f(x)=x-5 

Domain of a function f is a set of all values of x at which the given function is defind.

Here clearly the given function f(x)=x-5 is defind for every real value of x.so domain of f= R (set of real numbers)

Range of a function is a set of values of function f(x) at domain of f(x).

Range of f= R

In 60sec = 360revolutions 

    1 sec =6 revolution 

1 revolution =2 pie

6 revolution = 12 pie

if n is even answer is zero

if n is odd answer is one

if n is odd the answer will be zero

if n is even all cancel and answer is 1

We know i​​​​​​²= -1 

And {tex}i^4=(i^2)^2=(-1)^2= 1{/tex}

So all the terms having i will be cancelled out. We ll get answer as 1 

Let the two numbers are a and b.

Then A.T.Q.

a+b=C ------(1)      and  {tex}{a\over b}={p\over q}{/tex} ----(2)

From eq(2)

{tex}a={p\over q}×b{/tex} ---------(3)

Substituting this value in eq.(1) we get

{tex}{pb\over q}+b=C{/tex}

{tex}{pb+bq\over q}=C{/tex}

pb+bq=C×q

b(p+q)=Cq

         b={tex}{Cq\over p+q}{/tex}

On substituting value of b in eq.(3) we get

a={tex}{p\over q}×{Cq\over p+q}={pC\over p+q}{/tex}

Mehod 1:- In this method first of all we find value of a and  c from given equations in variable b. After that we find a+b and b+c and then divide them.

Given {tex}{a\over b} ={2\over 3}{/tex} and {tex}{b\over c}={4\over 5}{/tex}

     {tex}=> a={2\over 3}b{/tex} and {tex}{c\over b}={5\over 4}{/tex} {tex}=> c={5\over 4}b{/tex}

Now a+b={tex}{2b\over 3}+b={2b+3b\over 3}={5b\over 3}{/tex}

and b+c={tex}b+{5b\over 4}={4b+5b\over 4}={9b\over 4}{/tex}

So {tex}{a+b\over b+c}={{5b\over 3}\over {9b\over 4}}={5b\over 3}×{4\over 9b}{/tex}={tex}{20\over27}{/tex}

Method 2:-

Given {tex}{a\over b}={2\over 3}{/tex} and {tex}{b\over c}={4\over 5}{/tex}

Now {tex}{a+b\over b+c}={{a+b\over b}\over {b+c\over b}}{/tex}        [on dividing numerator and denominator by b]

                  ={tex}{{a\over b}+{b\over b}}\over{{b\over b} +{c\over b}}{/tex}={tex}{{a\over b}}+1\over{1+{c\over b}}{/tex}={tex}{{2\over 3}}+1\over {1+{5\over4}}{/tex}={tex}{2+3\over 3}\over {4+5\over 4}{/tex}={tex}{5\over 3}\over {9\over 4}{/tex}={tex}{5\over 3}×{4\over 9}{/tex}={tex}20\over 27{/tex}   [by substituting the values of a/b and c/b]

equation is{tex} ax^3+bx^2+cx+d{/tex}

when x=1

{tex} a+b+c+d=0{/tex}  when {tex} x=1{/tex}

{tex} -8a+4b-2c+d=0{/tex}  when {tex} x=-2{/tex}

{tex} -a+b-c+d=4{/tex}  when {tex} x=-1{/tex}

{tex} 8a+4b+2c+d=28{/tex}  when {tex} x=2{/tex}

{tex}\begin{pmatrix} 1 & 1 & 1 & 1 \\ -8 & 4 & -2 & 1 \\ -1 & 1 & -1& 1\\ 8 & 4 &2 & 1\end{pmatrix} \ \begin{pmatrix} a \\ b \\ c\\ d\end{pmatrix} \ =\begin{pmatrix} 0 \\ 0 \\ 4\\ 28\end{pmatrix} \ {/tex}

If a - b is a factor of given expression, then a - b = 0 => a = b

Putting a = b, in the given expression, we get

b(b^{<font size="2">2</font>}-c^{<font size="2">2</font>}) +b(c^{<font size="2">2</font>}- b^{<font size="2">2</font>}) +c(b^{<font size="2">2</font>}- b^{<font size="2">2</font>})

= b³ - bc^{<font size="2">2</font>} + bc^{<font size="2">2</font>}  - b^{<font size="2">3</font>} + c(0)

= 0

Therefore, (a - b) is a factor of given expression.

Again if (b - c) is a factor of given expression, then

Putting b - c = 0 => b = c in the given expression, we get

a(c^{<font size="2">2 </font>}- c^{<font size="2">2</font>}) +c(c^{<font size="2">2 </font>}- a^{<font size="2">2</font>}) + c(a^{<font size="2">2 </font>}- c^{<font size="2">2</font>})

= a(0) + c³ - ca² + ca² - c³ 

= 0

Therefore, (b - c) is a factor of given expression.

Again if (c - a) is a factor of given expression, then

Putting c - a = 0 => c = a in the given expression, we get

a(b^{<font size="2">2 </font>}- a^{<font size="2">2</font>}) + b(a^{<font size="2">2 </font>}- a^{<font size="2">2</font>}) + a(a^{<font size="2">2 </font>}- b^{<font size="2">2</font>})

= ab² - a^{<font size="2">3</font>} + b(0) + a³ - ab²

= 0

Therefore, (c - a) is a factor of given expression

||) {tex}{x^4} + 4 = {x^2} + {2^2} = {\left( {x + 2} \right)^2} - 2 \times x \times 2{/tex}         [Since {tex}{a^2} + {b^2} = {\left( {a + b} \right)^2} - 2ab{/tex}]

= {tex}{\left( {x + 2} \right)^2} - 4x{/tex}

= {tex}{\left( {x + 2} \right)^2} - {\left( {2\sqrt x } \right)^2}{/tex}

= {tex}\left( {x + 2 - 2\sqrt x } \right)\left( {x + 2 + 2\sqrt x } \right){/tex}

= {tex}\left( {x + 2\sqrt x + 2} \right)\left( {x - 2\sqrt x + 2} \right){/tex}

|) 1+x^{<font size="2">4</font>}+x^{<font size="2">8</font>}<font size="2"> = x⁸ + x⁴ + 1</font>

Adding and subtracting 2(x⁴)(1)

{tex}{\left( {{x^4}} \right)^2} + 2\left( {{x^4}} \right)\left( 1 \right) + {\left( 1 \right)^2} - 2\left( {{x^4}} \right) + {x^4} = {\left( {{x^4} + 1} \right)^2} - {x^4}{/tex}

= {tex}\left( {{x^4} + 1 + {x^2}} \right)\left( {{x^4} + 1 - {x^2}} \right) = \left( {{x^4} + {x^2} + 1} \right)\left( {{x^4} - {x^2} + 1} \right){/tex}

= {tex}\left[ {{{\left( {{x^2}} \right)}^2} + 2\left( {{x^2}} \right)\left( 1 \right) + 1 - 2\left( {{x^2}} \right)\left( 1 \right) + {x^2}} \right]\left( {{x^4} - {x^2} + 1} \right){/tex}

= {tex}\left[ {{{\left( {{x^2} + 1} \right)}^2} - {x^2}} \right]\left( {{x^4} - {x^2} + 1} \right){/tex}

= {tex}\left( {{x^2} + 1 + x} \right)\left( {{x^2} + 1 - x} \right)\left( {{x^4} - {x^2} + 1} \right){/tex}

= {tex}\left( {{x^2} + x + 1} \right)\left( {{x^2} - x + 1} \right)\left( {{x^4} - {x^2} + 1} \right){/tex}

{tex} \cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta \\ take \theta =20 \\ \frac{1}{2} = 4{x^3} - 3x \\ 8{x^3} - 6x-1=0 {/tex}

solve the equation

In <a href="https://googleweblight.com/i?u=https://en.m.wikipedia.org/wiki/Set_theory&hl=en-IN">set theory</a> (and, usually, in other parts of <a href="https://googleweblight.com/i?u=https://en.m.wikipedia.org/wiki/Mathematics&hl=en-IN">mathematics</a>), a Cartesian product is a <a href="https://googleweblight.com/i?u=https://en.m.wikipedia.org/wiki/Mathematical_operation&hl=en-IN">mathematical operation</a> that returns a <a href="https://googleweblight.com/i?u=https://en.m.wikipedia.org/wiki/Set_(mathematics)&hl=en-IN">set</a> (or product set or simply product) from multiple sets. That is, for sets Aand B, the Cartesian product A × B is the set of all <a href="https://googleweblight.com/i?u=https://en.m.wikipedia.org/wiki/Ordered_pair&hl=en-IN">ordered pairs</a> (a, b) where a ∈ A and b ∈ B. Products can be specified using <a href="https://googleweblight.com/i?u=https://en.m.wikipedia.org/wiki/Set-builder_notation&hl=en-IN">set-builder notation</a>, e.g.

{\displaystyle A\times B=\{\,(a,b)\mid a\in A\ {\mbox{ and }}\ b\in B\,\}.}

<a href="https://googleweblight.com/i?u=https://en.m.wikipedia.org/wiki/Cartesian_product%23cite_note-1&hl=en-IN">[1]</a>

A table can be created by taking the Cartesian product of a set of rows and a set of columns. If the Cartesian product rows × columns is taken, the cells of the table contain ordered pairs of the form (row value, column value).

More generally, a Cartesian product of nsets, also known as an n-fold Cartesian product, can be represented by an array of n dimensions, where each element is an n-<a href="https://googleweblight.com/i?u=https://en.m.wikipedia.org/wiki/Tuple&hl=en-IN">tuple</a>. An ordered pair is a <a href="https://googleweblight.com/i?u=https://en.m.wikipedia.org/wiki/Tuple%23Names_for_tuples_of_specific_lengths&hl=en-IN">2-tuple or couple</a>.

The Cartesian product is named after <a href="https://googleweblight.com/i?u=https://en.m.wikipedia.org/wiki/Ren%25C3%25A9_Descartes&hl=en-IN">René Descartes</a>,<a href="https://googleweblight.com/i?u=https://en.m.wikipedia.org/wiki/Cartesian_product%23cite_note-2&hl=en-IN">[2]</a> whose formulation of <a href="https://googleweblight.com/i?u=https://en.m.wikipedia.org/wiki/Analytic_geometry&hl=en-IN">analytic geometry</a> gave rise to the concept, which is further generalized in terms of <a href="https://googleweblight.com/i?u=https://en.m.wikipedia.org/wiki/Direct_product&hl=en-IN">direct product</a>.

Examples

A deck of cards

<a href="https://googleweblight.com/i?u=https://en.m.wikipedia.org/wiki/File:Piatnikcards.jpg&hl=en-IN"></a>

Standard 52-card deck

An illustrative example is the <a href="https://googleweblight.com/i?u=https://en.m.wikipedia.org/wiki/Standard_52-card_deck&hl=en-IN">standard 52-card deck</a>. The <a href="https://googleweblight.com/i?u=https://en.m.wikipedia.org/wiki/Playing_cards%23Anglo-American&hl=en-IN">standard playing card</a>ranks {A, K, Q, J, 10, 9, 8, 7, 6, 5, 4, 3, 2} form a 13-element set. The card suits {♠, ♥, ♦, ♣} form a four-element set. The Cartesian product of these sets returns a 52-element set consisting of 52 <a href="https://googleweblight.com/i?u=https://en.m.wikipedia.org/wiki/Ordered_pairs&hl=en-IN">ordered pairs</a>, which correspond to all 52 possible playing cards.

Ranks × Suits returns a set of the form {(A, ♠), (A, ♥), (A, ♦), (A, ♣), (K, ♠), ..., (3, ♣), (2, ♠), (2, ♥), (2, ♦), (2, ♣)}.

Suits × Ranks returns a set of the form {(♠, A), (♠, K), (♠, Q), (♠, J), (♠, 10), ..., (♣, 6), (♣, 5), (♣, 4), (♣, 3), (♣, 2)}.

Both sets are distinct, even disjoint.

Let {tex}f(x)=8x^{ 3 }+1001x+2008=0\Rightarrow f(x)=8x^{ 3 }+0{ x }^{ 2 }+1001x+2008=0{/tex}

Given the roots of this equation are r,s,t.

Now using the relationship between the zeroes and coefficients of a cubic polynomial we have, 

 sum of the roots= {tex}r+s+t=\frac { -\left( 0 \right) }{ 8 } =0{/tex}.........................(i)

Also product of the roots={tex}rst=\frac { -\left( 2008 \right) }{ 8 } =-251{/tex}............(ii)

Now {tex}(r+s)^{ 3 }+(s+t)^{ 3 }+(t+r)^{ 3 }=(-t)^{ 3 }+(-r)^{ 3 }+(-s)^{ 3 }=-\left( t^{ 3 }+r^{ 3 }+s^{ 3 } \right) {/tex}.............(iii)          [using (i)]

But we can get  {tex}r^{ 3 }+s^{ 3 }+t^{ 3 }=\left( r+s+t \right) \left( r^{ 2 }+s^{ 2 }+t^{ 2 }-rs-st-tr \right) +3rst{/tex}

{tex}\Rightarrow r^{ 3 }+s^{ 3 }+t^{ 3 }=0+3\left( -251 \right) =-753{/tex}   [using (i) and (ii)]

Therefore from (iii) we have {tex}(r+s)^{ 3 }+(s+t)^{ 3 }+(t+r)^{ 3 }=-\left( t^{ 3 }+r^{ 3 }+s^{ 3 } \right) =-\left( -753 \right) =753{/tex}

If 1,w,w² are the cube roots of unity then we have {tex}{ \omega }^{ 3 }=1,1+\omega +{ \omega }^{ 2 }=0{/tex}

w²(1+w)³-(1+w²)w= {tex}{ \omega }^{ 2 }{ \left( -{ \omega }^{ 2 } \right) }^{ 3 }-\left( -\omega \right) .\omega ={ \omega }^{ 2 }.{ -\omega }^{ 6 }+{ \omega }^{ 2 }=-{ \omega }^{ 8 }+{ \omega }^{ 2 }=-{ \omega }^{ 3 }.{ \omega }^{ 3 }.{ \omega }^{ 2 }+{ \omega }^{ 2 }=-{ \omega }^{ 2 }+{ \omega }^{ 2 }=0{/tex}   {tex}\\ \left[ \because { \left( { a }^{ m } \right) }^{ n }={ a }^{ mn },{ a }^{ m }.{ a }^{ n }={ a }^{ m+n } \right] {/tex}

Let {tex}{{\rm{S}}_n} = \sum {{{\left( {{\rm{odd\ number}}} \right)}^3}} {/tex}

=>     {tex}{{\rm{S}}_n} = \sum\limits_{k = 1}^n {\left( {2k - 1} \right)} {/tex}                      [S_n = Sum of n odd natural numbers]

=>     {tex}{{\rm{S}}_n} = \sum\limits_{k = 1}^n {\left( {8{k^3} - 12{k^2} + 6x - 1} \right)} {/tex}

=>     {tex}{{\rm{S}}_n} = 8\left( {\sum\limits_{k = 1}^n {{k^3}} } \right) - 12\left( {\sum\limits_{k = 1}^n {{k^3}} } \right) + 6\left( {\sum\limits_{k = 1}^n k } \right) - n{/tex}       [1 + 1 + 1 + ...... to n terms = n]

=>     {tex}{{\rm{S}}_n} = 8.{1 \over 4}{n^2}{\left( {n + 1} \right)^2} - 12.{1 \over 6}n\left( {n + 1} \right)\left( {2n + 1} \right) + 6.{1 \over 2}n\left( {n + 1} \right) - n{/tex}

=>     {tex}{{\rm{S}}_n} = 2{n^2}{\left( {n + 1} \right)^2} - 2n\left( {n + 1} \right)\left( {2n + 1} \right) + 3n\left( {n + 1} \right) - n{/tex}

=>     {tex}{{\rm{S}}_n} = n\left( {n + 1} \right)\left[ {2n\left( {n + 1} \right) - 2\left( {2n + 1} \right) + 3} \right] - n{/tex}

=>     {tex}{{\rm{S}}_n} = n\left( {n + 1} \right)\left[ {2{n^2} - 2n + 1} \right] - n{/tex}

=>     {tex}{{\rm{S}}_n} = n\left[ {\left( {n + 1} \right)\left( {2{n^2} - 2n + 1} \right) - 1} \right]{/tex}

=>     {tex}{{\rm{S}}_n} = n\left[ {2{n^3} - 2{n^2} + n + 2{n^2} - 2n + 1 - 1} \right]{/tex}

=>     {tex}{{\rm{S}}_n} = n\left[ {2{n^3} - n} \right]{/tex}

=>     {tex}{{\rm{S}}_n} = {n^2}\left[ {2{n^2} - 1} \right]{/tex}

In Relation, each input has only one output whereas a Function is a relation in which no input relates to more than one output.

Let the linear function from Z into Z be {tex}f\left( x \right) =ax+b{/tex}

Given {tex}f\left( 1 \right) =1,f\left( 2 \right) =3,f\left( 0 \right) =-1,f\left( -1 \right) =-3{/tex}

Now {tex}f\left( 1 \right) =1\Rightarrow a+b=1\\ f\left( 0 \right) =-1\Rightarrow 0+b=-1\Rightarrow b=-1{/tex}

Substituting {tex}b=-1{/tex} in a+b=1 we get a-1 =1 {tex}\Rightarrow{/tex}a=1+1=2

So we have  a = 2 and {tex}b=-1{/tex}

Hence  the linear function from Z into Z is {tex}f\left( x \right) =2x-1{/tex}

Given: {tex}3\tan {\rm{A}}\tan {\rm{B}} = 1{/tex}

=>     {tex}{{3\sin {\rm{A}}\sin {\rm{B}}} \over {\cos {\rm{A}}\cos {\rm{B}}}} = 1{/tex}

=>     {tex}3\sin {\rm{A}}\sin {\rm{B}} = \cos {\rm{A}}\cos {\rm{B}}{/tex}

=>     {tex}3\left[ {{1 \over 2}\left\{ {\cos \left( {{\rm{A}} - {\rm{B}}} \right) - \cos \left( {{\rm{A}} + {\rm{B}}} \right)} \right\}} \right] = {1 \over 2}\left[ {\cos \left( {{\rm{A}} - {\rm{B}}} \right) + \cos \left( {{\rm{A}} + {\rm{B}}} \right)} \right]{/tex}

=>     {tex}3\cos \left( {{\rm{A}} - {\rm{B}}} \right) - 3\cos \left( {{\rm{A}} + {\rm{B}}} \right) = \cos \left( {{\rm{A}} - {\rm{B}}} \right) + \cos \left( {{\rm{A}} + {\rm{B}}} \right){/tex}

=>     {tex} - 4\cos \left( {{\rm{A}} + {\rm{B}}} \right) = - 2\cos \left( {{\rm{A}} - {\rm{B}}} \right){/tex}

=>     {tex}2\cos \left( {{\rm{A}} + {\rm{B}}} \right) = \cos \left( {{\rm{A}} - {\rm{B}}} \right){/tex}

Hence proved.

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