What is TIFR

3(b) Prove that ( + ) is an algebraic number. Is every irrational number, an algebraic number? Justify your answer.

Ans :-

Algebraic Number :- A number is called an algebraic number if it satisfies a polynomial equation

a₀xⁿ + a₁x^n-1 + a₂x^n-2 + …………. a_n-1x + a_n = 0

where the coefficients a₀₁ a₁₁ a₁₂ .......... a_1n  are integers, n>1.

Given number         X =  +

X² = ( + )²

X² = ()² + ()²  + 2()()

X²  = 2 + 5 + 2

X²  = 7 + 2

X² – 7 = 2

(X² – 7)² = (2²

X² + 49 – 14X² = 40

\[ Ram\quad is\quad going\quad to\quad school.\\ x=\frac { -b+\sqrt { b^{ 2 }-4ac } }{ 2a } \\ 4x^{ 2 }-8x+9=0\\ \int _{ 4 }^{ 5 }{ 5x } dx=K \]

\[ \begin{array}{l}4x^2-8x+9=0\\\end{array} \]

\[ \left. \begin{array} { l } { \text { If } u = e ^ { x y z } \text { , find the value of } \frac { \partial ^ { 3 } u } { \partial x \partial y \partial z } } \\ { \qquad \left. \begin{array}{l}{ u = e ^ { x y z } }\\{ \frac { \partial u } { \partial z } = e ^ { x y z } ( x y ) }\\{ \frac { \partial ^ { 2 } u } { \partial y \partial z } = e ^ { x y z } ( x ) + e ^ { x y z } ( x z ) ( x y ) = e ^ { x y z } ( x + x ^ { 2 } y z ) }\\{ \frac { \partial ^ { 3 } u } { \partial x \partial y \partial z } = e ^ { x y z } ( 1 + 2 x y z ) + e ^ { x y z } ( y z ) \cdot ( x + x ^ { 2 } y z ) }\\{ = e ^ { x y z } [ 1 + 2 x y z + x y z + x ^ { 2 } y ^ { 2 } z ^ { 2 } ] }\\{ = e ^ { x y z } [ 1 + 3 x y z + x ^ { 2 } y ^ { 2 } z ^ { 2 } ] }\end{array} \right. } \end{array} \right. \]

\[ \left. \begin{array} { | c | c | c | c | c | c | c | c | c | c | } \hline \text { Iteration-1 } & { } & { C _ { j } } & { 5 } & { 10 } & { 8 } & { 0 } & { 0 } & { 0 } & { } \\ \hline B & { C _ { B } } & { X _ { B } } & { x _ { 1 } } & { x _ { 2 } } & { x _ { 3 } } & { S _ { 1 } } & { S _ { 2 } } & { S _ { 3 } } & { } \\ \hline \end{array} \right. \]

Problem is

The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate

1. As the constraint-1 is of type '≤' we should add slack variable S1

2. As the constraint-2 is of type '≤' we should add slack variable S2

3. As the constraint-3 is of type '≤' we should add slack variable S3

\[ \int _{\:0}^{\pi }\sin \left(x\right)dx \]

\[ \begin{pmatrix}5&4&5&r\\ g&t&y&5\\ 6&6&6&5\\ 4&3&5&t\\ v&t&t&5\end{pmatrix} \]

\[ \left. \begin{array} { l } { \text { Taylor series of function } f ( x ) \text { at a is defined as: } } \\ { \qquad f ( x ) = f ( a ) + \frac { f ( a ) } { 1 ! } ( x - a ) + \frac { f ^ { \prime \prime } ( a ) } { 2 ! } ( x - a ) ^ { 2 } + \frac { f ^ { \prime \prime } ( a ) } { 3 ! } ( x - a ) ^ { 3 } + \ldots } \\ { \text { Maclaurin series of function } f ( x ) \text { is a taylor series of function } f ( x ) \text { at: } a = 0 } \\ { \qquad f ( x ) = f ( 0 ) + \frac { f ( 0 ) } { 1 ! } ( x ) + \frac { f ( 0 ) } { 2 ! } ( x ) ^ { 2 } + \frac { f ^ { \prime \prime } ( 0 ) } { 3 ! } ( x ) ^ { 3 } + \ldots } \end{array} \right. \]

\[ \left. \begin{array} { | c | c | c | c | c | c | c | c | c | c | } \hline \text { Iteration-1 } & { } & { C _ { j } } & { 5 } & { 10 } & { 8 } & { 0 } & { 0 } & { 0 } & { } \\ \hline B & { C _ { B } } & { X _ { B } } & { x _ { 1 } } & { x _ { 2 } } & { x _ { 3 } } & { S _ { 1 } } & { S _ { 2 } } & { S _ { 3 } } & { } \\ \hline \end{array} \right. \]

\[ \left. \begin{array} { l } { 3 x _ { 1 } + 5 x _ { 2 } + 2 x _ { 3 } \leq 60 } \\ { 4 x _ { 1 } + 4 x _ { 2 } + 4 x _ { 3 } \leq 72 } \\ { 2 x _ { 1 } + 4 x _ { 2 } + 5 x _ { 3 } \leq 100 } \\ { \text { and } x _ { 1 } , x _ { 2 } , x _ { 3 } \geq 0 } \end{array} \right. \]