(101)⁵ = (100 + 1)⁵
= (100)⁵ + 5C₁ (100)⁴ + 5C₂ (100)³ + 5C₃ (100)² + 5C₄ (100) + 5C₅
= 10000000000+ 5(100000000) + 10(1000000) + 10(10000) + 5(100) + 1
= 10000000000 + 500000000 + 10000000 + 100000 + 500 + 1
= 10510100501

B = \(\left[ \begin{matrix} 0.8 & 0.2 \\ 0.9 & 0.7 \end{matrix} \right] \)
I - B=\(\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)-\(\left[ \begin{matrix} 0.8 & 0.2 \\ 0.9 & 0.7 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 0.2 & -0.2 \\ - 0.9 & 0.3 \end{matrix} \right] \)
|I - B|=  \(\left[ \begin{matrix} 0.2 & -0.2 \\ - 0.9 & 0.3 \end{matrix} \right] \)
= (0.2)(0.3) - (-0.2)(-0.9)
= 0.06 - 0.18
= 0.12 < 0
Since |I - B| is negative, Hawkins – Simon conditions are not satisfied.
Therefore, the given system is not viable.

\(\left| \begin{matrix} 1 & a & { a }^{ 2 } \\ 1 & b & { b }^{ 2 } \\ 1 & c & { c }^{ 2 } \end{matrix} \right| =\left| \begin{matrix} 0 & a-b & { a }^{ 2 }-{ b }^{ 2 } \\ 0 & b-c & { b }^{ 2 }-{ c }^{ 2 } \\ 1 & c & { c }^{ 2 } \end{matrix} \right| \begin{matrix} { R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2 } \\ { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 3 } \end{matrix}\)
\(=\left| \begin{matrix} 0 & a-b & (a-b)(a+b) \\ 0 & b-c & (b-c)(b+c) \\ 1 & c & { c }^{ 2 } \end{matrix} \right| \)
\(=(a-b)(b-c)\left| \begin{matrix} 0 & 1 & a+b \\ 0 & 1 & b+c \\ 1 & c & { c }^{ 2 } \end{matrix} \right| \)
= (a - b)(b - c)[0 - 0 + {b + c - (a + b)}]
= (a - b)(b - c)(c - a)

u = log(x²+y²)
\(\frac { \partial u }{ \partial x } =\frac { 1 }{ { x }^{ 2 }+{ y }^{ 2 } } (2x)=\frac { 2x }{ { x }^{ 2 }+{ y }^{ 2 } } \)
\(\frac { { \partial }^{ 2 }u }{ \partial { x }^{ 2 } } =\frac { \left( { x }^{ 2 }+{ y }^{ 2 } \right) . 2-2x.2x }{ \left( { x }^{ 2 }+{ y }^{ 2 } \right) ^{ 2 } } =\frac { 2\left( { y }^{ 2 }-{ x }^{ 2 } \right) }{ \left( { x }^{ 2 }+{ y }^{ 2 } \right) ^{ 2 } } \)
\(\frac { \partial u }{ \partial y } =\frac { 1 }{ { x }^{ 2 }+{ y }^{ 2 } } (2y)=\frac { 2y }{ { x }^{ 2 }+{ y }^{ 2 } } \)
\(\frac { { \partial }^{ 2 }u }{ \partial y^{ 2 } } =\frac { \left( { x }^{ 2 }+{ y }^{ 2 } \right) . 2-2y.2y }{ \left( { x }^{ 2 }+{ y }^{ 2 } \right) ^{ 2 } } =\frac { 2\left( { y }^{ 2 }-{ x }^{ 2 } \right) }{ \left( { x }^{ 2 }+{ y }^{ 2 } \right) ^{ 2 } } \)
\(\therefore \frac { { \partial }^{ 2 }u }{ { 2x }^{ 2 } } +\frac { { \partial }^{ 2 }u }{ \partial { y }^{ 2 } } =0\)

Given C = x (x²-2) = x³-2x
(i) Average Cost  = \(\frac { C }{ x } =\frac { { x }^{ 3 }-2x }{ x } ={ x }^{ 2 }-2\)
(ii) Marginal Cost (MC) = \(\frac { dc }{ dx } =\frac { d }{ dx } \left( { x }^{ 2 }-2x \right) ={ 3x }^{ 2 }-2\)
(iii) R = 14x - x²
Average Revenue (AR) = \(\frac { R }{ x } =14-x\)
(iv) Marginal Revenue (MR) \(=\frac { d R}{ dx } =14-2x\)

(i) Variables: Let x₁ be the number of days required to run plant C₁ and x₂ be the number of days required to run plant C₂
Objective function: Minimize Z = 600 x₁ + 400 x₂
(ii) Constraints: 3000 x₁ + 1000 x₂ ≥ 24000 (since there is a demand of 24000 bottles of drink A, production should not be less than 24000)
1000 x₁ + 1000 x₂ ≥ 16000
2000 x₁ + 6000 x₂ ≥ 48000
(iii) Non-negative restrictions: Since be the number of days required of a firm are non-negative, we have x₁, x₂ ≥ 0
Thus we have the following LP model.
Minimize Z = 600 x₁ + 400 x₂
subject to 3000 x₁ + 1000 x₂ ≥ 24000
1000 x₁ + 1000 x₂ ≥ 16000
2000 x₁ + 6000 x₂ ≥ 48000 and x₁, x₂ ≥ 0

Let P (x₁, y₁) be any point on the locus.
A (2, 2), B (1, 5) are the given points
Since P, A, B are collinear
Slope of AP = slope of AB
\(\frac{y_1-2}{x_1-2}=\frac{5-2}{1-2}\)
\(y_1-2=-3\left(x_1-2\right)\)
\(y_1-2=-3 x_1+6\)
\(3 x_1+y_1-8=0\)
\(\therefore \text {Locus of } P\left(x_1, y_1\right) \text {is } 3 x+y-8=0\)

x = a cos \(\theta\), y = a sin \(\theta\)
\(\frac {dx }{ d\theta } =-a\sin { \theta } \) ; \(\frac {dy }{ d\theta } =a\cos { \theta } \)    
\(= { \frac { dy }{ dx} } =\frac { a\cos { \theta } }{ -a\sin { \theta } } =-\cot { \theta } \)
\(\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}(-\cot \theta)\)
\(=\frac{d}{d \theta}(-\cot \theta) \frac{d \theta}{d x}\)
\(=\operatorname{cosec}^2 \theta\left(\frac{-1}{a \sin \theta}\right)=\frac{-1}{a} \operatorname{cosec}^3 \theta\)

\(\lim _{ x\rightarrow 2 }{ f(x) } \)= \(\lim _{ x\rightarrow 2 }{ \frac { { x }^{ 7 }-128 }{ { x }^{ 5 }-32 } }\)
\(=\lim _{x \rightarrow 2} \frac{\frac{x^7-2^7}{x-2}}{\frac{x^5-2^5}{x-2}}=\frac{7(2)^6}{5(2)^4}=\frac{7(4)}{5}=\frac{28}{5}\)
 

(n + 2)! = 60(n - 1)!
(n + 2)(n + 1)(n)(n - 1)! = 60(n - 1)!
(n + 2)(n + 1)(n) = 60
(n + 2)(n + 1)(n) = 5 \(\times \) 4  \(\times \) 3
n = 3
(Counting 60 as product of 3 consecutive natural numbers)

\(AB=\begin{bmatrix} 3&-1\\2&1 \end{bmatrix}\begin{bmatrix} 3&0\\1&-2 \end{bmatrix}=\begin{bmatrix} 9-1&0+2\\6+1&0-2 \end{bmatrix}=\begin{bmatrix} 8&2\\7&-2 \end{bmatrix}\)
= -16 - 14 = -30
\(\therefore\) |AB| = -30

\(r =\frac{N \Sigma X Y-(\Sigma X)(\Sigma Y)}{\sqrt{N \Sigma X^2-(\Sigma X)^2} \sqrt{N \Sigma Y^2-(\Sigma Y)^2}} \)
\(=\frac{10(-115)-(50)(-30)}{\sqrt{10(290)-(50)^2} \sqrt{10(300)-(-30)^2}} \)
\(=\frac{-1150+1500}{\sqrt{400 \times 2100}}=\frac{350}{916.52}=0.382\)

Here a = 64,000, n = 12 and i = \(\frac{10}{100}=0.1\)
Amount of Ordinary annuity (A) = \(\frac{a}{i}[(1+i)^n-1]\)
= \(\frac{64000}{0.1}[(1+0.1)^{12}-1]\)
= 6,40,000 [(1.1)¹² – 1]
= 6,40,000[3.3184 – 1]
= 6,40,000 [2.3184]
= 64 x 23184
∴ A  = Rs. 14,83,776.

Here n = 11 observations are arranged into ascending order
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22
D₂ = size of 2 \(\left( \frac { n+1 }{ 10 } \right) \)^th value
D₆ = size of 6  \(\left( \frac { n+1 }{ 10 } \right) \)^th value
D₂ = size of 2.4^th value ≈ size of 2^nd value = 4
D₆ = size of 7.2^th value ≈ size of 7^th value = 14

 \(\sin { \frac { \pi }{ 4 } } \cos { \frac { \pi }{ 12 } } +\cos { \frac { \pi }{ 4 } } \sin { \frac { \pi }{ 12 } } \)
\(=\sin { \left( \frac { \pi }{ 4 } +\frac { \pi }{ 12 } \right) } \)  
\(=\sin { \left( \frac { 3\pi +\pi }{ 12 } \right) } =\sin { \left( \frac { \pi }{ 3 } \right) } =\frac { \sqrt { 3 } }{ 2 } \)

x = log t, y = sin t
\(\frac { dx }{ dt } =\frac { 1 }{ t } ;\frac { dy }{ dt } =cost\)
\(\frac { dy }{ dx } =\frac { \frac { dy }{ dt } }{ \frac { dx }{ dt } } =\frac { cost }{ \frac { 1 }{ t } } =t.cos \ t\)

The equation of the circle is \({ x }^{ 2 }+{ y }^{ 2 }-4x+4y-8=0\) 
\({ PT }^{ 2 }={ x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }-{ 4x }_{ 1 }+{ 4y }_{ 1 }-8\) 
At \(P(0,1)\) 
\({ PT }^{ 2 }=0+1+0+4-8=-3<0\)
\(At \ Q(5,9)\)
\({ QT }^{ 2 }=25+81-20+36-8=114>0\)
\(At \ R(-2,3)\)
\({ RT }^{ 2 }=4+9+12-8=25>0\)
\(At \ S(2,2)\)
\(ST^{ 2 }=4+4-8+8-8=0\) 
\(\therefore \) The point P lies inside the circle. The points Q and R lie outside the circle and the point S lies on the circle.