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 × 2 ]

i) Determine whether the binary operations \ast on R defined by a \ast b = | a - b | is commutative. Also, find the value of (-3) \ast 2 .

ii) Prove that

\tan^2( \sec^{-1} 2) + \cos^2 ( \mathrm{cosec}^{-1} 3) = 11

iii) Without expanding at any stage, find the value of the determinant:

\triangle = \left| \begin{array}{ccc} 20 & a & b+c \\ 20 & b & a+c \\ 20 & c & a+b    \end{array} \right|

iv) If \begin{bmatrix}  2 & 3 \\ 5 & 7 \end{bmatrix}  \begin{bmatrix}  1 & -3 \\ -2 & 4  \end{bmatrix} =   \begin{bmatrix}  -4 & 6 \\ -9 & x \end{bmatrix} , find x

v)  Find \frac{dy}{dx} if x^3 + y^3 = 3axy

vi) The edge of a variable cube is increasing at a rate of 10 cm/sec. How fast is the volume of the cube increasing when the edge is 5 cm long?

vii) Evaluate \int \limits_{4}^{5}  |x-5| dx

viii) Form a differential equation of the family of the curves y^2 = 4ax

ix) A bag contains 5 white, 7 red and 4 black balls. If four balls are drawn one by one with replacements, what is the probability that none is white?

x) Let A and B be two events such that P(A) = \frac{1}{2} , P(B) = p and P(A \cup B) = \frac{3}{5} . Find 'p' and A and B are independent events.

Answer:

i)  For commutative

a \ast b = b \ast a for all a, b \in A

Given a \ast b = | a - b |

b \ast a = |b - a | = |a - b | = a \ast b . Hence proved.

(-3) \ast 2 = |-3 - 2 | = |-5| = 5 .

ii)  Given \tan^2( \sec^{-1} 2) + \cos^2 ( \mathrm{cosec}^{-1} 3) = 11

Let \sec^{-1} 2 = \theta \Rightarrow 2 = \sec \theta

And let \mathrm{cosec}^{-1} 3 = \alpha \Rightarrow 3 = \mathrm{cosec} \alpha

\therefore LHS = \tan^2 \theta + \cot^2 \alpha = ( \sec^2 \theta - 1 ) +  ( \mathrm{cosec}^2 \alpha -1)

= (2^2 -1 ) + ( 3^2 - 1 )

= (4-1) + ( 9-1) = 3 + 8 = 11 = RHS. Hence proved.

iii) Given \triangle = \left| \begin{array}{ccc} 20 & a & b+c \\ 20 & b & a+c \\ 20 & c & a+b    \end{array} \right|

\Rightarrow \triangle = 20 \left| \begin{array}{ccc} 1 & a & b+c \\ 1 & b & a+c \\ 1 & c & a+b    \end{array} \right|

C_2 \rightarrow C_2+C_3

\Rightarrow \triangle = 20 \left| \begin{array}{ccc} 1 & a+b+c & b+c \\ 1 & b+c+a & a+c \\ 1 & c+a+b & a+b    \end{array} \right|

\Rightarrow \triangle = 20(a+b+c) \left| \begin{array}{ccc} 1 & 1 & b+c \\ 1 & 1 & a+c \\ 1 & 1 & a+b    \end{array} \right| = 0

Note: Determinant of a Matrix with two Identical rows or columns is equal to 0. It is one of the property of determinants. If, we have any matrix with two identical rows or columns then its determinant is equal to zero.

iv) Given

\begin{bmatrix}  2 & 3 \\ 5 & 7 \end{bmatrix}  \begin{bmatrix}  1 & -3 \\ -2 & 4  \end{bmatrix} =   \begin{bmatrix}  -4 & 6 \\ -9 & x \end{bmatrix}

\Rightarrow  \begin{bmatrix}  2-6 & -6+12 \\ 5-14 & -15+28  \end{bmatrix} =   \begin{bmatrix}  -4 & 6 \\ -9 & x \end{bmatrix}

\Rightarrow  \begin{bmatrix}  -4 & 6 \\ -9 & 13  \end{bmatrix} =   \begin{bmatrix}  -4 & 6 \\ -9 & x \end{bmatrix}

\Rightarrow x = 13

v)  x^3 + y^3 = 3axy

Differentiating on both sides

3x^2 + 3y^2  \frac{dy}{dx} = 3a \Big[  x \frac{dy}{dx} + y \Big]

\Rightarrow x^2 + y^2  \frac{dy}{dx} = ax \frac{dy}{dx} + a y

\Rightarrow x^2 - ay = \Big( \frac{dy}{dx} \Big) (ax - y^2)

\Rightarrow \frac{dy}{dx} = \frac{x^2 - ay}{ax - y^2}

vi)  Let the edge be x

Given \frac{dx}{dt} = 10 cm/sec

We need to find \frac{dV}{dt} at x = 5 cm

We know V = x^3

\Rightarrow \frac{dV}{dt} = 3x^2 \frac{dx}{dt} = 3 \times 25 \times 10 = 750 \ cm^3/sec

vii) \int \limits_{4}^{5}  |x-5| dx

= - \int \limits_{4}^{5}  (x-5) dx

= - \Big[ \frac{x^2}{2} - 5x   \Big]_{4}^{5}

= - \Big[  \Big( \frac{5^2}{2} - 5\times 5  \Big) - \Big( \frac{4^2}{2} - 5\times 4  \Big)    \Big]

= - \Big[ \frac{25}{2} - 25 - \frac{16}{2} + 20 \Big]

= \frac{1}{2}

viii) Consider the given equation.

y^2=4ax      … … … … … (i)

On differentiating both sides w.r.t x ,  we get

2y \frac{dy}{dx} = 4a     … … … … … (ii)

From equations (i) and (ii), we get

y^2 = 2y \frac{dy}{dx} x

2x \frac{dy}{dx} = y

\frac{dy}{dx} = \frac{y}{2x}

ix) Balls are drawn one by one with replacements and each try is independent

Probability of not drawing a White ball = \frac{11}{16}

Therefore the probability of not drawing  white ball in four consecutive draws = \Big( \frac{11}{16} \Big)^4 

x)      P(A \cup B) + P(A \cap B) = P(A) + P(B) 

\Rightarrow \frac{3}{5} + P(A) .P(B) = \frac{1}{2} +p 

\Rightarrow \frac{3}{5} + \frac{1}{2} p = \frac{1}{2} +p 

\Rightarrow \frac{1}{2} p = \frac{1}{10}

\Rightarrow p = \frac{1}{5}

\\

Question 2:  If the function

f : R \rightarrow R be defined as  f(x) = \frac{3x+4}{5y-7} , ( x \neq \frac{7}{5} ) and

g :   R \rightarrow R be defined as g(x) = \frac{7x+4}{5x-3} , ( x \neq \frac{3}{5} ) .

Show that (gof) = (fog) (x)                                                                         [ 4 ]

Answer:

If f : A \rightarrow B and g : B \rightarrow A

(gof) (x) = g(f(x)) = g \Big( \frac{3x+4}{5y-7} \Big)

= \frac{7\Big( \frac{3x+4}{5y-7} \Big)+4}{5\Big( \frac{3x+4}{5y-7} \Big)-3}   = \frac{21x+ 28 + 20x - 28}{15x + 20 - 15x + 21}   = \frac{41x}{41} = x

Similarly,

(fog) (x) = f(g(x)) = f \Big( \frac{7x+4}{5x-3} \Big)

= \frac{3\Big( \frac{7x+4}{5x-3} \Big)+4}{5\Big( \frac{7x+4}{5x-3} \Big)-7}   = \frac{21x + 12 + 20x - 12}{35x+20-35x+21}    = \frac{41x}{41} = x

Hence proved.

\\

Question 3:

a)  If  \cos^{-1} \frac{x}{2} + \cos^{-1} \frac{y}{3} = \theta , then prove that 9x^2 - 12 xy \cos \theta + 4 y^2 = 36 \sin^2 \theta

OR

b) Evaluate \cos( 2 \cos^{-1} x + \sin^{-1} x) at x = \frac{1}{5}                                               [ 4 ]

Answer:

a)  \cos^{-1} \frac{x}{2} + \cos^{-1} \frac{y}{3} = \theta

\cos^{-1} \Big[  \Big( \frac{x}{2} \Big)  \Big( \frac{y}{2} \Big)  - \sqrt{1 - \frac{x^2}{4}} \sqrt{1 - \frac{y^2}{4}} \Big] = \theta

\Rightarrow \frac{xy}{6} - \frac{1}{6} \sqrt{(4-x^2)(9-y^2)} = \cos \theta

\Rightarrow  xy - \sqrt{(4-x^2)(9-y^2)} = 6 \cos \theta

\Rightarrow  xy -6 \cos \theta =  \sqrt{(4-x^2)(9-y^2)}

Squaring both sides

\Rightarrow  x^2y^2 + 36 \cos^2 \theta - 12 xy \cos \theta = 36 - 9x^2 -4y^2 + x^2 y^2

\Rightarrow  9x^2 + 4y^2 - 12xy \cos \theta = 36 ( 1 - \cos \theta)

\Rightarrow 9x^2 + 4y^2 - 12xy \cos \theta = 36 \sin^2 \theta

OR

b)  \cos ( 2 \cos^{-1} x + \sin^{-1} x)

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

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

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

Let \cos^{-1} x = \theta

\therefore x = \cos \theta

= - \sin \theta

= -\sqrt{1 - \cos^2 \theta}

= - \sqrt{1 - x^2}

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

= - \sqrt{\frac{24}{25}}

=- \frac{\sqrt{24}}{5}

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Question 4:  Using the properties of determinants, show that

\left| \begin{array}{ccc} x & p & q \\ p & x & q \\ q & p & x    \end{array} \right| = ( x-p) (x^2 +px - 2q^2)                                                        [ 4 ]

Answer:

LHS = \left| \begin{array}{ccc} x & p & q \\ p & x & q \\ q & q & x    \end{array} \right|

C_1 \rightarrow C_1 - C_2

= \left| \begin{array}{ccc} x-p & p & q \\ p-x & x & q \\ 0 & q & x    \end{array} \right|

= (x-p) \left| \begin{array}{rrr} 1 & p & q \\ -1 & x & q \\ 0 & q & x    \end{array} \right|

R_2 \rightarrow R_2+R_1

= (x-p) \left| \begin{array}{ccc} 1 & p & q \\ 0 & x+p & 2q \\ 0 & q & x    \end{array} \right|

=  ( x-p) [ (x+p) x - 2q^2]

=  ( x-p) [ x^2 +px - 2q^2] . Hence proved.

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Question 5:  Verify Rolle’s theorem for the function  f(x) = - 1 + \cos x in the interval [0, 2\pi ]                                                               [ 4 ]

Answer:

Step I: The function is continuous in [0, 2\pi ]

Step II: The function is derivable in (0, 2\pi )

Step III: f(0) = - 1 + \cos 0 = -1 + 1 = 0

f(2\pi) = - 1 + \cos 2\pi = -1 + 1 = 0

\therefore f(0) = f( 2\pi)

Therefore there must be at least one point c in (0, 2\pi ) at which f'(c) = 0

f'(x) = - \sin x

\Rightarrow f'(c) = - \sin c = 0

\Rightarrow c = \pi                 \{ c \neq 0, 2\pi \}

Hence verified.

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Question 6:  If y = e^{m \sin^{-1}x} , prove that (1-x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = m^2y            [ 4 ]

Answer:

y = e^{m \sin^{-1}x}

Differentiating w.r.t x

\Rightarrow y' = e^{m \sin^{-1}x}. m. \frac{1}{\sqrt{1-x^2}}

\Rightarrow y' = \frac{my}{\sqrt{1-x^2}}

\Rightarrow y' \sqrt{1-x^2} = my

Squaring both sides

\Rightarrow (y')^2 ( 1- x^2) = m^2 y^2

Differentiating again

\Rightarrow (y')^2 ( - 2x) + ( 1- x^2) 2y'y'' = 2 m^2yy'

\Rightarrow -x(y')^2 + ( 1 - x^2)y'y'' = m^2 y y'

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

\Rightarrow (1-x^2) y'' - xy' = m^2y

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Question 7:

(a) The equation of a tangent at (2, 3) on the curve y^2 = px^3+q is y = 4x - 7 . Find the value of 'p' and 'q' .

OR

(b) Using L’Hospital’s rule, evaluate \lim \limits_{x \to 0} \frac{xe^x - \log (1+x)}{x^2}                                   [ 4 ]

Answer:

(a)     y^2 = px^3 + q

Differentiating  w.r.t x

2 y y' = 3 px^2 + 0

\Rightarrow \frac{dy}{dx} = y' = \frac{3px^2}{2y}

Now as y = 4x - y , slope of tangent = 4

\therefore \frac{dy}{dx} = 4

\therefore \frac{3p(2)^2}{2(3)} = 4

\Rightarrow p = 2

Since (2, 3) is lying on the curve,

3^2 = ( 2) 2^3 + q

\Rightarrow q = 9 - 16 = - 7

OR

(b) \lim \limits_{x \to 0} \frac{xe^x - \log (1+x)}{x^2}

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

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

= \frac{0+e^0+e^0+ \frac{1^2}{1}}{2} = \frac{1+1+1}{2} = \frac{3}{2}

\\

Question 8:

(a) Evaluate: \int \limits_{}^{} \frac{dx}{\sqrt{5x-4x^2}}

OR

(b) Evaluate: \int \limits_{}^{} \sin^3 x \cos^4 x dx                                                               [ 4 ]

Answer:

(a)

\int \limits_{}^{} \frac{dx}{\sqrt{5x-4x^2}}

Using the method of completing the square first

5x - 4x^2 = - 4 \Big[ x^2 - \frac{5}{4} x \Big]

\Rightarrow 5x - 4x^2 = -4 \Big[ x^2 - \frac{5}{4} x + \Big( \frac{5}{8} \Big)^2 - ( \frac{5}{8} )^2 \Big]

\Rightarrow 5x - 4x^2 = -4 \Big[ \Big(x- \frac{5}{8} \Big)^2 - \frac{25}{64} \Big]

\Rightarrow 5x - 4x^2 = 4 \Big[ \frac{25}{64} - \Big(x- \frac{5}{8} \Big)^2  \Big]

Therefore

\int \limits_{}^{} \frac{dx}{\sqrt{5x-4x^2}}

= \int \limits_{}^{} \frac{dx}{\sqrt{4[ ( \frac{5}{8})^2 - ( x - \frac{5}{8})^2  ]}}

= \frac{1}{2} \int \limits_{}^{} \frac{dx}{\sqrt{( \frac{5}{8})^2 - ( x - \frac{5}{8})^2 }}

Since   \int \limits_{}^{} \frac{dx}{\sqrt{a^2 - x^2}}  = \sin^{-1} \Big(  \frac{x}{a} \Big) + c

= \frac{1}{2} \sin \Big[  \frac{x - \frac{5}{8}}{\frac{5}{8}} \Big] + c

OR

(b)

\int \limits_{}^{}   \sin^3 x \cos^4 x dx

=    \int \limits_{}^{} \sin^2 x ( \cos^2 x)^2 \sin x dx

= \int \limits_{}^{}   (1 - \cos^2 x) \cos^4  \sin x dx

Let  \cos x = t    \therefore - \sin x dx = dt

= - \int \limits_{}^{} (1 - t^2) t^4 dt

= - \int \limits_{}^{} (t^4 - t^6) dt

= - \Big[ \frac{t^5}{5}   - \frac{t^7}{7}   \Big] + c

= - \frac{t^5}{5}   + \frac{t^7}{7}   + c

= - \frac{\cos^5}{5}   + \frac{\cos^7}{7}   + c

\\

Question 9:  Solve the differential equation (1+x^2) \frac{dy}{dx} = 4x^2 - 2xy    [ 4 ]

Answer:

(1+x^2) \frac{dy}{dx} = 4x^2 - 2xy

\Rightarrow \frac{dy}{dx} = \frac{4x^2}{1+x^2} - \Big( \frac{2x}{1+x^2} \Big) y

\Rightarrow \frac{dy}{dx} + \Big( \frac{2x}{1+x^2} \Big) y= \frac{4x^2}{1+x^2}

\Rightarrow \frac{dy}{dx} +Py = Q

Integration Factor = e^{ \int \limits_{}^{} p dx } = e^{ \int \limits_{}^{} \frac{2x}{1+x^2} dx } = e^{ \log |1+x^2| } = (1+x^2)

Now y ( 1+x^2) =  \int \limits_{}^{} \Big( \frac{4x}{1+x^2} \Big) (1 + x^2) dx + c

\Rightarrow y(1+x^2) = 4 . \frac{x^3}{3} + c

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Question 10:  Three persons A, B and C shoot to hit a target. Their probability of hitting the target are \frac{5}{6} , \frac{4}{3} and \frac{3}{4} respectively. Find the probability that

i) Exactly two persons hit the target

ii) At least one person hits the target                                                          [ 4 ]

Answer:

Given:

P(A) = \frac{5}{6}           P(\overline{A}) = \frac{1}{6}

P(B) = \frac{4}{3}           P(\overline{B}) = \frac{1}{5}

P(C) = \frac{3}{4}           P(\overline{C}) = \frac{1}{4}

i) P( exactly two persons hit the target )

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

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

= \frac{1}{6} + \frac{1}{10} + \frac{1}{8} = \frac{188}{480} = 0.39

ii) P( none of the three persons hit the target ) = \frac{1}{6} \times \frac{1}{5} \times \frac{1}{4} = \frac{1}{120}

Therefore Probability of at least one hitting the target = 1 - \frac{1}{120} = \frac{119}{120}

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Question 11: Solve the following system of linear equations using matrices: x - 2y = 10, 2x - y - z = 8 and -2y + z = 8                                                       [6]

Answer:

Given equations

x-2y=10, 2x-y-z=8 and -2y+z = 7

AX = B

where

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

|A| = 1( -1-2) + 2(2) = -3 + 4 = 1 \neq 0

\because A^{-1} exists means there is a unique solution

X = A^{-1}B

A^{-1} = \frac{1}{|A|} . adj A

= \frac{1}{1} \begin{bmatrix} +(-3) & -2 & +(-4) \\  -(-2) & +1 & -(-2) \\ +2 & -(-1) & +(3) \end{bmatrix}'   = \begin{bmatrix} -3 & -2 & -4 \\  2 & 1 & 2 \\ 2 & 1 & 3 \end{bmatrix}'   = \begin{bmatrix} -3 & 2 & 2 \\  -2 & 1 & 1 \\ -4 & 2 & 3 \end{bmatrix}

\therefore X = \begin{bmatrix} -3 & 2 & 2 \\  -2 & 1 & 1 \\ -4 & 2 & 3 \end{bmatrix}  \begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix}   = \begin{bmatrix} -30+16+14 \\  -20+8+7 \\ -40+16+21 \end{bmatrix} = \begin{bmatrix} 0 \\  -5 \\ -3 \end{bmatrix}

\therefore x = 0, y = -5 , z = -3

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Question 12:

(a) Show that the radius of a closed right circular cylinder of a given surface area and maximum volume to is equal to half of its height.

OR

(b) Prove that the area of a right angled triangle of given hypotenuse is maximum when the triangle is isosceles.                                                  [6]

Answer:

(a) A = 2\pi r h + 2 \pi r^2 \Rightarrow h = \frac{A - 2\pi r^2}{2 \pi r}

V = \pi r^2 h = \pi r^2 \Big( \frac{A - 2\pi r^2}{2 \pi r} \Big) = \frac{r}{2} [A - 2\pi r^2]

\therefore V = \frac{Ar}{2} - \pi r^3

Differentiating w.r.t. r

\frac{dV}{dr} = \frac{A}{2} - 3\pi r^2

Now \frac{dV}{dr} = 0

\therefore \frac{A}{2} - 3\pi r^2 = 0

\Rightarrow \frac{A}{2} = 3 \pi r^2

\Rightarrow \frac{A}{6\pi} = r^2

\Rightarrow r = \sqrt{\frac{A}{6\pi}}

Differentiating again w.r.t. r

\frac{d^2V}{dr^2} = 0 - 6 \pi r   which is negative. Hence volume maximum at r = \sqrt{\frac{A}{6\pi}}

Substituting

A = 2\pi \Big( \sqrt{\frac{A}{6\pi}} \Big)  h + 2 \pi \Big( \sqrt{\frac{A}{6\pi}} \Big)^2

\Rightarrow A = 2\pi \Big( \sqrt{\frac{A}{6\pi}} \Big) h + 2 \pi \Big( \frac{A}{6\pi} \Big) 

\Rightarrow \frac{2}{3} A = 2 \pi \Big( \sqrt{\frac{A}{6\pi}} \Big)   h

\Rightarrow h = \frac{\frac{2}{3}A}{2\pi \sqrt{\frac{A}{6\pi}}} = \sqrt{\frac{2A}{3\pi}}

Now r = \sqrt{\frac{A}{6\pi}}    and h = \sqrt{\frac{2A}{3\pi}}

\therefore \frac{r}{h} = \sqrt{\frac{A}{6\pi}} \times \sqrt{\frac{3\pi}{2A}} = \frac{1}{2}

\therefore r = \frac{h}{2} . Hence proved.

OR

(b)

A = \frac{1}{2} xy

We know x^2 + y^2 = H^2 \Rightarrow y = \sqrt{H^2 - x^2}

A = \frac{1}{2} x \sqrt{H^2 - x^2}

Differentiating w.r.t. x

\frac{dA}{dx} = \frac{1}{2} \Big[  2 \Big( \frac{-2x}{2(\sqrt{H^2-x^2})} \Big) +  \sqrt{H^2 - x^2} (1)     \Big]

= \frac{1}{2} \Big[  \frac{-x^2+H^2-x^2}{\sqrt{H^2 - x^2}} \Big] = 0

\Rightarrow H^2 = 2x^2

\frac{H}{\sqrt{2}} = x

Differentiating again w.r.t. x we get less than 0 at x = \frac{H}{\sqrt{2}} . Hence it’s maximum.

\therefore x^2 + y^2 = H^2

\frac{H^2}{2} + y^2 = H^2

y = \frac{H}{\sqrt{2}} = x

Therefore the triangle is an isosceles triangle.

\\

Question 13:

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

OR

(b) Evaluate: \int \limits_{}^{} \frac{2x+7}{x^2 - x - 2} dx                                                                              [6]

Answer:

(a)

I = \int \limits_{}^{} \tan^{-1} \sqrt{\frac{1-x}{1+x}}  dx

Let x = \cos 2\theta  \Rightarrow dx = - 2 \sin 2 \theta d\theta

\therefore = -2 \int \limits_{}^{} \tan^{-1} \sqrt{\frac{1-\cos 2\theta}{1+\cos 2\theta}}  \sin 2 \theta d\theta

= -2 \int \limits_{}^{} \tan^{-1} \sqrt{\frac{2 \sin^2 \theta}{2\cos^2 \theta}}  \sin 2 \theta d\theta

= -2 \int \limits_{}^{} \tan^{-1} (\tan \theta) \sin 2\theta d\theta

= -2 \int \limits_{}^{} \theta \sin 2\theta d\theta

Let 2\theta = y  \Rightarrow 2 d\theta = dy

= - \frac{1}{2} \int \limits_{}^{} y \sin y dy

= - \frac{1}{2} \Big[ y \int \limits_{}^{} \sin y dy - \int \limits_{}^{} \frac{d(y)}{dy} \int \limits_{}^{} \sin y dy + c \Big]

= - \frac{1}{2} \Big[ -y \cos y + \int \limits_{}^{} \cos y dy + c \Big]

= - \frac{1}{2} \Big[ -y \cos y + \sin y + c \Big]

= \frac{1}{2} y \cos y - \frac{1}{2}  \sin y + c'

= \frac{1}{2} (2\theta) \cos 2\theta - \frac{1}{2}  \sin 2\theta + c'

Where \theta = \frac{1}{2} \cos^{-1}x

OR

(b)  \int \limits_{}^{} \frac{2x+7}{x^2 - x - 2} dx

= \int \limits_{}^{} \frac{2x-1}{x^2 - x - 2} dx + \int \limits_{}^{} \frac{8}{x^2 - x - 2} dx

= \int \limits_{}^{} \frac{2x-1}{x^2 - x - 2} dx + 8 \int \limits_{}^{} \frac{1}{x^2 - x - 2} dx

= \int \limits_{}^{} \frac{2x-1}{x^2 - x - 2} dx + 8 \int \limits_{}^{} \frac{1}{(x-2)(x+1)} dx

= \int \limits_{}^{} \frac{2x-1}{x^2 - x - 2} dx + \frac{8}{3} \int \limits_{}^{} \Big[ \frac{1}{(x-2)} - \frac{1}{(x+1)} \Big] dx 

Now, if u = x^2 - x + 1 \Rightarrow \frac{du}{dx} = 2x-1 \Rightarrow dx = \frac{1}{2x-1} du

\therefore \int \limits_{}^{} \frac{2x-1}{x^2 - x - 2} = \int \limits_{}^{} \frac{1}{u} du = \log |u| = \log |x^2 - x - 2  |

Now, if u = x-2 \Rightarrow \frac{du}{dx} = 1 \Rightarrow dx = du

\therefore \int \limits_{}^{} \frac{1}{x- 2} = \int \limits_{}^{} \frac{1}{u} du = \log |u| = \log |x - 2  |

Now, if u = x+1 \Rightarrow \frac{du}{dx} = 1 \Rightarrow dx =  du

\therefore \int \limits_{}^{} \frac{1}{x+1} = \int \limits_{}^{} \frac{1}{u} du = \log |u| = \log |x +1|

\therefore \int \limits_{}^{} \frac{2x+7}{x^2 - x - 2} dx = \log |x^2 - x - 2  | + \frac{8}{3} [ \log |x - 2  | - \log |x +1| ]

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Question 14:  The probability that a bulb produced in a factory will fuse after 150 days is 0.05 . Find the probability that out of five such bulbs:

i) None will fuse after 150 days of use

ii) Not more than one will fuse after 150 days of use

iii) More than one will fuse after 150 days of use

iv) At least one will fuse after 150 days of use

Answer:

Let X be the number of bulbs that will fuse after 150 days of use in an experiment of 5 trials.

p = 0.05     q = 1 - 0.05 = 0.95     n = 5

\therefore p(X=x) = {^nC_x}  q^{n-x} p^x = {^5C_x}  (0.95)^{5-x}(0.05)^x

i) P(none) = P(X=0) = {^5C_0}  (0.95)^{5-0}(0.05)^0 = (0.95)^5

ii) P(X \leq 1) = P(X=0) + P(X=1)

= {^5C_0}  (0.95)^{5-0}(0.05)^0 + {^5C_1}  (0.95)^{5-1}(0.05)^1 = (0.95)^4 \times 1.2

iii) P(X > 1) = 1 - P(X \leq 1) = 1 - (0.95)^4 \times 1.2

iv) P(X = \ at \ least \ one \ will \ fuse ) = 1 - P(X =0) = 1 -(0.95)^5

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SECTION B (20 Marks)

Question 15:                                                                                              [ 3 × 2]

(a) Write a vector of magnitude 18 units in the direction of the vector \hat{i}-2\hat{j} - 2\hat{k} .

(b) Find the angle between the two lines

\frac{x+1}{2} = \frac{y-2}{5} = \frac{z+3}{4}    and   \frac{x-1}{5} = \frac{y+2}{2} = \frac{z-1}{-5}

(c) Find the equation of the plane passing through the point (2, -3, 1) and perpendicular to the line joining the points (4, 5, 0) and (1, -2, 4) .

Answer:

(a)    \overrightarrow{a} = \hat{i}-2\hat{j} - 2\hat{k}

\hat{a} = \frac{\overrightarrow{a}}{|\overrightarrow{a}|} = \frac{\hat{i}-2\hat{j} - 2\hat{k}}{\sqrt{1+4+4}} = \frac{\hat{i}-2\hat{j} - 2\hat{k}}{3}

\overrightarrow{b} = |\overrightarrow{b}| \hat{a}

\Rightarrow \overrightarrow{b} = 18 [ \frac{1}{3} ( \hat{i}-2\hat{j} - 2\hat{k}) ]

\Rightarrow \overrightarrow{b} =  6\hat{i}-12\hat{j} - 12\hat{k}

(b)    L_1 : \frac{x+1}{2} = \frac{y-2}{5} = \frac{z+3}{4}

\Rightarrow a_1 = 2, b_1 = 5 , c_1 = 4 and a_2 = 5, b_2 = 2 , c_2 = -5

L_2 :  \frac{x-1}{5} = \frac{y+2}{2} = \frac{z-1}{-5}

Angle between L_1 and L_2 is given by

\cos \theta = \Big| \frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{  {a_1}^2+ {b_1}^2 + {c_1}^2  } \sqrt{  {a_2}^2+ {b_2}^2 + {c_2}^2  } } \Big|

= \Big| \frac{2 \times 5+ 5 \times 2 + 4 \times (-5) }{\sqrt{  4+25+16  } \sqrt{  25+4+16 }} \Big|

= \Big| \frac{10+10-20}{45} \Big| = 0

\Rightarrow \cos \theta = 0 \Rightarrow \theta = 90^{\circ}   or \frac{\pi}{2}

(c)    Given point ( 2, -3, 1 ). d.rs of line joining the points A( 4, 5, 0) and B ( 1, -2, 4) is < 1-4, -2-5, 0-4 > i.e. < -3, -7, -4> or <3, 7, 4 >

Equation of plant passing through (2, -3, 1) is

A(x-x_1) + B ( y - y_1) + c( z-z_1) = 0

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

\Rightarrow 3x + 7y + 4z + 11=0

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Question 16:                                                                                                       [ 4]

(a) Prove that \overrightarrow{a} . [ (\overrightarrow{b}+\overrightarrow{c}) \times (\overrightarrow{a} + 3\overrightarrow{b} + 4 \overrightarrow{c})] = [\overrightarrow{a}\  \overrightarrow{b}\  \overrightarrow{c} ] 

OR

(b) Using vectors, find the area of the triangle whose vertices are: A ( 3, -1, 2) , B ( 1, -1, -3)  and C( 4, -3, 1)  .

Answer:

(a)    We have \overrightarrow{a} . [ (\overrightarrow{b}+\overrightarrow{c}) \times (\overrightarrow{a} + 3\overrightarrow{b} + 4 \overrightarrow{c})]

= \overrightarrow{a} . [ \overrightarrow{b} \times \overrightarrow{a}  + 3( \overrightarrow{b} \times \overrightarrow{b}) + 4 (\overrightarrow{b} \times \overrightarrow{c})  + \overrightarrow{c} \times \overrightarrow{a}  + 3( \overrightarrow{c} \times \overrightarrow{b}) + 4 (\overrightarrow{c} \times \overrightarrow{c})  ]

= \overrightarrow{a} . [ \overrightarrow{b} \times \overrightarrow{a}  +  4( \overrightarrow{b} \times \overrightarrow{c})  + \overrightarrow{c} \times \overrightarrow{a}  + 3( \overrightarrow{c} \times \overrightarrow{b})   ]

= \overrightarrow{a} . [ \overrightarrow{b} \times \overrightarrow{a}  +  4( \overrightarrow{b} \times \overrightarrow{c})  + \overrightarrow{c} \times \overrightarrow{a}  - 3( \overrightarrow{b} \times \overrightarrow{c})   ]

= \overrightarrow{a} . [ \overrightarrow{b} \times \overrightarrow{a}  +   \overrightarrow{b} \times \overrightarrow{c} + \overrightarrow{c} \times \overrightarrow{a}    ]

=  \overrightarrow{a}( \overrightarrow{b} \times \overrightarrow{a})  +  \overrightarrow{a}( \overrightarrow{b} \times \overrightarrow{c}) + \overrightarrow{a} (\overrightarrow{c} \times \overrightarrow{a})    ]

= 0 + [ \overrightarrow{a}\  \overrightarrow{b}\  \overrightarrow{c}] + 0

= [ \overrightarrow{a}\  \overrightarrow{b}\  \overrightarrow{c}]  = RHS. Hence proved.

OR

(b)   Given A ( 3, -1, 2) , B ( 1, -1, -3)  and C( 4, -3, 1)  .

\overrightarrow{AB} = (1-3) \hat{i}+ ( -1 + 1 ) \hat{j}+ (-3-2) \hat{k} = 2\hat{i}+ 0\hat{j} -5\hat{k}

\overrightarrow{AC} = (4-3) \hat{i}+ ( -3 + 1 ) \hat{j}+ (1-2) \hat{k} = \hat{i}-2\hat{j} -\hat{k}

Area of \triangle ABC = \frac{1}{2} | \overrightarrow{AB} \times \overrightarrow{AC} |

\overrightarrow{AB} \times \overrightarrow{AC} = \left| \begin{array}{rrr} \hat{i} & \hat{j} & \hat{j}  \\ -2 & 0 & -5 \\ 1 & -2 & -1 \end{array} \right|

= \hat{i} ( 0 - 10) - \hat{j} (2+5) + \hat{k} (4-0)

= -10\hat{i} -7\hat{j} + 4\hat{k}

\therefore Area of \triangle ABC = \frac{1}{2} \sqrt{100+49+16} = \frac{1}{2} \sqrt{165} sq. units.

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Question 17:                                                                                                           [ 4 ]

(a) Find the image of the point (3, -2, 1) on the plant 3x-y+4z=2

OR

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

Answer:

(a)    The directional ratios of normal to plane are <3, -1, 4>

\therefore Equation of AB = \frac{x-3}{3} = \frac{y+2}{-1} = \frac{z-1}{4} = k

Coordinate of a general point on line AB is ( 3k+3, -k-2, 4k+1)

This point lies on the plane 3x -y + 4z=2

Therefore 3(3k+3)-(-k-2)+4(4k+1) =2

\Rightarrow 9k + 9 + k + 2 + 16k + 4 = 2

\Rightarrow 26k+15=2

\Rightarrow k = - \frac{13}{26} = - \frac{1}{2}

\therefore coordinate of M are

\Big( 3 \times (- \frac{1}{2} ) + 3, \frac{1}{2} - 2, 4 \times (- \frac{1}{2} ) + 1 \Big) = \Big( \frac{3}{2} , - \frac{3}{2} , -1 \Big)

OR

(b)    Equation of line passing through the point ( -1, 3, -2) is

L:  \frac{x+1}{a} = \frac{y-3}{b} = \frac{z+2}{c} where a, b, c are direction ratios of the line.

Line L is \perp to lines  \frac{x}{1} = \frac{y}{2} = \frac{z}{3}   and \frac{x+2}{-3} = \frac{y-1}{2} = \frac{z+1}{5}

The directional ratios of like \perp can be found by solving equation

\frac{a}{10-6} = \frac{b}{-9-5} = \frac{c}{2+6}

\Rightarrow \frac{a}{4} = \frac{b}{-14} = \frac{c}{8}

\Rightarrow \frac{a}{2} = \frac{b}{-7} = \frac{c}{4}

Therefore the equation of line L is  \frac{x+1}{2} = \frac{y-3}{-7} = \frac{z+2}{4}

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Question 18: Draw a rough sketch of the curves y^2 = x and y^2 = 4 - 3x and find the area enclosed between them.                                                    [ 6 ]

Answer:

y^2 = x      … … … … … i) (equation of a parabola)

y^2 = 4 - 3x      … … … … … ii) (equation of a parabola)

Points of intersection:

4-3x = x \Rightarrow x = 1

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

Hence the two points of intersection are ( 1, 1) and ( 1, -1)

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

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

2020-08-16_16-36-23

Let PQ cut the x-axis at  R

Total area = POQLP = 2 area of PORLP

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

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

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

= 2    \Big[  \frac{2}{3} + \frac{2}{9}    \Big]

= \frac{16}{9} sq. units

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SECTION C (20 Marks)

Question 19:                                                                                                           [3 × 2]

(a) The selling price of a commodity is fixed at Rs. 60   and its cost function is C(x) = 35x + 250

i) Determine it’s profit margin

ii) Find the break even point

(b) The Revenue function is given by R(x) = 100x - x^2 -x^3 . Find

i) The demand function

ii) Marginal revenue function

(c) For the lines of regression 4x - 2y = 4 and 2x-3y+6 = 0 , find the value of 'x' and the mean of 'y'

Answer:

(a)    C(x) = 35x + 250 and S(x) = 60x

i) P(x) = S(x) - C(x) = 60x-35x-250 = 25x-250 = 25( x - 10)

ii)At break even point P(x) = 0

\therefore 60x = 35x + 250

\Rightarrow 25x = 250  \Rightarrow x = 10

(b)

i) R(x) = P(x).x

P(x) = \frac{R(x)}{x} = \frac{100x - x^2 -x^3}{x} = 100 - x - x^2

\therefore P(x) = 100 - x - x^2

ii) MR = \frac{dR}{dx} = 100-2x-3x^2

\therefore MR = 100-2x-3x^2

(c)    4x - 2y = 4      … … … … … i)

2x-3y=-6      … … … … … ii)

Solving i) and ii)  we get y = 4 and x = 3

\therefore \overline{x} = 3 and \overline{y} = 4

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Question 20:                                                                                                           [4]

(a) The correlation coefficient between x and y is 0.6 . If the variance of x = 225 and the variance of y is 400 , mean of x is 10 and the mean of y = 20 , find:

i) the equations of two regression lines

ii) the expected value of y when x = 2

OR

(b) Find the regression coefficient  b_{yx}, b_{xy} and the correlation coefficient 'r' for the following data: ( 2, 8), ( 6, 8), ( 4, 5), ( 7, 6), (5, 2)

Answer:

(a)    Given r = 0.6     \overline{x} =10     \overline{y} =10

Var(x) = 225       Var(y) = 400

\sigma_x = 15               \sigma_y = 20

b_{yx} = r \frac{\sigma_y}{\sigma_x} = 0.6 \times \frac{20}{15} = \frac{4}{5}

b_{xy} = r \frac{\sigma_x}{\sigma_y} = 0.6 \times \frac{15}{20} = \frac{9}{20}

i)       Line of regression of y on x

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

\Rightarrow y - 20 = \frac{4}{5} (x - 10)

\Rightarrow 5y - 100 = 4x - 40

\Rightarrow 4x - 5y + 60 = 0

         Line of regression of x on y

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

\Rightarrow y - 10 = \frac{9}{20} (y - 20)

\Rightarrow 20x-200 = 9y - 180

\Rightarrow 20x - 9 y - 20 = 0

ii) At x = 2, 4(2) - 5y +60 = 0  \Rightarrow 5y = 68 \Rightarrow y = \frac{68}{5}

OR

(b)

x x^2 y y^2 xy
2 4 8 64 16
6 36 8 64 48
4 14 5 25 20
7 49 6 36 42
5 25 2 4 10
\Sigma x = 24 \Sigma x^2 = 130 \Sigma y = 29 \Sigma y^2 = 193 \Sigma xy = 136

 

b_{yx} = \frac{\Sigma xy - \frac{1}{n} \Sigma x \Sigma y}{\Sigma x^2 - \frac{1}{n} ( \Sigma x)^2}   = \frac{136 - \frac{1}{5} (24)(29)}{130 - \frac{1}{5} (24)^2}   = - \frac{3.2}{14.8} = -0.22

b_{xy} = \frac{\Sigma xy - \frac{1}{n} \Sigma x \Sigma y}{\Sigma y^2 - \frac{1}{n} ( \Sigma y)^2}   = \frac{136 - \frac{1}{5} (24)(29)}{193 - \frac{1}{5} (29)^2}   = - \frac{3.2}{24.8} = -0.13

|x| =\sqrt{b_{yx}b_{xy}} = \sqrt{(-0.22)(-0.13)} = \pm \sqrt{0.0286} = \pm 0.17

Since r has the same sign as regressions coefficient \therefore r(x, y) = -0.17

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Question 21:                                                                                                       [4]

(a)  The marginal cost of the production of the commodity is 30+2x , it is know that fixed cost are Rs. 200 , find;

i) The total cost

ii) The cost of increasing output from 100 to 200 units 

OR

(b)  The total cost function of the firm is given by C(x) = \frac{1}{3} x^3 - 5x^2 + 30 x - 15 where the selling price per unit is given as Rs. 6 . Find for what value of x will the profit be maximum.

Answer:

(a) 

i)      MC = \frac{dc}{dx} =30 + 2x

Integrating both sides

C = \int \limits_{}^{} ( 30+2x) dx

\Rightarrow C = 30x + x^2 + k

When x = 0, C = 200

\therefore 200 = 0 + 0 + k

\Rightarrow k = 200

\therefore cost function C(x) = 30x + x^2 + 200

ii) Cost of increasing the output from 100 to 200 units

= \int \limits_{100}^{200} (30+2x) dx

= {\Big[ 30x + x^2  \Big]}_{100}^{200}

= [ 30(200) + 200^2] - [ 30(100)+100^2]

= 33000

\therefore cost of increasing output = 33000 Rs.

OR

(b)   

C(x) = \frac{1}{3}   x^3 - 5x^2 + 30 x - 15

S(x) = 6x

\therefore P(x) = S(x) - C(x) = 5x^2 - \frac{1}{3} x^3 - 24 x + 15

\frac{dP}{dx}   = 10 x - x^2 - 24

Put \frac{dP}{dx}   = 0

\Rightarrow x^2 - 10x + 24 = 0

\Rightarrow (x-4)(x-6) = 0

\Rightarrow x = 4 \ or \  6

\frac{d^2P}{dx^2}   = 10- 2x

\frac{d^2P}{dx^2}   \Big|_{x=4} = 10 - 2(4) = 2 > 0 [ Therefore P is minimum at x = 4 ]

\frac{d^2P}{dx^2}   \Big|_{x=6} = 10 - 2(6) = -2 < 0 [ Therefore P is maximum at x = 6 ]

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Question 22:  A company uses three machines to manufacture two types of shirts, half sleeves shirts and full sleeves shirt. The number of hours required per week on machines M_1, M_2 and M_3 for one shirt of each type is given in the following table:

M_1 M_2 M_3
Half Sleeves Shirt 1 2 8/5
Full Sleeves Shirt 2 1 8/5

None of the machines can be in operations for more than 40 hours per week. The profit on each half sleeves shirt is Rs. 1 and the profit on each full sleeves shirt is Rs. 1.50 .

How many of each type of shirts should be made per week to maximize the company’s profit.                                                                                                [6]

Answer:

Let the number of half sleeves shirts produced = x

Let the number of full sleeves shirts produced = x

Maximize Z = x + 1.5 y

Constraints:

x+2y \leq 40

2x+y \leq 40

\frac{8}{5} x + \frac{8}{5} y \leq 40 \Rightarrow x + y \leq 25

Now draw the three lines:

x+2y=40

2x+y=40

x+y=25

x 0 40 x 0 20 x 0 25
y 20 0 y 40 0 y 25 0

2020-08-16_18-32-55

Z = x + 1.5 y
A( 0, 20) 30
B(10, 15) 32.5
C( 15, 10) 30
D( 20, 0) 20

Therefore the maximum value is at B(10, 15) . Hence the company should make 10 Half Sleeves Shirts and 15 Full Sleeves Shirts to maximize profit.