ISC Class 12 Mathematics Board Exam Question Paper 2013 with Solutions

ISC Class 12 Mathematics Board Exam Question Paper with Solutions

 

SECTION A

 

Question 1 [3×10 Marks]
(i) If \((A - 2I)(A - 3I) = 0\), where \(A = \begin{pmatrix} 4 & 2 \\ -1 & x \end{pmatrix}\) and \(I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\), find the value of \(x\).
(ii) Find the value(s) of \(k\) so that the line \(2x + y + k = 0\) may touch the hyperbola \(3x^2 - y^2 = 3\).
(iii) Prove that: \(\tan^{-1} \frac{1}{4} + \tan^{-1} \frac{2}{9} = \frac{1}{2} \sin^{-1} \frac{4}{5}\)
(iv) Using L'Hospital's Rule, evaluate: \(\lim_{x \to 0} \left( \frac{e^x - e^{-x} - 2x}{x - \sin x} \right)\)
(v) Evaluate: \(\int \frac{1}{x + \sqrt{x}} \, dx\)
(vi) Evaluate: \(\int_{0}^{1} \log \left(\frac{1}{x} - 1\right) dx\)
(vii) Two regression lines are represented by \(4x + 10y = 9\) and \(6x + 3y = 4\). Find the line of regression of \(y\) on \(x\).
(viii) If \(1\), \(w\) and \(w^2\) are the cube roots of unity, evaluate \((1 - w^4 + w^8)(1 - w^8 + w^{16})\).
(ix) Solve the differential equation: \(\log \left(\frac{dy}{dx}\right) = 2x - 3y\)
(x) If two balls are drawn from a bag containing three red balls and four blue balls, find the probability that:
(a) They are of the same colour.
(b) They are of different colours.

Answer:
(i) Given \(A = \begin{pmatrix} 4 & 2 \\ -1 & x \end{pmatrix}\), \(I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)
\(A - 2I = \begin{pmatrix} 4 - 2 & 2 \\ -1 & x - 2 \end{pmatrix} = \begin{pmatrix} 2 & 2 \\ -1 & x - 2 \end{pmatrix}\)
\(A - 3I = \begin{pmatrix} 4 - 3 & 2 \\ -1 & x - 3 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ -1 & x - 3 \end{pmatrix}\)
\((A - 2I)(A - 3I) = \begin{pmatrix} 2 & 2 \\ -1 & x - 2 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ -1 & x - 3 \end{pmatrix} = \begin{pmatrix} 0 & 2x - 2 \\ -x + 1 & x^2 - 5x + 4 \end{pmatrix}\)
Since \((A - 2I)(A - 3I) = 0\), equating the corresponding elements gives \(2x - 2 = 0 \implies x = 1\).

Teacher's Note:
a) Ensure matrix multiplication is performed row-by-column carefully before equating to the zero matrix.
b) Verify that the obtained value of \(x\) satisfies all elements of the resulting matrix equation simultaneously.

Answer:
(ii) The given equation of the hyperbola is \(3x^2 - y^2 = 3\), which can be written in standard form as \(\frac{x^2}{1} - \frac{y^2}{3} = 1\).
Here, \(a^2 = 1\) and \(b^2 = 3\).
The given line is \(2x + y + k = 0\), which can be written as \(y = -2x - k\). Here slope \(m = -2\) and intercept \(c = -k\).
The condition for a line \(y = mx + c\) to be a tangent to the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) is \(c^2 = a^2m^2 - b^2\).
Substituting the values: \((-k)^2 = (1)(-2)^2 - 3 \implies k^2 = 4 - 3 = 1 \implies k = \pm 1\).

Teacher's Note:
a) Remind students to first reduce the conic equation to standard form to correctly identify \(a^2\) and \(b^2\).
b) Watch out for sign errors when applying the condition of tangency for hyperbolas versus ellipses.

Answer:
(iii) \(\text{LHS} = \tan^{-1} \frac{1}{4} + \tan^{-1} \frac{2}{9} = \tan^{-1} \left( \frac{\frac{1}{4} + \frac{2}{9}}{1 - \frac{1}{4} \cdot \frac{2}{9}} \right) = \tan^{-1} \left( \frac{\frac{9 + 8}{36}}{\frac{36 - 2}{36}} \right) = \tan^{-1} \left( \frac{17}{34} \right) = \tan^{-1} \left( \frac{1}{2} \right)\)
Let \(\theta = \tan^{-1} \left( \frac{1}{2} \right) \implies \tan \theta = \frac{1}{2}\).
We know that \(\sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta} = \frac{2 \left(\frac{1}{2}\right)}{1 + \left(\frac{1}{2}\right)^2} = \frac{1}{1 + \frac{1}{4}} = \frac{4}{5}\).
Therefore, \(2\theta = \sin^{-1} \left(\frac{4}{5}\right) \implies \theta = \frac{1}{2} \sin^{-1} \left(\frac{4}{5}\right) = \text{RHS}\).

Teacher's Note:
a) Standard inverse trigonometric identities for addition of tangents must be applied accurately.
b) Converting \(\tan^{-1}\) to \(\sin^{-1}\) using double angle formulae is a standard technique that requires clear step-by-step substitution.

Answer:
(iv) \(\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{x - \sin x}\) is of the indeterminate form \(\left[\frac{0}{0}\right]\).
Applying L'Hospital's Rule by differentiating numerator and denominator with respect to \(x\):
\(= \lim_{x \to 0} \frac{e^x + e^{-x} - 2}{1 - \cos x}\), which is again of form \(\left[\frac{0}{0}\right]\).
Applying L'Hospital's Rule again:
\(= \lim_{x \to 0} \frac{e^x - e^{-x}}{\sin x}\), which is again of form \(\left[\frac{0}{0}\right]\).
Applying L'Hospital's Rule a third time:
\(= \lim_{x \to 0} \frac{e^x + e^{-x}}{\cos x} = \frac{1 + 1}{1} = 2\).

Teacher's Note:
a) Students should check for indeterminate forms at each stage before reapplying L'Hospital's Rule.
b) Be careful with the differentiation of exponential terms, especially the chain rule sign changes for \(e^{-x}\).

Answer:
(v) Let \(I = \int \frac{1}{x + \sqrt{x}} dx = \int \frac{1}{\sqrt{x}(\sqrt{x} + 1)} dx\)
Put \(\sqrt{x} + 1 = t \implies \frac{1}{2\sqrt{x}} dx = dt \implies \frac{1}{\sqrt{x}} dx = 2 dt\).
\(I = \int \frac{2}{t} dt = 2 \log |t| + c = 2 \log |\sqrt{x} + 1| + c\).

Teacher's Note:
a) Substitution method simplifies irrational algebraic integrands effectively.
b) Always remember to substitute back the original variable and include the constant of integration \(c\).

Answer:
(vi) Let \(I = \int_{0}^{1} \log \left(\frac{1}{x} - 1\right) dx = \int_{0}^{1} \log \left(\frac{1 - x}{x}\right) dx\)  --- (1)
Using the property \(\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx\):
\(I = \int_{0}^{1} \log \left(\frac{1 - (1 - x)}{1 - x}\right) dx = \int_{0}^{1} \log \left(\frac{x}{1 - x}\right) dx\)  --- (2)
Adding (1) and (2):
\(2I = \int_{0}^{1} \left[ \log \left(\frac{1 - x}{x}\right) + \log \left(\frac{x}{1 - x}\right) \right] dx = \int_{0}^{1} \log(1) dx = 0 \implies I = 0\).

Teacher's Note:
a) Definite integral properties such as \(\int_{a}^{b} f(x)dx = \int_{a}^{b} f(a+b-x)dx\) often simplify seemingly complex logarithmic integrals instantly.
b) Adding the original integral and its transformed version leverages logarithm addition properties (\(\log a + \log b = \log ab\)).

Answer:
(vii) Let the first equation be \(4x + 10y = 9 \implies 10y = -4x + 9 \implies y = -\frac{2}{5}x + \frac{9}{10}\). Thus, \(b_{yx} = -\frac{2}{5}\).
Let the second equation be \(6x + 3y = 4 \implies 3x = -3y + 4 \implies x = -y + \frac{4}{3}\). Thus, \(b_{xy} = -1\).
Now, \(r^2 = b_{yx} \cdot b_{xy} = \left(-\frac{2}{5}\right)(-1) = \frac{2}{5} = 0.4 < 1\), which is valid.
Since the product of regression coefficients is positive and less than 1, our assumption is correct. Hence, the line of regression of \(y\) on \(x\) is \(4x + 10y = 9\).

Teacher's Note:
a) To identify which equation is which, test the magnitude of regression coefficients or compute \(r^2 < 1\).
b) The line of regression of \(y\) on \(x\) is expressed in the form \(y = a_1 x + b_1\).

Answer:
(viii) Given that \(1, w, w^2\) are cube roots of unity, we know \(w^3 = 1\) and \(1 + w + w^2 = 0\).
Expression = \((1 - w^4 + w^8)(1 - w^8 + w^{16})\)
Since \(w^4 = w\), \(w^8 = w^2\), \(w^{16} = w\):
\(= (1 - w + w^2)(1 - w^2 + w)\)
Since \(1 + w^2 = -w\), the first term is \((-w - w) = -2w\).
Since \(1 + w = -w^2\), the second term is \((-w^2 - w^2) = -2w^2\).
Product \(= (-2w)(-2w^2) = 4w^3 = 4(1) = 4\).

Teacher's Note:
a) Always reduce higher powers of \(w\) using \(w^3 = 1\) before substitution.
b) Make effective use of the fundamental relation \(1 + w + w^2 = 0\).

Answer:
(ix) Given \(\log \left(\frac{dy}{dx}\right) = 2x - 3y\)
\(\frac{dy}{dx} = e^{2x - 3y} = e^{2x} \cdot e^{-3y}\)
Separating variables:
\(e^{3y} dy = e^{2x} dx\)Integrating both sides:
\(\int e^{3y} dy = \int e^{2x} dx\)
\(\frac{e^{3y}}{3} = \frac{e^{2x}}{2} + c \implies 2e^{3y} - 3e^{2x} = k\) (where \(k = 6c\)).

Teacher's Note:
a) Convert logarithmic differential equations into exponential form first before attempting variable separation.
b) Ensure constants of integration are handled neatly when multiplying across equations.

Answer:
(x) Total balls = 3 red + 4 blue = 7 balls.
Total number of ways to draw 2 balls out of 7 is \({^7}C_2 = \frac{7 \times 6}{2} = 21\).
(a) Probability of same colour = P(both red) + P(both blue)
\(= \frac{{^3}C_2}{{^7}C_2} + \frac{{^4}C_2}{{^7}C_2} = \frac{3}{21} + \frac{6}{21} = \frac{9}{21} = \frac{3}{7}\).
(b) Probability of different colours = 1 - P(same colour) = \(1 - \frac{3}{7} = \frac{4}{7}\) (or \(\frac{{^3}C_1 \times {^4}C_1}{{^7}C_2} = \frac{3 \times 4}{21} = \frac{12}{21} = \frac{4}{7}\)).

Teacher's Note:
a) Mutually exclusive cases for "same colour" must be added up correctly using combinations.
b) Complementary probability can provide a quick cross-check for the "different colours" case.

 

Question 2 [5 Marks]
(a) Using properties of determinants, prove that:
\begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ y+z & z+x & x+y \end{vmatrix} = (x - y)(y - z)(z - x)(x + y + z)

Answer:
Let \(\Delta = \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ y+z & z+x & x+y \end{vmatrix}\)
Applying \(R_3 \to R_3 + R_1\):
\(\Delta = \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x+y+z & x+y+z & x+y+z \end{vmatrix}\)
Taking \((x + y + z)\) common from \(R_3\):
\(\Delta = (x + y + z) \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ 1 & 1 & 1 \end{vmatrix}\)
Applying \(C_1 \to C_1 - C_2\) and \(C_2 \to C_2 - C_3\):
\(\Delta = (x + y + z) \begin{vmatrix} x - y & y - z & z \\ x^2 - y^2 & y^2 - z^2 & z^2 \\ 0 & 0 & 1 \end{vmatrix}\)
Taking \((x - y)\) from \(C_1\) and \((y - z)\) from \(C_2\):
\(\Delta = (x + y + z)(x - y)(y - z) \begin{vmatrix} 1 & 1 & z \\ x+y & y+z & z^2 \\ 0 & 0 & 1 \end{vmatrix}\)
Expanding along \(R_3\):
\(\Delta = (x + y + z)(x - y)(y - z)(1) \begin{vmatrix} 1 & 1 \\ x+y & y+z \end{vmatrix}\)
\(= (x + y + z)(x - y)(y - z) [ (y + z) - (x + y) ]\)
\(= (x + y + z)(x - y)(y - z)(z - x)\).

Teacher's Note:
a) Row and column operations should be chosen to create zeros and extract common linear factors progressively.
b) Verify the sign of each factor (especially cyclic ones like \(z - x\)) during extraction.

 

Question 2 [5 Marks]
(b) Find \(A^{-1}\), where \(A = \begin{pmatrix} 4 & 2 & 3 \\ 1 & 1 & 1 \\ 3 & 1 & -2 \end{pmatrix}\).
Hence, solve the following system of linear equations:
\(4x + 2y + 3z = 2\)
\(x + y + z = 1\)
\(3x + y - 2z = 5\)

Answer:
Determinant of \(A\):
\(|A| = 4(-2 - 1) - 2(-2 - 3) + 3(1 - 3) = 4(-3) - 2(-5) + 3(-2) = -12 + 10 - 6 = -8 \neq 0\).
Cofactors of elements of \(A\):
\(C_{11} = -3\), \(C_{12} = 5\), \(C_{13} = -2\)
\(C_{21} = 7\), \(C_{22} = -17\), \(C_{23} = 2\)
\(C_{31} = -1\), \(C_{32} = -1\), \(C_{33} = 2\)
\(\text{Adj } A = \begin{pmatrix} -3 & 7 & -1 \\ 5 & -17 & -1 \\ -2 & 2 & 2 \end{pmatrix}\)
\(A^{-1} = \frac{1}{|A|} \text{Adj } A = -\frac{1}{8} \begin{pmatrix} -3 & 7 & -1 \\ 5 & -17 & -1 \\ -2 & 2 & 2 \end{pmatrix} = \frac{1}{8} \begin{pmatrix} 3 & -7 & 1 \\ -5 & 17 & 1 \\ 2 & -2 & -2 \end{pmatrix}\)
The matrix equation is \(AX = B\), where \(X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\) and \(B = \begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix}\).
\(X = A^{-1}B = \frac{1}{8} \begin{pmatrix} 3 & -7 & 1 \\ -5 & 17 & 1 \\ 2 & -2 & -2 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix} = \frac{1}{8} \begin{pmatrix} 6 - 7 + 5 \\ -10 + 17 + 5 \\ 4 - 2 - 10 \end{pmatrix} = \frac{1}{8} \begin{pmatrix} 4 \\ 12 \\ -8 \end{pmatrix} = \begin{pmatrix} 1/2 \\ 3/2 \\ -1 \end{pmatrix}\).
Thus, \(x = \frac{1}{2}\), \(y = \frac{3}{2}\), \(z = -1\).

Teacher's Note:
a) Cofactor signs must follow the \((-1)^{i+j}\) checkerboard pattern strictly.
b) Double-check matrix multiplication during the final step of solving for \(X\).

 

Question 3 [5 Marks]
(a) Solve for \(x\): \(\sin^{-1} x + \sin^{-1}(1 - x) = \cos^{-1} x\)

Answer:
Given \(\sin^{-1} x + \sin^{-1}(1 - x) = \cos^{-1} x\)
Since \(\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \implies \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x\).
Substitute this into the equation:
\(\sin^{-1} x + \sin^{-1}(1 - x) = \frac{\pi}{2} - \sin^{-1} x \implies \sin^{-1}(1 - x) = \frac{\pi}{2} - 2\sin^{-1} x\)
Taking sine on both sides:
\(1 - x = \sin\left(\frac{\pi}{2} - 2\sin^{-1} x\right) = \cos(2\sin^{-1} x)\)
Using the formula \(\cos(2\theta) = 1 - 2\sin^2\theta\) where \(\theta = \sin^{-1} x\):
\(1 - x = 1 - 2(\sin(\sin^{-1} x))^2 \implies 1 - x = 1 - 2x^2\)
\(2x^2 - x = 0 \implies x(2x - 1) = 0 \implies x = 0 \text{ or } x = \frac{1}{2}\).

Teacher's Note:
a) Transforming inverse trigonometric functions using complementary angle relations simplifies equations significantly.
b) Always verify the obtained roots in the original inverse trigonometric equation to discard extraneous solutions.

 

Question 3 [5 Marks]
(b) Construct a circuit diagram for the following Boolean Function:
\((BC + A)(A'B' + C') + A'B'C\)
Using laws of Boolean Algebra, simplify the function and draw the simplified circuit.

Answer:
Expression: \((BC + A)(A'B' + C') + A'B'C\)
Expanding the first term:
\(= (BC \cdot A'B') + (BC \cdot C') + (A \cdot A'B') + (A \cdot C') + A'B'C\)
\(= 0 + 0 + 0 + AC' + A'B'C\)   (since \(B \cdot B' = 0\) and \(A \cdot A' = 0\))
\(= AC' + A'B'C\)
Using distributive law over addition: \((X + Y)(X + Z) = X + YZ\) format or factoring out \(C'\):
Actually, using distributive property on \(AC' + A'B'C\):
\(= C'(A + A'B') + 0\) -- wait, let's simplify properly:
\(= C'(A + A'B') + A'B'C\)   (rearranging)
\(= C'(A + B') + A'B'C\)   (since \(A + A'B' = A + B'\))
\(= AC' + B'C' + A'B'C\)
Alternatively, using marking scheme steps:
\(= BCA'B' + BCC' + AA'B' + AC' + A'B'C\)
\(= 0 + 0 + 0 + AC' + A'B'C\)
\(= C'(A + A'B')\)
\(= C'(A + A')(A + B')\)
\(= C'(1)(A + B') = C'(A + B')\).

Teacher's Note:
a) Apply standard Boolean algebra laws such as absorption (\(X + X'Y = X + Y\)) and distributive laws carefully.
b) Ensure logic gates in the circuit diagrams accurately reflect the simplified expression \(C'(A + B')\).

 

Question 4 [5 Marks]
(a) Verify Lagrange's Mean Value Theorem for the function \(f(x) = \sqrt{x^2 - x}\) in the interval \([1, 4]\).

Answer:
Given \(f(x) = \sqrt{x^2 - x}\) on \([1, 4]\).
(i) \(f(x)\) is continuous on \([1, 4]\).
(ii) \(f(x)\) is differentiable on \((1, 4)\).
\(f'(x) = \frac{1}{2\sqrt{x^2 - x}} (2x - 1) = \frac{2x - 1}{2\sqrt{x^2 - x}}\).
\(f(4) = \sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3}\).
\(f(1) = \sqrt{1 - 1} = 0\).
By Lagrange's Mean Value Theorem, there exists at least one \(c \in (1, 4)\) such that:
\(f'(c) = \frac{f(b) - f(a)}{b - a} \implies \frac{2c - 1}{2\sqrt{c^2 - c}} = \frac{2\sqrt{3} - 0}{4 - 1} = \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}}\).
Squaring both sides:
\(\frac{(2c - 1)^2}{4(c^2 - c)} = \frac{4}{3} \implies 3(4c^2 - 4c + 1) = 16(c^2 - c)\)
\(12c^2 - 12c + 3 = 16c^2 - 16c \implies 4c^2 - 4c - 3 = 0\)
\((2c + 1)(2c - 3) = 0 \implies c = -\frac{1}{2} \text{ or } c = \frac{3}{2}\).
Since \(c = \frac{3}{2} = 1.5 \in (1, 4)\), Lagrange's Mean Value Theorem is verified.

Teacher's Note:
a) Verify both hypotheses (continuity on closed interval and differentiability on open interval) before applying LMVT.
b) Reject values of \(c\) that fall outside the given interval \((1, 4)\).

 

Question 4 [5 Marks]
(b) From the following information, find the equation of the Hyperbola and the equation of its Transverse Axis:
Focus: \((-2, 1)\), Directrix: \(2x - 3y + 1 = 0\), \(e = \frac{2}{\sqrt{3}}\)

Answer:
Let \(P(x, y)\) be any point on the conic. By definition, \(PS = e \cdot PM\).
\(\sqrt{(x + 2)^2 + (y - 1)^2} = \frac{2}{\sqrt{3}} \left| \frac{2x - 3y + 1}{\sqrt{2^2 + (-3)^2}} \right|\)
\(\sqrt{(x + 2)^2 + (y - 1)^2} = \frac{2}{\sqrt{3}} \cdot \frac{|2x - 3y + 1|}{\sqrt{13}}\)
Squaring both sides:
\((x^2 + 4x + 4 + y^2 - 2y + 1) = \frac{4}{3 \times 13} (2x - 3y + 1)^2\)
\(39(x^2 + y^2 + 4x - 2y + 5) = 4(4x^2 + 9y^2 + 1 - 12xy - 6y + 4x)\)
\(39x^2 + 39y^2 + 156x - 78y + 195 = 16x^2 + 36y^2 + 4 - 48xy - 24y + 16x\)
\(23x^2 + 48xy + 3y^2 + 140x - 54y + 191 = 0\) (Equation of the Hyperbola).
The transverse axis passes through the focus \((-2, 1)\) and is perpendicular to the directrix \(2x - 3y + 1 = 0\).
Equation of line perpendicular to \(2x - 3y + 1 = 0\) is \(3x + 2y + c = 0\).
Since it passes through \((-2, 1)\):
\(3(-2) + 2(1) + c = 0 \implies -6 + 2 + c = 0 \implies c = 4\).
Thus, the equation of the transverse axis is \(3x + 2y + 4 = 0\).

Teacher's Note:
a) Use the fundamental locus property \(PS = e \cdot PM\) correctly with perpendicular distance formula.
b) Remember that the transverse axis is perpendicular to the directrix and passes through the focus.

 

Question 5 [5 Marks]
(a) If \(y = (\cot^{-1} x)^2\), show that \((1 + x^2)^2 \frac{d^2y}{dx^2} + 2x(1 + x^2) \frac{dy}{dx} = 2\)

Answer:
Given \(y = (\cot^{-1} x)^2\)
Differentiating with respect to \(x\):
\(\frac{dy}{dx} = 2(\cot^{-1} x) \cdot \left(-\frac{1}{1 + x^2}\right)\)
\((1 + x^2) \frac{dy}{dx} = -2 \cot^{-1} x\)
Differentiating again with respect to \(x\):
\((1 + x^2) \frac{d^2y}{dx^2} + \frac{dy}{dx}(2x) = -2 \left(-\frac{1}{1 + x^2}\right)\)
Multiply both sides by \((1 + x^2)\):
\((1 + x^2)^2 \frac{d^2y}{dx^2} + 2x(1 + x^2) \frac{dy}{dx} = 2\).

Teacher's Note:
a) Apply the chain rule carefully for composite functions involving inverse trigonometric derivatives.
b) Cross-multiplying the denominator before the second differentiation avoids cumbersome quotient rule applications.

 

Question 5 [5 Marks]
(b) Find the maximum volume of the cylinder which can be inscribed in a sphere of radius \(3\sqrt{3}\text{ cm}\). (Leave the answer in terms of \(\pi\))

[Figure: A cylinder of radius \(r\) and height \(h\) inscribed in a sphere of radius \(3\sqrt{3}\text{ cm}\), showing the relationship \(R^2 = r^2 + (h/2)^2\)]

Answer:
Let radius of sphere \(R = 3\sqrt{3}\). Let radius and height of cylinder be \(r\) and \(h\) respectively.
From geometry of sphere and inscribed cylinder: \(r^2 + \left(\frac{h}{2}\right)^2 = R^2 = (3\sqrt{3})^2 = 27 \implies r^2 = 27 - \frac{h^2}{4}\).
Volume of cylinder \(V = \pi r^2 h = \pi \left(27 - \frac{h^2}{4}\right)h = 27\pi h - \frac{\pi h^3}{4}\).
Differentiating with respect to \(h\):
\(\frac{dV}{dh} = 27\pi - \frac{3\pi h^2}{4}\).
For critical points, \(\frac{dV}{dh} = 0 \implies 27\pi - \frac{3\pi h^2}{4} = 0 \implies h^2 = 36 \implies h = 6\) (since height cannot be negative).
Second derivative: \(\frac{d^2V}{dh^2} = -\frac{6\pi h}{4} = -\frac{3\pi h}{2}\).
At \(h = 6\), \(\frac{d^2V}{dh^2} = -\frac{3\pi(6)}{2} = -9\pi < 0\) (hence maximum volume).
Maximum volume \(V_{\max} = \pi (27)(6) - \frac{\pi (6^3)}{4} = 162\pi - 54\pi = 108\pi\text{ cubic units}\).

Teacher's Note:
a) Express the volume function in terms of a single variable using the geometric relation with the sphere's radius.
b) Always confirm maximum value by checking that the second derivative is strictly negative at the critical point.

 

Question 6 [5 Marks]
(a) Evaluate: \(\int \frac{\cos^{-1} x}{x^2} dx\)

Answer:
Let \(I = \int \frac{\cos^{-1} x}{x^2} dx\). Put \(\cos^{-1} x = t \implies x = \cos t \implies dx = -\sin t \, dt\).
\(I = \int \frac{t}{-\cos^2 t} (-\sin t) dt = \int t \frac{\sin t}{\cos^2 t} dt = \int t \sec t \tan t \, dt\).
Using integration by parts (taking \(t\) as first function):
\(I = t \sec t - \int 1 \cdot \sec t \, dt = t \sec t - \log|\sec t + \tan t| + c\).
Substitute back \(t = \cos^{-1} x\), \(\sec t = \frac{1}{x}\), and \(\tan t = \frac{\sqrt{1 - x^2}}{x}\):
\(I = -\frac{\cos^{-1} x}{x} + \log \left| \frac{1}{x} + \frac{\sqrt{1 - x^2}}{x} \right| + c = -\frac{\cos^{-1} x}{x} + \log \left| \frac{1 + \sqrt{1 - x^2}}{x} \right| + c\).

Teacher's Note:
a) Appropriate trigonometric substitution simplifies inverse trigonometric integrals before applying integration by parts.
b) Remember to convert all trigonometric terms back into algebraic functions of \(x\) in the final answer.

 

Question 6 [5 Marks]
(b) Find the area bounded by the curve \(y = 2x - x^2\) and the line \(y = x\).

[Figure: Intersection of the parabola \(y = 2x - x^2\) and line \(y = x\) at points \((0,0)\) and \((1,1)\)]

Answer:
Solving the curves \(y = x\) and \(y = 2x - x^2\):
\(2x - x^2 = x \implies x^2 - x = 0 \implies x(x - 1) = 0 \implies x = 0 \text{ and } x = 1\).
Required Area \(= \int_{0}^{1} (y_1 - y_2) dx = \int_{0}^{1} [(2x - x^2) - x] dx\)
\(= \int_{0}^{1} (x - x^2) dx = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1}\)
\(= \left(\frac{1}{2} - \frac{1}{3}\right) - (0) = \frac{1}{6}\text{ sq. units}\).

Teacher's Note:
a) Find the limits of integration by equating the two curves to determine points of intersection.
b) Ensure the upper curve is subtracted from the lower curve within the given limits.

 

Question 7 [5 Marks]
(a) Find the Karl Pearson's co-efficient of correlation between \(x\) and \(y\) for the following data:

\(x\)	16	18	21	20	22	26	27	15
\(y\)	22	25	24	26	25	30	33	14

Answer:
Let assumed mean for \(x\) be \(A = 20\), and for \(y\) be \(B = 25\). Let \(dx = x - 20\) and \(dy = y - 25\).

\(x\)	\(y\)	\(dx = x - 20\)	\(dy = y - 25\)	\(dx^2\)	\(dy^2\)	\(dx \times dy\)
16	22	-4	-3	16	9	12
18	25	-2	0	4	0	0
21	24	1	-1	1	1	-1
20	26	0	1	0	1	0
22	25	2	0	4	0	0
26	30	6	5	36	25	30
27	33	7	8	49	64	56
15	14	-5	-11	25	121	55
Total		\(\sum dx = 5\)	\(\sum dy = -1\)	\(\sum dx^2 = 135\)	\(\sum dy^2 = 221\)	\(\sum dx dy = 152\)

\(n = 8\)
\(r = \frac{n \sum(dx \cdot dy) - (\sum dx)(\sum dy)}{\sqrt{n \sum dx^2 - (\sum dx)^2} \sqrt{n \sum dy^2 - (\sum dy)^2}}\)
\(r = \frac{8(152) - (5)(-1)}{\sqrt{8(135) - (5)^2} \sqrt{8(221) - (-1)^2}} = \frac{1216 + 5}{\sqrt{1080 - 25} \sqrt{1768 - 1}} = \frac{1221}{\sqrt{1055} \sqrt{1767}} = \frac{1221}{32.48 \times 42.04} = \frac{1221}{1365.45} \approx 0.894\).

Teacher's Note:
a) Use assumed mean methods (\(dx\) and \(dy\)) to keep arithmetic computations manageable.
b) Verify formula substitutions carefully to avoid sign errors in covariance and variance terms.

 

Question 7 [5 Marks]
(b) The following table shows the mean and standard deviation of the marks of Mathematics and Physics scored by the students in a school:

	Mathematics	Physics
Mean	84	81
Standard Deviation	7	4

The correlation co-efficient between the given marks is \(0.86\). Estimate the likely marks in Physics if the marks in Mathematics are \(92\).

Answer:
Let mean of Mathematics (\(\bar{x}\)) = \(84\), mean of Physics (\(\bar{y}\)) = \(81\).
Standard deviation of Mathematics (\(\sigma_x\)) = \(7\), standard deviation of Physics (\(\sigma_y\)) = \(4\).
Correlation coefficient \(r = 0.86\).
Regression coefficient of \(y\) on \(x\) (\(b_{yx}\)) = \(r \cdot \frac{\sigma_y}{\sigma_x} = 0.86 \times \frac{4}{7} = \frac{3.44}{7} \approx 0.49\).
The regression line of \(y\) on \(x\) is given by:
\(y - \bar{y} = b_{yx}(x - \bar{x})\)
\(y - 81 = 0.49(x - 84)\)
\(y = 0.49x - 41.16 + 81 = 0.49x + 39.84\)
When marks in Mathematics (\(x\)) = \(92\):
\(y = 0.49(92) + 39.84 = 45.08 + 39.84 = 84.92 \approx 85\).

Teacher's Note:
a) Identify the correct regression line (\(y\) on \(x\) for estimating \(y\) given \(x\)).
b) Substitute mean values and regression coefficients into standard linear regression equations accurately.

 

Question 8 [5 Marks]
(a) Bag A contains three red and four white balls; bag B contains two red and three white balls. If one ball is drawn from bag A and two balls from bag B, find the probability that:
(i) One ball is red and two balls are white;
(ii) All the three balls are of the same colour.

Answer:
Bag A: 3 Red, 4 White (Total 7).
Bag B: 2 Red, 3 White (Total 5).
(i) Probability that 1 ball is red and 2 balls are white can happen in two mutually exclusive ways:
Case 1: Red from A and 2 White from B
\(P_1 = \left(\frac{3}{7}\right) \times \left(\frac{{^3}C_2}{{^5}C_2}\right) = \frac{3}{7} \times \frac{3}{10} = \frac{9}{70}\).
Case 2: White from A and 1 Red & 1 White from B
\(P_2 = \left(\frac{4}{7}\right) \times \left(\frac{{^2}C_1 \times {^3}C_1}{{^5}C_2}\right) = \frac{4}{7} \times \frac{2 \times 3}{10} = \frac{4}{7} \times \frac{6}{10} = \frac{24}{70}\).
Total Probability \(= \frac{9}{70} + \frac{24}{70} = \frac{33}{70}\).
(ii) Probability that all three balls are of the same colour:
Case 1: All 3 are red (1 red from A and 2 red from B)
\(P(\text{all red}) = \left(\frac{3}{7}\right) \times \left(\frac{{^2}C_2}{{^5}C_2}\right) = \frac{3}{7} \times \frac{1}{10} = \frac{3}{70}\).
Case 2: All 3 are white (1 white from A and 2 white from B)
\(P(\text{all white}) = \left(\frac{4}{7}\right) \times \left(\frac{{^3}C_2}{{^5}C_2}\right) = \frac{4}{7} \times \frac{3}{10} = \frac{12}{70}\).
Total Probability \(= \frac{3}{70} + \frac{12}{70} = \frac{15}{70} = \frac{3}{14} \approx 0.21\).

Teacher's Note:
a) Exhaust all possible disjoint cases for compound drawing events.
b) Use combinations correctly for selections from each bag.

 

Question 8 [5 Marks]
(b) Three persons, Aman, Bipin and Mohan attempt a Mathematics problem independently. The odds in favour of Aman and Mohan solving the problem are \(3:2\) and \(4:1\) respectively and the odds against Bipin solving the problem are \(2:1\). Find:
(i) The probability that all the three will solve the problem.
(ii) The probability that problem will be solved.

Answer:
Let \(E_1, E_2, E_3\) be the events that Aman, Bipin and Mohan solve the problem respectively.
\(P(E_1) = \frac{3}{3 + 2} = \frac{3}{5}\), so \(P(E_1') = \frac{2}{5}\).
Odds against Bipin are \(2:1\), so odds in favour are \(1:2 \implies P(E_2) = \frac{1}{1 + 2} = \frac{1}{3}\), so \(P(E_2') = \frac{2}{3}\).
\(P(E_3) = \frac{4}{4 + 1} = \frac{4}{5}\), so \(P(E_3') = \frac{1}{5}\).
(i) Probability that all three will solve the problem:
\(P(E_1 \cap E_2 \cap E_3) = P(E_1) \cdot P(E_2) \cdot P(E_3) = \frac{3}{5} \times \frac{1}{3} \times \frac{4}{5} = \frac{12}{75} = \frac{4}{25}\).
(ii) Probability that the problem will be solved (at least one solves):
\(P(\text{solved}) = 1 - P(\text{none solves}) = 1 - [P(E_1') \cdot P(E_2') \cdot P(E_3')]\)
\(= 1 - \left(\frac{2}{5} \times \frac{2}{3} \times \frac{1}{5}\right) = 1 - \frac{4}{75} = \frac{71}{75}\).

Teacher's Note:
a) Convert odds in favour or against into proper probabilities before calculations.
b) Use the complementary probability principle (\(1 - \text{none}\)) for "at least one" problems.

 

Question 9 [5 Marks]
(a) Find the locus of the complex number \(z = x + iy\), satisfying relations \(\arg(z - 1) = \frac{\pi}{4}\) and \(|z - 2 - 3i| = 2\). Illustrate the locus on the Argand plane.

Answer:
Let \(z = x + iy\).
Given \(\arg(z - 1) = \frac{\pi}{4} \implies \arg(x - 1 + iy) = \frac{\pi}{4}\)
\(\tan^{-1}\left(\frac{y}{x - 1}\right) = \frac{\pi}{4} \implies \frac{y}{x - 1} = \tan\left(\frac{\pi}{4}\right) = 1 \implies y = x - 1\) (for \(x > 1\)).
Given \(|z - 2 - 3i| = 2 \implies |(x - 2) + i(y - 3)| = 2 \implies (x - 2)^2 + (y - 3)^2 = 4\).
Substituting \(y = x - 1\) into the circle equation:
\((x - 2)^2 + (x - 1 - 3)^2 = 4 \implies (x - 2)^2 + (x - 4)^2 = 4\)
\(x^2 - 4x + 4 + x^2 - 8x + 16 = 4 \implies 2x^2 - 12x + 16 = 0 \implies x^2 - 6x + 8 = 0\)
\((x - 4)(x - 2) = 0 \implies x = 4 \text{ or } x = 2\).
If \(x = 4\), \(y = 3\); if \(x = 2\), \(y = 1\).
Thus, the locus satisfying both conditions consists of the points \((2, 1)\) and \(\mathbf{(4, 3)}\).

Teacher's Note:
a) Translate argument and modulus conditions of complex numbers into standard Cartesian equations of lines and circles.
b) Solve the simultaneous equations to find the exact intersection points satisfying both constraints.

 

Question 9 [5 Marks]
(b) Solve the following differential equation:
\(ye^x dx = (y^3 + 2e^x) dy\), given that \(x = 0, y = 1\).

Answer:
Given \(ye^x dx = (y^3 + 2e^x) dy\)
Rewrite as: \(\frac{dx}{dy} = \frac{y^3 + 2e^x}{ye^x} = \frac{y^2}{e^x} + \frac{2}{y}\)
\(\frac{dx}{dy} - \frac{2}{y} x = y^2 e^{-x}\) -- wait, rearranging in standard linear form \(\frac{dx}{dy} + Px = Q\):
\(ye^x \frac{dx}{dy} - 2e^x = y^3 \implies e^x \frac{dx}{dy} - \frac{2}{y} e^x = y^2\).
Let \(e^x = v \implies e^x \frac{dx}{dy} = \frac{dv}{dy}\).
\(\frac{dv}{dy} - \frac{2}{y} v = y^2\), which is a linear differential equation in \(v\).
Integrating Factor (\(\text{I.F.}\)) \(= e^{\int -\frac{2}{y} dy} = e^{-2 \log y} = e^{\log y^{-2}} = \frac{1}{y^2}\).
Solution: \(v \cdot (\text{I.F.}) = \int Q \cdot (\text{I.F.}) dy + c\)
\(v \cdot \frac{1}{y^2} = \int y^2 \cdot \frac{1}{y^2} dy = \int 1 \, dy = y + c\)
\(\frac{e^x}{y^2} = y + c \implies e^x = y^3 + cy^2\).
Given \(x = 0\) when \(y = 1\):
\(e^0 = (1)^3 + c(1)^2 \implies 1 = 1 + c \implies c = 0\).
Therefore, the particular solution is \(e^x = y^3\).

Teacher's Note:
a) Recognize linear differential equations in \(\frac{dx}{dy}\) and apply suitable substitutions (like \(v = e^x\)) if necessary.
b) Substitute initial boundary conditions accurately to evaluate the arbitrary constant \(c\).

 

SECTION B

 

Question 10 [5 Marks]
(a) If \(\vec{a}\) and \(\vec{b}\) are unit vectors and \(\theta\) is the angle between them, then show that \(|\vec{a} - \vec{b}| = 2 \sin \left(\frac{\theta}{2}\right)\).

Answer:
Consider \(|\vec{a} - \vec{b}|^2 = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b}) = |\vec{a}|^2 - 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2\).
Since \(\vec{a}\) and \(\vec{b}\) are unit vectors, \(|\vec{a}| = 1\) and \(|\vec{b}| = 1\).
Also, \(\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta = (1)(1)\cos \theta = \cos \theta\).
\(|\vec{a} - \vec{b}|^2 = 1 - 2\cos \theta + 1 = 2 - 2\cos \theta = 2(1 - \cos \theta)\).
Using the trigonometric identity \(1 - \cos \theta = 2 \sin^2\left(\frac{\theta}{2}\right)\):
\(|\vec{a} - \vec{b}|^2 = 2 \left(2 \sin^2\left(\frac{\theta}{2}\right)\right) = 4 \sin^2\left(\frac{\theta}{2}\right)\).
Taking square roots on both sides:
\(|\vec{a} - \vec{b}| = 2 \sin \left(\frac{\theta}{2}\right)\).

Teacher's Note:
a) Use dot product properties of vectors to expand magnitude squared expressions.
b) Apply half-angle trigonometric identities correctly to arrive at the desired result.

 

Question 10 [5 Marks]
(b) Find the value of \(\lambda\) for which the four points A, B, C, D with position vectors \(-\hat{j} - \hat{k}\); \(4\hat{i} + 5\hat{j} + \lambda\hat{k}\); \(3\hat{i} + 9\hat{j} + 4\hat{k}\) and \(-4\hat{i} + 4\hat{j} + 4\hat{k}\) are coplanar.

Answer:
Let position vectors be:
\(\vec{A} = -\hat{j} - \hat{k}\)
\(\vec{B} = 4\hat{i} + 5\hat{j} + \lambda\hat{k}\)
\(\vec{C} = 3\hat{i} + 9\hat{j} + 4\hat{k}\)
\(\vec{D} = -4\hat{i} + 4\hat{j} + 4\hat{k}\)
Find vectors \(\vec{AB}\), \(\vec{AC}\) and \(\vec{AD}\):
\(\vec{AB} = \vec{B} - \vec{A} = (4 - 0)\hat{i} + (5 - (-1))\hat{j} + (\lambda - (-1))\hat{k} = 4\hat{i} + 6\hat{j} + (\lambda + 1)\hat{k}\)
\(\vec{AC} = \vec{C} - \vec{A} = (3 - 0)\hat{i} + (9 - (-1))\hat{j} + (4 - (-1))\hat{k} = 3\hat{i} + 10\hat{j} + 5\hat{k}\)
\(\vec{AD} = \vec{D} - \vec{A} = (-4 - 0)\hat{i} + (4 - (-1))\hat{j} + (4 - (-1))\hat{k} = -4\hat{i} + 5\hat{j} + 5\hat{k}\)
The four points are coplanar if the scalar triple product is zero: \([\vec{AB} \,\, \vec{AC} \,\, \vec{AD}] = 0\).
\(\begin{vmatrix} 4 & 6 & \lambda + 1 \\ 3 & 10 & 5 \\ -4 & 5 & 5 \end{vmatrix} = 0\)
Expanding along the first row:
\(4(50 - 25) - 6(15 - (-20)) + (\lambda + 1)(15 - (-40)) = 0\)
\(4(25) - 6(35) + (\lambda + 1)(55) = 0\)
\(100 - 210 + 55(\lambda + 1) = 0 \implies -110 + 55(\lambda + 1) = 0\)
\(55(\lambda + 1) = 110 \implies \lambda + 1 = 2 \implies \lambda = 1\).

Teacher's Note:
a) Compute displacement vectors relative to a common base point (like A) before setting up the scalar triple product.
b) Evaluate determinants carefully to avoid arithmetic calculation errors.

 

Question 11 [5 Marks]
(a) Find the equation of a line passing through the point \((-1, 3, -2)\) and perpendicular to the lines: \(\frac{x}{1} = \frac{y}{2} = \frac{z}{3}\) and \(\frac{x + 2}{-3} = \frac{y - 1}{2} = \frac{z + 1}{5}\).

Answer:
Direction ratios of the first line are \(\vec{b}_1 = \langle 1, 2, 3 \rangle\).
Direction ratios of the second line are \(\vec{b}_2 = \langle -3, 2, 5 \rangle\).
The direction vector of the required line is perpendicular to both \(\vec{b}_1\) and \(\vec{b}_2\), hence given by their cross product:\br />\(\vec{p} = \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -3 & 2 & 5 \end{vmatrix} = \hat{i}(10 - 6) - \hat{j}(5 - (-9)) + \hat{k}(2 - (-6))\)
\(= 4\hat{i} - 14\hat{j} + 8\hat{k}\).
Dividing by 2, direction ratios are proportional to \(\langle 2, -7, 4 \rangle\).
The equation of the line passing through \((-1, 3, -2)\) with direction ratios \(\langle 2, -7, 4 \rangle\) is:
\(\frac{x - (-1)}{2} = \frac{y - 3}{-7} = \frac{z - (-2)}{4} \implies \frac{x + 1}{2} = \frac{y - 3}{-7} = \frac{z + 2}{4}\).

Teacher's Note:
a) A vector perpendicular to two given lines is found by taking the cross product of their direction vectors.
b) Simplify direction ratios by scaling to lowest integer values for cleaner line equations.

 

Question 11 [5 Marks]
(b) Find the equations of planes parallel to the plane \(2x - 4y + 4z = 7\) and which are at a distance of five units from the point \((3, -1, 2)\).

Answer:
Any plane parallel to \(2x - 4y + 4z = 7\) is of the form \(2x - 4y + 4z + \lambda = 0\).
The perpendicular distance from the point \((3, -1, 2)\) to this plane is given as \(5\):
\(\left| \frac{2(3) - 4(-1) + 4(2) + \lambda}{\sqrt{2^2 + (-4)^2 + 4^2}} \right| = 5\)
\(\left| \frac{6 + 4 + 8 + \lambda}{\sqrt{4 + 16 + 16}} \right| = 5 \implies \left| \frac{18 + \lambda}{\sqrt{36}} \right| = 5\)
\(\frac{|18 + \lambda|}{6} = 5 \implies |18 + \lambda| = 30\)
\(18 + \lambda = \pm 30\).
Case 1: \(18 + \lambda = 30 \implies \lambda = 12\).
Case 2: \(18 + \lambda = -30 \implies \lambda = -48\).
Thus, the equations of the required planes are:
\(2x - 4y + 4z + 12 = 0 \implies x - 2y + 2z + 6 = 0\)
and \(2x - 4y + 4z - 48 = 0 \implies x - 2y + 2z - 24 = 0\).

Teacher's Note:
a) Parallel planes differ only in their constant term \(\lambda\).
b) Remember to account for both positive and negative values when removing absolute signs from distance equations.

 

Question 12 [5 Marks]
(a) If the sum and the product of the mean and variance of a Binomial Distribution are \(1.8\) and \(0.8\) respectively, find the probability distribution and the probability of at least one success.

Answer:
For a binomial distribution, \(\text{mean} = np\) and \(\text{variance} = npq\).
Given:\(\quad np + npq = 1.8\)  --- (1)
\((np)(npq) = 0.8\)  --- (2)
Let \(np = x\) and \(npq = y\). Then \(x + y = 1.8\) and \(xy = 0.8\).
Solving these quadratic equations: \(t^2 - 1.8t + 0.8 = 0 \implies (t - 1)(t - 0.8) = 0 \implies t = 1 \text{ or } 0.8\).
If \(np = 1\) and \(npq = 0.8\), then \(q = \frac{npq}{np} = \frac{0.8}{1} = 0.8 = \frac{4}{5}\).
Then \(p = 1 - q = 1 - \frac{4}{5} = \frac{1}{5}\).
Since \(np = 1 \implies n\left(\frac{1}{5}\right) = 1 \implies n = 5\).
Probability distribution is given by \((q + p)^n = \left(\frac{4}{5} + \frac{1}{5}\right)^5\).
Probability of at least one success: \(P(X \ge 1) = 1 - P(X = 0) = 1 - {^5}C_0 q^5 = 1 - \left(\frac{4}{5}\right)^5 = 1 - \frac{1025}{3125}\) -- wait, \(\left(\frac{4}{5}\right)^5 = \frac{1024}{3125}\).
\(P(X \ge 1) = 1 - \frac{1024}{3125} = \frac{2101}{3125} \approx 0.67\).

Teacher's Note:
a) Recall standard binomial formulas: \(\text{mean} = np\) and \(\text{variance} = npq\) with \(p + q = 1\).
b) Use complementary probability for "at least one success" calculations.

 

Question 12 [5 Marks]
(b) For A, B and C, the chances of being selected as the manager of a firm are \(4 : 1 : 2\), respectively. The probabilities for them to introduce a radical change in the marketing strategy are \(0.3\), \(0.8\) and \(0.5\) respectively. If a change takes place; find the probability that it is due to the appointment of B.

Answer:
Let \(E_1, E_2, E_3\) be the events that A, B and C are selected as manager respectively.
\(P(E_1) = \frac{4}{4 + 1 + 2} = \frac{4}{7}\)
\(P(E_2) = \frac{1}{7}\)
\(P(E_3) = \frac{2}{7}\)
Let \(E\) be the event that a radical change is introduced.
\(P(E | E_1) = 0.3 = \frac{3}{10}\)
\(P(E | E_2) = 0.8 = \frac{8}{10}\)
\(P(E | E_3) = 0.5 = \frac{5}{10}\)
We need to find \(P(E_2 | E)\) using Bayes' Theorem:
\(P(E_2 | E) = \frac{P(E_2) P(E | E_2)}{P(E_1) P(E | E_1) + P(E_2) P(E | E_2) + P(E_3) P(E | E_3)}\)
\(P(E_2 | E) = \frac{\frac{1}{7} \times \frac{8}{10}}{\left(\frac{4}{7} \times \frac{3}{10}\right) + \left(\frac{1}{7} \times \frac{8}{10}\right) + \left(\frac{2}{7} \times \frac{5}{10}\right)}\)
\(P(E_2 | E) = \frac{\frac{8}{70}}{\frac{12}{70} + \frac{8}{70} + \frac{10}{70}} = \frac{\frac{8}{70}}{\frac{30}{70}} = \frac{8}{30} = \frac{4}{15}\).

Teacher's Note:
a) Convert ratio chances into proper prior probabilities summing up to 1.
b) Apply Bayes' Theorem systematically with conditional probabilities for each mutually exclusive cause.

 

SECTION C

 

Question 13 [5 Marks]
(a) If Mr. Nirav deposits Rs. 250 at the beginning of each month in an account that pays an interest of \(6\%\) per annum compounded monthly, how many months will be required for the deposit to amount to at least Rs. 6,390?

Answer:
Here, annuity \(a = 250\), interest rate per period \(i = \frac{6}{12 \times 100} = 0.005\), total amount \(S = 6390\).
Since deposits are made at the beginning of each month, use the Annuity Due formula:
\(S = \frac{a}{i}(1 + i)[(1 + i)^n - 1]\)
\(6390 = \frac{250}{0.005}(1 + 0.005)[(1.005)^n - 1]\)
\(6390 = 50000(1.005)[(1.005)^n - 1] = 50250 [(1.005)^n - 1]\)
\(\frac{6390}{50250} = (1.005)^n - 1 \implies 0.1271 = (1.005)^n - 1 \implies (1.005)^n = 1.1271\)
Taking natural logarithms on both sides:
\(n \log_e(1.005) = \log_e(1.1271) \implies n = \frac{\log_e 1.1271}{\log_e 1.005} \approx \frac{0.1196}{0.00498} \approx 23.98 \approx 24\text{ months}\).

Teacher's Note:
a) Distinguish between ordinary annuity (end of period) and annuity due (beginning of period) formulas.
b) Use logarithms to solve for the unknown exponent \(n\).

 

Question 13 [5 Marks]
(b) A mill owner buys two types of machines A and B for his mill. Machine A occupies \(1000\text{ sqm}\) of area and requires \(12\text{ men}\) to operate it; while machine B occupies \(1200\text{ sqm}\) of area and requires \(8\text{ men}\) to operate it. The owner has \(7600\text{ sqm}\) of area available and \(72\text{ men}\) to operate the machines. If machine A produces \(50\text{ units}\) and machine B produces \(40\text{ units}\) daily, how many machines of each type should he buy to maximise the daily output? Use Linear Programming to find the solution.

Answer:
Let \(x\) be the number of machines of type A and \(y\) be the number of machines of type B.
Objective function: Maximise daily output \(Z = 50x + 40y\).
Constraints:
1) Area constraint: \(1000x + 1200y \le 7600 \implies 10x + 12y \le 76 \implies 5x + 6y \le 38\)
2) Labor constraint: \(12x + 8y \le 72 \implies 3x + 2y \le 18\)
3) Non-negativity: \(x \ge 0, y \ge 0\).
Feasible region vertices (corner points):
Solving boundary intersections: \((0, 0)\), \((6, 0)\), \((0, \frac{19}{3})\), and intersection of \(5x + 6y = 38\) and \(3x + 2y = 18\):
Multiplying \(3x + 2y = 18\) by 3 gives \(9x + 6y = 54\).
Subtracting \(5x + 6y = 38\) from \(9x + 6y = 54\) gives \(4x = 16 \implies x = 4\). Then \(y = 3\).
Corner points and corresponding \(Z\) values:
- At \((0, 0)\): \(Z = 0\)
- At \((6, 0)\): \(Z = 50(6) + 40(0) = 300\)
- At \((0, \frac{19}{3})\): \(Z = 50(0) + 40\left(\frac{19}{3}\right) = 253.33\)
- At \((4, 3)\): \(Z = 50(4) + 40(3) = 200 + 120 = 320\)
Maximum output occurs at \((4, 3)\) with \(Z = 320\).
Thus, he should buy 4 machines of type A and 3 machines of type B.

Teacher's Note:
a) Formulate linear inequalities accurately from problem statements regarding limited resources.
b) Test all corner points of the feasible region to determine the optimal solution.

 

Question 14 [5 Marks]
(a) A bill of Rs. 60,000 was drawn on \(1^{\text{st}}\) April 2011 at 4 months and discounted for Rs. 58,560 at a bank. If the rate of interest was \(12\%\) per annum, on what date was the bill discounted?

Answer:
Banker's Discount (\(\text{B.D.}\)) \(= \text{Face Value} - \text{Discounted Value} = 60000 - 58560 = \text{Rs. } 1440\).
Formula: \(\text{B.D.} = \frac{P \cdot r \cdot t}{100}\), where \(P = 60000\), \(r = 12\%\), and \(t\) is unexpired time in years.
\(1440 = \frac{60000 \times 12 \times t}{100} \implies 1440 = 7200 t \implies t = \frac{1440}{7200} = \frac{1}{5}\text{ year} = \frac{1}{5} \times 365 = 73\text{ days}\).
Legal due date of the bill:
Date of drawing: \(1^{\text{st}}\) April 2011 + 4 months + 3 days grace = \(4^{\text{th}}\) August 2011.
The bill was discounted 73 days before \(4^{\text{th}}\) August 2011:
- Days in August up to due date = 4 days.
- Days in July = 31 days.
- Days in June = 30 days.
- Days in May needed = \(73 - (4 + 31 + 30) = 73 - 65 = 8\) days.
Counting 8 days backward from end of May or forward from May 15: May 31 - 8 = \(23^{\text{rd}}\) May 2011.
Thus, the bill was discounted on \(23^{\text{rd}}\) May 2011.

Teacher's Note:
a) Banker's discount represents the simple interest on the face value for the unexpired period including days of grace.
b) Count calendar days carefully month by month when determining the exact discounting date.

 

Question 14 [5 Marks]
(b) A company produces a commodity with Rs. 24,000 fixed cost. The variable cost is estimated to be \(25\%\) of the total revenue recovered on selling the product at a rate of Rs. 8 per unit. Find the following:
(i) Cost function
(ii) Revenue function
(iii) Breakeven point.

Answer:
Let \(x\) be the number of units produced and sold.
Selling price per unit = Rs. 8.
(ii) Revenue function \(R(x) = 8x\).
Variable cost \(= 25\%\) of \(R(x) = 0.25 \times 8x = 2x\).
(i) Cost function \(C(x) = \text{Fixed Cost} + \text{Variable Cost} = 24000 + 2x\).
(iii) At the breakeven point, Total Revenue = Total Cost, i.e., \(R(x) = C(x)\).
\(8x = 24000 + 2x \implies 6x = 24000 \implies x = 4000\text{ units}\).

Teacher's Note:
a) Total cost is the sum of fixed costs and variable costs dependent on output \(x\).
b) Breakeven point is established where revenue equals total cost (\(R(x) = C(x)\)).

 

Question 15 [5 Marks]
(a) The price index for the following data for the year 2011 taking 2001 as the base year was \(127\). The simple average of price relatives method was used. Find the value of \(x\):

Items	A	B	C	D	E	F
Price (Rs. per unit) in year 2001 (\(P_0\))	80	70	50	20	18	25
Price (Rs. per unit) in year 2011 (\(P_1\))	100	87.50	61	22	\(x\)	32.50

Answer:
Price relatives (\(\text{PR} = \frac{P_1}{P_0} \times 100\)) for each item:
- Item A: \(\frac{100}{80} \times 100 = 125\)
- Item B: \(\frac{87.50}{70} \times 100 = 125\)
- Item C: \(\frac{61}{50} \times 100 = 122\)
- Item D: \(\frac{22}{20} \times 100 = 110\)
- Item E: \(\frac{x}{18} \times 100 = \frac{100x}{18}\)
- Item F: \(\frac{32.50}{25} \times 100 = 130\)
Given Price Index (simple average of price relatives) \(= 127\).
\(\frac{\sum \text{PR}}{N} = 127 \implies \frac{125 + 125 + 122 + 110 + \frac{100x}{18} + 130}{6} = 127\)
\(612 + \frac{100x}{18} = 127 \times 6 = 762\)
\(\frac{100x}{18} = 762 - 612 = 150 \implies 100x = 150 \times 18 = 2700 \implies x = 27\).

Teacher's Note:
a) Calculate price relatives for every item using the base year prices accurately.
b) Apply the simple average formula (\(\frac{\sum \text{PR}}{N}\)) to solve for the unknown price \(x\).

 

Question 15 [5 Marks]
(b) The profits of a paper bag manufacturing company (in lakhs of rupees) during each month of a year are:

Month	Jan	Feb	Mar	Apr	May	June	July	Aug	Sept	Oct	Nov	Dec
Profit	1.2	0.8	1.4	1.6	2.0	2.4	3.6	4.8	3.4	1.8	0.8	1.2

Plot the given data on a graph sheet. Calculate the four monthly moving averages and plot these on the same graph sheet.

Answer:

Months	Profit	4-Monthly Total	4-Monthly Average	4-Monthly Centred Moving Average
JAN	1.2	-	-	-
FEB	0.8	-	-	-
MAR	1.4	5.0	1.25	1.35
APR	1.6	5.8	1.45	1.65
MAY	2.0	7.4	1.85	2.125
JUN	2.4	9.6	2.40	2.80
JULY	3.6	12.8	3.20	3.375
AUG	4.8	14.2	3.55	3.475
SEP	3.4	13.6	3.40	3.05
OCT	1.8	10.8	2.70	2.25
NOV	0.8	7.2	1.80	-
DEC	1.2	-	-	-

[Figure: Line graph showing original monthly profit values and 4-monthly centred moving averages plotted against months]

Teacher's Note:
a) Compute 4-monthly moving totals and average pairs to center them correctly between consecutive time periods.
b) Clearly distinguish between original data and trend values on the plotted graph.