company manufactures two types of toys A and B. A toy of type A requires 5 minutes for cutting and 10 minutes for assembling. A toy of type B requires 8 minutes for cutting and 8 minutes for assembling. There are 3 hours available for cutting and 4 hours available for assembling the toys in a day. The profit is Rs. 50 each on a toy of Type A and Rs. 60 each on a toy of type B. How many of each type should the company manufacture in a day to maximize the profit? Use linear programming to find the solution. [5 Marks]

Answer:
Let \( x \) be the number of toys of Type A and \( y \) be the number of toys of Type B manufactured per day.
Objective: Maximize Profit \( Z = 50x + 60y \)
Subject to constraints:
1. Cutting time constraint (3 hours \( = 180 \) minutes):
\( 5x + 8y \le 180 \)
2. Assembling time constraint (4 hours \( = 240 \) minutes):
\( 10x + 8y \le 240 \implies 5x + 4y \le 120 \)
3. Non-negative constraints:
\( x \ge 0, \, y \ge 0 \)
Finding corner points of the feasible region:
- Intersection of \( 5x + 8y = 180 \) and \( 5x + 4y = 120 \):
Subtracting equations: \( 4y = 60 \implies y = 15 \).
Substitute \( y = 15 \) into \( 5x + 4(15) = 120 \implies 5x + 60 = 120 \implies 5x = 60 \implies x = 12 \).
Corner point: \( (12, 15) \).
- Other corner points: \( (0, 0), \, (24, 0) \) [from \( 5x + 4(0) = 120 \)], and \( (0, 22.5) \) [from \( 5(0) + 8y = 180 \to y = 22.5 \)].
Evaluating \( Z = 50x + 60y \) at corner points:
- At \( (0, 0) \): \( Z = 0 \)
- At \( (24, 0) \): \( Z = 50(24) + 60(0) = 1200 \)
- At \( (0, 22.5) \): \( Z = 50(0) + 60(22.5) = 1350 \)
- At \( (12, 15) \): \( Z = 50(12) + 60(15) = 600 + 900 = 1500 \)
Maximum profit is Rs. 1,500 achieved by manufacturing 12 toys of Type A and 15 toys of Type B.

Teacher's Note:
a) Convert all time units to minutes consistently before setting up inequalities.
b) Evaluate the objective function at all feasible corner points to find the absolute maximum.

Question 14

(a) A firm has the cost function \( C = \frac{x^3}{3} - 7x^2 + 111x + 50 \) and demand function \( x = 100 - p \):
(i) Write the total revenue function in terms of \( x \).
(ii) Formulate the total profit function P in terms of \( x \).
(iii) Find the profit maximising level of output \( x \). [5 Marks]

Answer:
Given demand function \( x = 100 - p \implies p = 100 - x \).
(i) Total Revenue \( R = p \cdot x = (100 - x)x = 100x - x^2 \).
(ii) Total Profit function \( P(x) = R - C \):
\( P(x) = (100x - x^2) - \left(\frac{x^3}{3} - 7x^2 + 111x + 50\right) \)
\( P(x) = -\frac{x^3}{3} + 6x^2 - 11x - 50 \)
(iii) To find the profit maximizing level of output, differentiate \( P(x) \) with respect to \( x \) and set to zero:
\( P'(x) = -x^2 + 12x - 11 = 0 \)
\( x^2 - 12x + 11 = 0 \)
\( (x - 1)(x - 11) = 0 \implies x = 1 \) or \( x = 11 \)
Second derivative test: \( P''(x) = -2x + 12 \)
- At \( x = 1 \): \( P''(1) = -2(1) + 12 = 10 \gt 0 \) (minimum profit)
- At \( x = 11 \): \( P''(11) = -2(11) + 12 = -10 \lt 0 \) (maximum profit)
Therefore, the profit maximizing level of output is \( x = 11 \).

Teacher's Note:
a) Revenue is price multiplied by quantity (\( R = px \)), and profit is revenue minus cost (\( P = R - C \)).
b) Use the second derivative test to confirm that the critical point corresponds to a maximum profit.

(b) A bill of Rs. 5050 is drawn on \( 13^\text{th} \) April 2013. It was discounted on \( 4^\text{th} \) July 2013 at 5% per annum. If the banker's gain on the transaction is Rs. 0.50, find the nominal date of the maturity of the bill. [5 Marks]

Answer:
Face Value \( FV = \text{Rs. } 5050 \), Rate of interest \( r = 5\% = 0.05 \), Banker's Gain \( BG = \text{Rs. } 0.50 \).
We know the relation \( BG = \frac{TD^2}{PW} \) or \( BG = \frac{(TD)^2}{FV} \) and Banker's Discount \( BD = TD + BG \).
Also, \( BG = \frac{TD \cdot r \cdot t}{1} \) where \( TD \) is true discount. More directly, \( BG = \frac{BD^2}{FV} \) or use formula: \( BG = \frac{FV \cdot r^2 \cdot t^2}{1 + r \cdot t} \) or via interest on true discount:
\( BG = TD \cdot r \cdot t \), and \( BD = TD(1 + rt) \).
Let time in years be \( t \).
Banker's Discount \( BD = \frac{FV \cdot r \cdot t}{1 + r \cdot t} \) — wait, simpler formula for Banker's Gain: \( BG = \frac{BD \cdot r \cdot t}{1 + r \cdot t} \) or \( BG = \text{Interest on True Discount} \).
Let's use \( BG = \frac{TD^2}{PW} \). Alternatively, \( BD - TD = BG \).
We know \( BD = \frac{FV \cdot r \cdot t}{1 + r \cdot t} \) and \( TD = \frac{FV \cdot r \cdot t}{1} \) (Wait, \( TD = P \cdot r \cdot t \), where \( FV = PW + TD \)).
Let us use the standard relation: \( BD = TD + BG \), and \( TD = \frac{BG}{r \cdot t} \).
Let's determine time \( t \) from \( BG = \frac{(BD)^2}{FV} \) or directly calculate period from days.
Let unexpired time be \( t \) years. Discounted date is \( 4^\text{th} \) July 2013.
Number of days from \( 4^\text{th} \) July to maturity: July has 31 days (remaining 27 days), August 31, September 30, October 31, etc.
Using formula \( BG = \frac{TD \cdot r \cdot t}{1} \)... Let's use the standard textbook approach:
\( BG = \frac{FV \cdot r^2 \cdot t^2}{1 + r \cdot t} \implies 0.50 = \frac{5050 \cdot (0.05)^2 \cdot t^2}{1 + 0.05 t} \)
\( 0.50 = \frac{5050 \times 0.0025 \cdot t^2}{1 + 0.05t} = \frac{12.625 t^2}{1 + 0.05t} \)
\( 0.50(1 + 0.05t) = 12.625 t^2 \)
\( 0.50 + 0.025t = 12.625 t^2 \implies 12.625 t^2 - 0.025t - 0.50 = 0 \)
Multiply by 1000 and divide by 25 to simplify: approximately \( t = \frac{1}{5} \) year \( = \frac{365}{5} = 73 \) days.
Thus, the unexpired time is 73 days from \( 4^\text{th} \) July 2013.
Adding 73 days to \( 4^\text{th} \) July 2013:
- July remaining: \( 31 - 4 = 27 \) days
- August: 31 days
- September: \( 73 - (27 + 31) = 73 - 58 = 15 \) days.
So the legal date of maturity is \( 15^\text{th} \) September 2013.
Subtracting 3 days of grace, the nominal date of maturity is \( 12^\text{th} \) September 2013.

Teacher's Note:
a) Use the standard financial formula connecting Banker's Gain, Face Value, rate of interest, and time: \( BG = \frac{FV \cdot r^2 \cdot t^2}{1 + r \cdot t} \).
b) Add the calculated days to the discounting date and remember to subtract 3 days of grace to find the nominal maturity date.

Question 15

(a) The price of six different commodities for years 2009 and 2011 are as follows:

The Index number for the year 2011 taking 2009 as the base year for the above data was calculated to be 125. Find the values of \( x \) and \( y \) if the total price in 2009 is Rs. 360. [5 Marks]

Answer:
Let base year prices be \( p_0 \) and current year prices be \( p_1 \).
Given total price in 2009 (base year) \( \sum p_0 = 360 \):
\( 35 + 80 + 25 + 30 + 80 + x = 360 \)
\( 250 + x = 360 \implies x = 110 \).
The simple aggregate index number formula is:
\( I_{01} = \frac{\sum p_1}{\sum p_0} \times 100 \)
Given Index number \( I_{01} = 125 \) and \( \sum p_0 = 360 \):
\( 125 = \frac{\sum p_1}{360} \times 100 \)
\( \sum p_1 = \frac{125 \times 360}{100} = 1.25 \times 360 = 450 \)
Sum of prices in 2011 \( \sum p_1 = 50 + y + 45 + 70 + 120 + 105 = 450 \)
\( 390 + y = 450 \implies y = 60 \).
Therefore, \( x = 110 \) and \( y = 60 \).

Teacher's Note:
a) Use the simple aggregate price index formula \( I = \frac{\sum p_1}{\sum p_0} \times 100 \).
b) Sum up the given base year and current year prices to form linear equations in \( x \) and \( y \).

(b) The number of road accidents in the city due to rash driving, over a period of 3 years, is given in the following table:

Calculate four quarterly moving averages and illustrate them and original figures on one graph using the same axes for both. [5 Marks]

Answer:

[Figure: Graph showing the original quarterly accident figures alongside the centered 4-quarter moving average trend line plotted against time quarters.]

Teacher's Note:
a) For quarterly data, compute 4-quarter moving totals and center them by taking 2-period moving averages.
b) Plot both the original time series and the smoothed trend values on the same graph for comparison.