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 × 3]

 

(i) If A = \begin{bmatrix}  3 & 1 \\ 7 & 5 \end{bmatrix} , find the values of x \ and \  y such that A^2 +xI_2=yA

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

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

(iv) Using L’Hospital’s Rule, evaluate: \lim \limits_{x \to 0} (1+ \sin \ x)^{\cot \ x}

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

(vi) Using the properties of definite integrals, evaluate:

\int \limits_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{ \sin x}+\sqrt{ \cos x}} dx

(vii) For the given lines of regression, 3x-2y=5 and x-4y=7 , find:

(a) regression coefficient b_{yx} and b_{xy}

(b) coefficient of correlation r(x,y)

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

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

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

Answer:

(i)  Given, A = \begin{bmatrix}  3 & 1 \\ 7 & 5 \end{bmatrix}

\therefore A^2 = \begin{bmatrix}  3 & 1 \\ 7 & 5 \end{bmatrix} \times \begin{bmatrix}  3 & 1 \\ 7 & 5 \end{bmatrix}

= \begin{bmatrix}  9+7 & 3+5 \\ 21+35 & 7+25 \end{bmatrix}

= \begin{bmatrix}  16 & 8 \\ 56 & 32 \end{bmatrix}

Now, A^2 +xI_2=yA

\begin{bmatrix}  16 & 8 \\ 56 & 32 \end{bmatrix} + x \begin{bmatrix}  1 & 0 \\ 0 & 1 \end{bmatrix}= y \begin{bmatrix}  3 & 1 \\ 7 & 5 \end{bmatrix}

\Rightarrow \begin{bmatrix}  16+x & 8 \\ 56 & 32+x \end{bmatrix} = \begin{bmatrix}  3y & y \\ 7y & 5y \end{bmatrix}

\Rightarrow y = 8

and 16+x = 3y

\Rightarrow x = 3 \times 8 - 16 = 8

(ii)  25x^2 - 9y^2 = 225

\Rightarrow \frac{x^2}{9} - \frac{y^2}{25} = 1

\Rightarrow a = 3, b = 5

c = \sqrt{a^2 + b^2} = \sqrt{34}

eccentricity (e) = \frac{c}{a} = \frac{\sqrt{34}}{3} 

Foci = (\pm \frac{\sqrt{34}}{3} , 0), Vertices = (\pm 3, 0)

Latus rectum = \frac{2b^2}{a} = \frac{75}{3}

(iii) \tan \ [ 2 \tan^{-1} \frac{1}{2} - \cot^{-1} \ 3]

= \tan \Big[ \tan^{-1} \Big( \frac{2 \times \frac{1}{2}}{1 - (\frac{1}{2})^2} \Big) - \cot^{-1} 3 \Big]

= \tan \Big[ \tan^{-1} \Big( \frac{1}{1 - \frac{1}{4}} \Big) - \tan^{-1} \frac{1}{3} \Big]

= \tan \Big[ \tan^{1} \frac{4}{3} - \tan^{-1} \frac{1}{3} \Big]

= \tan \Big[ \tan^{-1} \Big( \frac{\frac{4}{3} - \frac{1}{3}}{1+\frac{4}{9}} \Big) \Big]

= \tan \Big[ \tan^{-1} \Big( \frac{9}{13} \Big) \Big]

= \frac{9}{13}

(iv) \lim \limits_{x \to 0} (1+ \sin \ x)^{\cot \ x}

= e^{ \ \lim \limits_{x \to 0} \sin x \cot x} 

= e^{ \ \lim \limits_{x \to 0} \frac{\sin x \cos x}{\sin x}} 

= e^{\lim \limits_{x \to 0} \cos x} 

= e^{\cos 0} 

= e

v)  I = \int \limits_{}^{} e^x \Big( \frac{(2+\sin 2x)}{\cos^2 x} \Big) \ dx

= \int \limits_{}^{} e^x \Big( \frac{2}{\cos^2x} + \frac{2 \sin x \cos x}{\cos^2 x} \Big) \ dx

= \int \limits_{}^{} e^x (2 \sec^2 x + 2 \tan x) \ dx

= 2 \int \limits_{}^{} e^x (\sec^2 x + \tan x) \ dx

= 2 \Big( \int \limits_{}^{} e^x \sec^2x \ dx + \int \limits_{}^{} e^x \tan x \ dx \Big)

= 2 \Big[ e^x \int \limits_{}^{} \sec^2 x \ dx - \int \limits_{}^{} \Big( \frac{d}{dx} e^x  \int \limits_{}^{} \sec^2 x \ dx \Big) \ dx + \int \limits_{}^{} e^x \tan x \ dx \Big]

= 2 \Big[ e^x \tan x - \int \limits_{}^{} e^x \tan x \ dx + \int \limits_{}^{} e^x \tan x \ dx \Big]

= 2 e^x \tan x + c

(vi)  I = \int \limits_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{ \sin x}+\sqrt{ \cos x}} dx    … … … … … i)

= \int \limits_{0}^{\pi/2} \frac{\sqrt{\sin (\frac{\pi}{2} - x)}}{\sqrt{ \sin (\frac{\pi}{2} - x)}+\sqrt{ \cos (\frac{\pi}{2} - x)}} dx

I = \int \limits_{0}^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt { \cos x}+\sqrt{ \sin x}} dx    … … … … … ii)

Adding i) and ii) we get

2I = \int \limits_{0}^{\pi/2} \frac{\sqrt{\sin x} + \sqrt{\cos x} }{\sqrt{ \sin x}+\sqrt{ \cos x}} dx

2I = \int \limits_{0}^{\pi/2} 1 \ dx = \Big[ x \Big]^{\frac{\pi}{2}}_0

I = \frac{1}{2} \Big[ \frac{\pi}{2} - 0 \Big]

(vii)   (a) The two regression lines are

3x-2y= 5      … … … … … i)

x-4y=7    … … … … … ii)

From equation i) 3x = 2y + 5

\Rightarrow x = \frac{2}{3} y + \frac{5}{3} 

\therefore b_{xy} = \frac{2}{3}  (Regression of x on y )

From equation ii) 4y = x-7

\Rightarrow y = \frac{1}{4} x - \frac{7}{4} 

\therefore b_{yx} = \frac{1}{4}   (Regression of y on x )

(b) \therefore b_{xy} \times b_{yx} = \frac{2}{3} \times \frac{1}{4} < 1 . Therefore our assumption is correct.

Hence i) is the regression line of x on y and ii) is the regression line of y on x .

(viii)  \frac{(1+\sqrt{3}i)^2}{\sqrt{3}-i}

= \frac{1-3 +2\sqrt{3} i}{\sqrt{3}-i}

= \frac{-2 +2\sqrt{3} i}{\sqrt{3}-i}

= \frac{-2 +2\sqrt{3} i}{\sqrt{3}-i} \times \frac{\sqrt{3}+i}{\sqrt{3}+i}

= \frac{-2\sqrt{3} + 6i -2i -2\sqrt{3}}{\sqrt{3} + 1}

= \frac{-4\sqrt{3}+4i}{\sqrt{3} + 1}

= \frac{-4\sqrt{3}}{\sqrt{3} + 1} + \frac{4i}{\sqrt{3} + 1}

(ix) Number of balls = 20 \Rightarrow n(S) = 20

Number of balls marked with multiple of 3 = 3, 6, 9, 12, 15, 18 \Rightarrow n(A) = 6

Number of balls marked with multiple of 4 = 4, 8, 12, 16, 20 \Rightarrow n(B) = 5

Number of ball multiple of both 3 and 4 = 12 \Rightarrow n(A \cap B) = 1

The required marked with probability of P (A \cup B) = P(A) + P(B) - P(A \cap B)

= \frac{6}{20} + \frac{5}{20} - \frac{1}{20} = \frac{10}{20} = \frac{1}{20}

(x)  (x+1) \ dy - 2xy \ dx = 0

\frac{dy}{dx} = \frac{2xy}{x+1}

\frac{1}{y} dy = \Big( \frac{2x}{x+1} \Big) dx

\frac{1}{y} dy = \frac{2x+2-2}{x+1} dx

\frac{1}{y} dy = \Big( \frac{2(x+1)}{(x+1)} - \frac{2}{(x+1)} \Big) dx

\int \limits_{}^{} \frac{1}{y} dy =  \int \limits_{}^{}2 \ dx - \int \limits_{}^{} \frac{2}{(x+1)} dx

\log |y| = 2x - 2 \log |x+1| + \log c

\log |y| + 2 \log |x+1|-\log c = 2x

\log \Big(  \frac{y(x+1)^2}{c} \Big) = 2x

\frac{y(x+1)^2}{c} = e^{2x}

y = \frac{c \ e^x}{(x+1)}

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Question 2:

(a) Using properties of determinants, prove that:

\left| \begin{array}{ccc}  a^2+1 & ab & ac \\ ba & b^2+1 & bc \\ ca & cb & c^2+1 \end{array} \right|= a^2 + b^2 + c^2 + 1     [5]

(b) Solve the following system of linear equations using matrix method:

x-2y=10, \ \ 2x+y+3z=8, \ \ -2y+z=7     [5]

Answer:

(a)   \Delta = \left| \begin{array}{ccc}  a^2+1 & ab & ac \\ ba & b^2+1 & bc \\ ca & cb & c^2+1 \end{array} \right|

Applying R_1 \rightarrow \frac{1}{a} R_1,    R_2 \rightarrow \frac{1}{b} R_2, R_3 \rightarrow \frac{1}{c} R_3

\Delta = abc \left| \begin{array}{ccc}  a+\frac{1}{a} & b & c \\ a & b+\frac{1}{b} & c \\ a & b & c+\frac{1}{c} \end{array} \right|

\Delta =  \left| \begin{array}{ccc}  a^2+1 & b^2 & c^2 \\ a^2 & b^2+1 & c^2 \\ a^2 & b^2 & c^2+1 \end{array} \right|

Applying C_1 \rightarrow C_1 + C_2 + C_3

\Delta =  \left| \begin{array}{ccc}  1+a^2+b^2+c^2 & b^2 & c^2 \\ 1+a^2+b^2+c^2 & b^2+1 & c^2 \\ 1+a^2+b^2+c^2 & b^2 & c^2+1 \end{array} \right|

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

Applying R_2\rightarrow R_2 - R_1 \ and \  R_3 \rightarrow R_3 - R_1

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

Expanding along C_1 , we get

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

= 1+a^2+b^2+c^2 = RHS. Hence proved.

(b)  Given system of equation is,

x - 2y = 10

2x+ y + 3z = 8

-2y + z = 7

\Rightarrow A = \begin{bmatrix}  1 & -2  & 0 \\ 2 & 1  & 3 \\ 0 & -2 & 1 \end{bmatrix},  X = \begin{bmatrix}  x \\ y \\ z \end{bmatrix}, B = \begin{bmatrix}  10 \\ 8 \\ 7 \end{bmatrix}

\left| \begin{array}{c}  A  \end{array} \right| = 1(1+6) + 2(2-0) + 0 = 7 + 4 + 11 \neq 0

\therefore A^{-1} exists

Therefore The system has the unique solution X = A^{-1}B

A_{11} = \left| \begin{array}{cc}  1 & 3 \\ -2 & 1  \end{array} \right| = 7    A_{12} = - \left| \begin{array}{cc}  2 & 3 \\ 0 & 1  \end{array} \right| = -2    A_{13} = \left| \begin{array}{cc}  2 & 1 \\ 0 & 2  \end{array} \right| = -4

A_{21} = - \left| \begin{array}{cc}  -2 & 0 \\ -2 & 1  \end{array} \right| = 2    A_{22} = \left| \begin{array}{cc}  1 & 0 \\ 0 & 1  \end{array} \right| = 1    A_{23} = - \left| \begin{array}{cc}  1 & -2 \\ 0 & -2  \end{array} \right| = 2

A_{31} = \left| \begin{array}{cc}  -2 & 0 \\ 1 & 3  \end{array} \right| = -6    A_{32} = - \left| \begin{array}{cc}  1 & 0 \\ 2 & 3  \end{array} \right| = -3    A_{33} = \left| \begin{array}{cc}  1 & -2 \\ 2 & 1  \end{array} \right| = 5

Therefore adj \ A = \left| \begin{array}{ccc}  7 & -2  & -4 \\ 2 & 1  & 2 \\ -6 & -3 & 5 \end{array} \right|^T

= \left| \begin{array}{ccc} 7 & 2  & -6 \\ -2 & 1  & -3 \\ -4 & 2 & 5 \end{array} \right|

A^{-1} = \frac{adj \ A}{\left| \begin{array}{c}  A  \end{array} \right|}

\Rightarrow A^{-1} = \frac{1}{11} \left| \begin{array}{ccc} 7 & 2  & -6 \\ -2 & 1  & -3 \\ -4 & 2 & 5 \end{array} \right|

Now, AX = B \Rightarrow X = A^{-1} B

\begin{bmatrix}  x \\ y \\ z \end{bmatrix} = \frac{1}{11} \left| \begin{array}{ccc} 7 & 2  & -6 \\ -2 & 1  & -3 \\ -4 & 2 & 5 \end{array} \right|  \begin{bmatrix}  10 \\ 8 \\ 7 \end{bmatrix}

\Rightarrow \begin{bmatrix}  x \\ y \\ z \end{bmatrix} = \frac{1}{11} \left| \begin{array}{c} 70+16-42 \\ -20+8-21 \\ -40+16+35 \end{array} \right|

\Rightarrow \begin{bmatrix}  x \\ y \\ z \end{bmatrix} = \frac{1}{11} \begin{bmatrix}  44 \\ -33 \\ 11 \end{bmatrix}

\Rightarrow x = 4, y = -3 \ \& z = 11

\\

Question 3:

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

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

Answer:

(a)  Given \cos^{-1}x + \cos^{-1}y+ \cos^{-1}z = \pi

\cos^{-1}x + \cos^{-1}y = \pi - \cos^{-1}z

\Rightarrow \cos^{-1} (xy - \sqrt{1-x^2} . \sqrt{1-y^2}) = \pi - \cos^{-1}z

\Rightarrow xy - \sqrt{1-x^2} . \sqrt{1-y^2} = \cos (\pi - \cos^{-1}z)

\Rightarrow xy - \sqrt{1-x^2} . \sqrt{1-y^2} = -\cos (\cos^{-1}z)

\Rightarrow xy - \sqrt{1-x^2} . \sqrt{1-y^2} = -z

\Rightarrow xy + z = \sqrt{1-x^2} . \sqrt{1-y^2}

Squaring both sides we get

\Rightarrow (xy + z)^2 = (1-x^2)(1-y^2)

\Rightarrow \Rightarrow x^2y^2 + z^2 + 2 xyz = 1- x^2 -y^2 + x^2 y^2

x^2 + y^2 + z^2 + 2xyz = 1

(b)  The given Boolean expression is PR + Q (Q'+R)(P+QR)

The switching circuit is2019-03-17_14-09-07

Now PR + Q (Q'+R)(P+QR)

= PR  + (QQ' + QR)(P + QR)

= PR+QR(P+QR)       (QQ'=1)2019-03-17_14-09-31

= PR+PQR+QR         (1+Q= 1)

= PR(1+Q)+QR

= PR+QR = (P+Q)R

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Question 4:

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

(b) Find the equation of the parabola with latus rectum joining points (4, 6) and (4, -2) .

Answer:

(a)  Given function is, f(x)=e^x ( \sin x- \cos x) on [\frac{\pi}{4}, \frac{5 \pi}{4} ]

We know, \cos and exponential functions are always continuous

Therefore given function is continuous in [\frac{\pi}{4}, \frac{5 \pi}{4} ]

Differentiating w.r.t to x , we get

f^{'} (x) = e^x (\cos x + \sin x) + e^x (\sin x - \cos x) 

= e^x (\cos x + \sin x + \sin x - \cos x)

= 2 e^x \sin x

Which exists for all x in [\frac{\pi}{4}, \frac{5 \pi}{4} ]

f( \frac{\pi}{4} ) = e^{\frac{\pi}{4}} \Big(  \frac{1}{\sqrt{2}} -  \frac{1}{\sqrt{2}} \Big) = 0

and f( \frac{5\pi}{4} ) = e^{\frac{5\pi}{4}} \Big( - \frac{1}{\sqrt{2}} +  \frac{1}{\sqrt{2}} \Big) = 0

f( \frac{\pi}{4} ) =  f( \frac{5\pi}{4} )

Therefore, the given function satisfies all three conditions of Rolle’s theorem. For maxima or minima,

f^{'} (c) = 0

\Rightarrow 2 e^x \sin c = 0

\Rightarrow \sin c = 0

\Rightarrow c = n \pi + (-1)^n (0)

\Rightarrow c = n \pi

\Rightarrow c = \pi

Since \pi lies between [\frac{\pi}{4}, \frac{5 \pi}{4} ] so Rolle’s theorem is verified.

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Question 5:

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

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

Answer:

(a)  Given y = \frac{x \sin^{-1} x}{\sqrt{1-x^2}}

y \sqrt{1-x^2} = x \sin^{-1} x

Differentiating both sides w.r.t x , we get

y \frac{-2x}{2 \sqrt{1-x^2}} + \sqrt{1-x^2} \frac{dy}{dx} = x \frac{1}{\sqrt{1-x^2}} + \sin^{-1}x

\Rightarrow -xy + (1-x^2) \frac{dy}{dx} = x + \sqrt{1-x^2} \sin^{-1} x

\Rightarrow -xy + (1-x^2) \frac{dy}{dx} = x + \sqrt{1-x^2} . \frac{y}{x} \sqrt{1-x^2}

\Rightarrow -xy + (1-x^2) \frac{dy}{dx} = x + \frac{y}{x} (1-x^2)

\Rightarrow -xy + (1-x^2) \frac{dy}{dx} =x + \frac{y}{x} -yx

\Rightarrow (1-x^2) \frac{dy}{dx} = x + \frac{y}{x} 

Hence proved.

(b)  Given length of the wire = 50 \ m

Let the length of the square = x \ m

Therefore the length of circle = (50- x) \ m

Let the side of the square = a

Therefore perimeter 4a = x \Rightarrow a = \frac{x}{4} 

Let the radius of the circle = r

Therefore circumference 2 \pi r = (50 - x) \Rightarrow r = \frac{50-x}{2\pi } 

Combined area (A)  = a^2 + \pi r^2

A = \frac{x^2}{16} + \pi \Big( \frac{50-x}{2 \pi} \Big)^2

A = \frac{x^2}{16} + \pi  \frac{(50-x)^2}{4 \pi^2} 

A = \frac{x^2}{16} +  \frac{(50-x)^2}{4 \pi} 

Differentiating w.r.t x we get

\frac{dA}{dx} = \frac{2x}{16} + \frac{2(50-x)(-1)}{4 \pi} 

\frac{dA}{dx} = \frac{x}{8} + \frac{(x-50)}{2 \pi} 

\frac{dA}{dx} = \frac{\pi x +4x - 200}{8 \pi} 

\frac{dA}{dx} = \frac{x(\pi +4) - 200}{8 \pi} 

For maxima and minima, \frac{dA}{dx} = 0

Therefore x(\pi + 4) - 200 = 0 \Rightarrow x = \frac{200}{\pi + 4} 

Since \frac{d^2 A}{dx^2} > 0

\Rightarrow A is minimum at x = \frac{200}{\pi + 4} 

Therefore the length of the square wire x = \frac{200}{\pi + 4} 

Therefore length of circle wire = 50 - x = 50 - \frac{200}{\pi + 4} = \frac{50 \pi}{\pi + 4} 

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Question 6:

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

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

Answer:

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

= \int \limits_{}^{} \frac{x + \sin x}{2 \cos^2 \frac{x}{2} } dx

= \int \limits_{}^{} \frac{x }{2 \cos^2 \frac{x}{2}} dx +   \int \limits_{}^{} \frac{ 2 \sin \frac{x}{2} . \cos \frac{x}{2} }{2 \cos^2 \frac{x}{2}} dx

= \frac{1}{2} \int \limits_{}^{} x \sec^2 \frac{x}{2} dx + \int \limits_{}^{} \tan \frac{x}{2}

= \frac{1}{2} \Big[ x.    \int \limits_{}^{} \sec^2 \frac{x}{2} - \int \limits_{}^{} \Big( \frac{d}{dx}   x . \int \limits_{}^{} \sec^2 \frac{x}{2} \ dx \Big)  \ dx \Big] + \int \limits_{}^{} \tan \frac{x}{2} \ dx

= \frac{1}{2} \Big [ x .  \int \limits_{}^{} \frac{\tan \frac{x}{2}}{\frac{1}{2}} -  \int \limits_{}^{} \frac{\tan \frac{x}{2}}{\frac{1}{2}} \Big] + \int \limits_{}^{} \tan \frac{x}{2} dx +c

= x \tan \frac{x}{2} - \int \limits_{}^{} \tan \frac{x}{2} \ dx + \int \limits_{}^{} \tan \frac{x}{2} \ dx + c

= x \tan \frac{x}{2} + c

(b)  Given curves are y^2= -3x + 4

y^2 = -3 \Big( x - \frac{4}{3} \Big)

2019-03-16_13-34-56

The vertex L of this parabola is (\frac{4}{3}, 0). It intercepts the y-axis at A(0,2) and B(0, -2) .

The points of intersection of these two parabolas are given by the equation x = -3x + 4 \Rightarrow x = 1

Then y^2 = 1 \Rightarrow y = \pm 1

Therefore the points of intesection are P(1,1) and Q(1, -1) . Let PQ intercepts the x-axis at R .

Therefore total area of POQLP = 2 area of OPLRO

= 2 \Big[ \int \limits_{0}^{1} \sqrt{x} \ dx + \int \limits_{1}^{\frac{4}{3}} \sqrt{4-3x} \ dx \Big]

= 2 \Big[  \Big(  \frac{x^{3/2}}{\frac{3}{2}} \Big)_{0}^{1} +  \Big(  \frac{2(4-3x)^{\frac{3}{2}}}{(-3) \times 3} \Big)_{1}^{\frac{4}{3}}  \Big]

= 2 \Big[ ( \frac{2}{3} - 0) - \frac{2}{9} (0-1) \Big]

= 2 \Big[ \frac{2}{3} + \frac{2}{9} \Big] = 2 \Big( \frac{6+2}{9} \Big) = 2( \frac{8}{9} ) = \frac{16}{9} sq. units.

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Question 7:

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

Pairs 1 2 3 4 5 6 7 8 9 10 11
Method A 24 29 19 14 30 19 27 30 20 28 11
Method B 37 35 16 26 23 27 19 20 16 11 21

Calculate Spearman’s Rank correlation.      [5]

(b) The coefficient of correlation between the values denoted by X and Y is 0.5 . The mean of X is 3 and that of Y is 5 . Their standard deviations are 5 and 4 respectively. Find:

(i) the two lines of regression

(ii) the expected value of Y , when X is given 14 .

(iii) the expected value of X , when Y is given as 9 .       [5]

Answer:

(a) 

Pairs A B Rank A Rank B D D²
1 24 37 6 1 5 25
2 29 35 3 2 1 1
3 19 16 8.5 9.5 1 1
4 14 26 10 4 6 36
5 30 23 1.5 5 -3.5 12.25
6 19 27 8.5 3 5.5 30.25
7 27 19 5 8 -3 9
8 30 20 1.5 7 -5.5 30.25
9 20 16 7 9.5 -2.5 6.25
10 28 11 4 11 -7 49
11 11 21 11 6 5 25
\Sigma \ d^2 = 225

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

= 1-  \frac{6 [ 225 + \frac{(2^3-2)}{12} + \frac{(2^3-2)}{12} + \frac{(2^3-2)}{12} ] }{11 \times 120}

= 1- \frac{6(225+1.5)}{1320}

= 1 - \frac{1350}{1320}

= 1 - 1.029

= - 0.029 , shows negative correlation.

(b)  Given \overline{x} = 3, \overline{y} = 5, \sigma_x=5, \sigma_y=4 and r=0.5

b_{yx} = r. \frac{\sigma_y}{\sigma_x} = 0.5 \times \frac{4}{5} = 0.4

b_{xy} = r. \frac{\sigma_x}{\sigma_y} = 0.5 \times \frac{5}{4} = 0.625

i) Regression equation of y on x is:

y - \overline{y} = b_{yx} (x - \overline{x})

y - 5 = 0.4 (x-3)

y = 0.4x + 3.8

Regression equation of x on y is:

x - \overline{x} = b_{xy} (y - \overline{y})

x - 3 = 0.625 (y-5)

x = 0.625y -0.125

ii) When x = 14 \Rightarrow y = 0.4 \times 14 + 3.8 = 9.4

iii) When y = 9 \Rightarrow x = 0.625 \times 9 - 0.125 = 5.5

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Question 8:

(a) In a college, 70\% students pass in Physics, 75\% pass in Mathematics and 10\% students fail in both. One student is chosen at random. What is the probability that:

(i) He passes in Physics and Mathematics

(ii) He passes in Mathematics given that he passes in Physics

(iii) He passes in Physics given that he passes in Mathematics.       [5]

(b) A bag contains 5 white and 4 black balls and another bag contains 7 white and 9 black balls. A ball is drawn from the first bag and two balls are drawn from the second bag. What is the probability of drawing 1 white and 2 black balls?       [5]

Answer:

(a)  2019-03-16_18-13-28.jpgLet x\% students pass in both Mathematics and Physics

Students who pass in Physics = 70\% \Rightarrow P(P) = \frac{70}{100}

Students who pass in Mathematics = 75\% \Rightarrow P(M) = \frac{75}{100}

Students who fail in Both = P(\overline{P} \cap \overline{M}) = 10\%

70\% - x + x + 75\% - x = 90\%

x = 55\% \Rightarrow P(M \cap P) = \frac{55}{100}

i) P (Pass in Physics and Mathematics) = \frac{55}{100} = \frac{11}{20}

ii) P (M/P) = \frac{P (M \cap P}{P(P)} = \frac{55/100}{70/100} = \frac{55}{70} = \frac{11}{14}

iii) P (P/M) = \frac{P (M \cap P}{P(M)} = \frac{55/100}{75/100} = \frac{55}{75} = \frac{11}{15}

(b) Let A : the event of getting 1 black ball from Bag I

B : the event of getting 1 black and 1 white ball from Bag II

C : the event of getting 1 white ball from Bag I

D : the event of getting 2 black balls from Bag II

Then, P(A) = \frac{^4C_1}{^9C_1} = \frac{4}{9}

P(B) = \frac{^9C_1 \times ^7C_1}{^{16}C_2} = \frac{21}{40}

P(C) = \frac{^5C_1}{^9C_1} = \frac{5}{9}

P(D) = \frac{^9C_2}{^{16}C_2} = \frac{4}{9}

Therefore required probability = P(A) P(B) + P(C) P(D)

= \frac{4}{9} \times \frac{21}{40} + \frac{5}{9} \times \frac{4}{9} = \frac{21+15}{90} \frac{2}{5}

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Question 9:

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

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

Answer:

(a) We have

\frac{2i}{1+i} = \frac{2i}{1+i} \times \frac{1-i}{1-i} = \frac{2(1+i)}{2} = 1+i

\frac{2i}{1+i} = \sqrt{2 \Big( \frac{1}{\sqrt{2}}+ \frac{1}{\sqrt{2}} i \Big) }

\frac{2i}{1+i} = \sqrt{2 \Big( \cos {\frac{\pi}{4}}+ i \sin {\frac{\pi}{4}} \Big) }

\Big( \frac{2i}{1+i} \Big)^n= \Bigg( \sqrt{2 \Big( \cos {\frac{\pi}{4}}+ i \sin {\frac{\pi}{4}} } \Bigg)^n

\Big( \frac{2i}{1+i} \Big)^n = 2^{\frac{n}{2}} \Big(  \cos {\frac{n\pi}{4}} + i \sin {\frac{n\pi}{4}}  \Big)

Which is positive integer if  {\frac{n\pi}{4}} = 0, 2\pi , 4\pi , 6\pi ...

\Rightarrow n = 0, 8, 16, 24

Therefore the least positive value of n = 8

(b)  (3xy+y^2)dx + (x^2+xy)dy=0

\frac{dy}{dx} = - \Big( \frac{3xy + y^2}{x^2+xy} \Big)

Put y = vx \Rightarrow \frac{dy}{dx} = v + x \frac{dy}{dx} 

\frac{dv}{dx} . x + v = -\Big( \frac{3x.vx+v^2x^2}{x^2 + x . vx} \Big)

\frac{dv}{dx} .x = \frac{-3v-v^2}{1+v} - v

x \frac{dv}{dx} = \frac{-3v-v^2-v-v^2}{1+v} 

x \frac{dv}{dx} = \frac{-2v^2-4v}{1+v} 

Integrating both sides

\int \limits_{}^{} \frac{1+v}{2v^2+4v} dv = \int \limits_{}^{} - \frac{1}{x} dx

\Rightarrow \frac{1}{4} \int \limits_{}^{} \frac{2 + 2v}{2v + v^2} dv = - \int \limits_{}^{} \frac{1}{x} dx

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

\frac{1}{4} \log \Big( \frac{y^2}{x^2} + 2 \frac{y}{x} \Big) = \log \frac{c}{x} 

\log \Big( \frac{y^2}{x^2} + 2 \frac{y}{x} \Big) = 4 \log \frac{c}{x} 

\log \Big( \frac{y^2}{x^2} + 2 \frac{y}{x} \Big) =  \log \frac{c^4}{x^4} 

 \frac{y^2}{x^2} + 2 \frac{y}{x}   =  \frac{c^4}{x^4} 

x^2y^2 + 2x^3y = c^4

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Section – B (20 Marks)

Question 10:

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

(b) Prove that:

\overrightarrow{a}.(\overrightarrow{b}+\overrightarrow{c}) \times (\overrightarrow{a}+ 2 \overrightarrow{b} + 3 \overrightarrow{c}) = [ \overrightarrow{a} \overrightarrow{b} \overrightarrow{c} ]        [5]

Answer:

(a)  \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = \overrightarrow{0}

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

Squaring both sides

(\overrightarrow{a} + \overrightarrow{b}) (\overrightarrow{a} + \overrightarrow{b}) = (- \overrightarrow{c})(- \overrightarrow{c})

b^2 + a^2 + 2ab \cos (\pi - C) = c^2

b^2 + a^2 + 2ab (- \cos C) = c^2

c^2 = a^2 + b^2 -2ab \cos C

(b)  LHS = \overrightarrow{a}.(\overrightarrow{b}+\overrightarrow{c}) \times (\overrightarrow{a}+ 2 \overrightarrow{b} + 3 \overrightarrow{c})

= \overrightarrow{a} . ( \overrightarrow{b} \times \overrightarrow{a} + \overrightarrow{b} \times 2\overrightarrow{b} + \overrightarrow{b} \times 3 \overrightarrow{c} + \overrightarrow{c} \times \overrightarrow{a} + \overrightarrow{c} \times 2\overrightarrow{b} + \overrightarrow{c} \times 3\overrightarrow{c})

= \overrightarrow{a} . ( \overrightarrow{b} \times \overrightarrow{a} + 3 \overrightarrow{b} \times \overrightarrow{c} + \overrightarrow{c} \times  \overrightarrow{a} + 2 \overrightarrow{c} \times \overrightarrow{b} )

= \overrightarrow{a} . (\overrightarrow{b} \times \overrightarrow{a}) + \overrightarrow{a} . (3 \overrightarrow{b} \times \overrightarrow{c}) + \overrightarrow{a} . (\overrightarrow{c} \times  \overrightarrow{a}) + \overrightarrow{a} . (2 \overrightarrow{c} \times \overrightarrow{b})

= 0 + 3 [ \overrightarrow{a} . (\overrightarrow{b} \times \overrightarrow{c}) + 0 + 2 ( \overrightarrow{a} . (\overrightarrow{c} \times \overrightarrow{b}) ]

= 3 [ \overrightarrow{a} \overrightarrow{b} \overrightarrow{c} ] - 2[ \overrightarrow{a} \overrightarrow{b} \overrightarrow{c} ]

= [ \overrightarrow{a} \overrightarrow{b} \overrightarrow{c} ] = RHS

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Question 11:

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

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

Answer:

(a)  Equation of the line passing through point P(-1, 3, 2) and Q (-4, 2, -2) is

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

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

Since Point R(5, 5, \lambda) lies on it

\frac{5+1}{3} = \frac{5-3}{1} = \frac{\lambda-2}{4} 

\Rightarrow \lambda = 10

(b) The required plan is perpendicular to the given plane

x-2y+5z+1 = 0

Therefore required plan is parallel to the line which is perpendicular to the given plan. Direction ratio of line a=1, b=-2, c=5

Hence the required plane is

\left| \begin{array}{ccc}  x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a & b & c \end{array} \right| = 0

\Rightarrow \left| \begin{array}{ccc}  x-2 & y+3 & z-1 \\ -1-2 & 1+3 & -7-1 \\ 1 & -2 & 5 \end{array} \right| = 0

\Rightarrow \left| \begin{array}{ccc}  x-2 & y+3 & z-1 \\ -3 & 4 & -8 \\ 1 & -2 & 5 \end{array} \right| = 0

\Rightarrow (x-2)(20-16) - (y-3)(-15+8)+(z-1)(6-4) = 0

\Rightarrow (x-2)4-(y+3)(-7) + (z-1)2 = 0

\Rightarrow 4x+7y+2z+11=0

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Question 12:

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

(b) On dialing certain telephone numbers, assume that on an average, one telephone number out of five is busy, ten telephone numbers are randomly selected and dialed. Find the probability that at least three of them will be busy.       [5]

Answer:

(a)  Let E_1 : the event that the bolt is produced by machine A

E_2 : the event that the bolt is produced by machine B

E_3 : the event that the bolt is produced by machine C

E_1, E_2 and E_3 are mutually exclusive and exhaustive events

We have P(E_1) = 25\% = \frac{25}{100}

P(E_2) = 35\% = \frac{35}{100}

P(E_3) = 40\% = \frac{40}{100}

Let E be the event that bolt chosen is found to be defective

Therefore P(E/E_1) = 5\% = \frac{5}{100}

P(E/E_2) = 4\% = \frac{4}{100}

P(E/E_3) = 2\% = \frac{2}{100}

P (defective bolt produced by machine C )

P(E_3/E) = \frac{P(E_3) . P(E/E_3)}{P(E_1).P(E/E_1) + P(E_2).P(E/E_2) + P(E_3).P(E/E_3)}

\Rightarrow P(E_3/E) = \frac{\frac{40}{100} \times \frac{2}{100}}{\frac{25}{100} \times \frac{5}{100} + \frac{35}{100} \times \frac{4}{100} + \frac{40}{100} \times \frac{2}{100}}

\Rightarrow P(E_3/E) = \frac{80}{125+140+80} = \frac{80}{345} = 0.23

(b)  The probability of one phone number out of five is busy is p = \frac{1}{5} 

Then probability of phone not busy is q = 1 - \frac{1}{5} = \frac{4}{5}

Given n = 10

Therefore required probability (at least three phones are busy)

= 1 - (Probability maximum two phones busy)

= 1 - [ P(0) + P(1) + P(2) ]

= 1 - \Big[ {^{10}C_{0}} \Big( \frac{1}{5} \Big)^0 . \Big( \frac{4}{5} \Big)^{10} + ^{10}C_{1} \Big( \frac{1}{5} \Big)^1 . \Big( \frac{4}{5} \Big)^{9} + ^{10}C_{2} \Big( \frac{1}{5} \Big)^2 . \Big( \frac{4}{5} \Big)^{8} \Big]

= 1 - \Big[ 1 \times 1 \times \Big( \frac{4}{5} \Big)^{10} + 10 \times \frac{1}{5} . \Big( \frac{4}{5} \Big)^{9} + 45 \times \Big( \frac{1}{5} \Big)^2 . \Big( \frac{4}{5} \Big)^{8} \Big]

= 1 - \Big( \frac{4}{5} \Big)^{8} \Big[ \frac{16}{25} + \frac{8}{9} + \frac{9}{5} \Big]

= 1 - \Big( \frac{4}{5} \Big)^{8} \Big[ \frac{101}{25} \Big]

= 1 - 0.1678 \times 4.04 = 1 - 0.678 = 0.322

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Section – C (20 Marks)

Question 13:

(a) A person borrows Rs. 68962 on the condition that he will repay the money with compound interest at 5\%   per annum in 4 equal annual installments, the first one being payable t the end of the first year. Find the value of each installment.       [5]

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

Answer:

(a)  Let the installment = a

Money borrowed (P)  = Rs. 68962      Time period (n) = 4      Rate of Interest (I) = 5\% = 0.05

P = \frac{a}{i} \{ 1 - (1+i)^{-n} \}

68962 = \frac{a}{0.05} \{ 1 - (1.05)^{-4} \}

68962 = \frac{a}{0.05} \times 0.1773

\Rightarrow a = \frac{68962 \times 0.05}{0.1773} 

\Rightarrow a = 19447.82 Rs

(b)

Toy A Toy B Time in a Day
Cutting Time 5 minutes 8 minutes 180 minutes
Assembling Time 10 minutes 8 minutes 240 minutes
Profit 50 60
Assumed Quantity x y

Profit function z = 50x + 60y

x \geq 0, y \geq 0

Now plot the line 5x+ 8y = 180

A B
x 0 36
y 22.5 0

Now plot the line 10x+ 8y = 240

C D
x 0 24
y 30 0

2019-03-17_11-03-38

Corner Point Objective Function: z = 50x + 60y
O (0,0) z = 50 × 0 + 60 × 0 = 0
D (24, 0) z = 50 × 24 + 60 × 0 = 1200
E (12, 15) z = 50 × 12 + 60 × 15 = 1500
A (0, 22.5) z = 50 × 0 + 60 × 22.5 = 1350

Hence the maximum profit is Rs. 1500 at E (12, 15)

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Question 14:

(a) A firm has the cost function C = \frac{x^3}{3} - 7x^2+111x+50 and demand function x = 100-p .

(i) Write the total revenue function in terms of x

(ii) Formulate the total profit function P in terms of x

(iii) Find the profit maximizing level of output.       [5]

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

Answer:

(a)  Cost function C(x) = \frac{x^3}{3} - 7x^2 + 11x + 50

Demand function x = 100 - p \Rightarrow p = 100 - x

i) Revenue function R(x) = p.x = x (100- x) = 100x - x^2

ii) Profit function, P(x) = Revenue function – Cost function

P(x) = R(x) - C(x) = 100x - x^2 - \frac{x^3}{3} + 7x^2 - 111x - 50

P(x) = - \frac{x^3}{3} + 6x^2 - 11x - 50

iii) For profit function to be maximum

Differentiating P(x) w.r.t  x

\frac{dP}{dx} = - x^2 + 12x - 11

Now,  \frac{dP}{dx} =0 \Rightarrow - x^2+12x-11= 0 \Rightarrow x^2- 12x+11=0

\Rightarrow x^2 - 11x -x + 11 = 0

\Rightarrow x (x-11) -(x-11) = 0

\Rightarrow (x-1)(x-11) = 0 \Rightarrow  x = 1, 11

Again differentiating equation iii) w.r.t x we get

\frac{d^2P}{dx^2} = 12 - 2x

At x = 1, \ \ \ \frac{d^2P}{dx^2} = 10 \Rightarrow \frac{d^2P}{dx^2} > 0 (minimum value)

at x = 11, \ \ \ \frac{d^2P}{dx^2} = -10 \Rightarrow \frac{d^2P}{dx^2} < 0 (maximum value)

Hence, the profit maximizing level of output at x = 11

(b) Let the unexpired period of the bill at the time of discounting be t years

B.G = \frac{A(ni)^2}{1+ni}   where A is the face value of the bill.

Here A = 5050, i = 0.05

Therefore 0.05 = \frac{5050(0.05t)^2}{1+0.05t}

505t^2 - t - 20 = 0

\Rightarrow t = \frac{1 \pm \sqrt{1+40400}}{1010}

\Rightarrow t = \frac{1 \pm 201}{1010}

\Rightarrow t = \frac{1}{5} = 73 days

Therefore , the legal due date of maturity is 73 days after 4th July which comes to 15th September .

Therefore the legal due date is 15th September and the nominal due date  is 12th September.

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Question 15:

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

Commodities A B C D E F
Price in 2009 (Rs.) 35 80 25 30 80 x
Price in 2011 (Rs.) 50 y 45 70 120 105

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

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

Year Jan-Mar April-June July-Sept Oct-Dec
2010 70 60 45 72
2011 79 56 46 84
2012 90 64 45 82

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

Answer:

(a)

Commodities Price in 2009 (Rs.) Price in 2011 (Rs)
A 35 50
B 80 y
C 25 45
D 30 70
E 80 120
F x 105
\Sigma P_0 = 250+x \Sigma P_1 = 390+y

Given P_0  = 360

\Rightarrow 250 + x = 360  \Rightarrow x = 110

Price Index = \frac{\Sigma P_1 }{\Sigma P_0 } \times 100

\Rightarrow 125 = \frac{390+y}{360} \times 100

\Rightarrow 4500 = 3900 + 10y

\Rightarrow y = 60

(b) Calculations for trends by 4 quarterly moving average:

Year Quarter Values 4- Quarterly moving total 4-Quarterly moving average 4-Quarterly moving average centered
2010 1 70
2 60
247 247/4=61.75
3 45 125.75/2=62.875
256 256/4=64
4 72 127/2=63.5
252 252/4/=63
2011 1 79 126.25/2=63.125
253 253/4=63.25
2 56 129.50/2=64.750
265 265/4=66.25
3 46 135.25/2=67.625
276 276/4=69
4 84 140/2=70
284 284/4=71
2012 1 90 141.75/2=70.875
283 283/4=70.75
2 64 141/2=70.50
281 281/4=70.25
3 45
4 82

2019-03-17_21-23-24

 

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