MATHEMATICS

(Maximum Marks: 100)

(Time Allowed: Three Hours)

(Candidates are allowed additional 15 minutes for only reading the paper. 

They must NOT start writing during this time)

The Question Paper consists of three sections A, B and C. 

Candidates are required to attempt all questions from Section A and all question EITHER from Section B OR Section C

Section A: Internal choice has been provided  in three questions  of four marks each and two questions of six marks each.

Section B: Internal choice has been provided in two question of four marks each.

Section C: Internal choice has been provided in two question of four marks each.

All working, including rough work, should be done on the same sheet as, and adjacent to, the rest of the answer. 

The intended marks for questions or parts of questions are given in brackets [ ].

Mathematical tables and graphs papers are provided.

Section – A (80 Marks)

Question 1:                                                                                       [10 \times 3]

 

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

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

(iii) Solve: \cos^{-1} (\sin \cos^{-1} x) = \frac{\pi}{6}

(iv) Using L’Hospital’s rule, evaluate: \lim \limits_{x \to 0} \frac{x- \sin x}{x^2 \sin x}

(v) Evaluate: \int \limits_{}^{}\frac{2y^2}{y^2+4} \ dy

(vi) Evaluate: \int \limits_{0}^{3} f(x) \ dx , where f(x) = \Bigg \{ \begin{matrix}  cos \ 2x, 0 \leq x \leq \frac{\pi}{2} \\  \\ 3, \ \ \ \ \ \ \frac{\pi}{2} \leq x \leq 3 \end{matrix}

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

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

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

(x) Solve the differential equation: \sin^{-1} (\frac{dy}{dx}) = x + y

Answer:

(i)   M^2-kM-I_2=0

\Rightarrow \begin{bmatrix}  1 & 2 \\ 2 & 3 \end{bmatrix} \times \begin{bmatrix}  1 & 2 \\ 2 & 3 \end{bmatrix} - k \times \begin{bmatrix}  1 & 2 \\ 2 & 3 \end{bmatrix} - \begin{bmatrix}  1 & 0 \\ 0 & 1 \end{bmatrix} = 0

\Rightarrow \begin{bmatrix}  5 & 8 \\ 8 & 13 \end{bmatrix} - \begin{bmatrix}  k & 2k \\ 2k & 3k \end{bmatrix} - \begin{bmatrix}  1 & 0 \\ 0 & 1 \end{bmatrix} = 0

\Rightarrow \begin{bmatrix}  5-k & 8-2k \\ 8-2k & 13-3k \end{bmatrix} = \begin{bmatrix}  1 & 0 \\ 0 & 1 \end{bmatrix} 

\Rightarrow 5 - k = 1 \Rightarrow k = 4 

Similarly, 13-3k = 1 \Rightarrow k = 4 

(ii)  \frac{2b^2}{a} =8

e = \frac{1}{3}

b^2 = 4a

b^2 = a^{2(1-e^2)}

\Rightarrow 4a = a^{2(1-\frac{1}{9})}

\Rightarrow a = \frac{9}{2}

b^2 = 18

Equation of ellipse: \frac{4x^2}{81} + \frac{y^2}{18} = 1 \Rightarrow 8x^2 + 9y^2  = 162

(iii)  \cos^{-1} (\sin \cos^{-1} x) = \frac{\pi}{6}

\sin (\cos^{-} x) = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}

\sqrt{1-x^2} =  \frac{\sqrt{3}}{2}

1-x^2 = \frac{3}{4}

\Rightarrow x^2 = \frac{1}{4}

x = \pm \frac{1}{2}

(iv)  \lim \limits_{x \to 0} \frac{x- \sin x}{x^2 \sin x}

Apply L’Hospital’s rule

=\lim _{x\to \:0} \left(\frac{1-\cos x}{2x\sin x+\cos x \ x^2}\right)

=\lim _{x\to \:0} \left(\frac{1-\cos x}{x \left(x\cos x+2\sin x \right)}\right)

Apply L’Hospital’s rule

=\lim _{x\to \:0} \left(\frac{\sin x}{-x^2\sin x+4x\cos x+2\sin x}\right)

=\lim _{x\to \:0} \left(\frac{-\sin x}{\sin x \ x^2-4\cos x \ x-2\sin x}\right)

Apply L’Hospital’s rule

=\lim _{x\to \:0} \left(\frac{-\cos x}{x^2\cos x+6x\sin \left(x\right)-6\cos x}\right)

\mathrm{Plug\:in\:the\:value}\:x=0

= \frac{-\cos 0}{0^2\cos 0+6\cdot \:0\cdot \sin 0-6\cos 0}

= \frac{1}{6}

(v)  \int \limits_{}^{}\frac{2y^2}{y^2+4} \ dy

= 2 \int \limits_{}^{} \Big(1 - \frac{4}{y^2+4} \Big)

= 2y - 8 .  \frac{1}{2} \tan^{-1} \frac{y}{2} + c

= 2y - 4 \tan^{-1} \frac{y}{2} + c

(vi)  \int \limits_{0}^{3} f(x) \ dx , where f(x) = \Bigg \{ \begin{matrix}  cos \ 2x, 0 \leq x \leq \frac{\pi}{2} \\  \\ 3, \ \ \ \ \ \ \frac{\pi}{2} \leq x \leq 3 \end{matrix}

Therefore 

= \int \limits_{0}^{\frac{\pi}{2}} \cos 2x dx + \int \limits_{\frac{\pi}{2}}^{3} 3 dx

= \Big[ \frac{\sin 2x}{2} \Big]_{0}^{\frac{\pi}{2}} + \Big[ 3x \Big]_{\frac{\pi}{2}}^{3}

= (0-0) + 3(3 - \frac{\pi}{2} )

= 9 - \frac{3\pi}{2} 

(vii)  Let the line of regression of x on y   be

4x+2y -3 = 0  \Rightarrow x = \frac{-1}{2} y + \frac{3}{4} 

\Rightarrow b_{xy}  = \frac{-1}{2} 

Let the line of regression of y on x be

3x+6y+5 = 0 \Rightarrow y = \frac{-1}{2} x - \frac{5}{6} 

\Rightarrow b_{yx} = \frac{-1}{2} 

Therefore r^2 = b_{yx} \times b_{xy} = \frac{-1}{2} \times \frac{-1}{2} = \frac{1}{4} 

Hence r = - \frac{1}{2}   since both b_{xy} and b_{yx} are negative.

(viii) Probability P(E) = Probability of drawing a spade + Probability of drawing an ace – Probability of drawing ace of spade

= \frac{13}{52} + \frac{4}{52} - \frac{1}{52} 

= \frac{16}{52} = \frac{4}{13} 

(ix)  \frac{a + b\omega + c\omega^2}{c+a\omega + b\omega^2}

Multiplying numerator and denominator by \omega^2

= \frac{\omega^2(a + b\omega + c\omega^2)}{\omega^2(c+a\omega + b\omega^2)}

= \frac{\omega^2(a + b\omega + c\omega^2)}{(c\omega^2+a\omega^3 + b\omega^4)}

We know \omega^3 = 1, \omega^4 = \omega^3.\omega = \omega

= \frac{\omega^2(a + b\omega + c\omega^2)}{(c\omega^2+a + b\omega)}

= \omega^2

(x)  \sin^{-1} (\frac{dy}{dx}) = x + y

\frac{dy}{dx} = \sin (x+y)

Let x + y = v

\frac{d(v-x)}{dx} = \sin v

\frac{dv}{dx} -1 = \sin v

\frac{dv}{dx} = 1+ \sin v

\int \limits_{}^{} \frac{dv}{1+ \sin x} = \int \limits_{}^{} dx 

\int \limits_{}^{} \frac{(1- \sin x) dv}{1- \sin^2 x} = \int \limits_{}^{} dx 

\int \limits_{}^{} \frac{(1- \sin x) dv}{\cos^2 x} = \int \limits_{}^{} dx 

\int \limits_{}^{} \sec^2 v - \tan v \sec v \ dx = \int \limits_{}^{} dx 

\tan v - \sec v  = x + c 

\Rightarrow \tan (x+y) - \sec (x+y)  = x + c

\\

Question 2: 

(a) Using properties of determinants, prove that:

\left| \begin{array}{ccc}  1+a^2-b^2 & 2ab & -2b \\ 2ab & 1-a^2+b^2 & 2a \\ 2b & -2a & 1-a^2-b^2 \end{array} \right|= (1+a^2+b^2)^3     [5]

(b) Given two matrices A \ and \  B

A = \begin{bmatrix} 1 & -2 & 3 \\ 1 & 4 & 1 \\ 1 & -3 & 2 \end{bmatrix} and B = \begin{bmatrix}  11 & -5 & -14 \\ -1 & -1 & 2 \\ -7 & 1 & 6 \end{bmatrix} . Find AB .

Using this result, solve the following system of equation:

x-2y+3z=6, \ x+4y+z=12, \ x-3y+2z=1     [5]

Answer:

(a)  LHS = \left| \begin{array}{ccc}  1+a^2-b^2 & 2ab & -2b \\ 2ab & 1-a^2+b^2 & 2a \\ 2b & -2a & 1-a^2-b^2 \end{array} \right|

C_1 \rightarrow C_1 - b C_3

= \left| \begin{array}{ccc}  1+a^2+b^2 & 2ab & -2b \\ 0 & 1-a^2+b^2 & 2a \\ b(1+a^2+b^2) & -2a & 1-a^2-b^2 \end{array} \right|

= (1+a^2+b^2) \left| \begin{array}{ccc}  1 & 2ab & -2b \\ 0 & 1-a^2+b^2 & 2a \\ b & -2a & 1-a^2-b^2 \end{array} \right|

C_2 \rightarrow C_2+ a C_3

= (1+a^2+b^2) \left| \begin{array}{ccc}  1 & 0 & -2b \\ 0 & 1+a^2+b^2 & 2a \\ b & -a(1+a^2+b^2) & 1-a^2-b^2 \end{array} \right|

= (1+a^2+b^2)^2 \left| \begin{array}{ccc}  1 & 0 & -2b \\ 0 & 1 & 2a \\ b & -a & 1-a^2-b^2 \end{array} \right|

R_3 \rightarrow R_3 - bR_1

= (1+a^2+b^2)^2 \left| \begin{array}{ccc}  1 & 0 & -2b \\ 0 & 1 & 2a \\ 0 & -1 & 1-a^2+b^2 \end{array} \right|

= (1+a^2+b^2)^2 (1 - a^2 + b^2 + 2a^2)

= (1+a^2+b^2)^3

= RHS. Hence proved.

(b)  A = \begin{bmatrix} 1 & -2 & 3 \\ 1 & 4 & 1 \\ 1 & -3 & 2 \end{bmatrix}

B = \begin{bmatrix}  11 & -5 & -14 \\ -1 & -1 & 2 \\ -7 & 1 & 6 \end{bmatrix}

AB = \begin{bmatrix} 1 & -2 & 3 \\ 1 & 4 & 1 \\ 1 & -3 & 2 \end{bmatrix} \times \begin{bmatrix}  11 & -5 & -14 \\ -1 & -1 & 2 \\ -7 & 1 & 6 \end{bmatrix} = \begin{bmatrix}  -8 & 0 & 0 \\ 0 & -8 & 0 \\ 0 & 0 & -8 \end{bmatrix} = -8 \begin{bmatrix}  1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

AB = -8I

Now,

A = \begin{bmatrix} 1 & -2 & 3 \\ 1 & 4 & 1 \\ 1 & -3 & 2 \end{bmatrix}      X = \begin{bmatrix} x  \\ y  \\ z  \end{bmatrix}      C = \begin{bmatrix} 6  \\ 12  \\ 1  \end{bmatrix}

X = A^{-1} C

\begin{bmatrix} x  \\ y  \\ z  \end{bmatrix} = - \frac{1}{8} \begin{bmatrix}  11 & -5 & -14 \\ -1 & -1 & 2 \\ -7 & 1 & 6 \end{bmatrix} \times \begin{bmatrix} 6  \\ 12  \\ 1  \end{bmatrix} = \begin{bmatrix} 1  \\ 2  \\ 3  \end{bmatrix}

Hence x = 1, y = 2 \ and \ z = 3

\\

Question 3:

(a) Solve the equation for x :

\sin^{-1} \frac{5}{x} + \sin^{-1} \frac{12}{x} = \frac{\pi}{2} , x \neq 0.      [5]

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

Answer:

(a)  \sin^{-1} \frac{5}{x} + \sin^{-1} \frac{12}{x} = \frac{\pi}{2}

\sin^{-1} \frac{5}{x}   = \frac{\pi}{2} - \sin^{-1} \frac{12}{x}

\frac{5}{x} = \sin \Big( \frac{\pi}{2} - \sin^{-1} \frac{12}{x} \Big)

\frac{5}{x} =\cos \Big( \sin^{-1} \frac{12}{x} \Big)

\frac{5}{x} = \cos \Big( \cos^{-1} \sqrt{1 - (\frac{12}{x})^2} \Big)

\frac{5}{x} = \sqrt{1 - ( \frac{12}{x})^2}

\Rightarrow \frac{25}{x^2} = 1 - ( \frac{12}{x} )^2

\Rightarrow \frac{25}{x^2} = 1 - \frac{144}{x^2} 

\Rightarrow \frac{169}{x^2} = 1

\Rightarrow x = \pm 13

(b)  ABC+AB'C+A'B'C

= ACB + ACB' + A'B'C

= AC (B+B') + A'B'C Since (B+B') = 1

= AC + A'B'C

= (A+A')(A+B')C

= (A+B')C

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Question 4:

(a) Verify Lagrange’s Mean Value Theorem for the following function:

f(x) = 2 \sin x + \sin 2x \ on \ [0, \pi]      [5]

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

Answer:

(a)  f(x) = (2 \sin x + \sin 2x) is continuous in [0, \pi]

f(0) = 0, f(\pi) = 0

f'(x) exists in (0, \pi)

f'(x) = 2 \cos x + 2 \cos 2x

All the conditions  of Lagrange’s Mean Value theorem satisfied  there exists 'c' in (0, \pi) such that  f'(c)  = \frac{f(b) - f(a)}{b-a} 

2 \cos c  + 2 \cos 2c = 0

2 \cos^2 c + \cos c - 1 = 0

\cos c = -1 \Rightarrow \cos c = \cos \pi \Rightarrow c = \pi (not possible)

or \cos c = \cos \frac{\pi}{3} \Rightarrow c = \frac{\pi}{3} \in (0, \frac{\pi}{3} ) which lies between 0 and \pi .

Hence Lagrange’s Mean Value theorem is verified.

(b)  Foci = (0, \pm \sqrt{10})

be = \sqrt{10}

a^2 = b^2 (e^2 - 1) = b^2 e^2 - b^2

a^2 = 10 - b^2

Let the equation be : \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1

\Rightarrow \frac{9}{10-a^2} - \frac{4}{a^2} = 1

\Rightarrow 9a^2 - 40  + 4a^2  = 10a^2  - a^4

\Rightarrow a^4 + 3a^2  - 40 = 0

\Rightarrow (a^2+8) (a^2 - 5)  = 0  

\Rightarrow a^2 = -8 ( not possible as a^2 cannot be negative )   or a^2 = 5

Therefore a^2 = 5 and b^2 = 10-5 = 5

Hence the equation is  \frac{y^2}{5} - \frac{x^2}{5} = 1 \Rightarrow y^2 -x^2 = 5

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Question 5:

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

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

Answer:

(a)   y = e^{m \cos^{-1} x}

Differentiating with respect to x

\frac{dy}{dx} = e^{m \cos^{-1} x} \times \frac{-m}{\sqrt{1-x^2}} 

\sqrt{1-x^2} \frac{dy}{dx} = -me^{m \cos^{-1} x} 

\sqrt{1-x^2} \frac{dy}{dx} = -my

Differentiating again with respect to x

\Rightarrow \sqrt{1-x^2} \frac{d^2y}{dx^2} - \frac{2x}{2\sqrt{1-x^2}} \frac{dy}{dx} = -m \frac{dy}{dx} 

(1-x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = - m \sqrt{1-x^2} \frac{dy}{dx} = -m (-my) = m^2y

(b)  Please refer to the adjoining figure2018-07-21_21-00-56

AB = 2x and BC = 2y

(2x)^2+(2y)^2 = (20)^2 \Rightarrow x^2 + y^2 = 100

P = 4x + 4y = 4x + 4\sqrt{100-x^2}

Differentiating with respect to x

\frac{dP}{dx} = 4 - \frac{4x}{\sqrt{100-x^2}} = 0

\Rightarrow 100 - x^2 = x^2 

\Rightarrow x^2 = 50

\Rightarrow x = 5\sqrt{2}

Differentiating with respect to x

\frac{d^2P}{dx^2} = -4 \Big[  \frac{\sqrt{100-x^2} - \frac{x(-x)}{ \sqrt{100-x^2}} }{100-x^2}    \Big] = \frac{-4 \times 100}{(100-x^2)^{\frac{3}{2}}} < 0

Hence the perimeter is maximum when x = 5\sqrt{2}

Therefore y = 5\sqrt{2}  \Rightarrow x = y

Therefore ABCD is a square with side  of 10\sqrt{2}

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Question 6:

(a) Evaluate: \int \frac{ \sec x}{1+ cosec \ x} \ dx      [5]

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

Answer:

(a)  I = \int \frac{ \sec x}{1+ \mathrm{cosec} x} \ dx

I = \int \frac{ \sin x}{\cos x (1+ \sin x)} \ dx

I = \int \frac{ \sin x \cos x}{\cos^2 x (1+ \sin x)} \ dx

Now put \sin x = t  \Rightarrow dt = \cos x dx

I = \int \frac{ t }{(1-t^2)(1+t)} dt

I = \int \frac{ t }{(1-t)(1+t)^2} dt

Now

\frac{ t }{(1-t^2)(1+t)} = \frac{A}{1-t} + \frac{B}{1+t} + \frac{C}{(1+t)^2} 

\Rightarrow t = A(1+t)^2 + B(1-t)(1+t) + C(1-t)

\Rightarrow t = A(1+t^2 + 2t) + B(1-t^2) + (C(1-t)

\Rightarrow t = t^2(A-B) + t(2A-C) + (A+B+C)

\Rightarrow A-B = 0, 2A-C = 1 \ and \  A+B+C = 0

\Rightarrow A = \frac{1}{4} , B = \frac{1}{4} , C = - \frac{1}{2} 

Thus

I = \int   \frac{A}{1-t} + \frac{B}{1+t} + \frac{C}{(1+t)^2}  dt

I = \frac{1}{4}\int \Big[ \frac{1}{1-t} + \frac{1}{1+t} - \frac{2}{1+t^2} \Big] \ dt

= \frac{1}{4} | - \log |1-t| + \log |1+t| + \frac{2}{1+t} | + c

= \frac{1}{4} | \log | \frac{1+t}{1-t} | + \frac{2}{1+t}  | + c

Therefore

= \frac{1}{4} | \log | \frac{1+\sin x}{1-\sin x} | + \frac{2}{1+\sin x}  | + c

 (b)  We have to calculate

\int \limits_{0}^{2} \sqrt{4-x^2} \ dx - \int \limits_{0}^{2} 2-x \ dx

First Integral:2018-07-21_21-07-35

\int \limits_{0}^{2} \sqrt{4-x^2} \ dx

Let x = 2 \sin u \Rightarrow dx = 2 \cos u \ du

= \int \limits_{0}^{2} \sqrt{4- 4\sin^2 u} \ 2 \cos u \ du   = \int \limits_{0}^{2} 4 \sqrt{1- \sin^2 u} \  \cos u \ du   = 4 \int \limits_{0}^{2} \cos^2 u \ du

Now \cos^2 u = \frac{1+\cos 2u}{2} , Therefore

= 4 \int \limits_{0}^{2} \frac{1+\cos 2u}{2} \ du   = 2 \int \limits_{0}^{2} 1 + \cos 2u \ du   = \Big[ 2(u + \frac{1}{2} \sin 2u) \Big]_{0}^{2}

Now substituting back

x = 2 \sin u \Rightarrow u = \sin^{-1} (\frac{x}{2})

= \Big[ 2(\sin^{-1} (\frac{x}{2}) + \frac{1}{2} \sin (2 \sin^{-1} (\frac{x}{2})) \Big]_{0}^{2}

=  \Big[ 2(\sin^{-1} (1) + \frac{1}{2} \sin (2 \sin^{-1} (1)) \Big] -  \Big[ 2(\sin^{-1} (0) + \frac{1}{2} \sin (2 \sin^{-1} (0)) \Big]   = \pi

Second Integral:

\int \limits_{0}^{2} 2-x \ dx   = \Big[ 2x - \frac{1}{2} x^2 \Big]_{0}^{2}   = 4 - 2  = 2

Hence \int \limits_{0}^{2} \sqrt{4-x^2} \ dx - \int \limits_{0}^{2} 2-x \ dx = \pi - 2

\\

Question 7:

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

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

Competitors A B C D E F G H I J
Judge A 2 11 11 18 6 5 8 16 13 15
Judge B 6 11 16 9 14 20 4 3 13 17

Answer:

(a) 

x y xy x^2 y^2
9 -4 -36 81 16
10 -3 -30 100 9
11 -1 -11 121 1
12 0 0 144 0
13 1 13 169 1
14 3 42 196 9
15 5 75 225 25
16 8 128 256 64
\Sigma x = 100 \Sigma y = 9 \Sigma xy = 181 \Sigma x^2 = 1292 \Sigma y^2 = 125

\overline{x} = \frac{100}{8} = 12.5

\overline{y} = \frac{9}{8} = 1.125

b_{yx} = \frac{\Sigma xy - \frac{1}{n} \Sigma x . \Sigma y }{\Sigma x^2 - \frac{1}{n} (\Sigma x)^2 }   = \frac{181 - \frac{1}{8} 100 . 9 }{1292 - \frac{1}{8} (100)^2 } = \frac{68.5}{42}   = 1.63

b_{xy} = \frac{\Sigma xy - \frac{1}{n} \Sigma x . \Sigma y }{\Sigma y^2 - \frac{1}{n} (\Sigma y)^2 }   = \frac{181 - \frac{1}{8} 100 . 9 }{125 - \frac{1}{8} (9)^2 } = \frac{68.5}{114.875}   = 0.596

Line of regression of y \ on \  x

y - \frac{9}{8}   = 1.63 (x - 12.5) \Rightarrow y = 1.63x - 19.25

Line of regression of x \ on \  y 

x - 12.5 = 0.596 (y - \frac{9}{8} ) \Rightarrow x = 0.596y + 11.83

When x = 13.5, y = 1.63 \times 13.5 - 19.25 = 2.76

(b) 

Judge A Judge B Rank (R_x) Rank (R_y) d = R_x-R_y d^2
2 6 10 8 2 4
11 11 5.5 6 -0.5 0.25
11 16 5.5 3 2.5 6.25
18 9 1 7 -6 36
6 14 8 4 4 16
5 20 9 1 8 64
8 4 7 9 -2 4
16 3 2 10 -8 64
13 13 4 5 -1 1
15 17 3 2 1 1
\Sigma d^2 = 196.5

r = 1 - 6 \Big[ \frac{ \Sigma d^2 + \frac{1}{12} \Sigma (m^3 - m)}{n(n^2-1)} \Big]

r = 1 - 6 \Big[ \frac{ 196.5 + \frac{1}{12} \Sigma (2^3 - 2)}{10(99)} \Big] = 1 - \frac{6 \times 197}{990} = -0.194

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Question 8:

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

(b) Three person A, B \ and \  C shoot to hit a target. If A hits the target 4 times out of 5 trials, B hits it 3 times in 4 trials and C hits is 2 times in 3 trials, find the probability that:

(i) Exactly two person hit the target.

(ii) At least two person hit the target.

(iii) Non hit the target.     [5]

Answer:

(a) Please note: The number of balls in the urn change based on what color is drawn.

Every time you get a Black ball, not only the ball is put back in the urn, another black ball is put in the urn. So the number of balls increase every time you pull a black ball.

In case you draw a white ball, it is not put back in the urn. So the number of balls decrease when you pull white ball.

Therefore the probability that the third ball is black is:

P(E) = P (WWB) + P(WBB) + P(BWB) + P(BBB)

= \frac{2}{4} \times \frac{1}{3} \times \frac{2}{2} +  \frac{2}{4} \times \frac{2}{3} \times \frac{3}{4} +  \frac{2}{4} \times \frac{2}{5} \times \frac{3}{4} +  \frac{2}{4} \times \frac{3}{5} \times \frac{4}{6}

= \frac{1}{6} + \frac{1}{4} + \frac{3}{20} + \frac{1}{5}

= \frac{23}{30}

(b)  Given: P(A) = \frac{4}{5} , P(B) = \frac{3}{4} and P(C) = \frac{2}{3}

Therefore P(\overline{A}) = \frac{1}{5} , P(\overline{B}) = \frac{1}{4}  and P(\overline{C}) = \frac{1}{3}

(i) Probability of at least two hitting the target

= P(AB\overline{C})+P(A\overline{B}C)+P(\overline{A}BC)

=   \frac{4}{5} \times  \frac{3}{4} \times  \frac{1}{3} +  \frac{4}{5} \times  \frac{1}{4} \times  \frac{2}{3} +  \frac{1}{5} \times  \frac{3}{4} \times  \frac{2}{3}   = \frac{26}{60} = \frac{13}{30}

(ii) At least two people hit the target

= P(AB\overline{C})+P(A\overline{B}C)+P(\overline{A}BC) + P(ABC)

=   \frac{4}{5} \times  \frac{3}{4} \times  \frac{1}{3} +  \frac{4}{5} \times  \frac{1}{4} \times  \frac{2}{3} +  \frac{1}{5} \times  \frac{3}{4} \times  \frac{2}{3} + \frac{4}{5} \times  \frac{3}{4} \times  \frac{2}{3}   = \frac{26}{60} + \frac{24}{60} = \frac{50}{60} = \frac{5}{6}

(iii) None hit the target

= P(\overline{A}.\overline{B}.\overline{C}) =  \frac{1}{5} . \frac{1}{4} . \frac{1}{3} = \frac{1}{60}

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Question 9:

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

(b) Solve the differential equation:

e^{x/y}(1 - \frac{x}{y} ) + (1 + e^{x/y}) \frac{dx}{dy} = 0 \ when \ x=0, y=1      [5]

Answer:

(a)  |\omega| = 1

| \frac{2 - iz}{2z-i} | = 1

\Rightarrow |2 - iz| = |2z-i|

Given z = x + iy 2018-07-21_21-09-59

|2 - i(x+iy)| = |2(x+iy)-i|

|2-ix+y| = |2x+2iy-i|

|(2+y)-ix| = |2x+i(2y-i)|

\sqrt{ (-x)^2 + (2+y)^2 } = \sqrt{(2x)^2+(2y-1)^2}

4 + y^2 + 4y + x^2 = 4x^2 + 4y^2 + 1 - 4y

\Rightarrow 3x^2 + 3y^2 -8y -3 = 0

\Rightarrow x^2 + y^2 - \frac{8}{3} y -1 = 0 … … … … … (i)

Now we know that the equation of a circle with center (h,k) and radius r is of the form (x-h)^2 + (y - k)^2 = r^2

Therefore (i) can be written as

(x-0)^2 + (y- \frac{4}{3} )^2 = ( \frac{5}{3} )^2

Therefore center is (0, - \frac{4}{3} ) and radius is \frac{5}{3} .

(b)  e^{\frac{x}{y}} \Big( 1 - \frac{x}{y} \Big) + \Big( 1 + e^{\frac{x}{y}} \Big) \frac{dx}{dy} = 0

Substitute x = vy \Rightarrow  \frac{dx}{dy} = v + y \frac{dv}{dy}

e^{v}(1-v) + (1+e^v).\Big(v + y \frac{dv}{dy} \Big) = 0

(v+e^v)+(1-e^v) y . \frac{dv}{dy} = 0

\Big( \frac{1+e^v}{v+e^v} \Big) dv + \frac{dy}{y} = 0

\int \limits_{}^{}  \Big( \frac{1+e^v}{v+e^v} \Big) dv = - \int \limits_{}^{}  \frac{dy}{y}

\log (v+e^v) = - \log y + \log c

\log \{(v+e^v).y \} = \log c

\log \Big[ \Big( \frac{x}{y} + e^{\frac{x}{y}} \Big).y \Big] = \log c

\Big[ \Big( \frac{x}{y} + e^{\frac{x}{y}} \Big).y \Big] = c

x + y e^{\frac{x}{y}} = c

When x = 0 and y = 1 we get c = 1

Hence x + y e^{\frac{x}{y}} = 1

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Section – B (20 Marks)

Question 10:

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

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

Answer:

(a)   In the adjoining diagram, we have a circle with center O (origin) and the vertices of the triangle are A, B and C. AB is the diameter.2018-07-21_21-11-56

Let \overrightarrow{OA} = \overrightarrow{a} \ and \  \overrightarrow{OC} = \overrightarrow{c}

Therefore \overrightarrow{CA} = \overrightarrow{OA} - \overrightarrow{OC} = \overrightarrow{a} - \overrightarrow{c}

\overrightarrow{CB} = \overrightarrow{OB} - \overrightarrow{OC} = -\overrightarrow{a} - \overrightarrow{c}

\overrightarrow{CA}. \overrightarrow{CB} = ( \overrightarrow{a} - \overrightarrow{c})(-\overrightarrow{a} - \overrightarrow{c})

= -\overrightarrow{a}.\overrightarrow{a}- \overrightarrow{a}.\overrightarrow{c} + \overrightarrow{c}.\overrightarrow{a} + \overrightarrow{c}.\overrightarrow{c}

= -|\overrightarrow{a}|^2 + |\overrightarrow{c}|^2

= - r^2 + r^2 = 0   (since r is the radius)

Therefore \angle ACB is a right angle.

(b)  The volume of a parallelopiped:

\begin{bmatrix} 2 & - 3 & -4 \\ 1 & 2 & -1 \\ 3 & 1 & 2 \end{bmatrix}

= 2\begin{bmatrix} 2 & - 1  \\ 1 & 2   \end{bmatrix} + 3\begin{bmatrix} 1 & - 1  \\ 3 & 2   \end{bmatrix}  - 4 \begin{bmatrix} 1 & 1  \\ 3 & 1   \end{bmatrix}

= 2 \times 5 + 3 \times 5  - 4 \times (-5) = 45 \ cubic \ units.

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Question 11:

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

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

Answer:

(a)  Given planes x+y+z+1 = 0 and 2x-3y+5z-2=0 and the point (-1, 2, 1)

The equation of a line passing through the intersection of the plans is given as

(x+y+z+1) + k (2x-3y+5z-2) = 0 

\Rightarrow 5k = 3 \Rightarrow k = \frac{3}{5}

Hence the equation of line will be

(x+y+z+1) + \frac{3}{5} (2x-3y+5z-2) = 0 

\Rightarrow 5x + 5 y + 5 z + 5 + 6x -9y+15z-6 = 0

\Rightarrow 11x - 4y + 20 z -1 = 0

(b)  \overrightarrow{a_1} = \hat{i} + 2 \hat{j} + 3 \hat{k}

\overrightarrow{a_2} = 2\hat{i} + 4 \hat{j} + 5 \hat{k}

\overrightarrow{b} = 2\hat{i} + 3 \hat{j} + 4 \hat{k}

\overrightarrow{a_2} - \overrightarrow{a_1} = 2\hat{i} + 4 \hat{j} + 5 \hat{k} - (\hat{i} + 2 \hat{j} + 3 \hat{k}) = \hat{i} + 2 \hat{j} + 2 \hat{k}

|\overrightarrow{b} |= \sqrt{4+9+16} = \sqrt{29}

(\overrightarrow{a_2} - \overrightarrow{a_1}) \times \overrightarrow{b} = \begin{bmatrix} \hat{i} &  \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 3 & 4 \end{bmatrix} = 2\hat{i} - 0 \hat{j} - \hat{k}

|(\overrightarrow{a_2} - \overrightarrow{a_1}) \times \overrightarrow{b} | = \sqrt{4+1}

\frac{|(\overrightarrow{a_2} - \overrightarrow{a_1}) \times \overrightarrow{b} |}{|\overrightarrow{b} |} = \sqrt{\frac{5}{29}} 

Hence the shortest distance = 0.415 units

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Question 12:

(a) Box \  I contains 2 white and 3 black balls. Box \ II contains 4 white and 1 black balls and Box \ III contains 3 white and 4 black balls. A dice having 3 red, 2 yellow and 1 green face, is thrown to select a box. If red face turns up we pick Box \  I , if yellow face turns up we pick Box \ II otherwise we pick Box \ III . The we draw a ball from a selected box. If the ball drawn is white, what is the probability that the dice had turned up with a Red face.     [5]

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

Answer:

(a)  Given: Probability of a red face when the dice is rolled : P(A) = \frac{3}{6}

Probability of a yelloq face when the dice is rolled : P(B) = \frac{2}{6}

Probability of a green face when the dice is rolled : P(C)  = \frac{1}{6}

Let D be the probability of drawing a White ball. Therefore

Probability of drawing a white ball from Bag I: P(D/A) = \frac{2}{5}

Probability of drawing a white ball from Bag II: P(D/B) = \frac{4}{5}

Probability of drawing a white ball from Bag III: P(D/C) = \frac{3}{7}

Therefore the probability of white ball drawn when red face show up:

P(A/D) = \frac{P(A) \times P(D/A)}{P(A) \times P(D/A) + P(B) \times P(D/B) + P(C) \times P(D/C)}

=  \frac{\frac{3}{6} \times \frac{2}{5}}{\frac{3}{6} \times \frac{2}{5} + \frac{2}{6} \times \frac{4}{5} + \frac{1}{6} \times \frac{3}{7}}

(b)  Number of dice : 5

P(odd number) = \frac{1}{2} 

P(even number) =  \frac{1}{2} 

 P(x \leq 3) = 1 - P(x =4, 5)

 = 1 - ^5C_4 \Big(\frac{1}{2}\Big)^2.\Big(\frac{1}{2}\Big) - ^5C_5 \Big(\frac{1}{2}\Big)^5

 = 1 - \frac{5}{32} - \frac{1}{32} 

 = \frac{26}{32} = \frac{13}{16} = 0.81

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Section – C (20 Marks)

Question 13:

(a) Mr. Nirav borrowed Rs. 50000 from the bank for 5 years. The rate of interest is 9\% per annum compounded monthly. Find the payment he makes if he pays back at the beginning of each month.     [5]

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

Food Vitamin A Vitamin B Vitamin C
X 1 unit 2 units 3 units
Y 2 units 2 units 1 unit

One kg of food X costs Rs. \ 24 and one kg of food Y costs Rs. \ 36 . Using linear programming, find the least cost of the total mixture which will contain the required vitamins.     [5]

Answer:

(a)  Principal = 50000 \ Rs.

Annual rate of interest = 9\% \Rightarrow Monthly rate of interest = 0.0075\%

Now P = \frac{A}{i} (1+i) [1- (1+i)^{-n}]

50000 = \frac{A}{0.0075} (1+0.0075) [1- (1+0.0075)^{-60}]

50000 = \frac{A}{0.0075} (1.0075) [1- (1+0.0075)^{-60}]

A = \frac{50000 \times 0.0075}{(1.0075) [1- (1+0.0075)^{-60}]} = 1032.19

Hence monthly installment = 1032.19 \ Rs.

(b) Let  us say we have x kgs of food X and y kgs of food Y .

Therefore  we have the following inequations:

x + 2y \geq 10

2x + 2y \geq 12

3x + y \geq 8

Now lets plot these lines. We got the following graph

2018-07-21_18-43-31

We have to minimize z = 24x + 36y 

Hence the four point of intersection from the graph are:

(10, 0) \Rightarrow Cost = 24 \times 10 + 36 \times 0 = 240 \ Rs. 

(2, 4) \Rightarrow Cost = 24 \times 2 + 36 \times 4 = 192 \ Rs. 

(1, 5) \Rightarrow Cost = 24 \times 1 + 36 \times 5 = 204 \ Rs. 

(0, 8) \Rightarrow Cost = 24 \times 0 + 36 \times 8 = 288 \ Rs. 

Hence, the 2 \ kg \ of \ X  and 4 kg \ of \ Y  should be bought.

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Question 14:

(a) A bill of Rs. \ 7650 was drawn on 8th March 2013, at 7 months. It was discounted on 18th May, 2013 and the holder of the bill received Rs. \ 7497 . What is the rate of interest charged by the bank?     [5]

(b) The average cost function, AC for a commodity is given by AC = x + 5 + \frac{36}{x} in terms of output x . Find:

(i) The total cost, C and marginal cost, MC as a function of x

(ii) The outputs for which AC increases.     [5]  

Answer:

(a)  Face value of the bill (FV) = 7650 \ Rs.

Discounted value of the bill (DV) = 7497 \ Rs.

Banker’s discount = (7650 -7497) = 153 \ Rs.

Nominal date due  is 8th October

Legal due date  of the bill is 11th October (3 days of grace period)

Number of unexpired days  from 8th May  to 11th October  is 146  days \Rightarrow n = \frac{146}{365} = \frac{2}{5} 

Banker’s discount = Ani 

\Rightarrow 153 = 7650 \times \frac{2}{5} \times r

\Rightarrow r = \frac{1}{20} = 5\%

(b)  AC = x + 5 + \frac{36}{x}

Cost function C = (x + 5 + \frac{36}{x} ) x = x^2 + 5x + 36

Marginal cost = \frac{dC}{dx} = 2x + 5

\frac{dAC}{dx} = 1- \frac{36}{x^2} 

For AC to be increasing, \frac{dAC}{dx} > 0

\Rightarrow 1- \frac{36}{x^2} > 0

\Rightarrow (x-6)(x+6) >0

Case 1: x-6 > 0 \ and \ x+6 > 0 \Rightarrow x > 6 \ and \ x > -6 \Rightarrow x > 6

Case 2: x-6 < 0 \ and \ x+6 < 0 \Rightarrow x < 6 \ and \ x < -6 \Rightarrow x < -6

But x cannot be negative. Therefore x > 6

Hence the average cost function increase  if the output x > 6

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Question 15:

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

Commodity Price in Rs. Weight
2010 2014
A 2 4 8
B 5 6 10
C 4 5 14
D 2 2 19

(b) The quarterly profits of a small scale industry (in thousands of Rupees) is as follows: Calculate the quarterly moving averages. Display these and the original figures graphically on the same graph sheet.     [5]

Year Quarter 1 Quarter 2 Quarter 3 Quarter 4
2012 39 47 20 56
2013 68 59 66 72
2014 88 60 60 67

Answer:

(a)

Commodity

2010

2014

p_ow p_1w
p_o w p_1 w
A 2 8 4 8 16 32
B 5 10 6 10 50 60
C 4 14 5 14 56 70
D 2 19 2 19 38 38
\Sigma p_ow \\ = 160 \Sigma p_1w \\ = 200

Index number for the year 2014 with respect to 2010 as the base year

= \frac{200}{160} \times 100  = 125

(b) 

Year Quarter Quarterly Profits 4 years moving total 4 year average
2012 1 39
2 47 162 40.2
3 20 191 47.75
4 56 203 50.75
2013 1 68 249 62.25
2 59 265 66.25
3 66 285 71.25
4 72 286 71.5
2014 1 88 280 70
2 60 275 68.75
3 60
4 67

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