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 - 2I)(A - 3I) = 0 , where A = \begin{bmatrix}  4 & 2 \\ -1 & x \end{bmatrix} and I = \begin{bmatrix}  1 & 0 \\ 0 & 1 \end{bmatrix} , find the value of x .

(ii) Find the value(s) of k so that the line 2x+y+k=0 may touch the hyperbola 3x^2-y^2=3 .

(iii) Prove that: \tan^{-1} \frac{1}{4} + \tan^{-1} \frac{2}{9} = \frac{1}{2} \sin^{-1} \frac{4}{5}

(iv) Using L’Hospital’s Rule, evaluate:

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

(v) Evaluate: \int \limits_{}^{} \frac{1}{x+\sqrt{x}} dx

(vi) Evaluate: \int \limits_{0}^{1} \log (\frac{1}{x} -1) \ dx

(vii) Two regression lines are represented by 4x+10y=9 and 6x+3y=4 . Find the line of regression of y \ on \ x .

(viii) If 1, \ \omega \ and \  \omega^2 are the cube roots of unity, evaluate:

(1-\omega^4+ \omega^8)(1-\omega^8+\omega^{16})

(ix) Solve the differential equation:

\log (\frac{dy}{dx}) = 2x-3y

(x) If two balls are drawn from a bag containing three red and four blue balls, find the probability that:

(a) They are the same color

(b) They are of different colors

Answer:

(i)   Given, (A - 2I)(A - 3I) = 0

\Bigg( \begin{bmatrix}  4 & 2 \\ -1 & x \end{bmatrix} - 2 \begin{bmatrix}  1 & 0 \\ 0 & 1 \end{bmatrix} \Bigg)  \Bigg( \begin{bmatrix}  4 & 2 \\ -1 & x \end{bmatrix} -  3 \begin{bmatrix}  1 & 0 \\ 0 & 1 \end{bmatrix}  \Bigg) = \begin{bmatrix}  0 & 0 \\ 0 & 0 \end{bmatrix}

\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}

\begin{bmatrix}  0 & 2x-2 \\ -x+1 & x^2-5x+4 \end{bmatrix} = \begin{bmatrix}  0 & 0 \\ 0 & 0 \end{bmatrix}

Therefore

-x+1 = 0 \Rightarrow x = 1

2x-2 = 0 \Rightarrow x = 1

x^2-5x+4 = 0 \Rightarrow (x-1)(x-4) = 0 \Rightarrow x = 1, 4

However, 4 does not satisfy all the equations. Hence x = 1

(ii)  3x^2 - y^2 = 3

\Rightarrow \frac{x^2}{1} - \frac{y^2}{3} = 1

Therefore a^2 = 1, b^2 = 3

y = -2x -k

m=-2 , c = -k

Condition for tangent

c^2 = a^2 m^2 - b^2

\Rightarrow k^2 = 4-3= 1

k= \pm 1

(iii)  LHS = \tan^{-1} \frac{1}{4} + \tan^{-1} \frac{2}{9}

= \tan^{-1} \Big( \frac{\frac{1}{4}+\frac{2}{9} }{1- \frac{1}{4} . \frac{2}{9}} \Big)

= \tan^{-1} \Big( \frac{\frac{17}{36}}{\frac{17}{18}} \Big)

= \tan^{-1} \frac{1}{2}

= \frac{1}{2} \Big( 2 \tan^{-1} \frac{1}{2} \Big)

= \frac{1}{2} \sin^{-1} \Big( \frac{2 . \frac{1}{2} }{1 + (\frac{1}{2})^2} \Big)

= \frac{1}{2} \sin^{-1} \frac{4}{5} = RHS.  Hence Proved.

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

Using L’Hospital’s Rule

= \lim \limits_{x \to 0} ( \frac{e^x + e^{-x} - 2}{1 - \cos x})

Again applying L’Hospital’s Rule

= \lim \limits_{x \to 0} ( \frac{e^x - e^{-x}}{\sin x})

Once again applying L’Hospital’s Rule,

= \lim \limits_{x \to 0} ( \frac{e^x + e^{-x}}{\cos x})

= \frac{1+1}{1} = 2

(v)  Given, \int \limits_{}^{} \frac{1}{x+\sqrt{x}} dx

Put \sqrt{x} = t \Rightarrow x = t^2  \Rightarrow dx = 2 dt

Therefore

\int \limits_{}^{} \frac{dx}{x+\sqrt{x}}   = \int \limits_{}^{} \frac{2t \ dt}{t^2 + t} =  2 \int \limits_{}^{} \frac{1}{t+1} dt

= 2 \log |t+1 | + c

= 2 \log \ (\sqrt{x} + 1) + c

(vi) I = \int \limits_{0}^{1} \log (\frac{1}{x} -1) \ dx

\Rightarrow I = \int \limits_{0}^{1} \log ( \frac{1-x}{x} )  \ dx

Using property,  \int \limits_{0}^{a} f(x) \ dx = \int \limits_{0}^{a} f(a-x) \ dx

I = \int \limits_{0}^{1} \log ( \frac{1-(1-x)}{1-x} )  \ dx

\Rightarrow I = \int \limits_{0}^{1} \log ( \frac{x}{1-x} )  \ dx

\Rightarrow I = \int \limits_{0}^{1} \log ( (\frac{1-x}{x})^{-1} )  \ dx

\Rightarrow I = \int \limits_{0}^{1} (-1) \log ( \frac{1-x}{x} )  \ dx

\Rightarrow I =  - \int \limits_{0}^{1} \log ( \frac{1-x}{x} ) \ dx

\Rightarrow I=-I \Rightarrow 2I = 0 \Rightarrow I = 0

Therefore  \int \limits_{0}^{1} \log (\frac{1}{x} -1) \ dx = 0

(vii)  Let us assume that 4x+10y=9 be the regression line of y on x

10y = -4x + 9

\Rightarrow y = - \frac{4}{10} x + \frac{9}{10} 

\Rightarrow b_{yx} = - \frac{2}{5} 

Then 6x + 3y = 4

\Rightarrow 6x = -3y + 4

\Rightarrow  x = - \frac{3}{6} y + \frac{4}{6} 

\Rightarrow b_{xy} = - \frac{1}{2} 

Therefore b_{yx}.b_{xy} = (- \frac{2}{5} ).(- \frac{1}{2} ) = \frac{1}{5} < 1

Hence our assumption is true i.e. regression line y on x is 4x+10y=9

(viii)  (1-\omega^4+ \omega^8)(1-\omega^8+\omega^{16})

= (1-\omega+ \omega^2)(1-\omega^2+\omega)

= (-2 \omega) (-2 \omega^2)

= 4 \omega^3

= 4

(ix)  \log (\frac{dy}{dx}) = 2x-3y

\Rightarrow \frac{dy}{dx} = e^{2x-3y}

\Rightarrow \frac{dy}{dx} = \frac{e^{2x}}{e^{3y}} 

\Rightarrow e^{3y} dy = e^{2x} dx

Integrating both sides

\int \limits_{}^{} e^{3y} dy = \int \limits_{}^{} e^{2x} dx

\frac{e^{3y}}{3} = \frac{e^{2x}}{2} +c

2e^{3y}-3 e^{2x} = k

(x)  Total number of ways of drawing 2 balls = ^7C_2

(a) P (Balls are of same color) = \frac{^3C_2}{^7C_2} +  \frac{^4C_2}{^7C_2} =  \frac{^3C_2+^4C_2}{^7C_2}  = \frac{ \frac{3!}{2!.1!} +\frac{4!}{2!.2!} }{\frac{7!}{2!.5!}} = \frac{3+6}{7 \times 3} = \frac{3}{7} 

(b) P (Balls are of different color) = \frac{^3C_1 \times ^4C_1}{^7C_2} = \frac{ \frac{3!}{1!.2!} + \frac{4!}{1!.3!} }{\frac{7!}{2!.5!}} = \frac{3 \times 4}{7 \times 3} = \frac{4}{7} 

\\

Question 2:

(a) Using properties of determinants, prove that:

\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)     [5]

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

Hence, solve the system of linear equations:

4x+2y+3z = 2

x+y+z=1

3x+y-2z=5                   [5]

Answer:

(a)   \Delta = \left| \begin{array}{ccc}  x & y & z \\ x^2 & y^2 & z^2 \\ y+z & z+x & x+y \end{array} \right|

Applying R_3 \rightarrow R_3 + R_1 , we get

\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|

\Rightarrow \Delta = (x+y+z)\left| \begin{array}{ccc}  x & y & z \\ x^2 & y^2 & z^2 \\ 1 & 1 & 1 \end{array} \right|

Applying C_1 \rightarrow C_1 - C_2 and C_2 \rightarrow C_2 - C_3 , we get

\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|

\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|

Expanding along R_3 , we get

= (x+y+z) (x-y)(y-z) (y+z-x-y)

= (x-y)(y-z)(z-x)(x+y+z)

= RHS. Hence proved.

(b)  A = \left| \begin{array}{ccc}  4 & 2 & 3 \\ 1 & 1 & 1 \\ 3 & 1 & -2 \end{array} \right|

= 4(-2-1) -2(-2-3)+3(1-3) = -12+10-6 = -8 \neq 0

\Rightarrow A^{-1} exists

A^{-1} = \frac{1}{|A|} adj. \  A

A_{11} = \left| \begin{array}{cc}  1 & 1  \\ 1 & -2 \end{array} \right| = -3 A_{12} = - \left| \begin{array}{cc}  1 & 1  \\ 3 & -2 \end{array} \right| = 5 A_{13} = \left| \begin{array}{cc}  1 & 1  \\ 3 & 1 \end{array} \right| = -2

A_{21} = - \left| \begin{array}{cc}  2 & 3  \\ 1 & -2 \end{array} \right| = 7 A_{22} = \left| \begin{array}{cc}  4 & 3  \\ 3 & -2 \end{array} \right| = -17 A_{23} = - \left| \begin{array}{cc}  4 & 2  \\ 3 & 1 \end{array} \right| = 2

A_{31} = \left| \begin{array}{cc}  2 & 3  \\ 1 & 1 \end{array} \right| = -1 A_{32} = - \left| \begin{array}{cc}  4 & 3  \\ 1 & 1 \end{array} \right| = -1 A_{33} = \left| \begin{array}{cc}  4 & 2  \\ 1 & 1 \end{array} \right| = 2

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}

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}

Given system of equation is

4x+2y+3z = 2

x+y+z=1

3x+y-2z=5

This can be written as AX = B

Where A = \begin{bmatrix}  4 & 2 & 3 \\ 1 & 1 & 1 \\ 3 & 1 & -2 \end{bmatrix}   X = \begin{bmatrix}  x  \\ y  \\ z  \end{bmatrix}   and B = \begin{bmatrix}  2  \\ 1  \\ 5  \end{bmatrix}

Since |A| \neq 0 , the given system of equations have a unique solution X = A^{-1}B

\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}

\Rightarrow X = \frac{1}{2} \begin{bmatrix}  1 \\ 3 \\ -2 \end{bmatrix}

\Rightarrow x = \frac{1}{2} , y = \frac{3}{2} and z = -1 which is the solution to the system of the equations.

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Question 3:

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

(b) Construct a circuit diagram for the following Boolean function:

(BC+A)(A'B'+C')+A'B'C'

Using laws of Boolean Algebra, simplify the function and draw the simplified circuit.  [5]

Answer:

(a)  \sin^{-1} x + \sin^{-1} (1-x) = \cos^{-1} x   

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

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

Let \sin^{-1} x = y \Rightarrow x = \sin y

Therefore we get

\sin^{-1} (1-x) = \frac{\pi}{2} - 2y

\Rightarrow 1- x = \sin \Big( \frac{\pi}{2} - 2y \Big)

\Rightarrow 1- x = \cos 2y

\Rightarrow 1- x = 1 - 2 \sin^2 y

\Rightarrow 1- x = 1 - 2 x^2

\Rightarrow 2x^2 - x = 0

\Rightarrow x (2x-1) = 0

\Rightarrow x = 0, \frac{1}{2}

Hence the solution of the given equations are x = 0, \frac{1}{2}

(b)

2019-03-21_15-06-42

(BC+A).(A'B'+C') + A'B'C'

= BCA'B'+BCC' + AA'B + AC' + A'B'C' 2019-03-21_15-07-39

= 0+0+0+AC'+A'B'C'

= C'(A+A'B')

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

= C'(A+B')

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Question 4:

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

(b) From the following information, find the equation of the Hyperbola and the equation of its Transverse Axis:

Focus: (-2, 1) , Directrix: 2x-3y+1 = 0 , e = \frac{2}{\sqrt{3}}      [5]

Answer:

(a)  Given, f(x) = \sqrt{x^2 - x} , x \in [1, 4]

Since x^2 - x is continuous on R and \sqrt{x^2-x} exists in [1, 4]

\Rightarrow f(x) is continuous in [1, 4]

Differentiating the given function w.r.t x

f'(x) = \frac{1}{2} (x^2 - x)^{-\frac{1}{2}}. (2x-1) = \frac{2x-1}{2\sqrt{x^2 - x}} 

which exists for all x \in R

Thus, both the conditions of Lagrange’s mean value theorem is satisfied therefore at least one c exists in [1, 4] .

Such that f'(c) = \frac{f(4)-f(1)}{4-1} 

\Rightarrow \frac{2c-1}{2\sqrt{c^2-c}} = \frac{\sqrt{12}-0}{3} 

3(2c-1) = 2 \sqrt{c^2 - c} . \sqrt{12}

Squaring both sides we get

9(4c^2 + 1 - 4 c) = 48(c^2 - c)

\Rightarrow 12c^2 - 12c + 3 = 16c^2 - 16c

\Rightarrow 4c^2 - 4c - 3 = 0

\Rightarrow (2c-3)(2c+1) = 0

\Rightarrow c = \frac{3}{2} \ or \ \frac{-1}{2} 

Clearly, c = \frac{3}{2}  lies in the interval [1, 4]

Hence, Lagrange’s mean value theorem is verified and c = \frac{3}{2}

(b)  Let P(x, y) be the point on the conic, then

PS = e .PM

i.e. \sqrt{(x+2)^2 + (y-1)^2} = \frac{2}{\sqrt{3}} \Big( \frac{2x-3y+1}{\sqrt{4+9}} \Big)

\Rightarrow 39(x^2 + y^2 + 4x -2y + 5) = 4 (4x^2 + 9y^2 + 1 - 12xy + 4x -6y)

i.e. 23x^2 + 48xy + 3y^2 + 140 x - 54 y + 191 = 0 is the required hyperboloa.

Transverse axis  passes through (-2, 1) and is perpendicular  to Directrix, 2x-3y+1 = 0

TA: 3x+2y +c=0 where -6+2+c=0 \Rightarrow c = 4

Therefore TA: 3x+2y+4=0

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Question 5:

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

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

Answer:

(a)  Given y = (\cot^{-1} x)^2

Now, \frac{dy}{dx} = 2 \cot^{-1} x \frac{d}{dx} (\cot^{-1}x)

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

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

Differentiating once again w.r.t x

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

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

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

(b)  Let the height of the cylinder = h and the radius of the cylinder = r

Given Radius of the sphere = 3\sqrt{3} 2019-03-21_18-34-10

If V is the volume of the cylinder, then V = \pi r^2h

Let O be the center of the sphere and COC' \perp AB

For \triangle OCA ,

(3\sqrt{3})^2 = \Big( \frac{h}{2} \Big)^2 + r^2

\Rightarrow r^2 = 27 - \frac{h^2}{4} 

Therefore  V = \pi \Big( 27 - \frac{h^2}{4} \Big). h

V = 27 \pi h - \pi \frac{h^3}{4} 

Therefore \frac{dV}{dh} = 27 \pi - \frac{3 \pi h^2}{4} 

\frac{d^2V}{dh^2} = - \frac{6 \pi h}{4} 

For maxima and minima,

\frac{dV}{dh} = 0

\therefore 27 \pi = \frac{3 \pi h^2}{4} 

\Rightarrow h^2 = \frac{4 \times 27}{3} = 36

\Rightarrow h = \pm 6

and \Big( \frac{d^2V}{dh^2} \Big)_{h=6} = \frac{-6\pi \times 6}{4} < 0

\therefore V is maximum when h = 6 , putting h = 6

r^2 = 27 - \frac{36}{4} 

r^2 = 18

\therefore V = \pi r^2 h = \pi \times 18 \times 6 = 108 \pi \ cm^3

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Question 6:

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

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

Answer:

(a)  Given, \int \limits_{}^{} \frac{ \cos^{-1} x}{x^2} dx

Put \cos^{-1} x = t

\Rightarrow x = \cos t

\Rightarrow dx = - \sin t \ dt

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

= - \int \limits_{}^{} t (\sec t  \tan t) \ dt

= - t \sec t - \int \limits_{}^{} \sec t \ dt

= -t \sec t + \log | \sec t  + \tan t| + c

= - \frac{t}{\cos t} + \log  | \frac{1+ \sin t}{\cos t} | + c

Now putting the value of t back in the expression

\int \limits_{}^{} \frac{ \cos^{-1} x}{x^2} \ dx = - \frac{\cos^{-1} x}{x} + \log | \frac{1+ \sin (\cos^{-1}x)}{x} | + c

\Rightarrow \int \limits_{}^{} \frac{ \cos^{-1} x}{x^2} \ dx = - \frac{\cos^{-1} x}{x} + \log | \frac{1+\sqrt{1-x^2}}{x} | + c

(b)  Given: y = 2x - x^2 and line of intersection y = x 2019-03-22_6-53-45

-y = x^2 - 2x

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

-(y-1) = (x-1)^2 which represents a downward parabola with vertex at (1, 1)

Line of intersection y = x

Putting y = x in the above equation

-(x-1)= (x-1)^2

\Rightarrow -x + 1 = x^2 - 2x + 1

\Rightarrow x^2 - x = 0

\Rightarrow x (x-1) = 0

\Rightarrow x = 0, 1

Therefore the point of intersections are (0, 0) and (1, 1)

Therefore the area enclosed between the curve y = 2x - x^2 and the line y = x is

\int \limits_{0}^{1} (2x - x^2 - x) \ dx = \int \limits_{0}^{1} (x - x^2) \ dx = \Big[ \frac{x^2}{2} - \frac{x^3}{3} \Big]_{0}^{1} 

= ( \frac{1}{2} - \frac{1}{3} ) - (0-0)

= \frac{1}{6} sq. units.

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Question 7:

(a) Find the Karl Pearson’s coefficient of correlation between x and y for the following data.     [5]

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

(b) The following table shows the mean and standard deviation of the marks of Mathematics and Physics scored by students in a school:

Mathematics Physics
Mean 84 81
Standard Deviation 7 4

The correlation co-efficient between the given marks is 0.86 . Estimate the likely marks in Physics in the marks in Mathematics at 92 .     [5]

Answer:

(a) 

x y dx = x - 20 dy = y - 25 dx^2 dy^2 dy^2
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
5 -1 135 221 152

r = \frac{ n \Sigma (dx.dy) - \Sigma dx \times \Sigma dy} { \sqrt{n \Sigma dx^2 - (\Sigma dx)^2 }  . \sqrt{n \Sigma dy^2 - (\Sigma dy)^2} }

\Rightarrow r = \frac{ 8 \times 152 - (5 \times -1)}{ \sqrt{8 \times 135 - 5^2 }  . \sqrt{8 \times 221 - (-1)^2} }

\Rightarrow r = \frac{1216+5}{ \sqrt{1080 - 25}. \sqrt{1768-1} }

= 0.894

(b)  Mean marks in Mathematics \overline{x} = 84

Mean marks in Physics  \overline{y} = 81

\sigma_x = 7, \sigma_y = 4

r = 0.86

Therefore b_{yx} = r. \frac{\sigma_y}{\sigma_x} = 0.86 \times \frac{4}{7} = 0.49

Therefore regression equation of y on x

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

\Rightarrow y = 81 = 0.49 (x-84)

\Rightarrow y = 81 = 0.49x - 41.16

\Rightarrow y = 0.49x + 39.84

Putting x = 92, y = 45.08 + 39.84 = 84.92

Hence the likely marks in Physics are 84.92

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Question 8:

(a) Bag \ A contains three red and four white balls. Bag \ B contains two red and three white balls. If one ball is drawn from Bag \ A and two balls are drawn from Bag \ B , find the probability that:

(i) One ball is red and two balls are white

(ii) All the three balls are of the same color    [5]

(b) Three persons, Aman, Bipin and Mohan attempt a Mathematics problems independently. The odds in favor of Aman and Mohan solving the problem are 3:2 and 4:1 respectively and the odds against Bipin solving the problem are 2:1 . Find:

(i) The probability that all the three will solve the problem

(ii) the probability that the problem will be solved.    [5]

Answer:

(a) Possible selection are as follows:

i)     1 Red ball from Bag A, 2 white ball from Bag B

1 white ball from bag A, 1 white ball from bag B, 1 red ball from Bag B

Therefore P(one ball is red and two balls are white)

= \frac{^3C_1}{^7C_1} \times \frac{^3C_2}{^5C_2} + \frac{^4C_1}{^7C_1} \times \frac{^3C_1 \times ^2C_1}{^5C_2}

= \frac{3}{7} \times \frac{3}{10} + \frac{4}{7} \times \frac{3 \times 2}{10}

= \frac{9}{70} + \frac{24}{70} = \frac{33}{70}

ii)   Possible selections are as follows:

1 red ball from Bag A, 2 red balls from Bag B

1 white ball from Bag A and 2 white balls from bag B

Therefore P (all the three balls are of the same color)

= \frac{^3C_1}{^7C_1} \times \frac{^2C_2}{^5C_2} + \frac{^4C_1}{^7C_1} \times \frac{^3C_2}{^5C_2}

= \frac{3}{7} \times \frac{1}{10} + \frac{4}{7} \times \frac{3 }{10}

= \frac{3}{70} + \frac{12}{70} = \frac{15}{70}

(b)  Odd against = \frac{1-p}{p}

Odds in favor = \frac{p}{1-p}

A : Event Aman solves the problem

B : Event Bipin solves the problem

C : Event Mohan solves the problem

Aman Bipin Mohan
\frac{3}{2} = \frac{p}{1-p}

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

\Rightarrow P(A) = \frac{3}{5}

\frac{2}{1} = \frac{1-p}{p}

\Rightarrow p = \frac{1}{3}

\Rightarrow P(B) = \frac{1}{3}

\frac{4}{1} = \frac{p}{1-p}

\Rightarrow p = \frac{4}{5}

\Rightarrow P(C) = \frac{4}{5}

P( A \cap B \cap C) = P(A). P(B). P(C)

i) Probability that all three will solve the problem = \frac{3}{5} \times \frac{1}{3} \times \frac{4}{5} = \frac{4}{25}

ii) Probability that the problem is not solved = probability that all three fail to solve the problem

\Rightarrow P ( \overline{A} \cap \overline{B} \cap \overline{C}) = P( \overline{A}).P( \overline{B}).P( \overline{C})

= \Big( 1 - \frac{3}{5} \Big) . \Big( 1 - \frac{1}{3} \Big) . \Big( 1 - \frac{4}{5} \Big)

= \frac{2}{5} . \frac{2}{3} . \frac{1}{5} = \frac{4}{75}

Therefore the probability that the problem will be solved = 1 - \frac{4}{75} = \frac{71}{75}

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Question 9:

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

(b) Solve the following differential equation:

y e^y \ dx = (y^3 + 2 x e^y) \ dy , given that x =0, y = 1 .     [5]

Answer:

(a)   Let z = x + iy

arg(z-1) = \frac{\pi}{4}

arg(x+iy-1) = \frac{\pi}{4}

\tan^{-1} \frac{y}{x-1} = \frac{\pi}{4}

\frac{y}{x-1} = 1 2019-03-23_11-04-34

Therefore y = x-1

|z-2-3i| = 2

\Rightarrow |x+iy -2-3i| = 2

\Rightarrow (x-2)^2 + (y-3)^2 = 4

\Rightarrow (x-2)^2 + (x-1-3)^2= 4

\Rightarrow x^2 - 4x +4 +x^2-8x+16=4

\Rightarrow  2x^2-12x+16=0

(x-4)(x-2)=0 \Rightarrow  x = 4, 2

When x = 4, y = 3 and when x = 2, y = 1

Therefore the locus will be (2, 1) and (4, 3)

(b)  Given, y \ e^y \ dx = (y^3 + 2 x \ e^y) \ dy

\Rightarrow y \ e^y \frac{dx}{dy} = y^3 + 2x \ e^y

\Rightarrow \frac{dx}{dy} = \frac{2}{y} . x + y^2 \ e^{-y}

\Rightarrow \frac{dx}{dy}   - \frac{2}{y} . x = y^2 \ e^{-y}

It is a linear differential equation in x .

I.F. = e^{ \int -\frac{2}{y} . dy}

= e^{-2 \log |y|}

= e^{\log |y|^{-2}}

= |y|^{-2} = \frac{1}{y^2}

Therefore, general solution is

x. \frac{1}{y^2} = \int y^2 e^{-y} . \frac{1}{y^2} dy+c

x. \frac{1}{y^2} = \int  e^{-y} dy +c

\Rightarrow x. \frac{1}{y^2} = \frac{e^{-y}}{-1} +c

\Rightarrow Given that x = 0 and y = 1

\Rightarrow 0 = - e^{-1} +c

\Rightarrow c = e^-1

Substituting the value of c   we get

\frac{x}{y^2} = -e^{-y} + e^{-1}

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

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Question 10:

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

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

Answer:

a)   \overrightarrow{a}.\overrightarrow{b} = |\overrightarrow{a}| |\overrightarrow{b}| \cos \theta

\Rightarrow \overrightarrow{a}.\overrightarrow{b} = 1.1 \cos \theta = \cos \theta

Now, |\overrightarrow{a} - \overrightarrow{b} |^2 = (\overrightarrow{a} - \overrightarrow{b})^2

= |\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 - 2 \overrightarrow{a}.\overrightarrow{b}

= 1 + 1 - 2 \cos \theta

= 2 ( 1 - \cos \theta)

\Rightarrow |\overrightarrow{a} - \overrightarrow{b} |^2 = 2.2 \sin^2 \frac{\theta}{2} 

\Rightarrow |\overrightarrow{a} - \overrightarrow{b} | = 2 \sin \frac{\theta}{2} 

b)   Let A, B, C, D 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} and -4\hat{i}+4\hat{j}+4\hat{k}

\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}

\overrightarrow{AC} = 3\hat{i}+9\hat{j} +4\hat{k} - (-\hat{j} - \hat{k}) = 3\hat{i}+10\hat{j} +5\hat{k}

\overrightarrow{AD} = -4\hat{i}+4\hat{j}+4\hat{k} - (-\hat{j} - \hat{k}) = -4\hat{i}+5\hat{j}+5\hat{k}

Since the points A, B, C and D are coplanar, vectors \overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD} and coplanar.

\Rightarrow \begin{bmatrix}  \overrightarrow{AB} &  \overrightarrow{AC} & \overrightarrow{AD} \end{bmatrix} = 0

\Rightarrow \begin{bmatrix}  4 & 6 & \lambda + 1 \\ 3 & 10 & 5 \\ -4 & 5 & 5 \end{bmatrix} = 0

\Rightarrow 4 (50-25) - 3 (30 - 5 - 5 \lambda) - 4 (30-10-10\lambda) = 0

\Rightarrow 100 - 75 + 15\lambda - 80 +40\lambda = 0

\Rightarrow -55 + 55 \lambda = 0

\Rightarrow -1 + \lambda = 0

\Rightarrow \lambda = 1

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Question 11:

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

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

Answer:

a)    Any line passing through the point (-1, 3, -2) is

\frac{x-(-1)}{a} = \frac{y-3}{b} = \frac{z-(-2)}{c} where a, b and c are its direction vectors

It is perpendicular to the lines \frac{x}{1} = \frac{y}{2} = \frac{z}{3}

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

then 1.a + 2.b + 3.c=0

\Rightarrow a +2b+3c=0 … … … … …  i)

and -3.a + 2.b + 5.c=0

\Rightarrow -3a+2b+5c=0 … … … … …  ii)

Subtracting ii) from i) we get

4a-2c=0

\Rightarrow c = 2a … … … … …  iii)

and hence from i) we get

a+ 2b + 3.2a =0

\Rightarrow 2b=-7a

\Rightarrow b = \frac{-7a}{2} … … … … …  iv)

From iv) and iii) we get

a:b:c =a: \frac{-7a}{2} :2a

\Rightarrow a:b:c = 2: -7: 4

Putting these values in the equation of the line

\frac{x+1}{2} = \frac{y-3}{-7} = \frac{z+2}{4}

b)   The given plane is 2x-4y+4z=7 … … … … … i)

Equation of any plane parallel to above equation is

2x - 4y + 4z + k = 0 … … … … … ii)

Now ii) is at a distance of 5 units from the point (3, -1, 2) of

\frac{| 2 \times 3 - 4 (-1) + 4 \times 2 +k|}{\sqrt{2^2 + (-4)^2 + 4^2}} = 5

\Rightarrow \frac{|18+k|}{6} = 5

\Rightarrow |18+k| = 30

\Rightarrow 18+k = 30 or 18+k=-30

\Rightarrow k = 12 or k = -48

Substituting these values in ii) the equations of the required planes are

2x-4y+4z+12=0 and 2x-4y+4z-48=0

or x-2y+2z+6=0 and x-2y+2z-24=0

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Question 12:

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

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

Answer:

a)    Given,

np + npq = 1.8 \Rightarrow np(1+q) = 1.8 … … … … … i)

np.npq = 0.8 \Rightarrow n^2p^2q = 0.8 … … … … … ii)

Dividing the square of i) by ii) we get

\frac{n^2p^2(1+q^2)}{n^2p^2q} = \frac{1.8 \times 1.8}{0.8} 

\Rightarrow \frac{(1+q)^2}{q} = \frac{3.24}{0.8} 

\Rightarrow \frac{(1+q)^2}{q} = \frac{81}{20} 

\Rightarrow 20+2 \times 20q + 20q^2 = 81q

\Rightarrow 20q^2-41q+20=0

\Rightarrow 20q^2 - 25q - 16q +20 = 0

\Rightarrow 5q(4q-5)-4(4q-5) = 0

\Rightarrow (4q-5)(5q-4) = 0

Therefore q = \frac{5}{4} \ or \   \frac{4}{5}

But q \neq \frac{5}{4} \ as \   \frac{5}{4} > 1

Therefore q = \frac{4}{5} 

\Rightarrow p = \frac{1}{5} 

From i), n \times \frac{1}{5} \Big( 1 + \frac{4}{5} \Big) = 1.8

\Rightarrow 9n = 25 \times 1.8

\Rightarrow n = \frac{25 \times 1.8}{9} 

\Rightarrow n = \frac{45}{9} 

Therefore n = 5

Hence, the binomial distribution is (p+q)^n = \Big( \frac{1}{5} + \frac{4}{5} \Big)^5

Probability of getting at least 1 success = P(X \geq 1)

= 1- P(X = 0)

= 1- ^5C_0  \Big( \frac{1}{5} \Big)^0 \Big( \frac{4}{5} \Big)^5

= 1 - \frac{1024}{3125} = 0.67

b)    Let E_1, E_2 and E_3 and E be the events as defined below

E_1 = A is selected as a manager

E_2 = B is selected as a manager

E_3 = C is selected as a manager

and E = radical change occurs in marketing strategy

P(E_1) = \frac{4}{4+1+2} = \frac{4}{7}

P(E_2) = \frac{1}{4+1+2} = \frac{1}{7}

P(E_3) = \frac{2}{4+1+2} = \frac{2}{7}

Given, P(E/E_1) = 0.3 P(E/E_2) = 0.8 , P(E/E_3) = 0.5

We want to find the probability that the radical change in marketing strategy occurred due to the appointment of B i.e

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)}

= \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}

= \frac{0.8}{1.2+0.8+1} = \frac{4}{15}

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Question 13:

(a)  If Mr. Nirav deposits Rs. \ 250 at the beginning of each month in an account that pays an interest of 6\% per annum compounded monthly, how many months will be required for the deposit to at least Rs. \ 6390?   [5]

(b) A mill owner buys two types of machines A \ and \ B for his mill. Machine \ A occupies 1000 sqm of area and requires 12 men to operate it.; while Machine \ B occupies 1200 sqm of area and requires 8 men to operate it. The owner has 7600 sqm of area available and 72 men to operate the machines. If Machine \ A produces 50 units and Machine \ B produces 40 units daily, how many machines of each type should be buy to maximize the daily output? Use linear programming to find the solution.   [5]

Answer:

a)   a = 250

i = \frac{6}{100 \times 12} = 0.005

n = m

S = \frac{a}{i} (1+i) [(1+i)^m - 1]

6390 = \frac{250}{0.005} (1.005) [(1.005)^m - 1 ]

0.1271 = [ (1.005)^{m}  - 1]

1.1271 = 1.005^{m}

\log 1.1271 = m \log 1.005

m = \frac{\log 1.1271}{\log 1.005}

m = 23.98

\Rightarrow m = 24 months

b) Data given

Machine A Machine B Machine C
Area Needed (Sq. m) 1000 1200 7600
Labor Force 12 8 72
Daily Output (units) 50 40

Let x and y be the number of Machines A and Machine B respectively.

Constraints

1000x + 1200y \leq 7600   \Rightarrow 5x + 6y \leq 38

12x + 8 y  \leq 72  \Rightarrow 3x + 2y \leq 18

x\geq 0, y \geq 0

Total output Z = 50x + 40y

So minimize Z = 50x + 40y

Subject to 5x + 6y \leq 38 And 3x + 2y \leq 18  x\geq 0, y \geq 0

Now for 3x + 2y = 18

$latex x $ 6 0
y 0 9

Now for 5x + 6y = 38

x 7.2 0
y 0 (19/3)

2019-03-23_22-51-05

The vertices of the feasible region OAPD are O(0, 0), A (6, 0), P(4, 3) and D (0, 6.33)

Corner Points Object Function Z = 50x + 40 y
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

Thus we see that Z is maximum at P (4, 3) . Therefore number of Machine A = 4 and number of Machine B = 3 .

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Question 14:

(a) A bill of Rs. 60000 was drawn on 1st April 2011 at 4 months and discounted for Rs. 58560 at a bank. If the rate of interest was 12\% per annum, on what date was the bill discounted?   [5]

(b) A company produces a commodity with Rs. 24000 fixed cost. The variable cost is estimated to be 25\% of the total revenue recovered on selling the product at a rate of Rs. 8 per unit. FInd the following:

(i) Cost function

(ii) Revenue function

(iii) Break even point   [5]

Answer:

a)   Banker’s discount  (BD) = 60000 - 58560 = 1440

Let t be the expired period in years

BD = Ani

1440 =60000 \times \frac{12}{100} \times t \Rightarrow t = \frac{1}{5} \ years = 73 days

Legal due date of the bill

1st April 2011 + 4 months + 3 days of grace  = 4th August  2011

The bill was cashed 73 days before 4th August (4 days in August, 31 days in July, 30 days in June, 8 days in May) which comes to 23rd May 2011 as the date

b)    Supposed that x number of units are produced and sold.

Given fixed cost = 24000 Rs.

Revenue R(x) = 8x

i) As each unit’s variable cost of x units is 25\% of revenue,

Therefore Variable cost  of x units = 0.25 \times 8x = 2

Therefore total cost of producing x units = 24000 + 2x

ii) Price of one unit = 8 Rs.

Therefore revenue on selling x units = R(x) = 8x

ii) At break even values

C(x) = R(x)

24000 + 2x = 8 x

\Rightarrow 24000 = 6x

\Rightarrow x = 4000

Hence the break even point is 4000 units produced.

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Question 15:

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

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

(b) The profit of a paper bag manufacturing company (in lakhs / millions of Rs.) during each month pf a year are:

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

Plot the given data on a graph sheet. Calculate the four monthly moving averages and plot these on the same graph sheet.   [5]

Answer:

a) Using simple averages of price relatives, the price index of 2011 taking 2001 as the base year was 127 . From the following data we find x

P_0 80 70 50 20 18 25
P_1 100 87.50 61 22 x 32.5
PR 125 125 122 110 \frac{100x}{18} 130

\frac{\Sigma PR}{N} = 127

\Rightarrow 612 +   \frac{100x}{18} = 127 \times 6 \Rightarrow x = 27

b) 

Month Profit 4 monthly totals 4 monthly average 4 monthly centered moving average
Jan 1.2      
         
Feb 0.8      
    5 1.25  
Mar 1.4     1.35
    5.8 1.45  
Apr 1.6     1.65
    7.4 1.85  
May 2.0     2.125
    9.6 2.4  
Jun 2.4     2.8
    12.8 3.2  
July 3.6     3.375
    14.2 3.55  
Aug 4.8     3.475
    13.6 3.4  
Sep 3.4     3.05
    10.8 2.7  
Oct 1.8     2.25
    7.2 1.8  
Nov 0.8      
         
Dec 1.2      
         

2019-03-24_10-07-11

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