\displaystyle \textbf{MATHEMATICS} \ \

\displaystyle \textit{Maximum Marks: 80} \ \
\displaystyle \textit{Time Allowed: Three Hours} \ \
\displaystyle \text{(Candidates are allowed additional 15 minutes for only reading the paper.} \ \
\displaystyle \text{They must NOT start writing during this time.)} \ \

\displaystyle \text{This Question Paper consists of three sections A, B and C.} \ \
\displaystyle \text{Candidates are required to attempt all questions from Section A and all questions} \ \
\displaystyle \text{EITHER from Section B OR Section C.} \ \
\displaystyle \text{Section A: Internal choice has been provided in two questions of two marks each, two questions} \ \
\displaystyle \text{of four marks each and two questions of six marks each.} \ \
\displaystyle \text{Section B: Internal choice has been provided in one question of two marks and} \ \
\displaystyle \text{one question of four marks.} \ \
\displaystyle \text{Section C: Internal choice has been provided in one question of two marks and} \ \
\displaystyle \text{one question of four marks.} \ \
\displaystyle \text{All working, including rough work, should be done on the same sheet as, and adjacent to the rest} \ \
\displaystyle \text{of the answer.} \ \
\displaystyle \text{The intended marks for questions or parts of questions are given in brackets [ ].} \ \
\displaystyle \text{Mathematical tables and graph papers are provided.} \ \

\displaystyle \textbf{SECTION A - 80 MARKS} \ \

\displaystyle \textbf{Question 1: } \hspace{10.0cm} [10 \text{ times } 3]
\displaystyle \text{(i) If } (A - 2I)(A - 3I) = 0 ,\text{ where } A = \begin{bmatrix} 4 & 2 \\ -1 & x \end{bmatrix} \text{ and } I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} , \text{ find the value of } x .
\displaystyle \text{Answer:}
\displaystyle \text{(i) } \text{Given, } (A - 2I)(A - 3I) = 0
\displaystyle \left( \begin{bmatrix} 4 & 2 \\ -1 & x \end{bmatrix} - 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right)\left( \begin{bmatrix} 4 & 2 \\ -1 & x \end{bmatrix} - 3 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right) = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}
\displaystyle \begin{bmatrix} 2 & 2 \\ -1 & x-2 \end{bmatrix} . \begin{bmatrix} 1 & 2 \\ -1 & x-3 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}
\displaystyle \begin{bmatrix} 0 & 2x-2 \\ -x+1 & x^2-5x+4 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}
\displaystyle \text{Therefore}
\displaystyle -x+1 = 0 \Rightarrow x = 1
\displaystyle 2x-2 = 0 \Rightarrow x = 1
\displaystyle x^2-5x+4 = 0 \Rightarrow (x-1)(x-4) = 0 \Rightarrow x = 1, 4
\displaystyle \text{However, } 4 \text{ does not satisfy all the equations. Hence } x = 1
\displaystyle \\
\displaystyle \text{(ii) Find the value(s) of } k \text{ so that the line } 2x+y+k=0 \text{ may touch the hyperbola} \\ 3x^2-y^2=3 .
\displaystyle \text{Answer:}
\displaystyle \text{(ii) } 3x^2 - y^2 = 3
\displaystyle \Rightarrow \frac{x^2}{1} - \frac{y^2}{3} = 1
\displaystyle \text{Therefore } a^2 = 1, b^2 = 3
\displaystyle y = -2x -k
\displaystyle m = -2 , c = -k
\displaystyle \text{Condition for tangent}
\displaystyle c^2 = a^2 m^2 - b^2
\displaystyle \Rightarrow k^2 = 4-3= 1
\displaystyle k= \pm 1
\displaystyle \\
\displaystyle \text{(iii) Prove that: } \tan^{-1} \frac{1}{4} + \tan^{-1} \frac{2}{9} = \frac{1}{2} \sin^{-1} \frac{4}{5}
\displaystyle \text{Answer:}
\displaystyle \text{(iii) LHS } = \tan^{-1} \frac{1}{4} + \tan^{-1} \frac{2}{9}
\displaystyle = \tan^{-1} \left( \frac{\frac{1}{4}+\frac{2}{9}}{1- \frac{1}{4} \cdot \frac{2}{9}} \right)
\displaystyle = \tan^{-1} \left( \frac{\frac{17}{36}}{\frac{17}{18}} \right)
\displaystyle = \tan^{-1} \frac{1}{2}
\displaystyle = \frac{1}{2} \sin^{-1} \left( \frac{2 \cdot \frac{1}{2}}{1 + \left(\frac{1}{2}\right)^2} \right)
\displaystyle = \frac{1}{2} \sin^{-1} \frac{4}{5} = \text{ RHS. Hence Proved. }
\displaystyle \\
\displaystyle \text{(iv) Using L'Hospital's Rule, evaluate: } \lim_{x \to 0} \left( \frac{e^x - e^{-x} - 2x}{x - \sin x} \right)
\displaystyle \text{Answer:}
\displaystyle \text{(iv) } \lim_{x \to 0} \left( \frac{e^x - e^{-x} - 2x}{x - \sin x} \right)
\displaystyle \text{Using L'Hospital's Rule}
\displaystyle = \lim_{x \to 0} \left( \frac{e^x + e^{-x} - 2}{1 - \cos x} \right)
\displaystyle \text{Again applying L'Hospital's Rule}
\displaystyle = \lim_{x \to 0} \left( \frac{e^x - e^{-x}}{\sin x} \right)
\displaystyle \text{Once again applying L'Hospital's Rule,}
\displaystyle = \lim_{x \to 0} \left( \frac{e^x + e^{-x}}{\cos x} \right)
\displaystyle = \frac{1+1}{1} = 2
\displaystyle \\
\displaystyle \text{(v) Evaluate: } \int \frac{1}{x+\sqrt{x}} dx
\displaystyle \text{Answer:}
\displaystyle \text{(v) Given, } \int \frac{1}{x+\sqrt{x}} dx
\displaystyle \text{Put } \sqrt{x} = t \Rightarrow x = t^2 \Rightarrow dx = 2t dt
\displaystyle \text{Therefore}
\displaystyle \int \frac{dx}{x+\sqrt{x}} = \int \frac{2t dt}{t^2 + t} = 2 \int \frac{1}{t+1} dt
\displaystyle = 2 \log |t+1 | + c
\displaystyle = 2 \log (\sqrt{x} + 1) + c
\displaystyle \\
\displaystyle \text{(vi) Evaluate: } \int_{0}^{1} \log \left(\frac{1}{x} -1 \right) dx
\displaystyle \text{Answer:}
\displaystyle \text{(vi) } I = \int_{0}^{1} \log \left( \frac{1}{x} -1 \right) dx
\displaystyle \Rightarrow I = \int_{0}^{1} \log \left( \frac{1-x}{x} \right) dx
\displaystyle \text{Using property, } \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx
\displaystyle I = \int_{0}^{1} \log \left( \frac{1-(1-x)}{1-x} \right) dx
\displaystyle \Rightarrow I = \int_{0}^{1} \log \left( \frac{x}{1-x} \right) dx
\displaystyle \Rightarrow I = \int_{0}^{1} \log \left( \left(\frac{1-x}{x}\right)^{-1} \right) dx
\displaystyle \Rightarrow I = \int_{0}^{1} (-1) \log \left( \frac{1-x}{x} \right) dx
\displaystyle \Rightarrow I = - \int_{0}^{1} \log \left( \frac{1-x}{x} \right) dx
\displaystyle \Rightarrow I = -I \Rightarrow 2I = 0 \Rightarrow I = 0
\displaystyle \text{Therefore } \int_{0}^{1} \log \left(\frac{1}{x} -1 \right) dx = 0
\displaystyle \\
\displaystyle \text{(vii) Two regression lines are represented by } 4x+10y=9 \text{ and } \\ 6x+3y=4 . \text{ Find the line of regression of } y \text{ on } x .
\displaystyle \text{Answer:}
\displaystyle \text{(vii) Let us assume that } 4x+10y=9 \text{ be the regression line of } y \text{ on } x
\displaystyle 10y = -4x + 9
\displaystyle \Rightarrow y = - \frac{4}{10} x + \frac{9}{10}
\displaystyle \Rightarrow b_{yx} = - \frac{2}{5}
\displaystyle \text{Then } 6x + 3y = 4
\displaystyle \Rightarrow 6x = -3y + 4
\displaystyle \Rightarrow x = - \frac{3}{6} y + \frac{4}{6}
\displaystyle \Rightarrow b_{xy} = - \frac{1}{2}
\displaystyle \text{Therefore } b_{yx}.b_{xy} = (- \frac{2}{5} ).(- \frac{1}{2} ) = \frac{1}{5} < 1
\displaystyle \text{Hence our assumption is true i.e. regression line } y \text{ on } x \text{ is } 4x+10y=9
\displaystyle \\
\displaystyle \text{(viii) If } 1, \omega \text{ and } \omega^2 \text{ are the cube roots of unity, evaluate: } \\ (1-\omega^4+ \omega^8)(1-\omega^8+\omega^{16})
\displaystyle \text{Answer:}
\displaystyle \text{(viii) } (1-\omega^4+ \omega^8)(1-\omega^8+\omega^{16})
\displaystyle = (1-\omega+ \omega^2)(1-\omega^2+\omega)
\displaystyle = (-2 \omega) (-2 \omega^2)
\displaystyle = 4 \omega^3
\displaystyle = 4
\displaystyle \text{(ix) Solve the differential equation: } \log \left(\frac{dy}{dx}\right) = 2x-3y
\displaystyle \text{Answer:}
\displaystyle \text{(ix) } \log \left(\frac{dy}{dx}\right) = 2x-3y
\displaystyle \Rightarrow \frac{dy}{dx} = e^{2x-3y}
\displaystyle \Rightarrow \frac{dy}{dx} = \frac{e^{2x}}{e^{3y}}
\displaystyle \Rightarrow e^{3y} dy = e^{2x} dx
\displaystyle \text{Integrating both sides}
\displaystyle \int e^{3y} dy = \int e^{2x} dx
\displaystyle \frac{e^{3y}}{3} = \frac{e^{2x}}{2} + c
\displaystyle 2e^{3y}-3 e^{2x} = k
\displaystyle \\
\displaystyle \text{(x) If two balls are drawn from a bag containing three red and four blue balls,} \\ \text{find the probability that: }
\displaystyle \text{Answer:}
\displaystyle \text{(x) Total number of ways of drawing } 2 \text{ balls } = {}^{7}C_{2}
\displaystyle \text{(a) } P \text{ (Balls are of same color) } = \frac{{}^{3}C_{2}}{{}^{7}C_{2}} + \frac{{}^{4}C_{2}}{{}^{7}C_{2}} = \frac{{}^{3}C_{2}+{}^{4}C_{2}}{{}^{7}C_{2}} = \frac{ \frac{3!}{2!.1!} + \frac{4!}{2!.2!} }{\frac{7!}{2!.5!}} = \frac{3+6}{7 \times 3} = \frac{3}{7}
\displaystyle \text{(b) } P \text{ (Balls are of different color) } = \frac{{}^{3}C_{1} \times {}^{4}C_{1}}{{}^{7}C_{2}} = \frac{ \frac{3!}{1!.2!} \times \frac{4!}{1!.3!} }{\frac{7!}{2!.5!}} = \frac{3 \times 4}{7 \times 3} = \frac{4}{7}
\\

\displaystyle \textbf{Question 2:}
\displaystyle \text{(a) Using properties of determinants, prove that:}
\displaystyle \left| \begin{array}{ccc} x & y & z \\ x^2 & y^2 & z^2 \\ y+z & z+x & x+y \end{array} \right|= (x-y)(y-z)(z-x)(x+y+z) \hspace{5.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(a) } \Delta = \left| \begin{array}{ccc} x & y & z \\ x^2 & y^2 & z^2 \\ y+z & z+x & x+y \end{array} \right|
\displaystyle \text{Applying } R_3 \rightarrow R_3 + R_1 , \text{ we get}
\displaystyle \Delta = \left| \begin{array}{ccc} x & y & z \\ x^2 & y^2 & z^2 \\ x+y+z & x+y+z & x+y+z \end{array} \right|
\displaystyle \Rightarrow \Delta = (x+y+z)\left| \begin{array}{ccc} x & y & z \\ x^2 & y^2 & z^2 \\ 1 & 1 & 1 \end{array} \right|
\displaystyle \text{Applying } C_1 \rightarrow C_1 - C_2 \text{ and } C_2 \rightarrow C_2 - C_3 , \text{ we get}
\displaystyle \Rightarrow \Delta = (x+y+z)\left| \begin{array}{ccc} x-y & y-z & z \\ x^2 - y^2 & y^2 - z^2 & z^2 \\ 0 & 0 & 1 \end{array} \right|
\displaystyle \Rightarrow \Delta = (x+y+z) (x-y)(y-z) \left| \begin{array}{ccc} 1 & 1 & z \\ x+y & y+z & z^2 \\ 0 & 0 & 1 \end{array} \right|
\displaystyle \text{Expanding along } R_3 , \text{ we get}
\displaystyle = (x+y+z) (x-y)(y-z) (y+z-x-y)
\displaystyle = (x-y)(y-z)(z-x)(x+y+z)
\displaystyle = \text{RHS. Hence proved.}
\\
\displaystyle \text{(b) Find } A^{-1} , \text{ where } A = \begin{bmatrix} 4 & 2 & 3 \\ 1 & 1 & 1 \\ 3 & 1 & -2 \end{bmatrix}
\displaystyle \text{Hence, solve the system of linear equations:}
\displaystyle 4x+2y+3z = 2
\displaystyle x+y+z=1
\displaystyle 3x+y-2z=5 \hspace{12.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(b) } A = \left| \begin{array}{ccc} 4 & 2 & 3 \\ 1 & 1 & 1 \\ 3 & 1 & -2 \end{array} \right|
\displaystyle = 4(-2-1) -2(-2-3)+3(1-3) = -12+10-6 = -8 \neq 0
\displaystyle \Rightarrow A^{-1} \text{ exists}
\displaystyle A^{-1} = \frac{1}{|A|} \ adj. \ A
\displaystyle A_{11} = \left| \begin{array}{cc} 1 & 1 \\ 1 & -2 \end{array} \right| = -3 \quad A_{12} = - \left| \begin{array}{cc} 1 & 1 \\ 3 & -2 \end{array} \right| = 5 \quad A_{13} = \left| \begin{array}{cc} 1 & 1 \\ 3 & 1 \end{array} \right| = -2
\displaystyle A_{21} = - \left| \begin{array}{cc} 2 & 3 \\ 1 & -2 \end{array} \right| = 7 \quad A_{22} = \left| \begin{array}{cc} 4 & 3 \\ 3 & -2 \end{array} \right| = -17 \quad A_{23} = - \left| \begin{array}{cc} 4 & 2 \\ 3 & 1 \end{array} \right| = 2
\displaystyle A_{31} = \left| \begin{array}{cc} 2 & 3 \\ 1 & 1 \end{array} \right| = -1 \quad A_{32} = - \left| \begin{array}{cc} 4 & 3 \\ 1 & 1 \end{array} \right| = -1 \quad A_{33} = \left| \begin{array}{cc} 4 & 2 \\ 1 & 1 \end{array} \right| = 2
\displaystyle adj. \ A = \begin{bmatrix} -3 & 5 & -2 \\ 7 & -17 & 2 \\ -1 & -1 & 2 \end{bmatrix}^T = \begin{bmatrix} -3 & 7 & -1 \\ 5 & -17 & -1 \\ -2 & 2 & 2 \end{bmatrix}
\displaystyle \text{Therefore } A^{-1} = \frac{1}{|A|} adj. \ A= - \frac{1}{8} \begin{bmatrix} -3 & 7 & -1 \\ 5 & -17 & -1 \\ -2 & 2 & 2 \end{bmatrix} = \frac{1}{8} \begin{bmatrix} 3 & -7 & 1 \\ -5 & 17 & 1 \\ 2 & -2 & -2 \end{bmatrix}
\displaystyle \text{Given system of equation is}
\displaystyle 4x+2y+3z = 2
\displaystyle x+y+z=1
\displaystyle 3x+y-2z=5
\displaystyle \text{This can be written as } AX = B
\displaystyle \text{Where } A = \begin{bmatrix} 4 & 2 & 3 \\ 1 & 1 & 1 \\ 3 & 1 & -2 \end{bmatrix} \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \quad \text{and } B = \begin{bmatrix} 2 \\ 1 \\ 5 \end{bmatrix}
\displaystyle \text{Since } |A| \neq 0 , \text{ the given system of equations have a unique solution } X = A^{-1}B
\displaystyle \Rightarrow X = \frac{1}{8} \begin{bmatrix} 3 & -7 & 1 \\ -5 & 17 & 1 \\ 2 & -2 & -2 \end{bmatrix} . \begin{bmatrix} 2 \\ 1 \\ 5 \end{bmatrix} = \frac{1}{8} \begin{bmatrix} 6-7+5 \\ -10+17+5 \\ 4-2-10 \end{bmatrix} = \frac{1}{8} \begin{bmatrix} 4 \\ 12 \\ -8 \end{bmatrix}
\displaystyle \Rightarrow X = \frac{1}{2} \begin{bmatrix} 1 \\ 3 \\ -2 \end{bmatrix}
\displaystyle \Rightarrow x = \frac{1}{2} , y = \frac{3}{2} \text{ and } z = -1 \text{which is the solution to the system of the equations.}
\\

\displaystyle \textbf{Question 3:}
\displaystyle \text{(a) Solve for } x: \sin^{-1} x + \sin^{-1} (1-x) = \cos^{-1} x \hspace{5.0cm} [5]
\displaystyle \text{Answer:} \displaystyle \text{(a) } \sin^{-1} x + \sin^{-1} (1-x) = \cos^{-1} x
\displaystyle \Rightarrow \sin^{-1} x + \sin^{-1} (1-x) = \frac{\pi}{2} - \sin^{-1} x
\displaystyle \Rightarrow \sin^{-1} (1-x) = \frac{\pi}{2} - 2 \sin^{-1} x
\displaystyle \text{Let } \sin^{-1} x = y \Rightarrow x = \sin y
\displaystyle \text{Therefore we get}
\displaystyle \sin^{-1} (1-x) = \frac{\pi}{2} - 2y
\displaystyle \Rightarrow 1- x = \sin \left( \frac{\pi}{2} - 2y \right)
\displaystyle \Rightarrow 1- x = \cos 2y
\displaystyle \Rightarrow 1- x = 1 - 2 \sin^2 y
\displaystyle \Rightarrow 1- x = 1 - 2 x^2
\displaystyle \Rightarrow 2x^2 - x = 0
\displaystyle \Rightarrow x (2x-1) = 0
\displaystyle \Rightarrow x = 0, \frac{1}{2}
\displaystyle \text{Hence the solution of the given equations are } x = 0, \frac{1}{2}
\displaystyle \\
\displaystyle \text{(b) Construct a circuit diagram for the following Boolean function:} \displaystyle (BC+A)(A'B'+C')+A'B'C'
\displaystyle \text{Using laws of Boolean Algebra, simplify the function and draw the simplified circuit.} \hspace{0.2cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(b) } (BC+A).(A'B'+C') + A'B'C'
\displaystyle = BCA'B'+BCC' + AA'B + AC' + A'B'C'
\displaystyle = 0+0+0+AC'+A'B'C'
\displaystyle = C'(A+A'B')
\displaystyle = C'(A+A') (A+B')
\displaystyle = C'(A+B')
\displaystyle \\

\displaystyle \textbf{Question 4:}
\displaystyle \text{(a) Verify Lagrange's Mean Value Theorem for the function }  \\ f(x) = \sqrt{x^2 - x} \text{ in the interval } [1,4] \hspace{2.2cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given, } f(x) = \sqrt{x^2 - x} , x \in [1, 4]
\displaystyle \text{Since } x^2 - x \text{ is continuous on } R \text{ and } \sqrt{x^2-x} \text{ exists in } [1, 4]
\displaystyle \Rightarrow f(x) \text{ is continuous in } [1, 4]
\displaystyle \text{Differentiating the given function w.r.t } x
\displaystyle f'(x) = \frac{1}{2} (x^2 - x)^{-\frac{1}{2}} \cdot (2x-1) = \frac{2x-1}{2\sqrt{x^2 - x}}
\displaystyle \text{which exists for all } x \in R
\displaystyle \text{Thus, both the conditions of Lagrange's mean value theorem is satisfied therefore} \\ \text{at least one c exists in } [1, 4].
\displaystyle \text{Such that } f'(c) = \frac{f(4)-f(1)}{4-1}
\displaystyle \Rightarrow \frac{2c-1}{2\sqrt{c^2-c}} = \frac{\sqrt{12}-0}{3}
\displaystyle 3(2c-1) = 2 \sqrt{c^2 - c} \cdot \sqrt{12}
\displaystyle \text{Squaring both sides we get}
\displaystyle 9(4c^2 + 1 - 4 c) = 48(c^2 - c)
\displaystyle \Rightarrow 12c^2 - 12c + 3 = 16c^2 - 16c
\displaystyle \Rightarrow 4c^2 - 4c - 3 = 0
\displaystyle \Rightarrow (2c-3)(2c+1) = 0
\displaystyle \Rightarrow c = \frac{3}{2} \ or \ \frac{-1}{2}
\displaystyle \text{Clearly, } c = \frac{3}{2} \text{ lies in the interval } [1, 4]
\displaystyle \text{Hence, Lagrange's mean value theorem is verified and } c = \frac{3}{2}
\displaystyle \\
\displaystyle \text{(b) From the following information, find the equation of the Hyperbola and the} \\ \text{equation of its Transverse Axis:}
\displaystyle \text{Focus: } (-2, 1) , \text{ Directrix: } 2x-3y+1 = 0 , e = \frac{2}{\sqrt{3}} \hspace{1.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(b) Let } P(x, y) \text{ be the point on the conic, then}
\displaystyle PS = e \cdot PM
\displaystyle \text{i.e. } \sqrt{(x+2)^2 + (y-1)^2} = \frac{2}{\sqrt{3}} \left( \frac{2x-3y+1}{\sqrt{4+9}} \right)
\displaystyle \Rightarrow 39(x^2 + y^2 + 4x -2y + 5) = 4 (4x^2 + 9y^2 + 1 - 12xy + 4x -6y)
\displaystyle \text{i.e. } 23x^2 + 48xy + 3y^2 + 140 x - 54 y + 191 = 0 \text{ is the required hyperboloa.}
\displaystyle \text{Transverse axis passes through } (-2, 1) \text{ and is perpendicular to Directrix, } 2x-3y+1 = 0
\displaystyle TA: 3x+2y +c=0 \text{ where } -6+2+c=0 \Rightarrow c = 4
\displaystyle \text{Therefore } TA: 3x+2y+4=0
\displaystyle \\

\displaystyle \textbf{Question 5:}
\displaystyle \text{(a) If } y = ( \cot^{-1} x)^2 , \text{ show that } (1+x^2)^2 \frac{d^2y}{dx^2} + 2x(1+x^2) \frac{dy}{dx} = 2 \hspace{2.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given } y = (\cot^{-1} x)^2
\displaystyle \text{Now, } \frac{dy}{dx} = 2 \cot^{-1} x \frac{d}{dx} (\cot^{-1}x)
\displaystyle \Rightarrow \frac{dy}{dx} = 2 \cot^{-1} x \left( \frac{-1}{1+x^2} \right)
\displaystyle \Rightarrow (1+x^2) \frac{dy}{dx} = -2 \cot^{-1} x
\displaystyle \text{Differentiating once again w.r.t } x
\displaystyle (1+x^2) \frac{d^2 y}{dx^2} + 2x \frac{dy}{dx} = -2 \left( \frac{-1}{1+x^2} \right)
\displaystyle \Rightarrow (1+x^2) \frac{d^2 y}{dx^2} + 2x \frac{dy}{dx} = \frac{2}{1+x^2}
\displaystyle \Rightarrow (1+x^2)^2 \frac{d^2 y}{dx^2} + 2x (1+x^2) \frac{dy}{dx} =2
\displaystyle \\
\displaystyle \text{(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 ) \hspace{4.2cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(b) Let the height of the cylinder } = h \text{ and the radius of the cylinder } = r
\displaystyle \text{Given Radius of the sphere } = 3\sqrt{3}
\displaystyle \text{If } V \text{ is the volume of the cylinder, then } V = \pi r^2h
\displaystyle \text{Let } O \text{ be the center of the sphere and } COC' \perp AB
\displaystyle \text{For } \triangle OCA ,
\displaystyle (3\sqrt{3})^2 = \left( \frac{h}{2} \right)^2 + r^2
\displaystyle \Rightarrow r^2 = 27 - \frac{h^2}{4}
\displaystyle \text{Therefore } V = \pi \left( 27 - \frac{h^2}{4} \right) h
\displaystyle V = 27 \pi h - \pi \frac{h^3}{4}
\displaystyle \text{Therefore } \frac{dV}{dh} = 27 \pi - \frac{3 \pi h^2}{4}
\displaystyle \frac{d^2V}{dh^2} = - \frac{6 \pi h}{4}
\displaystyle \text{For maxima and minima,}
\displaystyle \frac{dV}{dh} = 0
\displaystyle \therefore 27 \pi = \frac{3 \pi h^2}{4}
\displaystyle \Rightarrow h^2 = \frac{4 \times 27}{3} = 36
\displaystyle \Rightarrow h = \pm 6
\displaystyle \text{and } \left( \frac{d^2V}{dh^2} \right)_{h=6} = \frac{-6\pi \times 6}{4} < 0
\displaystyle \therefore V \text{ is maximum when } h = 6 , \text{ putting } h = 6
\displaystyle r^2 = 27 - \frac{36}{4}
\displaystyle r^2 = 18
\displaystyle \therefore V = \pi r^2 h = \pi \times 18 \times 6 = 108 \pi \ cm^3
\displaystyle \\

\displaystyle \textbf{Question 6:}
\displaystyle \text{(a) Evaluate: } \int \limits_{}^{} \frac{ \cos^{-1} x}{x^2} dx \hspace{9.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given, } \int \limits_{}^{} \frac{ \cos^{-1} x}{x^2} dx
\displaystyle \text{Put } \cos^{-1} x = t
\displaystyle \Rightarrow x = \cos t
\displaystyle \Rightarrow dx = - \sin t \ dt
\displaystyle \text{Therefore } \int \limits_{}^{} \frac{ \cos^{-1} x}{x^2} dx = \int \limits_{}^{} \frac{t}{\cos^2 t} (- \sin t) dt
\displaystyle = - \int \limits_{}^{} t (\sec t \tan t) dt
\displaystyle = - t \sec t - \int \limits_{}^{} \sec t dt
\displaystyle = -t \sec t + \log | \sec t + \tan t| + c
\displaystyle = - \frac{t}{\cos t} + \log \left| \frac{1+ \sin t}{\cos t} \right| + c
\displaystyle \text{Now putting the value of } t \text{ back in the expression}
\displaystyle \int \limits_{}^{} \frac{ \cos^{-1} x}{x^2} dx = - \frac{\cos^{-1} x}{x} + \log \left| \frac{1+ \sin (\cos^{-1}x)}{x} \right| + c
\displaystyle \Rightarrow \int \limits_{}^{} \frac{ \cos^{-1} x}{x^2} dx = - \frac{\cos^{-1} x}{x} + \log \left| \frac{1+\sqrt{1-x^2}}{x} \right| + c
\displaystyle \\
\displaystyle \text{(b) Find the area bounded by the curve } y = 2x-x^2 \text{ and the line } y = x \hspace{3.0cm} [5]
\displaystyle \text{Answer:} \displaystyle \text{(b) Given: } y = 2x - x^2 \text{ and line of intersection } y = x
\displaystyle -y = x^2 - 2x
\displaystyle -y + 1 = x^2 - 2x + 1
\displaystyle -(y-1) = (x-1)^2 \text{ which represents a downward parabola with vertex at } (1, 1)
\displaystyle \text{Line of intersection } y = x
\displaystyle \text{Putting } y = x \text{ in the above equation}
\displaystyle -(x-1)= (x-1)^2
\displaystyle \Rightarrow -x + 1 = x^2 - 2x + 1
\displaystyle \Rightarrow x^2 - x = 0
\displaystyle \Rightarrow x (x-1) = 0
\displaystyle \Rightarrow x = 0, 1
\displaystyle \text{Therefore the point of intersections are } (0, 0) \text{ and } (1, 1)
\displaystyle \text{Therefore the area enclosed between the curve } y = 2x - x^2 \text{ and the line } y = x \text{ is }
\displaystyle \int \limits_{0}^{1} (2x - x^2 - x) dx = \int \limits_{0}^{1} (x - x^2) dx = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1}
\displaystyle = \left( \frac{1}{2} - \frac{1}{3} \right) - (0-0)
\displaystyle = \frac{1}{6} \text{ sq. units.}
\displaystyle \\

\displaystyle \textbf{Question 7:}
\displaystyle \text{(a) Find the Karl Pearsons coefficient of correlation between } x \text{ and } y \\  \text{ for the following data. \hspace{10.0cm} [5] }
\displaystyle \begin{array}{c|cccccccc} x & 16 & 18 & 21 & 20 & 22 & 26 & 27 & 15 \\ y & 22 & 25 & 24 & 26 & 25 & 30 & 33 & 14 \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{(a)}
\displaystyle \begin{array}{c|c|c|c|c|c|c} x & y & dx = x - 20 & dy = y - 25 & dx^2 & dy^2 & dx.dy \\ \hline 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 \\ \hline & & 5 & -1 & 135 & 221 & 152 \end{array}
\displaystyle r = \frac{ n \Sigma (dx.dy) - \Sigma dx \times \Sigma dy} { \sqrt{n \Sigma dx^2 - (\Sigma dx)^2 } \cdot \sqrt{n \Sigma dy^2 - (\Sigma dy)^2} }
\displaystyle \Rightarrow r = \frac{ 8 \times 152 - (5 \times -1)}{ \sqrt{8 \times 135 - 5^2 } \cdot \sqrt{8 \times 221 - (-1)^2} }
\displaystyle \Rightarrow r = \frac{1216+5}{ \sqrt{1080 - 25} \cdot \sqrt{1768-1} }
\displaystyle = 0.894
\displaystyle \\
\displaystyle \text{(b) The following table shows the mean and standard deviation of the marks of Mathematics} \\ \text{and Physics scored by students in a school:}
\displaystyle \begin{array}{c|cc} & \text{Mathematics} & \text{Physics} \\ \hline \text{Mean} & 84 & 81 \\ \text{Standard Deviation} & 7 & 4 \end{array}
\displaystyle \text{The correlation co-efficient between the given marks is } 0.86 . \text{ Estimate the likely marks in} \\ \text{Physics in the marks in Mathematics at } 92 . \hspace{0.2cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(b)}
\displaystyle \text{Mean marks in Mathematics } \overline{x} = 84
\displaystyle \text{Mean marks in Physics } \overline{y} = 81
\displaystyle \sigma_x = 7, \sigma_y = 4
\displaystyle r = 0.86
\displaystyle \text{Therefore } b_{yx} = r \cdot \frac{\sigma_y}{\sigma_x} = 0.86 \times \frac{4}{7} = 0.49
\displaystyle \text{Therefore regression equation of } y \text{ on } x
\displaystyle y - \overline{y} = b_{yx} ( x - \overline{x})
\displaystyle \Rightarrow y = 81 = 0.49 (x-84)
\displaystyle \Rightarrow y = 81 = 0.49x - 41.16
\displaystyle \Rightarrow y = 0.49x + 39.84
\displaystyle \text{Putting } x = 92, y = 45.08 + 39.84 = 84.92
\displaystyle \text{Hence the likely marks in Physics are } 84.92
\displaystyle \\

\displaystyle \textbf{Question 8:}
\displaystyle \text{(a) } Bag \ A \text{ contains three red and four white balls. } Bag \ B \text{ contains two red} \\ \text{and three white balls. If one ball is drawn from } Bag \ A \text{ and two balls are drawn} \\ \text{from } Bag \ B , \text{ find the probability that:}
\displaystyle \text{(i) One ball is red and two balls are white}
\displaystyle \text{(ii) All the three balls are of the same color } \hspace{5.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(a) Possible selection are as follows:}
\displaystyle \text{i) 1 Red ball from Bag A, 2 white ball from Bag B}
\displaystyle \text{1 white ball from bag A, 1 white ball from bag B, 1 red ball from Bag B}
\displaystyle \text{Therefore P(one ball is red and two balls are white)}
\displaystyle = \frac{{}^{3}C_{1}}{{}^{7}C_{1}} \times \frac{{}^{3}C_{2}}{{}^{5}C_{2}} + \frac{{}^{4}C_{1}}{{}^{7}C_{1}} \times \frac{{}^{3}C_{1} \times {}^{2}C_{1}}{{}^{5}C_{2}}
\displaystyle = \frac{3}{7} \times \frac{3}{10} + \frac{4}{7} \times \frac{3 \times 2}{10}
\displaystyle = \frac{9}{70} + \frac{24}{70} = \frac{33}{70}
\displaystyle \text{ii) Possible selections are as follows:}
\displaystyle \text{1 red ball from Bag A, 2 red balls from Bag B}
\displaystyle \text{1 white ball from Bag A and 2 white balls from bag B}
\displaystyle \text{Therefore P (all the three balls are of the same color)}
\displaystyle = \frac{{}^{3}C_{1}}{{}^{7}C_{1}} \times \frac{{}^{2}C_{2}}{{}^{5}C_{2}} + \frac{{}^{4}C_{1}}{{}^{7}C_{1}} \times \frac{{}^{3}C_{2}}{{}^{5}C_{2}}
\displaystyle = \frac{3}{7} \times \frac{1}{10} + \frac{4}{7} \times \frac{3}{10}
\displaystyle = \frac{3}{70} + \frac{12}{70} = \frac{15}{70}
\displaystyle \\
\displaystyle \text{(b) Three persons, Aman, Bipin and Mohan attempt a Mathematics problems} \\ \text{independently. The odds in favor of Aman and Mohan solving the problem are} \\ 3:2 \text{ and } 4:1 \text{ respectively and the odds against Bipin solving the problem are } \\ 2:1 . \text{ Find:}
\displaystyle \text{(i) The probability that all the three will solve the problem}
\displaystyle \text{(ii) the probability that the problem will be solved. } \hspace{5.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(b) Odd against } = \frac{1-p}{p}
\displaystyle \text{Odds in favor } = \frac{p}{1-p}
\displaystyle A : \text{ Event Aman solves the problem}
\displaystyle B : \text{ Event Bipin solves the problem}
\displaystyle C : \text{ Event Mohan solves the problem}
\displaystyle \begin{array}{c|c|c} \text{Aman} & \text{Bipin} & \text{Mohan} \end{array}
\displaystyle \frac{3}{2} = \frac{p}{1-p}
\displaystyle \Rightarrow p = \frac{3}{5}
\displaystyle \Rightarrow P(A) = \frac{3}{5}
\displaystyle \frac{2}{1} = \frac{1-p}{p}
\displaystyle \Rightarrow p = \frac{1}{3}
\displaystyle \Rightarrow P(B) = \frac{1}{3}
\displaystyle \frac{4}{1} = \frac{p}{1-p}
\displaystyle \Rightarrow p = \frac{4}{5}
\displaystyle \Rightarrow P(C) = \frac{4}{5}
\displaystyle P( A \cap B \cap C) = P(A). P(B). P(C)
\displaystyle \text{i) Probability that all three will solve the problem } = \frac{3}{5} \times \frac{1}{3} \times \frac{4}{5} = \frac{4}{25}
\displaystyle \text{ii) Probability that the problem is not solved = probability that all three fail to solve the problem}
\displaystyle \Rightarrow P ( \overline{A} \cap \overline{B} \cap \overline{C}) = P( \overline{A}).P( \overline{B}).P( \overline{C})
\displaystyle = \left( 1 - \frac{3}{5} \right) \cdot \left( 1 - \frac{1}{3} \right) \cdot \left( 1 - \frac{4}{5} \right)
\displaystyle = \frac{2}{5} \cdot \frac{2}{3} \cdot \frac{1}{5} = \frac{4}{75}
\displaystyle \text{Therefore the probability that the problem will be solved } = 1 - \frac{4}{75} = \frac{71}{75}
\displaystyle \\

\displaystyle \textbf{Question 9:}
\displaystyle \text{(a) Find the locus of the complex number } z = x+iy , \text{ satisfying the relation}\\  \arg (z-1) = \frac{\pi}{4} \text{ and } |z-2-3i| = 2.
\displaystyle \text{Illustrate the locus on the Argand plane. } \hspace{5.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(a) Let } z = x + iy
\displaystyle \arg(z-1) = \frac{\pi}{4}
\displaystyle \arg(x+iy-1) = \frac{\pi}{4}
\displaystyle \tan^{-1} \frac{y}{x-1} = \frac{\pi}{4}
\displaystyle \frac{y}{x-1} = 1
\displaystyle \text{Therefore } y = x-1
\displaystyle |z-2-3i| = 2
\displaystyle \Rightarrow |x+iy -2-3i| = 2
\displaystyle \Rightarrow (x-2)^2 + (y-3)^2 = 4
\displaystyle \Rightarrow (x-2)^2 + (x-1-3)^2 = 4
\displaystyle \Rightarrow x^2 - 4x +4 + x^2 - 8x +16 = 4
\displaystyle \Rightarrow 2x^2 - 12x + 16 = 0
\displaystyle (x-4)(x-2)=0 \Rightarrow x = 4, 2
\displaystyle \text{When } x = 4, y = 3 \text{ and when } x = 2, y = 1
\displaystyle \text{Therefore the locus will be } (2, 1) \text{ and } (4, 3)
\displaystyle \\
\displaystyle \text{(b) Solve the following differential equation:}
\displaystyle y e^y dx = (y^3 + 2 x e^y) dy , \text{ given that } x =0, y = 1 . \hspace{6.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given, } y e^y dx = (y^3 + 2 x e^y) dy
\displaystyle \Rightarrow y e^y \frac{dx}{dy} = y^3 + 2x e^y
\displaystyle \Rightarrow \frac{dx}{dy} = \frac{2}{y} x + y^2 e^{-y}
\displaystyle \Rightarrow \frac{dx}{dy} - \frac{2}{y} x = y^2 e^{-y}
\displaystyle \text{It is a linear differential equation in } x.
\displaystyle \text{I.F. } = e^{\int -\frac{2}{y} dy}
\displaystyle = e^{-2 \log |y|}
\displaystyle = e^{\log |y|^{-2}}
\displaystyle = |y|^{-2} = \frac{1}{y^2}
\displaystyle \text{Therefore, general solution is}
\displaystyle x \cdot \frac{1}{y^2} = \int y^2 e^{-y} \cdot \frac{1}{y^2} dy + c
\displaystyle x \cdot \frac{1}{y^2} = \int e^{-y} dy + c
\displaystyle \Rightarrow x \cdot \frac{1}{y^2} = \frac{e^{-y}}{-1} + c
\displaystyle \Rightarrow \text{Given that } x = 0 \text{ and } y = 1
\displaystyle \Rightarrow 0 = - e^{-1} + c
\displaystyle \Rightarrow c = e^{-1}
\displaystyle \text{Substituting the value of } c \text{ we get}
\displaystyle \frac{x}{y^2} = -e^{-y} + e^{-1}
\displaystyle \Rightarrow x = y^2 (-e^{-y} + e^{-1})
\displaystyle \\

\displaystyle \textbf{Question 10:}
\displaystyle \text{(a) If } \overrightarrow{a} \text{ and } \overrightarrow{b} \text{ are unit vectors and } \theta \text{ is the angle between them,}
\displaystyle \text{then show that } |\overrightarrow{a} - \overrightarrow{b} | = 2 \sin \frac{\theta}{2}. \hspace{8.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(a)} \ \overrightarrow{a}.\overrightarrow{b} = |\overrightarrow{a}| |\overrightarrow{b}| \cos \theta
\displaystyle \Rightarrow \overrightarrow{a}.\overrightarrow{b} = 1.1 \cos \theta = \cos \theta
\displaystyle \text{Now, } |\overrightarrow{a} - \overrightarrow{b} |^2 = (\overrightarrow{a} - \overrightarrow{b})^2
\displaystyle = |\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 - 2 \overrightarrow{a}.\overrightarrow{b}
\displaystyle = 1 + 1 - 2 \cos \theta
\displaystyle = 2 ( 1 - \cos \theta)
\displaystyle \Rightarrow |\overrightarrow{a} - \overrightarrow{b} |^2 = 2.2 \sin^2 \frac{\theta}{2}
\displaystyle \Rightarrow |\overrightarrow{a} - \overrightarrow{b} | = 2 \sin \frac{\theta}{2}
\displaystyle \\
\displaystyle \text{(b) If the value of } \lambda \text{ for which the four points } A, B, C, D \text{ 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} \text{ and } -4\hat{i}+4\hat{j}+4\hat{k} \text{ are coplanar.} \hspace{5.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(b) Let } A, B, C, D \text{ be the given points whose position vectors are } -\hat{j} - \hat{k} ; 4\hat{i}+5\hat{j}+ \lambda \hat{k} ; 3\hat{i}+9\hat{j} +4\hat{k} \text{ and } -4\hat{i}+4\hat{j}+4\hat{k}
\displaystyle \overrightarrow{AB} = 4\hat{i}+5\hat{j}+ \lambda \hat{k} - (-\hat{j} - \hat{k}) = 4\hat{i} + 6\hat{j} + (\lambda + 1) \hat{k}
\displaystyle \overrightarrow{AC} = 3\hat{i}+9\hat{j} +4\hat{k} - (-\hat{j} - \hat{k}) = 3\hat{i}+10\hat{j} +5\hat{k}
\displaystyle \overrightarrow{AD} = -4\hat{i}+4\hat{j}+4\hat{k} - (-\hat{j} - \hat{k}) = -4\hat{i}+5\hat{j}+5\hat{k}
\displaystyle \text{Since the points A, B, C and D are coplanar, vectors } \overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD} \text{ are coplanar.}
\displaystyle \Rightarrow \begin{vmatrix} 4 & 6 & \lambda + 1 \\ 3 & 10 & 5 \\ -4 & 5 & 5 \end{vmatrix} = 0
\displaystyle \Rightarrow 4 (50-25) - 3 (30 - 5 - 5 \lambda) - 4 (30-10-10\lambda) = 0
\displaystyle \Rightarrow 100 - 75 + 15\lambda - 80 +40\lambda = 0
\displaystyle \Rightarrow -55 + 55 \lambda = 0
\displaystyle \Rightarrow -1 + \lambda = 0
\displaystyle \Rightarrow \lambda = 1
\displaystyle \\

\displaystyle \textbf{Question 11:}
\displaystyle \text{(a) Find the equation of a line passing through the point } (-1, 3, -2) \text{ and perpendicular} \\ \text{to the lines: } \frac{x}{1} = \frac{y}{2} = \frac{z}{-3} \text{ and } \frac{x+2}{2} = \frac{y-1}{5} = \frac{z+1}{3}. \hspace{0.2cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(a) Any line passing through the point } (-1, 3, -2) \text{ is}
\displaystyle \frac{x-(-1)}{a} = \frac{y-3}{b} = \frac{z-(-2)}{c} \text{ where a, b and c are its direction vectors}
\displaystyle \text{It is perpendicular to the lines } \frac{x}{1} = \frac{y}{2} = \frac{z}{3}
\displaystyle \frac{x+2}{-3} = \frac{y-1}{2} = \frac{z+1}{5}
\displaystyle \Rightarrow 1.a + 2.b + 3.c=0
\displaystyle \Rightarrow a +2b+3c=0
\displaystyle \Rightarrow -3.a + 2.b + 5.c=0
\displaystyle \Rightarrow -3a+2b+5c=0
\displaystyle \text{Subtracting ii) from i) we get}
\displaystyle 4a-2c=0
\displaystyle \Rightarrow c = 2a
\displaystyle \text{and hence from i) we get}
\displaystyle a+ 2b + 3.2a =0
\displaystyle \Rightarrow 2b=-7a
\displaystyle \Rightarrow b = \frac{-7a}{2}
\displaystyle \text{From iv) and iii) we get}
\displaystyle a:b:c =a: \frac{-7a}{2} :2a
\displaystyle \Rightarrow a:b:c = 2: -7: 4
\displaystyle \text{Putting these values in the equation of the line}
\displaystyle \frac{x+1}{2} = \frac{y-3}{-7} = \frac{z+2}{4}
\displaystyle \\
\displaystyle \text{(b) Find the equation of planes parallel to the plane } 2x-4y+4z=7 \text{ and which are at a} \\ \text{distance of five units from the point } (3, -1, 2). \hspace{0.2cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(b) The given plane is } 2x-4y+4z=7
\displaystyle \text{Equation of any plane parallel to above equation is}
\displaystyle 2x - 4y + 4z + k = 0
\displaystyle \text{Now ii) is at a distance of } 5 \text{ units from the point } (3, -1, 2) \text{ of}
\displaystyle \frac{| 2 \times 3 - 4 (-1) + 4 \times 2 +k|}{\sqrt{2^2 + (-4)^2 + 4^2}} = 5
\displaystyle \Rightarrow \frac{|18+k|}{6} = 5
\displaystyle \Rightarrow |18+k| = 30
\displaystyle \Rightarrow 18+k = 30 \text{ or } 18+k=-30
\displaystyle \Rightarrow k = 12 \text{ or } k = -48
\displaystyle \text{Substituting these values in ii) the equations of the required planes are}
\displaystyle 2x-4y+4z+12=0 \text{ and } 2x-4y+4z-48=0
\displaystyle \text{or } x-2y+2z+6=0 \text{ and } x-2y+2z-24=0
\displaystyle \\

\displaystyle \textbf{Question 12:}
\displaystyle \text{(a) If the sum and the product of the mean and variance of a Binomial Distribution} \\ \text{are } 1.8 \text{ and } 0.8 \text{ respectively, find the probability distribution and the probability of at} \\ \text{least one success.} \hspace{1.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given,}
\displaystyle np + npq = 1.8 \Rightarrow np(1+q) = 1.8
\displaystyle np.npq = 0.8 \Rightarrow n^2p^2q = 0.8
\displaystyle \text{Dividing the square of i) by ii) we get}
\displaystyle \frac{n^2p^2(1+q)^2}{n^2p^2q} = \frac{1.8 \times 1.8}{0.8}
\displaystyle \Rightarrow \frac{(1+q)^2}{q} = \frac{3.24}{0.8}
\displaystyle \Rightarrow \frac{(1+q)^2}{q} = \frac{81}{20}
\displaystyle \Rightarrow 20+2 \times 20q + 20q^2 = 81q
\displaystyle \Rightarrow 20q^2-41q+20=0
\displaystyle \Rightarrow 20q^2 - 25q - 16q +20 = 0
\displaystyle \Rightarrow 5q(4q-5)-4(4q-5) = 0
\displaystyle \Rightarrow (4q-5)(5q-4) = 0
\displaystyle \text{Therefore } q = \frac{5}{4} \ or \ \frac{4}{5}
\displaystyle \text{But } q \neq \frac{5}{4} \text{ as } \frac{5}{4} > 1
\displaystyle \text{Therefore } q = \frac{4}{5}
\displaystyle \Rightarrow p = \frac{1}{5}
\displaystyle \text{From i), } n \times \frac{1}{5} \left( 1 + \frac{4}{5} \right) = 1.8
\displaystyle \Rightarrow 9n = 25 \times 1.8
\displaystyle \Rightarrow n = \frac{25 \times 1.8}{9}
\displaystyle \Rightarrow n = \frac{45}{9}
\displaystyle \text{Therefore } n = 5
\displaystyle \text{Hence, the binomial distribution is } (p+q)^n = \left( \frac{1}{5} + \frac{4}{5} \right)^5
\displaystyle \text{Probability of getting at least 1 success } = P(X \geq 1)
\displaystyle = 1- P(X = 0)
\displaystyle = 1- {}^{5}C_{0} \left( \frac{1}{5} \right)^0 \left( \frac{4}{5} \right)^5
\displaystyle = 1 - \frac{1024}{3125} = 0.67
\displaystyle \\
\displaystyle \text{(b) For } A, \ B \text{ and } C , \text{ the chances of being selected as the manager of firm are } \\ 4:1:2 \text{ respectively.} \text{The probabilities for them to introduce a radical chance in the} \\ \text{marketing strategy are } 0.3, \ 0.8 \text{ and } 0.5 \text{ respectively. If a chance takes place; find} \\ \text{the probability that it is due to the appointment of B.} \hspace{3.2cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(b) Let } E_1, E_2 \text{ and } E_3 \text{ and } E \text{ be the events as defined below}
\displaystyle E_1 = A \text{ is selected as a manager}
\displaystyle E_2 = B \text{ is selected as a manager}
\displaystyle E_3 = C \text{ is selected as a manager}
\displaystyle \text{and } E = \text{ radical change occurs in marketing strategy}
\displaystyle P(E_1) = \frac{4}{4+1+2} = \frac{4}{7}
\displaystyle P(E_2) = \frac{1}{4+1+2} = \frac{1}{7}
\displaystyle P(E_3) = \frac{2}{4+1+2} = \frac{2}{7}
\displaystyle \text{Given, } P(E/E_1) = 0.3 , P(E/E_2) = 0.8 , P(E/E_3) = 0.5
\displaystyle \text{We want to find the probability that the radical change in marketing strategy occurred due to the appointment of B i.e}
\displaystyle 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)}
\displaystyle = \frac{\frac{1}{7} \times 0.8}{\frac{4}{7} \times 0.3+\frac{1}{7} \times 0.8+\frac{2}{7} \times 0.5}
\displaystyle = \frac{0.8}{1.2+0.8+1} = \frac{4}{15}
\displaystyle \\

\displaystyle \textbf{Question 13:}
\displaystyle \text{(a) If Mr. Nirav deposits Rs. 250 at the beginning of each month in an account that pays} \\ \text{an interest of } 6\% \text{ per annum compounded monthly, how many months will be required for} \\ \text{the deposit to at least Rs. 6390?} \hspace{5.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(a)}
\displaystyle a = 250
\displaystyle i = \frac{6}{100 \times 12} = 0.005
\displaystyle n = m
\displaystyle S = \frac{a}{i} (1+i) [(1+i)^m - 1]
\displaystyle 6390 = \frac{250}{0.005} (1.005) [(1.005)^m - 1 ]
\displaystyle 0.1271 = [ (1.005)^{m} - 1]
\displaystyle 1.1271 = 1.005^{m}
\displaystyle \log 1.1271 = m \log 1.005
\displaystyle m = \frac{\log 1.1271}{\log 1.005}
\displaystyle m = 23.98
\displaystyle \Rightarrow m = 24 \text{ months}
\displaystyle \\
\displaystyle \text{(b) A mill owner buys two types of machines } A \text{ and } B \text{ for his mill. } \text{Machine } A \\ \text{occupies } 1000 \text{ sqm of area and requires } 12 \text{ men to operate it.; while } \text{Machine } B \\ \text{occupies } 1200 \text{ sqm of area and requires } 8 \text{ men to operate it. } \text{The owner has } 7600 \text{ sqm} \\ \text{of area available and } 72 \text{ men to operate the machines. If } \text{Machine } A \text{ produces } 50 \\ \text{ units and Machine } B \text{ produces } 40 \text{units daily, how } \text{many machines of each type} \\ \text{should be buy to maximize the daily output? Use linear } \text{programming to find} \\ \text{the solution.} \hspace{5.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{(b) Data given}
\displaystyle \begin{array}{c|c|c|c} & \text{Machine A} & \text{Machine B} & \text{Machine C} \\ \hline \text{Area Needed (Sq. m)} & 1000 & 1200 & 7600 \\ \text{Labor Force} & 12 & 8 & 72 \\ \text{Daily Output (units)} & 50 & 40 & - \end{array}
\displaystyle \text{Let } x \text{ and } y \text{ be the number of Machines A and Machine B respectively.}
\displaystyle \text{Constraints}
\displaystyle 1000x + 1200y \leq 7600 \quad \Rightarrow 5x + 6y \leq 38
\displaystyle 12x + 8 y \leq 72 \quad \Rightarrow 3x + 2y \leq 18
\displaystyle x\geq 0, y \geq 0\displaystyle \text{Total output } Z = 50x + 40y
\displaystyle \text{So minimize } Z = 50x + 40y
\displaystyle \text{Subject to } 5x + 6y \leq 38 \text{ And } 3x + 2y \leq 18 \quad x\geq 0, y \geq 0
\displaystyle \text{Now for } 3x + 2y = 18
\displaystyle \begin{array}{c|cc} x & 6 & 0 \\ y & 0 & 9 \end{array}
\displaystyle \text{Now for } 5x + 6y = 38
\displaystyle \begin{array}{c|cc} x & 7.2 & 0 \\ y & 0 & \frac{19}{3} \end{array}
\displaystyle \text{The vertices of the feasible region OAPD are } O(0, 0), A (6, 0), P(4, 3) \text{ and } D (0, 6.33)
\displaystyle \begin{array}{c|c} \text{Corner Points} & \text{Object Function } Z = 50x + 40 y \\ \hline O (0, 0) & Z = 50 \times 0 + 40 \times 0 = 0 \\ A (6, 0) & Z = 50 \times 6 + 40 \times 0 = 300 \\ P (4, 3) & Z = 50 \times 4 + 40 \times 3 = 320 \\ D (0, 6.33) & Z = 50 \times 0 + 40 \times 6.33 = 253.33 \end{array}
\displaystyle \text{Thus we see that } Z \text{ is maximum at } P (4, 3). \text{ Therefore number of Machine A } = 4 \text{ and number of Machine } B = 3.
\displaystyle \\

\displaystyle \textbf{Question 14:}
\displaystyle \text{(a) A bill of Rs. } 60000 \text{ was drawn on 1st April 2011 at 4 months and discounted for} \\ \text{Rs. } 58560 \text{ at a bank. If the rate of interest was } 12\% \text{ per annum, on what date was the} \\ \text{bill discounted?} \hspace{1.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{a) Banker's discount } (BD) = 60000 - 58560 = 1440
\displaystyle \text{Let } t \text{ be the expired period in years}
\displaystyle BD = Ani
\displaystyle 1440 =60000 \times \frac{12}{100} \times t \Rightarrow t = \frac{1}{5} \text{ years } = 73 \text{ days}
\displaystyle \text{Legal due date of the bill}
\displaystyle \text{1st April 2011 + 4 months + 3 days of grace = 4th August 2011}
\displaystyle \text{The bill was cashed 73 days before 4th August (4 days in August, 31 days in July, 30 days} \\ \text{in June, 8 days in May) which comes to 23rd May 2011 as the date}
\displaystyle \\
\displaystyle \text{(b) A company produces a commodity with Rs. } 24000 \text{ fixed cost. The variable cost} \\ \text{is estimated to be } 25\% \text{ of the total revenue recovered on selling the product at a rate} \\ \text{of Rs. } 8 \text{ per unit. FInd the following:}
\displaystyle \text{(i) Cost function}
\displaystyle \text{(ii) Revenue function}
\displaystyle \text{(iii) Break even point} \hspace{5.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{b) Supposed that } x \text{ number of units are produced and sold.}
\displaystyle \text{Given fixed cost } = 24000 \text{ Rs.}
\displaystyle \text{Revenue } R(x) = 8x
\displaystyle \text{i) As each unit's variable cost of } x \text{ units is } 25\% \text{ of revenue,}
\displaystyle \text{Therefore Variable cost of } x \text{ units } = 0.25 \times 8x = 2
\displaystyle \text{Therefore total cost of producing } x \text{ units } = 24000 + 2x
\displaystyle \text{ii) Price of one unit } = 8 \text{ Rs.}
\displaystyle \text{Therefore revenue on selling } x \text{ units } = R(x) = 8x
\displaystyle \text{ii) At break even values}
\displaystyle C(x) = R(x)
\displaystyle 24000 + 2x = 8 x
\displaystyle \Rightarrow 24000 = 6x
\displaystyle \Rightarrow x = 4000
\displaystyle \text{Hence the break even point is } 4000 \text{ units produced.}
\displaystyle \\

\displaystyle \textbf{Question 15:}
\displaystyle \text{(a) The price index for the following data for the year 2011 taking 2001 as the} \\ \text{base year was } 127. \text{ The simple average of price relative method was used. Find} \\ \text{the value of } x . \hspace{5.2cm} [5]
\displaystyle \begin{array}{c|cccccc} \text{Items} & A & B & C & D & E & F \\ \hline \text{Price (Rs. Per unit) in year 2001} & 80 & 70 & 50 & 20 & 18 & 25 \\ \text{Price (Rs. Per unit) in year 2011} & 100 & 87.50 & 61 & 22 & x & 32.50 \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{a) Using simple averages of price relatives, the price index of 2011 taking 2001 as the} \\ \text{base year was } 127 . \text{ From the following data we find } x
\displaystyle \begin{array}{c|cccccc} P_0 & 80 & 70 & 50 & 20 & 18 & 25 \\ \hline P_1 & 100 & 87.50 & 61 & 22 & x & 32.5 \\ \hline PR & 125 & 125 & 122 & 110 & \frac{100x}{18} & 130 \end{array}
\displaystyle \frac{\Sigma PR}{N} = 127
\displaystyle \Rightarrow 612 + \frac{100x}{18} = 127 \times 6 \Rightarrow x = 27
\displaystyle \\
\displaystyle \text{(b) The profit of a paper bag manufacturing company (in lakhs / millions of Rs.) } \\ \text{during each month pf a year are:}
\displaystyle \begin{array}{c|cccccccccccc} \text{Month} & \text{Jan.} & \text{Feb.} & \text{Mar.} & \text{Apr.} & \text{May} & \text{Jun.} & \text{Jul.} & \text{Aug.} & \text{Sept.} & \text{Oct.} & \text{Nov.} & \text{Dec.} \\ \hline \text{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 \end{array}
\displaystyle \text{Plot the given data on a graph sheet. Calculate the four monthly moving} \\ \text{averages and plot these on the same graph sheet.} \hspace{2.0cm} [5]
\displaystyle \text{Answer:}
\displaystyle \text{b)}
\displaystyle \begin{array}{c|c|c|c|c} \text{Month} & \text{Profit} & \text{4 monthly totals} & \text{4 monthly average} & \text{4 monthly centered moving average} \\ \hline \text{Jan} & 1.2 & & & \\ \text{Feb} & 0.8 & 5 & 1.25 & \\ \text{Mar} & 1.4 & 5.8 & 1.45 & 1.35 \\ \text{Apr} & 1.6 & 7.4 & 1.85 & 1.65 \\ \text{May} & 2.0 & 9.6 & 2.4 & 2.125 \\ \text{Jun} & 2.4 & 12.8 & 3.2 & 2.8 \\ \text{July} & 3.6 & 14.2 & 3.55 & 3.375 \\ \text{Aug} & 4.8 & 13.6 & 3.4 & 3.475 \\ \text{Sep} & 3.4 & 10.8 & 2.7 & 3.05 \\ \text{Oct} & 1.8 & 7.2 & 1.8 & 2.25 \\ \text{Nov} & 0.8 & & & \\ \text{Dec} & 1.2 & & & \end{array}
\displaystyle \\

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