\(P =\frac{\frac{a}{i}}{k} \)
\(=\frac{\frac{2000}{0.1}}{12}=2,40,000\)

\(x=\frac{12 \times 96}{10}=115.20\)

\(\sec A(-\cos A)=\frac{1}{\cos A}(-\cos A)=-1\)

Equation of x axis y = 0, equation of y axis x = 0
\(\therefore\) combined equation xy = 0

\(a+b=0 \Rightarrow k-2=0\)

\(\text {Since }|A|=4-4=0\)

\(A=\left(\begin{array}{cc} \frac{4}{5} & \frac{-5}{12} \\ \frac{-2}{5} & \frac{1}{2} \end{array}\right)\)
\(|A|=\frac{2}{5}-\frac{1}{6}=\frac{12-5}{30}=\frac{7}{30}\)
\(A^{-1}=\frac{1}{|A|} \text { adjA }=\frac{30}{7}\left(\begin{array}{ll} \frac{1}{2} & \frac{5}{12} \\ \frac{2}{5} & \frac{4}{5} \end{array}\right)\)

Differentiating \(y=\frac { 1-3x }{ 1+3x } \) with respect to x.
\(\cfrac { dy }{ dx } =\cfrac { \left( 1+3x \right) \frac { d }{ dx } \left( 1-3x \right) -\left( 1-3x \right) \frac { d }{ dx } \left( 1+3x \right) }{ \left( 1+{ 3x }^{ 2 } \right) } \) 
= \(\cfrac { \left( 1+3x \right) \left( -3 \right) -\left( 1-3x \right) (3) }{ \left( 1+3x \right) ^{ 2 } } \) 
= \(\cfrac { -6 }{ \left( 1+3x \right) ^{ 2 } } \)

Using the precedence relationship and following the rules of network construction, the required network is shown in the following figure.

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

We have \(\cos 2x=\frac{1-\tan^2x}{1+\tan^2x}\)
\(\therefore LHS=\cos2x+\sin^2\phi=\frac{1-\tan^2x}{1+\tan^2x}+\sin^2\phi\)
\(=\frac{1-\left(2\tan^\phi+1\right)}{1+(2\tan^2\phi)+1}+\sin^2\phi\)      \([\because tan^{ 2 }x=2tan^{ 2 }\phi +1]\)
\(=\frac{-2\tan^2\phi}{2(1+\tan^2\phi)}+\sin^2\phi=\frac{-\tan^2\phi}{\sec^2\phi}+\sin^2\phi\)
\(=\frac{-\sin^2\phi}{\cos^2\phi.\frac{1}{\cos^2\phi}}+\sin^2\phi=-\sin^2\phi+\sin^2\phi=0=RHS\)
Hence Proved.

Let A = \(\begin{vmatrix}5&20\\ 0&-1 \end{vmatrix}\)
Minor of 5 = M₁₁ = -1
Minor of 20 = M₁₂ = 0
Minor of 0 = M₂₁ = 20
Minor of -1 = M₂₂ = 5
Co-factor of 5 = A₁₁ = -1
Co-factor of 20 = A₁₂ = 0
Co-factor of 0 = A₂₁ = -20
Co-factor of -1 = A₂₂ = 5

  \(cos2A=\cfrac { 1-{ tan }^{ 2 }A }{ 1+{ tan }^{ 2 }A } =\cfrac { 1-\frac { 1 }{ 49 } }{ 1+\frac { 1 }{ 49 } } =\cfrac { 48 }{ 49 } \times \cfrac { 49 }{ 50 } \) = \(\cfrac { 24 }{ 25 } \)   ..(1)
Now, sin4B = 2sin2B cos2B
\(\\ \\ \\ =2\cfrac { 2tanB }{ 1+{ tan }^{ 2 }B } \times \cfrac { 1-{ tan }^{ 2 }B }{ 1+{ tan }^{ 2 }B } \)   
\( =\cfrac { 4\times \frac { 1 }{ 3 } }{ 1+\frac { 1 }{ 9 } } \times \cfrac { 1-\frac { 1 }{ 9 } }{ 1+\frac { 1 }{ 9 } } =\cfrac { 24 }{ 25 } \) ..(2)
From(1) and (2) we get, cos2A = sin4B.

Compare \(\left( \cfrac { x }{ 3 } +9y \right) ^{ 9 }\) with \(\left( x+a \right) ^{ n }\) 
Since n = 9, we have 10 terms(even)
\(\therefore\) There are two middle terms namely \(\cfrac { { t }_{ n+1 } }{ 2 } ,\cfrac { { t }_{ n+3 } }{ 2 } i.e.,\cfrac { { t }_{ 9+1 } }{ 2 } ,\cfrac { { t }_{ 9+3 } }{ 2 } \) 
 General term in the expansion of \(\left( x+a \right) ^{ n }\) is
\({ t }_{ r+1 }=n{ C }_{ r }{ x }^{ n-r }{ a }^{ r }\) ..(1)
Here t₅ and  t₆ are middle terms.
put r = 4 in (1)
\({ t }_{ 4+1 }={ t }_{ 5 }=9{ C }_{ 4 }\left( \cfrac { x }{ 3 } \right) ^{ 5 }.\left( 9y \right) ^{ 4 }\)
\(=9{ C }_{ 4 }\cfrac { { x }^{ 5 } }{ { 3 }^{ 5 } } .{ 9 }^{ 4 }{ y }^{ 4 }\)
\(=\cfrac { 9\times 8\times 7\times 6 }{ 4\times 3\times 2\times 1 } .\cfrac { { x }^{ 5 } }{ { 3 }^{ 5 } } .{ 9 }^{ 4 }{ y }^{ 4 }=3402{ x }^{ 5 }{ y }^{ 4 }\) 
​​​​​​put r = 5 in (1),
\(\\ { t }_{ 5+1 }={ t }_{ 6 }=9{ C }_{ 5 }\left( \cfrac { x }{ 3 } \right) ^{ 4 }.\left( 9y \right) ^{ 5 }\)
\(={ 9C }_{ 5 }\cfrac { { x }^{ 4 } }{ { 3 }^{ 4 } } .{ 9 }^{ 5 }{ y }^{ 5 }\)
\(=91854{ x }^{ 4 }{ y }^{ 5 }\)

\(A=\left[ \begin{matrix} 2 & 4 \\ -3 & 2 \end{matrix} \right] \)
\(|A|=\left[ \begin{matrix} 2 & 4 \\ -3 & 2 \end{matrix} \right] \)
=16 ≠ 0
Since A is a nonsingular matrix, A ^-1 exists
Now  adj \(A=\left[ \begin{matrix} 2 & -4 \\3 & 2 \end{matrix} \right] \)
\({ A }^{ -1 }=\frac { 1 }{ \left| A \right| } adjA\)
\(=\frac { 1 }{ 16 } \left[ \begin{matrix} 2 & -4 \\ 3 & 2 \end{matrix} \right] \)

Using the immediate precedence relationship and following the rules of network construction, the required network is shown in the diagram.

Given a = Rs.500,i = 5% =0.05,n =10
A=\(\cfrac { a }{ i } \left[ 1-\left( 1+i \right) ^{ -n } \right] \)
=\(\cfrac { 500 }{ 0.05 } \left( 1.05 \right) \left[ \left( 1.05 \right) ^{ 10 }-1 \right] \)
=10,500 [1.629-1]
=10,500(0.629)
= Rs.6604.50
\(\therefore\) At the end of 10 years. I will receive Rs.6604.5
(1.05)¹⁰ = 10 log(1.05)
= 10(0.0212)
= 0.2120
Antilog of 0.2120 is 1.629

N=7

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
70	63	728	651	676

Karl Pearson correlation co-efficient
r(x, y) =\(\frac { N\sum { XY-(\sum { X)(\sum { Y) } } } }{ \sqrt { N.\sum { { X }^{ 2 }-({ \sum { X) } }^{ 2 } } } .\sqrt { N.{ \sum { Y } }^{ 2 }-({ \sum { Y) } }^{ 2 } } } \)
=\(\frac { 7(676)-(70)(63) }{ \sqrt { 7(728)-{ (70) }^{ 2 } } \sqrt { 7(651)-{ (63) }^{ 2 } } } \)
=\(\frac { 4732-4410 }{ (14)(24.2487) } =\frac { 322 }{ 229.4818 } \)=0.9485

S = {1, 2, 3,.....10}
n(S) = 10
Let B be the event of drawing a card greater than 4
B = {5, 6,.... 10}
\(P(B) =\frac { 6 }{ 10 } \)
Let A be the event of getting an even number
A = {2, 4, 6, 8, 10}
\(P(A) =\frac { 5 }{ 10 } \)
\(A\cap B\) = {6, 8, 10}  
\(P(A\cap B)=\frac { 3 }{ 10 } \)
\(P(A/B)=\frac {P(A\cap B)}{P(B)}=\frac{\frac{3}{10}}{\frac{6}{10}}=\frac{3}{6}=\frac{1}{2}\)

Harmonic mean is useful in calculating average speed
HM = \(\frac { n }{ 1/a+1/b+1/c+1/d } =\frac { 4 }{ 1/100+1/200+1/300+1/400 } \)
\(=\frac { 4 }{ \frac { 12+6+4+3 }{ 1200 } } =\frac { 4 \times 1200}{ 25 } \)
= 192 km/hr

Number of shares = 500
Face value of a share = Rs. 100
Discount = Rs. 14
Market value of a share = 100 – 14 = Rs. 86 (face value – discount)
Market value of 500 shares = Number of shares × market value of a share
= 500 × 86 = 43,000
Market value of 500 shares = Rs. 43,000

p = 200 - \(\frac { x }{ 100 } \) 
R = px \(=200x-{x^2\over100}\)
C = 40x + 120
(i) Profit function = R - C
\(=200x-{x^2\over100}-40x-120\)
\(={x^2\over100}-160x-120\)
(ii) Averange Profit \(=\mathrm{A} \cdot \mathrm{P}=\frac{p(x)}{x}=\frac{R(x)-c(x)}{x}\)
\(=\frac{200 x \frac{-x^2}{100}+40 x+1}{x}=160-\frac{x}{100}-\frac{120}{x}\)
At x = 10 units
\(\text {A.P } =160-\frac{1}{10}-\frac{120}{10} \)
\(=160-12.1= 147.90\)
(iii) Marginal profit \(=\mathrm{M} \cdot \mathrm{P}=\frac{d p}{d x}=\frac{d}{d x} p(x)\)
\(=160-\frac{2 x}{100} \)
\(=160-\frac{x}{50}\)
At x = 10 units
\(\text {M.P }=160-\frac{1}{5}=160-0.2= 159.80\)
(iv) Marginal average profit = M.A.P \( =\frac{d(A P)}{d x}\)
\(=-\frac{1}{100}+\frac{120}{x^2}\)
At x = 10 units
\(\text {M.A.P }=-\frac{1}{100}+\frac{120}{100}= 1.19\)

C(x) = 50 + 4x + 3√x
Marginal cost (MC) =\({Dc\over dx}-{d\over dx}[C(x)]\)
\(={d\over dx}\left[50+4x+3\sqrt3\right]=4{3\over 2\sqrt3}\)
When x = 9, \({dC\over dx}=4{3\over 2\sqrt9}\)\(=4{1\over 2}\)(or) Rs.4.50
∴ MC is Rs.4.50 , when the level of output is 9 units.

y = sin² x
dy/dx = 2 sin x cos x
= sin 2x

The given lines are
2x - 5y + 3 = 0
5x - 9y + \(\lambda \) = 0
x - 2y + 1 = 0
The condition for the lines to be concurrent is
\(\left| \begin{matrix} 2 & -5 & 3 \\ 6 & -9 & \lambda \\ 1 & -2 & 1 \end{matrix} \right| =0\quad \)
Expanding along R₁ , we get
\(2\left| \begin{matrix} -9 & \lambda \\ -2 & 1 \end{matrix} \right| +5\left| \begin{matrix} 5 & \lambda \\ 1 & 1 \end{matrix} \right| +3\left| \begin{matrix} 5 & -9 \\ 1 & -2 \end{matrix} \right| =0\)
\(\Rightarrow 2(-9+2\lambda )+5(5-\lambda )+3(-10+9)=0\)
\(\Rightarrow -18+4\lambda +25-5\lambda -30+27=0\)
\(\Rightarrow -\lambda +4=0\Rightarrow \lambda =+4\)

Y = -x² + 10 x - 15
\(\Rightarrow\) x² - 10x = - y - 15
\(\Rightarrow\) (x- 5)² = -y - 15 + 25
\(\Rightarrow\) (x- 5)² = -y + 10 
\(\Rightarrow\) (x - 5)² = (-y - 10)
The best time to end the project is when x = 5 months.

(x + 2) ( x - 5) + (y -2 ) ( y -1) = 0
\(\Rightarrow\) x² -5x + 2x - 10 + y² - y - 2y + 2 = 0
\(\Rightarrow\)  x² + y² - 3x - 3y - 8 = 0
here 2g  = -3    \(\Rightarrow\)  \(g=-\frac { 3 }{ 2 }\)
2f  = -3   \(\Rightarrow\)   \(f=-\frac { 3 }{ 2 } \)
and C = -8
Center of the circle (-g, -f) = \(\left( \frac { 3 }{ 2 } ,\frac { 3 }{ 2 } \right) \)
A radius of the circle is \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \)
Radius of the circle is  \(= \sqrt { \frac { 9 }{ 4 } +\frac { 9 }{ 4 } +8 }\)
\(=\sqrt{\frac{18}{4}+8}=\sqrt{\frac{9}{2}+8}\)
\(r=\sqrt{\frac{25}{2}}=\frac{5}{\sqrt{2}} \text { units }\)

To draw a line we need two points
No. of chords \(=21 C_2=\frac{21 \times 20}{2 \times 1}=210\)

2x + 3y = 5
x - 2y = -1
The given system can be written as
\(\begin{bmatrix} 2 & 3 \\ 1 & -2 \end{bmatrix}\)\(\begin{bmatrix} X\\Y \end{bmatrix}\)=\(\begin{bmatrix} 5\\-1 \end{bmatrix}\)
AX = B \(\Rightarrow \) \(X=A^{-1} B\)
Where A =\(\begin{bmatrix} 2 & 3 \\ 1 & -2 \end{bmatrix}\), \(X= \left[ \begin{matrix} x \\ y \end{matrix} \right] \),B=\(\begin{bmatrix} 5\\-1 \end{bmatrix}\)
\(|A|=-4-3=-7 \neq 0\)
\(\therefore\) A^-1 exists.
\(\mathrm{A}^{-1}=\frac{1}{|A|} \operatorname{adj} A=\frac{-1}{7}\left(\begin{array}{cc} -2 & -3 \\ -1 & 2 \end{array}\right)\)
\(X=A^{-1} B\)
\(=\frac{-1}{7}\left(\begin{array}{cc} -2 & -3 \\ -1 & 2 \end{array}\right)\left(\begin{array}{c} 5 \\ -1 \end{array}\right)\)
\(=\frac{-1}{7}\left(\begin{array}{cc} -10 & +3 \\ -5 & -2 \end{array}\right)=\frac{-1}{7}\left(\begin{array}{l} -7 \\ -7 \end{array}\right)\)
\(\left(\begin{array}{l} x \\ y \end{array}\right)=\left(\begin{array}{l} 1 \\ 1 \end{array}\right)\)
\(\text {Solution is } x=1, y=1\)

\(\frac{n !}{r !(n-r) !}=\frac{5 !}{2 ! 3 !}\)
\(=\frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 3 \times 2 \times 1}=10\)

\(f(x)=|x|=\left\{\begin{array}{cll} x & \text { if } & x \geq 0 \\ -x & \text { if } & x<0 \end{array}\right.\)
\(\mathrm{L}\left[f^{\prime}(0)\right]=\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}\)
\(=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{0-h-0}, x=0-h\)
\(=\lim _{h \rightarrow 0} \frac{f(-h)-f(0)}{-h}\)
\(=\lim _{h \rightarrow 0} \frac{|-h|-|0|}{-h}\)
\(=\lim _{h \rightarrow 0} \frac{|-h|}{-h}\)
\(=\lim _{h \rightarrow 0} \frac{h}{-h}\)
\(=\lim _{h \rightarrow 0}(-1)=-1\)
\(\mathrm{R}\left[f^{\prime}(0)\right]=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}\)
\(=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{0+h-0}, x=0+h\)
\(=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}\)
\(=\lim _{h \rightarrow 0} \frac{|h|-|0|}{h}\)
\(=\lim _{h \rightarrow 0} \frac{|h|}{h}\)
\(=\lim _{h \rightarrow 0} \frac{h}{h}\)
\(=\lim _{h \rightarrow 0} 1=1\)
\(\text { Here, } \mathrm{L}\left[f^{\prime}(0)\right] \neq \mathrm{R}\left[f^{\prime}(0)\right]\)
\(\therefore\) f(x) is not differentiable at x = 0

Given f(x)= x³-27x+108
f'(x) = = 3x² - 27
f'(x) = 0 ⇒ 3x² - 27 = 0
⇒ x² - 9 = 0 [Divided by 3]
⇒ x²=9
⇒ x= 3, - 3
Also,f"(x) = 6x
when x = 3, f"(x) = 6(3) = 18 > 0
∴ f is minimum at x = 3.
∴ Minimum Value = f(3) = 3³ - 27(3) + 108
=27-81+108=54 ....(1)
when x = -3,f"(x) = 6(-3) = -18 < 0
∴ f is maximum at x = - 3
Maximum value = 1(-3) = (-3)³ - 27(-3) + 108
= -27 + 81 + 108 = 162   ...(2)
From (1) and (2)
Maximum value - Minimum value = 162 - 54 = 108.
Hence, the maximum value of f(x) is 108 more than the minimum value

Let x₁ toys of type A and x₂ toys of type B are produced. Let Z be the maximum profit on two types of toys A and B.

Type	Machine I	Machine II	Machine III	Profit
A	12	18	6	Rs. 7.50
B	6	0	9	Rs. 5
Time available	6h = 360 min	6h = 360 min	6h = 360 min

Thus, the mathematical formulation of the LPP is maximize Z = 7.50x₁ + 5x₂
Subject to the constraints
12x₁ + 6x₂ ≤ 360,18x₁ ≤ 360,6x₁ + 9x₂ ≤ 360,x₁,x₂ ≥ 0
Consider the equations
12x₁ + 6x₂ ≤ 360

\({ x }_{ 1 }\)	0	30
\({ x }_{ 2 }\)	60	0

\(18{ x }_{ 1 }=360\)

x1

20

\({ 6x }_{ 1 }+9{ x }_{ 2 }=360\)

\({ x }_{ 1 }\)	0	60
\({ x }_{ 2 }\)	40	0

X₁ = 20 is a line parallel to x₂-axis at a distance of 20 units from it

 

The feasible region is OABCD and its co-ordinates are O(0, 0) A(20, 0) D(0, 40), B is the point of intersection ofthe lines x₁ = 20 and 12x₁+ 6x₂ = 360
\(\Rightarrow 2{ x }_{ 1 }+{ x }_{ 2 }=60\)
\(\Rightarrow 40+{ x }_{ 2 }=60\)
\( \Rightarrow { x }_{ 2 }=20\)
Verification of Band C:
\(\therefore \quad B\quad (20,20)\)
And C is the point of intersection of the lines
\(2{ x }_{ 1 }+{ x }_{ 2 }=60...(1)\)
\( (-)\quad \quad (-)\quad \quad (-)\)
\(2{ x }_{ 1 }+3{ x }_{ 2 }=120 \left[ \because \quad 6{ x }_{ 1 }+{ 9x }_{ 2 }=360 \right] ...(2)\)
\(-------------\)
\( -2x_{ 2 }=-60 \Rightarrow { x }_{ 2 }=30\)
\(2{ x }_{ 1 }+30=60 [\because \quad From\quad (1)] \Rightarrow { x }_{ 1 }=\frac { 30 }{ 2 } =15\)
\( \therefore \quad C\quad is\quad (15,30)\) 

Corner Points	Z = 7.5x1 + 5x2
O(0,0)	0
A(20,0)	150
B(20, 20)	250
C(15,30)	262.5
D(0, 40)	200

Maximum of Z occurs at (15,30)
Hence, the solution is x₁ = 15, x₂ = 30 and Z_max= 262.5

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

Let y=x³-6x²+7
Differentiating w.r.t. 'x' we get,
\({dy\over dx}=3x^2-12x\)
\({dy\over dx}=0\)
\(\Rightarrow3x^2-12x=0\)
\(\Rightarrow3x(x-4)=0\)
\(\Rightarrow x=0 \ or \ x=4\)
\({d^2y\over dx^2}=6x-12\)
when x=0 \({d^2y\over dx^2}=-12<0\)
\(\therefore \) y is maximum at x=0
\(\therefore \) maximum value=0³-6(0)²+7=7
when x=4, \({d^2y\over dx^2}=-6(4)-12=12>0\)
\(\therefore \) y is minimum at x=4
\(\therefore \) Minimum value =4⁴-6(4)²+7=64-96+7=-25
Hence maximum value is 7 and minimum value is -25.

Here the sample space of the experiment is
S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}
A = {Three heads or Three tails}
= {HHH, TTT}
B = {at least two heads}
= {HHH, HHT, HTH, THH} and
C = {at most two heads} = {HHT, HTH, HTT, THH, TTH, THT, TTT}
Also (A∩B) = {HHH}; (A∩C) = {TTT} and (B∩C) ={HHT, HTH, THH}
∴ P(A) = \(\frac { 2 }{ 8 } =\frac { 1 }{ 4 } \); P(B) =\(\frac{1}{2}\); P(C) = \(\frac{7}{8}\) and
P(A∩B)= \(\frac{1}{8}\), P(A∩C)=\(\frac{1}{8}\), P(B∩C)=\(\frac{3}{8}\)
Also P(A). P(B)=\(\frac { 1 }{ 4 } .\frac { 1 }{ 2 } =\frac { 1 }{ 8 } \)
P(A). P(C) =\(\frac { 1 }{ 4 } .\frac { 7 }{ 8 } =\frac { 7 }{ 32 } \)
and P(B). P(C) =\(\frac { 1 }{ 2 } .\frac { 7 }{ 8 } =\frac { 7 }{ 16 } \)
Thus, P(A∩B) = P(A). P(B)
P(A∩C) ≠ P(A) P(C) and
P(B∩C) ≠ P(B). P(C)
Hence, the events (A and B) are independent, and the events (A and C) and (B and C) are dependent.

X	Y	dx = X-48	dy = Y-52	dx2	dy2	dx dy
40	38	-8	-14	64	196	112
50	60	2	8	4	64	16
38	55	-10	3	100	9	-30
60	70	12	18	144	324	216
65	60	17	8	289	64	136
50	48	2	-4	4	16	-8
35	30	-13	-22	169	484	286
​\(\Sigma X\)​ = 338	​\(\Sigma Y\)​ = 361	​\(\Sigma dx\)​ = 2	​\(\Sigma dy\)​ = -3	​\(\Sigma dx^2\)​ = 774	​\(\Sigma dy^2\)​ = 1157	​\(\Sigma dxdy\)​ = 728

\(\bar{X} =\frac{\Sigma X}{N}=\frac{338}{7}=48.29 \)
\(\bar{Y} =\frac{\Sigma Y}{N}=\frac{361}{7}=51.57 \)
\(b_{y x} =\frac{N \Sigma d x d y-\Sigma d x \Sigma d y}{N \Sigma d x^2-(\Sigma d x)^2} \)
\(=\frac{7(728)+6}{7(774)-4} \)
\(=\frac{5102}{5414}=0.942\)
Regression equation of Y on X is
\(Y-\overset{-}{Y}=b_{yx}(X-\overset{-}{X})\)
Y - 51.57 = 0.942(X - 48.29)
Y = 0.942X - 45.49 + 51.57
= 0.942X + 6.08
Y = 0.942(30) + 6.08
Y = 34.34 (In Crores of rupees)

Earliest start time (EST) and latest finish time (LFT) of each activity are given in the following network.

E₁ = 0
E₂ = E₁ + t₁₂ = 0 + 8 = 8
E₃ = E₁ + t₁₃ = 0 + 4 = 4
E₄ = E₂ + t₂₄ or E₃ + t₃₄ = 8 + 10 = 18
(take E₂ + t₂₄ or E₃ + t₃₄ whichever is maximum)
E₅ = (E₂ + t₂₅ or E₄ + t₄₅) = 18 + 3 = 21
(take E₂ + t₂₅ or E₄ + t₄₅ whichever is maximum)
L₅ = 21
L₄ = L₅ – t₄₅ = 21 – 3 = 18
L₃ = L₄ – t₃₄ = 18 – 5 = 13
L₂ = L₅ – t₂₅ or L₄ – t₂₄ = 18 – 10 = 8
(take L₅ – t₂₅ or L₄ – t₂₄ whichever is minimum)
L₁ = L₂ – t₁₂ or L₃ - t₁₃ = 8 – 8 = 0
(take L₂ – t₁₂ or L₃ – t₁₃ whichever is minimum)
Here the critical path is 1-2-4-5, which is denoted by double lines.

Activity	Duration(tij)	EST	EFT=EST+tij	LST=LFT-tij	LFT
1-2	8	0	8	0	8
1-3	4	0	4	9	13
2-4	10	8	18	8	18
2-5	2	8	10	19	21
3-4	5	4	9	13	18
4-5	3	18	21	18	21

The longest duration to complete this project is 21 days.
The path connected by the critical activities is the critical path(the longest path).
Critical path is 1-2-4-5 and project completion time is 21 days.

RHS = tan 4x = tan2(2x)
= \(\frac { 2\quad tan\quad 2x }{ 1-{ tan }^{ 2 }2x } \left[ \because tan2x=\frac { 2\quad tan\quad x }{ 1-{ tan }^{ 2 }x } \right] \)
= \(\frac { 2.\frac { 2\quad tan\quad x }{ 1-{ tan }^{ 2 }x } }{ 1-\left( \frac { 2\quad tan\quad x }{ 1-{ tan }^{ 2 }x } \right) ^{ 2 } } =\frac { \frac { 4\quad tan\quad x }{ 1-{ tan }^{ 2 }x } }{ \frac { (1-{ tan }^{ 2 }x)^{ 2 }-4{ tan }^{ 2 }x }{ (1-{ tan }^{ 2 }x)^{ 2 } } } \)
= \(\frac { 4\quad tan\quad x }{ 1-{ tan }^{ 2 }\quad x } \times \frac { (1-{ tan }^{ 2 }\quad { x })^{ 2 } }{ 1+{ tan }^{ 4 }x-2\quad { tan }^{ 2 }x-4{ tan }^{ 2 }x } \)
=  \(\frac { 4 tanx(1-{ tan }^{ 2 }\quad x) }{ 1+{ tan }^{ 4 }x-6\ { tan }^{ 2 }x } \)  = LHS

Since the numerator degree is equal to the degree of the denominator, let us divide.

\(\therefore\)   \({{x^2+x+1}\over{x^2+2x+1}}=1-{{x}\over{x^2+2x+1}}\)  ...(1)
Consider    \({{x}\over{x^2+2x+1}}={{x}\over{(x+1)^2}}={{A}\over{x+1}}+{{B}\over{{(x+1)}^{2}}}\)
\(\therefore\ {{x}\over{x^2+2x+11}}={{A(x+1)+B}\over{{(x+1)}^{2}}}\)
\(\Rightarrow\)      x = A(x+ 1) + B  ..(2)
Putting x = - 1 in (2) we get
-1 = 0 + B   \(\Rightarrow\ \ \boxed{B=-1}\)
Putting x = 0 in (2) we get,
0 =A+B \(\Rightarrow\) A - 1 = 0  \(\Rightarrow\)  \(\boxed{A=1}\)
\(\therefore\)   \({{x}\over{{(x+1)}^{2}}}={{1}\over{x+1}}-{{1}\over{{(x+1)}^{2}}}\) ..(2)
Substituting (2) in (1) we get,
\({{x^2+x+1}\over{x^2+2x+1}}=1-{{1}\over{x+1}}+{{1}\over{{(x+1)}^{2}}}\)

\(\mathrm{LHS}= \tan (\pi+x) \cot (x-\pi)-\cos (2 \pi-x) \cos (2 \pi+x)-\tan (\pi+x) \cot (\pi-x) -\cos (2 \pi-x) \cos (2 \pi+x)\)
\(=-(\tan x)(-\cot x)-(\cos x)(\cos x)\)
\(=\tan \left(\frac{1}{\tan x}\right)-\cos ^2 x\)
\(=1-\cos ^{ 2 }{ x } =\sin ^{ 2 }{ x } \ \left[ \because \ 1-\cos ^{ 2 }{ x } =\sin ^{ 2 }{ x } \right] \)
= RHS  Hence Proved.