The given D.E. is:
\(xcos\left( \frac { y }{ x } \right) \frac { dy }{ dx } =ycos\left( \frac { y }{ x } \right) +x\) ..(i)
When, \(x=1, y=\frac { \pi }{ 4 } \)
\(cos\left( \frac { y }{ x } \right) \frac { dy }{ dx } =\frac { y }{ x } cos\left( \frac { y }{ x } \right) +1\)....(ii)
Put, y/x = v
\(\Rightarrow\) \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
From (ii), \(cosv\left( v+x\frac { dv }{ dx } \right) =v\quad cosv+1\)
\(\Rightarrow\) v cosv+x cosv dv/dx = v cos v + 1
\(\Rightarrow\) \(\\ xcos\frac { dv }{ dx } =1\)
\(\Rightarrow\) \(\int { cosv\quad dv=\int { \frac { dx }{ x } } } \)
Put \(y=\frac { \pi }{ 4 } \) , x = 1
\(\Rightarrow\) \(\frac { 1 }{ \sqrt { 2 } } =C\)
The particular solution is
\(sin\left( \frac { y }{ x } \right) =logx+\frac { 1 }{ \sqrt { 2 } } \)

3x + 2y + z = 1,600
4x + Y + 3z = 2,300
x + y + z = 900
\(\therefore \left[ \begin{matrix} 3 & 2 & 1 \\ 4 & 1 & 3 \\ 1 & 1 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 1,600 \\ 2,300 \\ 900 \end{matrix} \right] \ or\ AX=B\)
|A| = 3(-2)-2(1)+1(3)
= -5 \(\ne\)0
Cofactors are:
\({ A }_{ 11 }=-2, { A }_{ 12 }=-1, { A }_{ 13 }=3 \)
\({ A }_{ 21 }=-1, { A }_{ 22 }=2, { A }_{ 23 }=-1 \)
\({ A }_{ 31 }=5, { A }_{ 32 }=-5, { A }_{ 33 }=-5 \)
\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =-\frac { 1 }{ 5 } \left[ \begin{matrix} -2 & -1 & 5 \\ -1 & 2 & -5 \\ 3 & -1 & -5 \end{matrix} \right] \left[ \begin{matrix} 1,600 \\ 2,300 \\ 900 \end{matrix} \right] \)
\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 200 \\ 300 \\ 400 \end{matrix} \right] \)
x = 200, y = 300 and z = 400
i.e., Rs. 200 for sincerity, Rs. 300 for truthfullness and Rs. 400 for helpfulness.
Value: One more value like honesty, kindness etc.

Let the company manufactures sweaters of type A = x, type B = y, daily.
\(\therefore\)  LPP is maximize. P = 200x + 20y s.t.
360x + 120y \(\le \) 72000 
\(\Rightarrow\) 3x + y \(\le \) 300
x + y \(\le \) 300
x - y  \(\le \) 100
\(3x+y=600,\begin{cases} x=0,y=600 \\ y=0,x=200 \end{cases}\)
\(x+y=300,\begin{cases} x=0,y=300 \\ y=0,x=300 \end{cases}\)
\(x-y=100,\begin{cases} x=100,\quad y=0 \\ y=100,\quad x=200 \end{cases}\)
\(\\ x\ge 0\)
\(y\ge 0\)

Getting vertices of feasible region as, O(0, 0), A(100, 0), B(175, 75), C(150, 150) and D(0, 300)
Maximum profit is P = 200(175) + 20(75)
= 35000 + 1500 = Rs. 36500

Let x kg of fertilizer A and y kg of fertilizer B be used. As per the question:

Fertilizer	Nitrogen	Phosphoric acid	Cost
A	12%	5%	Rs 10/kg
B	4%	5%	Rs 8/kg
Minimum requirement	12 kg	12 kg

\(\therefore\) The LPP is given as
Min Z = 10x + 8y,
subject to the constants
\(\frac { 12x }{ 100 } +\frac { 4y }{ 100 } \ge 12\)
\(\Rightarrow 12x+4y\ge 1200\)
\(\Rightarrow 3x+y\ge 300\)
\(\frac { 5x }{ 100 } +\frac { 5y }{ 100 } \ge 12\)
\(\Rightarrow 5x+5y\ge 1200\)
\(\Rightarrow x+y\ge 240\)
\(x\ge 0,y\ge 0\)
L₁: 3x + y = 300

x	0	100	50
y	300	0	150

L₂: x + y = 240

x	0	2400	120
y	240	0	120

By Corner- point Method :

Corner Points	Z = 10x+8y
(0, 300)	2,400
(240, 0)	2,400
(30, 210)	1,980(Min)

Here cost is minimum at (30, 210) and is Rs. 1,980/. Since the region is unbounded, we have to draw
10x + 8y < 2340
L : 10x + 8y = 2340
Clearly open half plane has no common point with the feasible region is minimum value of Rs. 1,980. 
Farmer has to use 30 kg of fertilizer A and 210 kg of fertilizer B so that nutrient requirements are met at a minimum cost.

Total number of oranges = 25
number of good oranges = 20
number of bad oranges = 5
Probability of getting a bad orange,
\(P(B)=\frac { 5 }{ 25 } =\frac { 1 }{ 5 } \)
Now  \(p=\frac { 1 }{ 5 } \)
\(\\ q=1-p=\frac { 4 }{ 5 } \)
Let X be the random variable of "Number of bad oranges".
\(\Rightarrow X=0,1,2,3,4\)
\(P(X=0)={ n }_{ { C }_{ r } }{ \left( p \right) }^{ n-r }{ \left( q \right) }^{ r }\)
\(P(X=0)={ 4 }_{ { C }_{ 0 } }{ \left( \frac { 4 }{ 5 } \right) }^{ 4 }=\frac { 256 }{ 625 } \)
\(P(X=1)=4\times \frac { 64 }{ 125 } \times \frac { 1 }{ 5 } =\frac { 256 }{ 625 } \)
\(P(X=2)={ 4 }_{ { C }_{ 2 } }{ \left( \frac { 4 }{ 5 } \right) }^{ 2 }{ \left( \frac { 1 }{ 5 } \right) }^{ 2 }\)
\(=6\times \frac { 16 }{ 625 } =\frac { 96 }{ 625 } \)
\(P(X=3)={ 4 }_{ { C }_{ 3 } }\left( \frac { 4 }{ 5 } \right) { \left( \frac { 1 }{ 5 } \right) }^{ 3 }\)
\(=4\times \frac { 4 }{ 625 } =\frac { 16 }{ 625 } \)
\(P(X=4)={ 4 }_{ { C }_{ 4 } }{ \left( \frac { 4 }{ 5 } \right) }^{ 0 }{ \left( \frac { 1 }{ 5 } \right) }^{ 4 }\)
\(=\frac { 1 }{ 625 } \)
\(\therefore\)   Probability distribution is

X	P(X)	XP(X)	X2P(X)
0	256/625	0	0
1	256/625	256/625	256/625
2	96/625	192/625	384/625
3	16/625	48/625	144/625
4	1/625	4/625	16/625

\(\therefore Mean=\sum { XP(X) } \)
\(=\frac { 500 }{ 625 } =\frac { 20 }{ 25 } =\frac { 4 }{ 5 } \)
\(Var.\quad (X)=E({ X }^{ 2 })-[E(X){ ] }^{ 2 }\)
\(=\frac { 800 }{ 625 } -\frac { 16 }{ 25 } \)
\(Var.(X)=\frac { 32 }{ 25 } -\frac { 16 }{ 25 } =\frac { 16 }{ 25 } \)

Let  \((t^{ 2 },t)\) be any point on the curve  y² = x. Its distance (S) from the
line x - y + 1 = 0 is given by
\(S=\left| \frac { t-t^{ 2 }-1 }{ \sqrt { 1+1 } } \right| \)
\(=\frac { t^{ 2 }-t+1 }{ \sqrt { 2 } } \)
\(\left\{ \because t^{ 2 }-t+1=\left( t-\frac { 1 }{ 2 } \right) ^{ 2 }+\frac { 3 }{ 4 } >0 \right\} 1\)
\(\Rightarrow \frac { ds }{ dt } =\frac { 1 }{ \sqrt { 2 } } (2t-1)\)
\(\frac { d^{ 2 }S }{ dt^{ 2 } } =\sqrt { 2 } >0\)
Now , \(\frac { ds }{ dt } =0\)
\(\Rightarrow \frac { 1 }{ \sqrt { 2 } } (2t-1)=0\)
\(t=\frac { 1 }{ 2 } \)
This S is minimum at \(t=\frac { 1 }{ 2 } \)
So the required shortest distance is 
\(\frac { \left( \frac { 1 }{ 2 } \right) ^{ 2 }-\left( \frac { 1 }{ 2 } \right) +1 }{ \sqrt { 2 } } =\frac { 3 }{ 4\sqrt { 2 } } ,\frac { 3\sqrt { 2 } }{ 8 } \)

Let ABCD be a rectangle inscribed in a given circle with centre at 0 and radius a.
Let AB = 2x and BC = 2y

Then, OA² = OM² + AM² 
\(\Rightarrow a^2=y^2+x^2\)
\(\Rightarrow y=\sqrt{a^2-x^2}\)
Let A be the area of the rectangle.
\(\therefore A=4xy=4x\sqrt{x^2-x^2}\)
\(\Rightarrow \ \ \frac{dA}{dx}=4\{\frac{a^2-2x^2}{\sqrt{a^2-x^2}}\}\)
For maximum or minimum value of A,
\(\frac{dA}{dx}=0\)
\(=4\{\frac{a^2-2x^2}{\sqrt{a^2-x^2}}\}=0\Rightarrow\ \ x=\frac{a}{\sqrt{2}}\)
Now, \(\frac{d^2A}{dx^2}=4\frac{d}{dx}\{(a^2-2x^2)(a^2-x^2)^{-1/2}\}\)
\(\Rightarrow \frac{d^2A}{dx^2}=4[-4x(a^2-x^2)^{-1/2}+(a^2-2x^2)\times(-1/2)(a^2-x^2)^(-3/2)(-2x)]\)
\(=[\frac{-4x}{\sqrt{a^2-x^2}}+\frac{x(a^2-2x^2)}{(a^2-x^2)^{3/2}}]\)
\(\therefore\ (\frac{d^2A}{dx^2})_{x=\frac{a}{\sqrt2}}=-16<0\)
Thus A is maximum when \(x=\frac{a}{\sqrt2}\)
putting \(x=\frac{a}{\sqrt2}\) in (i) \(y=\frac{a}{\sqrt2}\)
Therefore \(x=y=\frac{a}{\sqrt2}\)
Hence area is maximum when x = y ⇒ 2x = 2y
i.e., the rectangle is a square.

Let equation of plane through (1, 2, -4) be a(x-1) + b(y-2) + c(z+4) = 0.
The plane is parallel to the given lines
\(\therefore\)  2a + 3b + 6c = 0; a + b - c = 0
Solving: \(\frac { a }{ -9 } =\frac { b }{ 8 } =\frac { c }{ -1 } =k\left( say \right) \)
\(\therefore\) a = -9k, b = 8k, c = -k
From (i), -9k(x-1) + 8k(y-2) - k(z+4) = 0
\(\therefore\) Equation of plane in cartesian form is 9x - 8y + z+11 = 0
Vector form of plane is: \(\Rightarrow \vec { r } .\left( 9\hat { i } -8\hat { j } +\hat { k } \right) =-11\)
Distance of (9, -8, -10) from the plane = \(\left| \frac { 9.9-8\left( -8 \right) +1\left( -10 \right) +11 }{ \sqrt { 81+64+1 } } \right| =\sqrt { 146 } \)

y = |x+1|

y	x
1	0
0	-1
2	+1
3	+2
1	-2
2	-3

Required area is given by integral of curve:
\(A\left( x \right) =\int _{ -4 }^{ 2 }{ \left| x+1 \right| } dx\)
 |x+1| = 0 at x = -1
At  x > -1, |x+1| = x + 1
At  x < -1, |x+1| = -(x+1)
So, \(A\left( x \right) =\int _{ -4 }^{ -1 }{ -\left( x+1 \right) dx+\int _{ -1 }^{ 2 }{ \left( x+1 \right) } dx } \)
\(=\left[ -\left( \frac { { x }^{ 2 } }{ 2 } +x \right) \right] _{ -4 }^{ -1 }+\left[ \left( \frac { { x }^{ 2 } }{ 2 } +x \right) \right] _{ -1 }^{ 2 }\)
\(=-\left[ \left( \frac { 1 }{ 2 } -1 \right) -\left( \frac { 16 }{ 2 } -4 \right) \right] +\left[ \left( \frac { 4 }{ 2 } +2 \right) -\left( \frac { 1 }{ 2 } -1 \right) \right] \)
\(=4\frac { 1 }{ 2 } +4\frac { 1 }{ 2 } =\ 9\ sq.units.\)

Let a, b \(\in A\), a " b = a + b + ab
Commutatively : for all a, b \(\in A\)
.a * b = a + b + ab
= b + a + ba
= b * a
Associatively: Let a, b, c \(\in A\)
(a * b) " c = (a + b + ab) " c
= (a + b + ab) + c + (a + b + ab)c
(a * b) " c = a + b + c + ab + bc + ac + abc
a * (b * c) = a * (b + c + bc)
= a + b + c + ab + bc + ac + abc
Clearly
(a * b) * c = a * (b "c)  \(\forall \) a, b, c  \(\in A\)}
 * is associative.
Identity: Let e \(\in A\)}such that
a * c = a
c + a + ea = a
c = 0
Identity element of A is e = O.
Inverse: Let b \(\in A\)such that
a*b = b*a = e
\(\Rightarrow \) a + b + ab = 0 and b + a + ba = 0
\(\Rightarrow \)  a = -b-ab
\(\Rightarrow \)  a = -b(l + a)
\(\Rightarrow \) b = \(\frac { -a }{ 1+a } \) \([\because \quad a\in A\therefore a\neq -1]\)
Invertible element of A is \(\frac { -a }{ 1+a } \) for all \(a\in A\)

Point of intersection of Circle and line is (3, 3).
\(\therefore \ Area=\int _{ 0 }^{ 3 }{ x\quad dx } +\int _{ 3 }^{ 3\sqrt { 2 } }{ \sqrt { 18-{ x }^{ 2 } } } dx\)
\(=\left[ \frac { { x }^{ 2 } }{ 2 } \right] _{ 0 }^{ 3 }+\left[ \frac { x }{ 2 } \sqrt { 18-{ x }^{ 2 } } +9{ sin }^{ -1 }\frac { x }{ 3\sqrt { 2 } } \right] _{ 3 }^{ 3\sqrt { 2 } }\)
\(=\frac { 9 }{ 2 } +\frac { 9\pi }{ 2 } -\frac { 9 }{ 2 } -\frac { 9\pi }{ 4 } \)
\(=\frac { 9\pi }{ 4 } sq.units.\)

\(\Delta=\left| \begin{matrix} { a }^{ 2 } &-{ (b-c) }^{ 2 } & bc \\ { b }^{ 2 } &-{ (c-a) }^{ 2 } & ca \\ { c }^{ 2 } &-{ (a-b) }^{ 2 } & ab \end{matrix} \right| \)
= \(-\left| \begin{matrix} { a }^{ 2 } &{ (b-c) }^{ 2 } & bc \\ { b }^{ 2 } &{ (c-a) }^{ 2 } & ca \\ { c }^{ 2 } &{ (a-b) }^{ 2 } & ab \end{matrix} \right| \)
= \(\begin{vmatrix} a^2&a^2+b^2+c^2&bc\\b^2&a^2+b^2+c^2&ca\\c^2&a^2+b^2+c^2&ab\end{vmatrix}\)
= \(-(a^2+b^2+c^2)\begin{vmatrix} a^2&1&bc\\b^2&1&ca\\c^2&1&ab\end{vmatrix}\)
= \(-(a^2+b^2+c^2)\begin{vmatrix}a^2-b^2&0&c(b-a)\\b^2-c^2&0&a(c-b)\\c^2&1&ab \end{vmatrix}\)
= \((a^2+b^2+c^2(a-b)(b-c)\begin{vmatrix}a+b&0&c\\b+c&0& -a\\c^2&1&ab \end{vmatrix}\)
\(=-(a^2+b^2+c^2)(a-b)(b-c)(-1)(a^2-ab+bc+c^2)\)
= \((a-b)(b-c)(a^2+b^2+c^2)(c-a)(c+a+b)\)
= \((b-c)(c-a)(a+b+c)({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 })\)

For \(a\in Z,a-a=0,\) which is divisible by 5.
\(\therefore \) \((a,a)\in R\forall a\in Z\).
Thus R is reflexive.
Now let \((a,b)\in R\)  \(\Rightarrow \) a-b is divisible by 5
\(\Rightarrow \) b-a is divisible by 5 \(\Rightarrow \) \((b,a)\in R\).
Thus R is symmetric.
Again let \((a,b)\in R,\ (b,c)\in R\)
\(\Rightarrow \) a-b and b-c are divisible by 5
\(\Rightarrow \) (a-b) + (b-c) = a-c is divisible by 5
\(\Rightarrow \) \((a,c)\in R\).
Thus R is transitive.
Hence, R is an equivalence relation.

\(y={ tan }^{ -1 }\frac { 3x+2x }{ 1-3x2x } \)
= \({ tan }^{ -1 }3x+{ tan }^{ -1 }2x\)
\(\Rightarrow \frac { dy }{ dx } =\frac { 3 }{ 1+9{ x }^{ 3 } } +\frac { 2 }{ 1+4{ x }^{ 2 } } \)

\(\frac { dx }{ d\theta } =\theta cos\theta +sin\theta \)
\(\frac { dy }{ d\theta } =-\theta sin\theta +cos\theta \)
\(\frac { dy }{ dx } =\frac { cos\theta -\theta sin\theta }{ \theta cos\theta +sin\theta } \)
dy/dx at \(\theta\)= \(\pi/4\)
\(=\frac { cos\frac { \pi }{ 4 } -\frac { \pi }{ 4 } sin\frac { \pi }{ 4 } }{ \frac { \pi }{ 4 } cos\frac { \pi }{ 4 } +sin\frac { \pi }{ 4 } } \)
\(=\frac { \frac { 1 }{ \sqrt { 2 } } -\frac { \pi }{ 4 } \times \frac { 1 }{ \sqrt { 2 } } }{ \frac { \pi }{ 4 } \times \frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } } } \)
=\(\frac { 1-\frac { \pi }{ 4 } }{ \frac { \pi }{ 4 } +1 } \)
\(\Rightarrow\)\(\frac { dy }{ dx } =\frac { 4-\pi }{ 4+\pi } \)

\(P(\bar { A } /\bar { B } )=\frac { P(\bar { A } \cap \bar { B } ) }{ P(\bar { B } ) } \)
\(=\frac { 1-P(A\cup B) }{ 1-P(B) } \)
\(=\frac { 1-[P(A)+P(B)-P(A\cap B)] }{ 1-P(B) } \)
\(=\frac { 7 }{ 10 } \)

\(\left|\overset\rightarrow a+\overset\rightarrow b \right|^{ 2 } +\left|\overset\rightarrow a-\overset\rightarrow b \right|^{ 2 }=2{(\left| \overset\rightarrow a \right| ^{2 }+\left|\overset\rightarrow b \right| ^{ 2 })}\)
\(\Rightarrow\left| \overset\rightarrow b \right| ^{ 2 }=2,116\)
\(\Rightarrow\left| \overset\rightarrow b \right| =46\)

Given, differential equation is 
\(\frac{d^2 y}{d x^2}+\sqrt[3]{\frac{d y}{d x}}+(1+x)=0 \Rightarrow \frac{d^2 y}{d x^2}+(1+x)=-\sqrt[3]{\frac{d y}{d x}}\)
On cubing both sides, we get
\(\left\{\frac{d^2 y}{d x^2}+(1+x)\right\}^3=-\frac{d y}{d x}\)
Hence, the order is 2 and degree is 3. So, the sum is 5.

Given integral \(=\int { \left( \frac { 1 }{ 1+{ x }^{ 2 } } -\frac { 2x }{ (1+{ x }^{ 2 })^{ 2 } } \right) } { e }^{ x }dx\)
\(=\frac { 1 }{ 1+{ x }^{ 2 } } { e }^{ x }+C \left[ as\frac { d }{ dx } \left( \frac { 1 }{ 1+{ x }^{ 2 } } \right) =\frac { -2x }{ (1+{ x }^{ 2 })^{ 2 } } \right] \)

We have, \(A=\left[ \begin{matrix} cos\theta & -sin\theta \\ sin\theta & cos\theta \end{matrix} \right] \)
\({ A }^{ T }=\left[ \begin{matrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{matrix} \right] \)
\(\Rightarrow A+{ A }^{ T }=\left[ \begin{matrix} 2cos\theta & 0 \\ 0 & cos\theta \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
\(\Rightarrow 2cos\theta =1\)
\(\Rightarrow cos\theta =\frac { 1 }{ 2 } \Rightarrow \theta =\cos ^{ -1 }{ \left( \frac { 1 }{ 2 } \right) } \)
\(\because \quad \theta =\frac { \pi }{ 3 } \)

The given function is f: X → Y and relation on X is R={(a, b): f(a) = f (b)}
Reflexive Since, for every x ∈ X, we have
f'(x) = f(x)
⇒ (xx) ∈ R, ∀ ∈ X Therefore, R is reflexive.
Symmetric Let (x, y) ∈ R
Then, f(x)=f(y)
⇒ f(y)=f(x)
⇒ (y, x) ∈ R
Thus, (x, y) ∈ R ⇒  (y, x)∈ R, ∀x, y∈ X
Therefore, R is symmetric.
Transitive Let x, y, z∈ X such that
(x, y) ∈ R and (y, z) ∈ R
Given a relation 5 in \(N \times N\), defined as
(a, b) S(c, d), if a+d=b+c.
Reflexive Let (a, b) be any arbitrary element of \(N \times N\)
i.e. \((a, b) \in N \times N\), where \(a, b \in N\)
Now, as a+b=b+a
[∴ addition is commutative ]
Therefore \quad(a, b) S(a, b)
So, S is reflexive.
Symmetric \(\operatorname{Let}(a, b),(c, d) \in N \times N\), such that (a, b)
S(c, d). Then, a+d=b+c
\( \Rightarrow  b+c=a+d \Rightarrow c+b=d+a \)
\(\Rightarrow (c, d) S(a, b)\)
So, S is symmetric.
Transitive Let \((a, b),(c, d),(e, f) \in N \times N\) such that (a, b) S(c, d) and (c, d) S(e, f).
Then, a+d=b+c and c+f=d+e
On adding the above equations, we get
a+d+c+f=b+c+d+e 
\( \Rightarrow  a+f=b+e \Rightarrow(a, b) S(e, f)\)
So, S is transitive.
Thus, S is reflexive, symmetric and transitive. Hence, S is an equivalence

= x-2 tan^-1x + c

Given \({ X }_{ m\times 3 }{ Y }_{ p\times 4 }={ Z }_{ 2\times b }\)
If XY is defined,
then \(3=p\Rightarrow p=3\)and 
\({ Z }_{ m\times 4 }={ Z }_{ 2\times b }\Rightarrow \) m = 2 and b = 4.
\(\therefore \ p=3,m=2\quad and\quad b=4.\)

2Let D₁ denote the event that the patient has disease d₁. The events D₂ and D₃ are defined similarly.
Then, \(P({ D }_{ 1 })=\frac { 3,200 }{ 10,000 } =0.32\)
\(P({ D }_{ 2 })=\frac { 3,500 }{ 10,000 } =0.35\)
and \(P({ D }_{ 3 })=\frac { 3,300 }{ 10,000 } =0.33\)
Let S be the event that the patient shows the symptoms S,
Then, \(P(S/{ D }_{ 1 })=\frac { P(S\cap D)) }{ P({ D }_{ 1 }) } =\frac { 3,100 }{ 3,200 } \)
= 0.97 (approx.)
\(P(S/{ D }_{ 3 })=\frac { 3,000 }{ 3,500 } =0.94\quad (approx.)\quad \)
\(P(S/{ D }_{ 3 })=\frac { 3,000 }{ 3,300 } =0.91\quad (approx.)\)
Using Bayes' theorem, we get
P(D₁/S) = The probability that the patient has disease d₁ knowing that he/she has symtoms S₁, S₂ ,....,S₆
P(D₁/S)
\(=\frac { P({ D }_{ 1 })|P(S/{ D }_{ 1 }) }{ P({ D }_{ 1 })P(S/{ D }_{ 1 })+P({ D }_{ 2 })P(S/{ D }_{ 2 })+P({ D }_{ 3 })P(S/{ D }_{ 3 }) } \)
\(=\frac { 0.32\times 0.97 }{ 0.32\times 0.97+0.35\times 0.94+0.33\times 0.91 } \)
\(=\frac { 0.3104 }{ 0.3104+0.329+0.3003 } \)
\(=\frac { 0.3104 }{ 0.9397 } =0.33\quad approx\)
Similarly,
\(P({ D }_{ 2 }/S)=\frac { 0.329 }{ 0.9397 } =0.35\quad approx\)
and \(P({ D }_{ 3 }/S)=\frac { 0.3003 }{ 0.9397 } =0.32\quad approx\)
Thus, knowing that the patient has symtoms S₁,S₂, ..., S₃, the probability that he has disease d₁ is 0.33, the probability that he has disease d₂ is 0.35, the probability that he has disease d₃ is 0·32. Therefore, the doctor should conclude that the patient is most likely to have disease d₂. 

\(\int { \frac { \sin ^{ 6 }{ x } +\cos ^{ 6 }{ x } }{ \sin ^{ 2 }{ x } \cos ^{ 2 }{ x } } } dx.\)
\(=\int { \frac { { (\sin ^{ 2 }{ x } +\cos ^{ 2 }{ x } ) }^{ 3 }-3\sin ^{ 2 }{ x } \cos ^{ 2 }{ x } (\sin ^{ 2 }{ x+ } \cos ^{ 2 }{ x } ) }{ \sin ^{ 2 }{ x } \cos ^{ 2 }{ x } } } \)
\([\because { a }^{ 3 }+{ b }^{ 3 }={ (a+b) }^{ 3 }-3ab(a+b)]\)
\(=\int { \frac { { (1) }^{ 3 }-3\sin ^{ 2 }{ x } \cos ^{ 2 }{ x } (1) }{ \sin ^{ 2 }{ x } \cos ^{ 2 }{ x } } } dx\)
\(=\int { \frac { 1-3\sin ^{ 2 }{ x } \cos ^{ 2 }{ x } }{ \sin ^{ 2 }{ x } \cos ^{ 2 }{ x } } } dx\)
\(=\int { \left( \frac { 1 }{ \sin ^{ 2 }{ x } \cos ^{ 2 }{ x } } -3 \right) } dx\)
\(=\int { \left( \frac { \sin ^{ 2 }{ x+ } \cos ^{ 2 }{ x } }{ \sin ^{ 2 }{ x } \cos ^{ 2 }{ x } } -3 \right) } dx\)
\(=\int { \left( \frac { 1 }{ \cos ^{ 2 }{ x } } +\frac { 1 }{ \sin ^{ 2 }{ x } } -3 \right) } dx\)
\(=\int { \sec ^{ 2 }{ x } } dx+\int { co\sec ^{ 2 }{ x } } dx-3\int { 1dx } \)
\(=\tan { x } -\cot { x } -3x+C.\)

Let 'x' and 'y' be the number of pieces of type A and type B respectively.
Then the problem is:
Maximize Z - 80x + 120y subject to:
\(9x+12y\le 180,\quad x+3y\le 30,\quad x\ge 0,\ge 0.\)

For solution set, we draw the lines:
x = 0, y = 0, 9x + 12y = 180 and x + 3y = 30.
The feasible region OAED is bounded whose vertices are:
O(0, 0), A(20, 0), D(0, 10) and E(12, 6)
[Solving 9x +12y = 180 and x + 3y = 30; x = 12, y = 6]
Applying Corner Point Method, we have:

Corner Point	Z = 80x + 120y
O: (0, 0)	0
A : (20, 0)	1600
E : (12, 6)	1680 (Maximum)
D : (0, 10)	1200

Hence, the maximum profit is Rs. 1680 when 12 pieces of Type A and 6 pieces of Type B are manufactured per week.

We have \(f(x)=2x^{ 3 }-9x^{ 2 }+12x+15\)
\(f'(x)=6x^{ 2 }-18x+12\)
\(=6(x^{ 2 }-3x+12)\)
\(=6(x-1)(x-2)\)
(I)For f(x) to be strictly increasing function of x,
\(f'(x)>0\)
i.e., (x-1)(x-2)>0
i.e., \(x\in (-\infty ,1)\cup (2,\infty )\)
Hence the function is strictly increasing in \((-\infty ,1)\cup (2,\infty )\)
(II)For f(x) to be strictly decreasing function of x,
\(f'(x)<0\)
i.e., 6(x-1)(x-2)<0
Hence the function is strictly decreasing in the interval (1,2)

The given equation is  \({ e }^{ x }\sqrt { 1-{ y }^{ 2 } } dx+\frac { y }{ x } =0,\)
\(\Rightarrow \) \(x{ e }^{ x }\quad dx+\frac { y }{ \sqrt { 1-{ y }^{ 2 } } } dy=0\)
|Variables Separable 
Integrating, \(\int { x\quad { e }^{ x }\quad dx+ } \int { \frac { y }{ \sqrt { 1-{ y }^{ 2 } } } dy= } c\)
\(\Rightarrow \) \(x{ e }^{ x }-\int { \left( 1 \right) } { e }^{ x }\quad dx-\frac { 1 }{ 2 } \int { { \left( 1-{ y }^{ 2 } \right) }^{ -\frac { 1 }{ 2 } } } \left( -2\quad y \right) dy=c\)
\(\Rightarrow \) \(x{ e }^{ x }- { e }^{ x }-\frac { 1 }{ 2 } \frac { { \left( 1-{ y }^{ 2 } \right) }^{ -\frac { 1 }{ 2 } } }{ \frac { 1 }{ 2 } } =c\)
\(\Rightarrow \) \(\left( x-1 \right) { e }^{ x }-\sqrt { 1-{ y }^{ 2 } } =c\) ...(1)
When \(x=0,y=1\)
\(\therefore \)    \(\left( -1 \right) \left( 1 \right) -0=c\Rightarrow c=-1\)
Putting in (1), \(\left( x-1 \right) { e }^{ x }-\sqrt { 1-{ y }^{ 2 } } =-1\)
\(\Rightarrow \)  \(\sqrt { 1-{ y }^{ 2 } } =\left( x-1 \right) { e }^{ x }+1,\)
which is the required solution

\({ \left( I+A \right) }^{ 2 }=(I+A)(I+A)\)
\(=II+IA+AI+AA\)
\(=I+A+A+{ A }^{ 2 }\)
\( =I+2A+A \quad \left[ \because \quad { A }^{ 2 }=A \right] \)
\(=I+3A ....(1)\)
\(\therefore \ { \left( I+A \right) }^{ 3 }={ (I+A) }^{ 2 }(I+A)\)
\(=(I+3A)(I+A)\ \left[ Using\quad (1) \right] \)
\( =II+IA+3AI+3AA\)
\(=I+A+3A+3{ A }^{ 2 }\)
\(=I+A+3A+3A\ \left[ \because \quad { A }^{ 2 }=A \right] \)
\(=I+7A...(2)\)
\(Hence,\ 7A-{ \left( I+A \right) }^{ 3 }=7A-(I=7A) = -I\) Using (2)

In g : \(2\rightarrow 5\) and in f, \(5\rightarrow 2\); etc.
fog = {(2, 2), (3, 3)}.

Since a + 0 = 0 + a = a and \(a\times 1=1\times a=a,\ \therefore \ 0\ and\ 1\) are the identity elements for operations '+' and '\(\times \)' respectively.
Further, there is no \(e\in R\) such that \(a-e=e-a=a\) and \(a\div e=e\div a=a\forall a\in R\).
Hence, '-' and '\(\div \)' do not have identity elements.

(i)\(\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}\)
\(\Rightarrow 2 \times 1-5 \times 4=2 x \times x-6 \times 4 \)
\(\Rightarrow 2-20=2 x^{2}-24 \)
\(\Rightarrow 2 x^{2}=6 \)
\(\Rightarrow x^{2}=3 \)
\(\Rightarrow x=\pm \sqrt{3} \)
(ii)We have:
\(\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}=\begin{vmatrix} x & 3 \\2x & 5 \end{vmatrix}\)
\(\Rightarrow 2 \times 5-3 \times 4=x \times 5-3 \times 2 x \)
\(\Rightarrow 10-12=5 x-6 x \)
\(\Rightarrow-2=-x \)
\(\Rightarrow x=2\)