\(\bar{x}=\frac{11+12+13+14+15}{5}=\frac{65}{5}=13\)

\(P=\frac{a}{i}=\frac{5000}{0.1}=50,000\)

Income \(=500 \times 100 \times \frac{15}{100}=7,500\)

R = px = 20x- 3x²
M.R = dR/dx = 20 - 6x

Since x₁ = 2.5, x₂ = 35 satisfies all the Constraints

f(5) = 5² - 4(5) = 5

\(\cos (\alpha+\beta)=\frac{12}{13} \Rightarrow \sin (\alpha+\beta)=\frac{5}{13} \)
\(\sin (\alpha-\beta)=3 / 5 \Rightarrow \cos (\alpha-\beta)=4 / 5 \)
\(\sin 2 \alpha=\sin ((\alpha+\beta)+(\alpha-\beta)) \)
\(= \sin (\alpha+\beta) \cos (\alpha-\beta) +\cos (\alpha+\beta) \sin (\alpha-\beta) \)
\(= \frac{5}{13} \cdot \frac{4}{5}+\frac{3}{5} \cdot \frac{12}{13}=\frac{56}{65}\)

\(\frac{1}{2}\left(2 \sin 15^{\circ} \cos 15^{\circ}\right) =\frac{1}{2} \sin 2\left(15^{\circ}\right)=\frac{1}{2} \sin 30^{\circ} =\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}\)

\(4 a=1 \Rightarrow a=\frac{1}{4}\)
Equation x = a
\(x=\frac{1}{4}\)

Let \(\frac { 1 }{ \left( { x }^{ 2 }-1 \right) } =\frac { A }{ x-1 } +\frac { B }{ x+1 } \)
Multiplying both sides by (x - 1)(x + 1), we get
1 = A(x +1) + ( Bx -1) ... (1)
Put x =1 in (1) we get, 1 = A(2)
 \(\therefore\) A = \(\frac{1}{2}\)
Put x = -1 in (1) we get, 1 = A(0) + B(-2)
\(\therefore\)  B = - \(\frac{1}{2}\)

(i) Variables:
Let x₁, x₂ represent the number of units of X and Y. 
(ii) Constraints:

	Labour	Capital
X	2	3
Y	3	1

 

∴ 2x₁ + 3x₂ ≤ 30 and 3x₁ + x₂ ≤ 17
(iii) Non-negative restrictions:
Since the number of units of X and Y cannot be negative,x₁, x₂ ≥ 0.
Hence, the mathematical formulation of the LPP is maximize \(Z=100{ x }_{ 1 }+120{ x }_{ 2 }\)
Subject to the constraints
\( { 2x }_{ 1 }+3{ x }_{ 2 }\le 30\)
\({ 3x }_{ 1 }+{ x }_{ 2 }\le 17\)
and x₁, x₂ ≥ 0.

a = 50, i = \(\frac{5}{100}\) = .05
A = \(\cfrac { a }{ i } =\cfrac { 50 }{ 0.05 } \) Rs.1000

We know that correlation coefficient
r= \(\frac { N\Sigma XY-(\Sigma X)(\Sigma Y) }{ \sqrt { N\Sigma X^{ 2 }-(\Sigma X)^{ 2 }\sqrt { N\Sigma { Y }^{ 2 }-(\Sigma Y)^{ 2 } } } } \)
= \(\frac { 9(597)-(45\times 108) }{ \sqrt { 9(285)-(45)^{ 2 }\times \sqrt { 9(1356)-(108^{ 2 }) } } } \)
r = +0.95

LHS=cos18^o-sin 18^o
=cos18^o-cos72^o[sin18^o=sin(90-72^o)=cos72^o]
\(=2\sin\left(\frac{18^o+72'}{2}\right).\sin\left(\frac{72^o-18'}{2}\right)\left[\because\cos C-\cos D=2\sin\left(\frac{C+D}{2}\right)\sin\left(\frac{D-C}{2}\right)\right]\)
\(=2\sin45^o\sin27^o=2\times\frac{1}{\sqrt2}\sin27^o=\sqrt{2}\sin27^o=RHS\)
Hence proved.

Let y = \(\left( \sqrt { x } +\frac { 1 }{ \sqrt { x } } \right) ^{ 2 }\)=\(x+\frac { 1 }{ x } +2\)
\(\frac{d y}{d x}=1-\frac{1}{x^2}\)

x = 3 cos\(\theta\) , y = 3 sin\(\theta\)
\(\cos \theta=\frac{x}{3}\ \sin \theta=\frac{y}{3}\)
We know that \(\cos ^2 \theta+\sin ^2 \theta=1\)
\(\frac {x^2}{9} + \frac {y^2}{9} = 1\)
\(\Rightarrow\) x² + y² = 9

f(x) = x + x²
f(-x) = (-x) + (-x)² 
f(-x) ≠ f(x) and f(-x) ≠ -f(x)
\(\therefore\) f is neither even nor odd function.

\({{x-2}\over{(x+2){(x-1)}^{2}}}={{A}\over{x+2}}+{{B}\over{x-1}}+{{C}\over{{(x-1)}^{2}}}\)
\(\frac{x-2}{(x+2)(x-1)^2}=\frac{A(x-1)^2+B(x+2)(x-1)+C(x+2)}{(x+2)(x-1)^2}\)
x - 2 = A(x - 1)²+ B(x + 2)(x - 1)+ C(x + 2)     ..(1)
x = -2 in (1) we get,
-2 - 2 =  A (-3)2 \(\Rightarrow\) - 4 = 9 A \(\Rightarrow\) A = \({{-4}\over{9}}\)
x = 1 in (1) we get,
1- 2 = C(1 + 2) \(\Rightarrow\) -1 = 3C \(\Rightarrow\) C = \({{-1}\over{3}}\)
Equate coefficient of x² on both sides of (1)
0 = A + B
\(B=-A=\frac{4}{9}\)  
\(\therefore\)   \({{x-2}\over{(x+1){(x-1)}^{2}}}-{{-{{4}\over{9}}}\over{x+2}}+{{{{4}\over{9}}}\over{x-1}}+{{-{{1}\over{3}}}\over{{(x-1)}^{2}}}+{{-4}\over{9(x+2)}}+{{4}\over{9(x-1)}}-{{1}\over{3{(x+1)}^{2}}}\)

We know that, domain set is R, range set is (0, \(\infty\)) and the curve passing through the point (0, 1)
Case (i) when a > 1
\(y=f(x)=a^x= \begin{cases}<1 & \text { if } x<0 \\ =1 & \text { if } x=0 \\ >1 & \text { if } x>0\end{cases}\) 
We noticed that as x increases, y is also increases and y > 0
\(\therefore \) The graph is as shown in the figure.

Case (ii) when  0 < a < 1
\(y=f(x)=a^x= \begin{cases}>1 & \text { if } x>0 \\ =1 & \text { if } x=0 \\ <1 & \text { if } x>0\end{cases}\) 
We noticed that as x increases, y decreases and y > 0
\(\therefore \) The graph is as shown in the figure

\(\cos 75^{\circ}+\cos 45^{\circ}=2 \cos \left(\frac{75^{\circ}+45^{\circ}}{2}\right) \cos \left(\frac{75^{\circ}-45^{\circ}}{2}\right)\)
\(=2 \cos \left(\frac{120^{\circ}}{2}\right) \cos \left(\frac{30^{\circ}}{2}\right)\)
\(=2 \cos 60^{\circ} \cos 15^{\circ}\)
\(=2 \times \frac{1}{2} \cos 15^{\circ}=\cos 15^{\circ}\)

A_ij = (–1)^i+jM_ij
A₁₁ = (–1)¹⁺¹M₁₁ = \(\left| \begin{matrix} 1 & 0 \\ 1 & 0 \end{matrix} \right| \) = 0
A₁₂ = (–1)¹⁺²M₁₂ = \(-\left| \begin{matrix} 0 & 0 \\ -4 & 0 \end{matrix} \right| \) = 0
A₁₃ = (–1)¹⁺³M₁₃ = \(\left| \begin{matrix} 0 & 1 \\ -4 & 1 \end{matrix} \right| \)= 0 - (-4) = 4
A₂₁ = (–1)²⁺¹M₂₁ = \(-\left| \begin{matrix} -2 & -3 \\ 1 & 0 \end{matrix} \right| \)= -(0 - (-3)) = -3
A₂₂ = (–1)²⁺²M₂₂ = \(\left| \begin{matrix} 1 & -3 \\ -4 & 0 \end{matrix} \right| \) = 0 - 12 = -12
A₂₃ = (–1)²⁺³M₂₃ = \(-\left| \begin{matrix} 1 & -2 \\ -4 & 1 \end{matrix} \right| \) = -(1-8) = 7
A₃₁ = (–1)³⁺¹M₃₁ = \(\left| \begin{matrix} -2 & -3 \\ 1 & 0 \end{matrix} \right| \) = 0 - (-3) = 3
A₃₂ = (–1)³⁺²M₃₂ = \(-\left| \begin{matrix} 1 & -3 \\ 0 & 0 \end{matrix} \right| \) = 0 - 0 = 0
A₃₃ = (–1)³⁺³M₃₃ = \(\left| \begin{matrix} 1 & -2 \\ 0 & 1 \end{matrix} \right| \) = 1 - 0 = 1
\(\left[ { A }_{ ij } \right] =\left[ \begin{matrix} 0 & 0 & 4 \\ -3 & -12 & 7 \\ 3 & 0 & 1 \end{matrix} \right] \)
Adj A = [A_ij]^T
\(=\left[ \begin{matrix} 0 & -3 & 3 \\ 0 & -12 & 0 \\ 4 & 7 & 1 \end{matrix} \right] \)

Minor of 1 = M₁₁ = 3
Minor of − 2 = M₁₂ = 4
Minor of 4 = M₂₁ = − 2
Minor of 3 = M₂₂ = 1
Cofactor of 1 = A₁₁ = (-1)¹⁺¹M₁₁ = 3
Cofactor of -2 = A₁₂ = (-1)¹⁺²M₁₂ = -4
Cofactor of 4 = A₂₁ = (-1)²⁺¹M₂₁ = 2
Cofactor of 3 = A₂₂ = (-1)²⁺²M₂₂ = 1

Since the decision variables are non-negative, the solution lies in the I quadrant of the plane.
Consider the equations 
\({ x }_{ 1 }-{ x }_{ 2 }=-1\)

\({ x }_{ 1 }\)	0	1
\({ x }_{ 2 }\)	1	2

\({ -x }_{ 1 }+{ x }_{ 2 }=0\)

\({ x }_{ 1 }\)	2	1
\({ x }_{ 2 }\)	2	1

 

The feasible region is not common. Thus, there is no maximum value of Z.

Given A = Rs.3783,i = 0.05,n = 3
A = \(\cfrac { a }{ i } \left[ \left( 1+i \right) ^{ n }-1 \right] \)
3783 = \(\cfrac { a }{ i } \) [(1.05)³-1]
3783 x 0.05 = a[1.1576-1]
a = \(\cfrac { 189.15 }{ 0.1576 } \) = 1200.19
\(\therefore\) a =Rs.1200
(1.05)³ = 3 log (1.05)
= 3(0.212)
= 0.636
Antilog of 0.636 is 1.1576

The regression co-efficient of X on Y is
b_xy= \(\frac { N\sum { XY-(\sum { X)(\sum { Y) } } } }{ N.\sum { { Y }^{ 2 }-(\sum { { Y) }^{ 2 } } } } =\frac { 5(300)-(20)(40) }{ 5(345)-({ 40) }^{ 2 } } \)=5.6
Also, \(\bar {X}\) =\(\frac { \sum { X } }{ N } =\frac { 20 }{ 5 } \)=4
and \(\bar {Y}\)=\(\frac { \sum { Y } }{ N } =\frac { 40 }{ 5 } \)=8
\(\therefore\)The regression equation of X on Y is
X- \(\bar {X}\)=b_xy(Y-\(\bar {Y}\) )
X-4=5.6(Y-8)
X-4=5.6Y-44.8
X=5.6Y-40.8
When Y=6, X=5.6(6)-40.8=33.6-40.8
\(\therefore\)X=-7.2

% Rise	x	log x
5	105	2.0212
8	108	2.0334
77	177	2.2480
		\(\sum logx\) = 6.3026

GM = Antilog\(\left( \frac { \sum { logx } }{ n } \right) =Anitlog\left( \frac { 6.3026 }{ 3 } \right) =Antilog(2.1009)\)
GM = 126.2
\(\therefore\) Average increase of the commodity from 2004 to 2007
= 126.1 - 100 = 26.1%

Weight (x)	f	log x	flog x
130	3	2.1139	6.3417
135	4	2.1303	8.5212
140	6	2.1461	12.8766
145	6	2.1614	12.9684
146	3	2.1644	6.4932
148	5	2.1703	10.8515
149	2	2.1732	4.3464
150	1	2.1761	2.1761
157	1	2.1959	2.1959
	N = 31		\(\sqrt{\sum flogx}\) = 66.771

GM = Antilog\(\left( \frac { \sum { f\ log\ x } }{ N } \right) \)
\(=Anitlog\left( \frac { 66.771 }{ 31 } \right) \)
= 142.5 Ibs

In both the series items are in small number. Therefore correlation coefficient can also be calculated without taking deviations from actual means or assumed mean.

X	Y	X2	Y2	XY
12	14	144	196	168
9	8	81	64	72
8	6	64	36	48
10	9	100	81	90
11	11	121	121	121
13	12	169	144	156
7	3	49	9	21
ΣX = 70	ΣY = 63	ΣX2 = 728	ΣY2 = 651	ΣXY = 676

r = \(\frac { N\Sigma XY-(\Sigma X)(\Sigma Y) }{ \sqrt { N\Sigma X^{ 2 }-(\Sigma X)^{ 2 }\times \sqrt { N\Sigma { Y }^{ 2 }-(\Sigma Y)^{ 2 } } } } \)
= \(\frac { 7(676)-(70)(63) }{ \sqrt { 7(728)-(70)^{ 2 }\times \sqrt { 7(651)-(63)^{ 2 } } } } \)
= \(\frac { 322 }{ 339.48 } \)
r = +0.95

Given that f(x) = 2x³ + 9x² + 12x + 1.
f'(x) = 6x² + 18x + 12
= 6(x² + 3x + 2)
= 6(x + 2)(x + 1)
f'(x) = 0 \(\Rightarrow\) 6 (x + 2)(x + 1) = 0
\(\Rightarrow\) x + 2 = 0 (or) x + 1 = 0.
x = –2 (or) x = –1
f(x) has stationary points at x = – 2 and x = – 1
Stationary values are obtained by putting x = – 2 and x = – 1
When x = – 2, f(–2) = 2(–8)  +9(4) + 12(–2) + 1 = –3
When x = – 1, f(–1) = 2(–1) + 9(1) + 12(–1) + 1 = –4
The stationary points are (–2, –3) and (–1,–4).

(i) Variables: Let x₁, x₂ and x₃ be the number of units of products P₁, P₂ and P₃ to be produced.
(ii) Objective function: Profit on x₁ units of the product P₁ = 20 x₁
Profit on x₂ units of the product P₂ = 25 x₂
Profit on x₃ units of the product P₃ = 15 x₃
Total profit = 20 x₁ + 25 x₂ + 15 x₃
Since the total profit is to be maximized, we have to maximize Z = 20 x₁ + 25 x₂ + 15 x₃
Constraints: 6x₁ + 3x₂ + 12x₃ ≤ 200
2x₁ + 5x₂ + 4x₃ ≤ 350
x₁ + 2x₂ + x₃ ≤ 100
Non-negative restrictions: Since the number of units of the products A, B and C cannot be negative, we have x₁, x₂, x₃ ≥ 0
Thus, we have the following linear programming model.
Maximize Z = 20 x₁ + 25 x₂ + 15 x₃
Subject to 6 x₁ + 3 x₂ + 12 x₃ ≤ 200
2x₁ + 5x₂ + 4x₃ ≤ 350
x₁ + 2x₂ + x₃ ≤ 100
x₁, x₂, x₃ ≥ 0

\(x={P\over p+5}\)
\({dx\over dp}={(p+5)-p\over (p+5)^2}\)
\(={5\over (p+5)^2}\)
Elasticity of supply: \(η_S={p\over x}.{dx\over dp}\)
\(=(p+5){5\over (p+5)^2} = {5\over (p+5)}\)
When p = 20, \(η_S={5\over 20+5}\) = 0.2
Interpretation:
i) If the price increases by 1% from p = Rs. 20, then the quantity of supply increases by 0.2% approximately.
ii) If the price decreases by 1% from p = Rs. 20, then the quantity of supply decreases by 0.2% approximately.

Given equation of pair of lines is
12x² + 7xy + ky² + 13x - y + 3 = 0
2h = 7   2g = 13   2f = -1
\(\Rightarrow a=12,\quad h=\frac { 7 }{ 2 } ,b=k,\quad g=\frac { 13 }{ 2 } ,f=\frac { -1 }{ 2 } ,c=3\)
The condition to represent pair of lines is abc + 2fgh - af² - bg² - ch² = 0
\((12)(k)(3)+2\left( \frac { -1 }{ 2 } \right) \left( \frac { 13 }{ 2 } \right) \left( \frac { 7 }{ 2 } \right) -12\left( \frac { 1 }{ 4 } \right) -k\left( \frac { 169 }{ 4 } \right) -3\left( \frac { 49 }{ 4 } \right) =0\)
\(\Rightarrow 36k-\frac { 91 }{ 4 } -3-\frac { 169k }{ 4 } -\frac { 147 }{ 4 } =0\)
\(\Rightarrow \frac { 144k-91-12-169k-147 }{ 4 } =0\)
\(\Rightarrow -25k-250=0\times 4=0\)
\(\Rightarrow -25k=250\)
\(\Rightarrow k=-10\)

\(\left(x-\frac{2}{x^2}\right)^{15}\)
\(t_{r+1}=n C_r x^{n-r} a^r\)
\(=15 C_r(x)^{15-r}\left(\frac{-2}{x^2}\right)^r\)
\(=(-1)^r 15 C_r 2^r x^{15-r}\left(\frac{1}{x^{2 r}}\right)\)
\(=(-1)^r 15 C_r 2^r x^{15-r-2 r}\)
\(=(-1)^r 15 C_r 2^r x^{15-3 r}\)
To find the term independent of x, equate the power of x to 0
15 - 3r = 0
\(3 r=15 \Rightarrow r=\frac{15}{3}=5\)
\(t_{r+1}=(-1)^5 15 C_5 2^5\)
\(=-32\left(15 C_5\right)\)

Let B = \(\begin{vmatrix} 1 &-3 &2 \\4 &-1&2\\3&5&2 \end{vmatrix}\)
Minor of 1 = M₁₁ = \(\begin{vmatrix} -1 & 2 \\ 5 & 2 \end{vmatrix}=-2-10=-12\)
Minor of -3 = M₁₂ = \(\begin{vmatrix}4 &2 \\ 3 & 2 \end{vmatrix}=8-6=2\)
Minor of 2 = M₁₃ = \(\begin{vmatrix} 4 & -1 \\ 3 & 5\end{vmatrix}=20+3=23\)
Minor of 4 = M₂₁ = \(\begin{vmatrix} -3 & 2 \\5 & 2 \end{vmatrix}=-6+10=-16\)
Minor of -1 = M₂₂ = \(\begin{vmatrix}1 & 2 \\ 3 & 2 \end{vmatrix}=2-6=-4\)
Minor of 2 = M₂₃ = \(\begin{vmatrix} 1& -3 \\3 &5 \end{vmatrix}=5+9=14\)
Minor of 3 = M₃₁ = \(\begin{vmatrix} -3 &2 \\ -1 & 2 \end{vmatrix}=-6+2=-4\)
Minor of 3 = M₃₂ = \(\left|\begin{array}{ll} 1 & 2 \\ 4 & 2 \end{array}\right|=2-8=-6\)
Minor of 2 = M₃₃ = \(\begin{vmatrix} 1 & -3 \\4 & -1 \end{vmatrix}=-1+12=11\)
Co-factor of 1 = A₁₁ = (-1)¹⁺¹ M₁₁ = -12
Co-factor of -3 = A₁₂ = (-1)¹⁺² M₁₂ = -2
Co-factor of 2 = A₁₃ = (-1)¹⁺³ M₁₃ = 23
Co-factor of 4 = A₂₁ = (-1)²⁺¹ M₂₁ = 16
Co-factor of -1 = A₂₂ = (-1)²⁺² M₂₂ = -4
Co-factor of 2 = A₂₃ = (-1)2+3 M₂₃ = -14
Co-factor of 3 = A₃₁ = (-1)³⁺¹ M₃₁ = -4
Co-factor of 5 = A₃₂ = (-1)³⁺² M₃₂ = 6
Co-factor of 2 = A₃₃ = (-1)³⁺³ M₃₃ = 11

Given y = - x³ + 3x² +  9x - 27    ...(1)
Differentiating w.r.t. 'x' we get,
\({dy\over dx}=-3x^2 + 6x + 9\)
∴ Slope of the tangent is - 3x² + 6x + 9
Let M = - 3x² + 6x + 9
\({dM\over dx}=-6x+6\)    ..(2)
Slope is maximum when \({dM\over dx}=0\) and \({d^2M\over dx^2}<0\)
\({dM\over dx}=0⇒-6x+6=0⇒ -6x =- 6⇒x=1\)
\({d^2M\over dx^2}=-6<0\)
∴ Slope is maximum when x = 1
∴ Maximum slope= -3(1)² + 6(1) + 9 = -3 + 15 = 12 [From (2)]
when x = 1, y = - (1)³ + 3(1)² + 9(1) - 27
-1+3+9-27=-16    [From (1)]
∴ Maximum slope is 12 and the required point is (1, -16)

E1 =0	L6 =22
E2 = 0+7=7	L5 =(22 - 3) = 19
E3 =(7 + 3) or (0 + 6)  Whichever is maximum = 0	L4 =22 - 6 = 16 .
E4 =(7 + 9) or (10 + 2)  Whichever is maximum=16	L3 =(22 - 9) or (16 - 8) Whichever is minimum=13
E5 =(0+ 11)or(10+4)  Whichever is maximum=14	L2 =16 - 9 = 7
E6 =(16 + 6) or (10 + 9) or (14 + 3)  Whichever is maximum = 22	L1 =7-7=0

Activity	Duration	EST	EFT=EST+tij	LST	LFT
1-2	7	0	7	7-7=0	7
1-3	6	0	6	13-6=7	13
1-5	11	0	11	19-11=8	19
2-3	3	7	10	13-3=10	13
2-4	9	7	16	16-9=7	16
3-4	2	13	15	19-4=15	16
3-5	4	13	17	19-4=15	19
3-6	9	10	19	22-9=13	22
4-6	6	16	22	22-6=16	22
5-6	3	14	17	22-3=19	22

EFT and LFT are same in the activity, 1 - 2, 2 - 4 and 4 - 6.
Hence the critical path is 1 - 2 - 4 - 6 and the project completion time is 22 Weeks.

Let the amount invested in each stock be Rs.x
For 12% Stock
Investment =Rs.x
Purchased Price = 89+1 = 90
Income=\(\frac {\text { Investment} }{ \text {Purchase Price } }\) x Dividend Rate
=\(\frac { x }{ 90 } \times 12=\frac { 2x }{ 15 } \)   ...(1)
For 8% Stock
Investment = Rs.x
market Price = 95+1= 96
Dividend Rate = 8%
Income=\(\cfrac { x }{ 96 } \times 8 =\cfrac { x }{ 12 } \)  ...(2)
Difference in income = Rs.120
From (1) and (2),  \(\cfrac { 2x }{ 15 } -\cfrac { x }{ 12 } \) = 120
\(x\left( \cfrac { 2 }{ 15 } -\cfrac { 1 }{ 12 } \right) \) = 120
\(x\left( \cfrac { 8-5 }{ 60 } \right) \) = 120
\(\cfrac { 3x }{ 60 } \) = 120
\(\Rightarrow\) x = \(\cfrac { 120\times 60 }{ 3 } \) = Rs.2400
Hence, investment in each stock is Rs.2400

Given C =\({1\over5}x^2-6x+100\)
Differentiating w.r.t. 'x' we get,
\({dC\over dx}={1\over5}(2x)-6\)
\({dC\over dx}=0\)
\(\Rightarrow {2x\over 5}-6=0\)
\(\Rightarrow {2x\over 5}=6\)
\(\Rightarrow x={6\times5\over2}=15\)
Now  \({d^2C\over dx^2}={2\over5}>0\)
\(\therefore\)Cost function is minimum when x = 15.

Let the events E₁, E₂, E₃ and A be defined as
E₁ - Cars in the service station A
E₂ - Cars in the service station B
E₃ - Cars in the service station C
A is Event of failing to clean the glass
\(P({ E }_{ 1 })=\frac { 50 }{ 100 } ,P({ E }_{ 2 })=\frac { 30 }{ 100 } ,P({ E }_{ 3 })=\frac { 20 }{ 100 } \)
\(P(A/E_{ 1 })=\frac { 5 }{ 100 } ,P(A/{ E }_{ 2 })=\frac { 7 }{ 100 } ,P(A/{ E }_{ 3 })=\frac { 3 }{ 100 } \)
Probability that the glass is not cleaned = P(A)
= P(E₁).P(A/E₁)+P(E₂).P(A/E₂)+P(E₃).P(A/E₃)
\(=\frac { 50 }{ 100 } \times \frac { 5 }{ 100 } +\frac { 30 }{ 100 } \times \frac { 7 }{ 100 } +\frac { 20 }{ 100 } \times \frac { 3 }{ 100 } \)
\(=\frac { 250+210+60 }{ 10,000 } \)
\(=\frac { 520 }{ 10,000 } \)
= 0.052
Probability that glass is cleaned
= P(A') = 1 - P(A)
= 1 - 0.052 = 0.948

Given \(sin\left( { sin }^{ -1 }\frac { 1 }{ 5 } +{ cos }^{ -1 }x \right) =1\)
⇒ \({ sin }^{ -1 }\left( \frac { 1 }{ 5 } \right) +{ cos }^{ -1 }(x)={ sin }^{ -1 }(1)=\frac { \pi }{ 2 } \)
⇒ \({ sin }^{ -1 }\left( \frac { 1 }{ 5 } \right) +{ cos }^{ -1 }(x)=\frac { \pi }{ 2 } \)
⇒ \({ sin }^{ -1 }\left( \frac { 1 }{ 5 } \right) =\frac { \pi }{ 2 } -{ cos }^{ -1 }(x)\)
⇒ \({ sin }^{ -1 }\left( \frac { 1 }{ 5 } \right) =sin^{ -1 }x\)  \(\left[ \because { sin }^{ -1 }(x)+{ cos }^{ -1 }(x)=\frac { \pi }{ 2 } \right] \)
⇒ \({ sin }^{ -1 }\left( \frac { 1 }{ 5 } \right) =sin^{ -1 }x\)
⇒ \(x=\frac { 1 }{ 5 } \)
Thus, \(x=\frac { 1 }{ 5 } \) is a root of the given equation.

\(\mathrm{LHS} =\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ} \)
\(=\cos 20^{\circ} \cos 40^{\circ}\left(\frac{1}{2}\right) \cos 80^{\circ}\)
\(=\frac{1}{2} \cos 20^{\circ}\left(\cos 40^{\circ} \cos 80^{\circ}\right)\)
\(=\frac{1}{2} \cos 20^{\circ}\left(\cos \left(60^{\circ}-20^{\circ}\right) \cos \left(60^{\circ}+20^{\circ}\right)\right)\)
\(=\frac{1}{2} \cos 20^{\circ}\left(\cos ^2 60^{\circ}-\sin ^2 20\right)\)
\(=\frac{1}{2} \cos 20^{\circ}\left(\frac{1}{4}-\left(1-\cos ^2 20^{\circ}\right)\right)\)
\(=\frac{1}{2} \cos 20^{\circ}\left(\frac{1-4+4 \cos ^2 20^{\circ}}{4}\right)\)
\(=\frac{1}{8}\left(4 \cos ^3 20^{\circ}-3 \cos 20^{\circ}\right)\)
\(=\frac{1}{8} \cos 3\left(20^{\circ}\right)=\frac{1}{8} \cos 60^{\circ}\)
\(=\frac{1}{8}\left(\frac{1}{2}\right)=\frac{1}{16}\)
Hence proved.

Let P (n) be the statement. 1.3+2.3²+3.3 + ... +n.3ⁿ = \({{(2n-1){3}^{n+1}+3}\over{4}} \) for all n \(\in\) N.
Step-1:
Put n = 1  \(\Rightarrow1.3(1){{(2-1){3}^{1+1}+3}\over{4}}={{3^2+3}\over{4}}={{12}\over{4}}\Rightarrow\ 3=3\)
\(\therefore\) P(1) is true.
Step-2:
Let us assume that P(k) is true
\(\therefore\)  1.3 + 2.3² + 3.3³ + ... + k.3k = \({{(2k-1){3}^{k+1}+3}\over{4}}\)  ....(1)
Step-3:
To prove that P (k + 1) is true i.e. to P.T. 1.3 + 2.3² + 3.3³ + ... + k.3^k + (k + 1)3^k+1
\(={{[2(k+1)-1]{3}^{k+2}+3}\over{4}}={{(2k+1){3}^{k+2}+3}\over{4}}\)
LHS = 1.3 + 2.3² + ... + k.3^k + (k+ 1)3^{k+ 1}
\(={{(2k-1){3}^{k+1}+3}\over{4}}+(k+1){3}^{k+1}={{(2k+1){3}^{k+1}+3+(4k+3){3}^{k+1}}\over{4}}\)
\(={{{3}^{k+1}(2k-1+4k+4)+3}\over{4}}={{{3}^{k+1}(6k+3)+3}\over{4}}={{{3}^{k+1}(2k+1)+3}\over{4}}={{{3}^{k+2}(2k+1)+3}\over{4}}\) = RHS
\(\therefore\)  P (k + 1) is true whenever P(k) is true.
\(\therefore\) By mathematical induction, P(n) is true for all values n.

Let x and y represent the number of units produced and the cost of production respectively. By the given data.
x₁ (500)  y₁ (6000)
x₂ (1000)  y₂ (9000)
Equation of straight line is
\(\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
\(\Rightarrow \frac { y-6000 }{ 9000-6000 } =\frac { x-500 }{ 1000-500 } \)
\( { y-6000 }{ } =\frac { 3000 }{ 500 } (x - 500)\)
\(\Rightarrow\) y - 6000 = 6(x - 500)
\(\Rightarrow\) y - 6000 = 6x - 3000
\(\Rightarrow\) y = 6x + 3000