Let godown A supply x and y quintals of grain to the shops D and E respectively. Then, (100 − x − y) will be supplied to shop F.
The requirement at shop D is 60 quintals since x quintals are transported from godown A. Therefore, the remaining (60 −x) quintals will be transported from godown B.
Similarly, (50 − y) quintals and 40 − (100 − x − y) = (x + y − 60) quintals will be transported from godown B to shop E and F respectively.
The given problem can be represented diagrammatically as follows.
\(60-x\ge 0\Leftrightarrow x\le 60\) ..(1)
\(50-y\ge 0\Leftrightarrow y\le 50\) .(2)
\(100-(x+y)\ge 0\Leftrightarrow x+y\le 100\) ..(3)
\(x+y-60\ge 0\Leftrightarrow x+y\ge 60\) ..(4)
and \(x,y\ge 0\)  ...(5)
The given problem can be formulated as
Minimize z = 2.5x + 1.5y + 410 … (1)
subject to the constraints,
The feasible region determined by the system of constraints is as follows.

The corner points are A (60, 0), B (60, 40), C (50, 50), and D (10, 50).
The values of z at these corner points are as follows.

Corner point	z = 2.5x + 1.5y + 41
A (60, 0)	560
B (60, 40)	620
C (50, 50)	610
D (10, 50)	510	→ Minimum

The minimum value of z is 510 at (10, 50).
Thus, the amount of grain transported from A to D, E, and F is 10 quintals, 50 quintals, and 40 quintals respectively and from B to D, E, and F is 50 quintals, 0 quintals, and 0 quintals respectively.
The minimum cost is Rs. 510.

We have : {1, 2, 3, 4, 5, 6}
Here 'X' is the larger of the two numbers obtained>
Now  X = 2, 3, 4, 5, 6
\(P(X=2)=\frac { 1 }{ ^{ 6 }{ C }_{ 2 } } =\frac { 1 }{ 15 } \)
\(P(X=3)=\frac { 2 }{ ^{ 6 }{ C }_{ 2 } } =\frac { 2 }{ 15 } \)
\(P(X=4)=\frac { 3 }{ ^{ 6 }{ C }_{ 2 } } =\frac { 3 }{ 15 } \)
\(P(X=5)=\frac { 4 }{ ^{ 6 }{ C }_{ 2 } } =\frac { 4 }{ 15 } \)
\(P(X=6)=\frac { 5 }{ ^{ 6 }{ C }_{ 2 } } =\frac { 5 }{ 15 } \)
Hence, probability distribution is :       

X:	2	3	4	5	6
P(X):	\(\frac { 1 }{ 15 } \)	\(\frac { 2 }{ 15 } \)	\(\frac { 3 }{ 15 } \)	\(\frac { 4 }{ 15 } \)	\(\frac { 5 }{ 15 } \)

(ii) we have

xi	pi	pixi	xi2	pixi2
2	\(\frac { 1 }{ 15 } \)	\(\frac { 2 }{ 15 } \)	4	\(\frac { 4 }{ 15 } \)
3	\(\frac { 2 }{ 15 } \)	\(\frac { 6 }{ 15 } \)	9	\(\frac { 18 }{ 15 } \)
4	\(\frac { 3 }{ 15 } \)	\(\frac { 12 }{ 15 } \)	16	\(\frac { 48 }{ 15 } \)
5	\(\frac { 4 }{ 15 } \)	\(\frac { 20 }{ 15 } \)	25	\(\frac { 100 }{ 15 } \)
6	\(\frac { 5 }{ 15 } \)	\(\frac { 30 }{ 15 } \)	36	\(\frac { 180 }{ 15 } \)
Total

(I) Mean \(\mu =\sum { { p }_{ i }-{ x }_{ i }=\frac { 70 }{ 15 } =\frac { 14 }{ 3 } } \)
(II) Variance \({ \sigma }^{ 2 }=\sum { { p }_{ i } } { x }_{ i }^{ 2 }-{ \mu }^{ 2 }\)
\(=\frac { 350 }{ 15 } -{ \left( \frac { 14 }{ 3 } \right) }^{ 2 }=\frac { 350 }{ 15 } -\frac { 196 }{ 9 } =\frac { 1050-980 }{ 45 } =\frac { 70 }{ 45 } =\frac { 14 }{ 9 } \)

(i) The given equation is: \(2y{ e }^{ \frac { x }{ y } }dx+\left( y-2x\quad { e }^{ \frac { x }{ y } } \right) dy=0\)
\(\Rightarrow\) \(\frac { dx }{ dy } =\frac { 2x{ e }^{ \frac { x }{ y } }-y }{ 2y{ e }^{ \frac { x }{ y } } } \) .................. (1)
Here  \(f\left( x,y \right) =\frac { 2x{ e }^{ \frac { x }{ y } }-y }{ 2y{ e }^{ \frac { x }{ y } } } .\)
\(\therefore \)  \(f\left( \lambda x,\lambda y \right) =\frac { 2x{ e }^{ \frac { \lambda x }{ \lambda y } }-\lambda y }{ 2x{ e }^{ \frac { \lambda x }{ \lambda y } } } \)
\(=\frac { 2x{ e }^{ \frac { x }{ y } }-y }{ 2y{ e }^{ \frac { x }{ y } } } ={ \lambda }^{ 0 }f\left( x,y \right) \)
Thus \(\\ f\left( x,y \right) \) is homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation.
To solve it, we make the substitution
x = vy ............... (2)
Differentiating equation (2) with respect to y, we get
\(\frac{d x}{d y}=v+y \frac{d v}{d y}\)
Substituting the value of and \(\frac{d x}{d y}\) in equation (1), we get
\(\Rightarrow \) \(y\frac { dv }{ dy } =\frac { 2v{ e }^{ v }-1 }{ 2\quad { e }^{ v }\quad } \quad \quad \)
\( y \frac{d v}{d y}=\frac{2 v e^v-1}{2 e^v}-v \\ y \frac{d v}{d y}=-\frac{1}{2 e^v} \)
or \( 2 e^v d v =\frac{-d y}{y} \) 
or \(\int 2 e^v \cdot d v =-\int \frac{d y}{y} \)
or \(2 e^v =-\log |y|+\mathrm{C} \)
\(2 e^{\frac{x}{y}}+\log |y|=\mathrm{C}\) ................. (3)
Substituting x = 0 and y = 1 in equation (3), we get
\(2 e^0+\log |1|=\mathrm{C} \Rightarrow \mathrm{C}=2\)
Substituting the value of C in equation (3), we get  
\(2 e^{\frac{x}{y}}+\log |y|=2\)
which is the particular solution of the given differential equation.

Given, Z = -3x + 4y
Subject to the constraints
x + 2y \( \leq\) 8; 3x + 2y \( \leq\) 12 and x \(\geq\)0, y \(\geq\) 0
Now, considering the inequations as equations, we get
  x + 2y = 8                  ...(i)
3x + 2y = 12                ...(ii)
Table for line x + 2y = 8 is

x	8	0
y	0	4

On putting (0, 0) in the inequality x + 2y \( \leq\) 8
0 \( \leq\) 8  (which is true)
So, half plane is towards the origin.
Table for line 3x + 2y = 12

x	4	0
y	0	6

On putting (0, 0) in the inequality 3x + 2y \( \leq\)12
0 \( \leq\) 12 (which is true)
So, half plane is towards the origin.
Also, x \(\geq\) 0 and y \(\geq\) 0, so the feasible region lies in the Ist quadrant.
The point of intersection of Eqs. (i) and (ii) is (2, 3).
The graphical representation of the above system of inequations is given below.

Corner points	Value of Z = -3x + 4y
A(0, 4)	16
B(2, 3)	6
C(4, 0)	-12 (Minimum)
O(0, 0)	0

Hence, Z = -12 is minimum at (4, 0).

Total number of outcomes=36
Favourable outcomes are:{(4, 6), (6, 4), (5, 5)}
\(\therefore \)  \(P(A)=P(B)=\frac { 3 }{ 36 } =\frac { 1 }{ 12 } \)
and \(P(\overset { \_ }{ A } )=P(\overset { \_ }{ B } )=1-\frac { 1 }{ 12 } =\frac { 11 }{ 12 } \)
\(\therefore \) P(A wins) = \(P(A)+P(\overset { \_ }{ A } \overset { \_ }{ B } A)+P(\overset { \_ }{ A } \overset { \_ }{ B } \overset { \_ }{ A } \overset { \_ }{ B } A)+........\)
=\(\frac { 1 }{ 12 } +\frac { 11 }{ 12 } \times \frac { 11 }{ 12 } \times \frac { 1 }{ 12 } +\frac { 11 }{ 12 } \times \frac { 11 }{ 12 } \times \frac { 11 }{ 12 } \times \frac { 11 }{ 12 } \times \frac { 1 }{ 12 } \times ........|G.P\)
=\(\frac { \frac { 1 }{ 12 } }{ 1-{ \left( \frac { 11 }{ 12 } \right) }^{ 2 } } =\frac { 1 }{ 12 } \times \frac { 144 }{ 23 } =\frac { 12 }{ 23 } \)
And  P(B wins) = \(P(\overset { \_ }{ A } B)+P(\overset { \_ }{ A } \overset { \_ }{ B } \overset { \_ }{ A } B)+.........\)
= \(\frac { 11 }{ 12 } \times \frac { 1 }{ 12 } +\frac { 11 }{ 12 } \times \frac { 11 }{ 12 } \times \frac { 11 }{ 12 } \times \frac { 1 }{ 12 } +.........|G.P\)
= \(\frac { \frac { 11 }{ 12 } \times \frac { 1 }{ 12 } }{ 1-{ \left( \frac { 11 }{ 12 } \right) }^{ 2 } } =\frac { 111 }{ 144 } \times \frac { 144 }{ 23 } =\frac { 11 }{ 23 } \)

We have \(y={ e }^{ ax }cos\quad bx\) ...(1)
\(\frac { dy }{ dx } ={ e }^{ ax }(-sin\quad bx)(b)+a{ e }^{ ax }cos\quad bx\)
\(={ e }^{ ax }(a\quad cos\quad bx-b\quad sin\quad bx)....(2)\)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ e }^{ ax }(-ab\quad sin\quad bx-{ b }^{ 2 }cos\quad bx)+a{ e }^{ ax }(a\quad cos\quad bx-b\quad sin\quad bx)....(3)\)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =-2a\frac { dy }{ dx } +\left( { a }^{ 2 }+{ b }^{ 2 } \right) y\)
\(={ e }^{ ax }\left[ -ab\quad sin\quad bx-{ b }^{ 2 }cos\quad bx+{ a }^{ 2 }cos\quad bx-ab\quad sin\quad bx \right] \)
\(-2a({ e }^{ ax }(a\quad cos\quad bx-b\quad sin\quad bx))+\left( { a }^{ 2 }+{ b }^{ 2 } \right) { e }^{ ax }cos\quad bx]\)

Here \(p=\frac { 90 }{ 100 } =\frac { 9 }{ 10 } \)
and \(q=1-p=1-\frac { 9 }{ 10 } =\frac { 1 }{ 10 } \) and n = 10
\(\therefore \)  Required Probability = \(P(X\le 6)\)
= \(1-P(7\le X\le 10)\)
= \(1-\sum _{ r=7 }^{ 10 }{ ^{ 10 }{ C }_{ r } } { \left( \frac { 9 }{ 10 } \right) }^{ r }{ \left( \frac { 1 }{ 10 } \right) }^{ 10-r }\)
= \(1-\sum _{ r=7 }^{ 10 }{ ^{ 10 }{ C }_{ r }{ 0.9 }^{ r }{ 0.1 }^{ 10-r } } \)

Part (A) is the correct answer
Let E₁ and E₂ be the events when speaks the truth or not respectively.
\(\therefore \)  \(P({ E }_{ 1 })=\frac { 4 }{ 5 } ,P({ E }_{ 2 })=1-\frac { 4 }{ 5 } =\frac { 1 }{ 5 } \)
Let A be the event when head appears.
\(\therefore \)  \(P(A/{ E }_{ 1 })=\frac { 1 }{ 2 } ,P(A/{ E }_{ 2 })=\frac { 1 }{ 2 } \)
By Bayes' Theorem,
\(P({ E }_{ 1 }/A)=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 }) } \)
\(=\frac { \left( \frac { 4 }{ 5 } \right) \left( \frac { 1 }{ 2 } \right) }{ \left( \frac { 4 }{ 5 } \right) \left( \frac { 1 }{ 2 } \right) +\left( \frac { 1 }{ 5 } \right) \left( \frac { 1 }{ 2 } \right) } =\frac { \frac { 4 }{ 10 } }{ \frac { 1 }{ 2 } } =\frac { 4 }{ 5 } \)

Let S denote the success (getting a ‘6’) and F denote the failure (not getting a ‘6’).
Thus \(\mathrm{P}(\mathrm{S})=\frac{1}{6}, \mathrm{P}(\mathrm{F})=\frac{5}{6}\)
P(A wins in the first throw) = P(S) = 1/6
A gets the third throw, when the first throw by A and second throw by B result into failures.
Therefore, P(A wins in the 3rd throw) = P(FFS) =  P(F)P(F)P(S) \(=\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}\)
\(=\left(\frac{5}{6}\right)^2 \times \frac{1}{6}\)
P(A wins in the 5th throw) = P (FFFFS) \(=\left(\frac{5}{6}\right)^4\left(\frac{1}{6}\right)\) and so on.
Hence,
P(A  wins ) \(=\frac{1}{6}+\left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right)+\left(\frac{5}{6}\right)^4\left(\frac{1}{6}\right)+\ldots \)
\(=\frac{\frac{1}{6}}{1-\frac{25}{36}}=\frac{6}{11}\)
P(B wins) = 1 – P (A wins) \(=1-\frac{6}{11}=\frac{5}{11}\)

\((fog)=f(g(x))\)
= {(2, 5), (5, 2), (1, 5)}.
 \([\therefore Under\ g:2\rightarrow 3\Rightarrow (2,5);\ etc.\ Under\ f:3\rightarrow 2]\) 

By the equation, \(\frac { 1 }{ 2 } \left| \begin{matrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{matrix} \right| =\pm 35\)
\({1\over2}[2(4-4)+6(5-k)+(1)920-4k)]=\le35\)
\({1\over2}[0+30-6k+20-4k]=\le35\)
\(50-10k=\le70\)
\(10k=50\mp 70\)
10k = -20, 120.
Hence k = -2, 12

(B) is the correct answer.
Reason: A= \(A=\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\).
\(\therefore\ A\prime =\begin{bmatrix} cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{bmatrix}.\)
\(\therefore \ A+A\prime =\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}+\begin{bmatrix} cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{bmatrix}\)
\(=\begin{bmatrix} cos\alpha +cos\alpha & -sin\alpha +sin\alpha \\ sin\alpha -sin\alpha & cos\alpha +cos\alpha \end{bmatrix}\)
\(=\begin{bmatrix} 2cos\alpha & 0 \\ 0 & 2cos\alpha \end{bmatrix}.\)
\(Hence,\ A+A\prime =I\)
\(\Rightarrow \ 2cos\alpha =1 \ \Rightarrow \ cos\alpha =\frac { 1 }{ 2 } \)
\(\Rightarrow \alpha =\frac { \pi }{ 3 } .\)

P is the set of all subsets of given set X.
Here \((A,B)\rightarrow (A\cup B)\in P\).
\(\because \) It is a binary operation on the set P.
Similarly, \((A,B)\rightarrow (A\cap B)\in P\)
Hence, it is a binary operation on the set P.

Let \({ cos }^{ -1 }\left( \frac { \sqrt { 3 } }{ 2 } \right) =y\), where \(y\in [0,\pi ]\)
\(\Rightarrow cosy=\frac { \sqrt { 3 } }{ 2 } \Rightarrow cosy=cos\frac { \pi }{ 6 } \)
\(\Rightarrow \ y=\frac { \pi }{ 6 } \)
Hence, the required Principal value = \(\frac { \pi }{ 6 } \)

\(2*3=\frac { 2 }{ 2 } +\frac { 3 }{ 2 } =1+1=2\)