Given equation is $3x^4-5x^3+3x^2-4x+5=0$ .... $(i)$

We have,

Given equation is $\dfrac {3}{12}=\dfrac {3}{2x}$

$\sqrt{1+(\dfrac{x^4}{-2x^2})^2}$

Given equation is $2x^4-9x^3+14x^2-9x+2=0$ .... $(i)$

Reciprocal is said to be divide $1$ by a number. Reciprocal equation involves reciprocal of the given number.

Given reciprocal equation can be written as

This is the reciprocal function:

$\dfrac{(2{x}-1)(x-1)^{4}(x-2)^{4}}{(x-2)(x-4)^{4}} \le 0\implies x\neq 2$

A reciprocal equation is an equation whose roots can be divided into pairs of numbers, each the reciprocal of the other. 

(Originally) an equation whose roots can be divided into pairs of numbers, each the reciprocal of the other is called Reciprocal equation

We can see that in the giving equation, the multiplication of roots is $1$ 

We know that in reciprocal equation, if one value of $x$ is $a$ then the other value of $x$  will be $\dfrac{1}{a}$

The range is defined as the set of all output values or the set of all those possible values which appear on Y axis in the graph of given function. 

Let ${ x }^{ \cfrac { 1 }{ 3 }  }=t\ { t }^{ 2 }+t-2=0\ { t }^{ 2 }+2t-t-2=0\ t(t+2)-(t+2)=0\ (t-1)(t+2)=0\ t=1\ x^{ \cfrac { 1 }{ 3 }  }=1\ x=1\ t=-2\ x^{ \cfrac { 1 }{ 3 }  }=-2\ x=-8\ x=\left( -8,1 \right)$

Reciprocal equation is the equation which have even numbers of roots and if one root is $x$ then the other root will be $\dfrac{1}{x}$ and the multiplication of all roots will be one.

To be reciprocal equation, the multiplication of the roots $(\dfrac{e}{a})$ should be $1$ 

Here, the coefficients are not palindromic because the first and and last coefficients are opposite in sign , same is the case for second last and second coefficient.

The highest power of $x $ in this equation is $4$, so this is a $4th$ order equation.

$y=\cfrac { { e }^{ x }-{ e }^{ -x } }{ { e }^{ x }+{ e }^{ -x } }$

If $ax^3+bx^2+cx+d$ is a reciprocal equation of the first type,

We know that

$a _{r}=a _{n-r}$ where $a _{n}$ are the coefficient of the equation $f(x)$

So, as $f(x)=ax^3+bx^2+cx+d$

$a=d$ and $ b=c$

Let A & B be the part of intersection of line with X & Y axis respectively.

The equation will have 2 solutions, 1 and -1.
The solution will be x+1 and x-1,
$(x-1)(x+1)=x^2-1$

$2x^2-3x-5=0$

An equation whose roots can be divided into pairs of numbers, each the reciprocal of the other or an equation which is unchanged if the variable is replaced by its reciprocal is known as reciprocal euation.

When the reciprocal equation is of an odd degree, $x=-1$ is always a solution.
So, $x+1$ is a factor.

When the reciprocal equation is of an odd degree, second type, $x=1$ is always a solution.

$x^4-3x^3+4x^2-3x+1=0$
This equation is resiprocal equation of first type as $a _{n-i}=a _i$
Dividing equation by $x^2$:
$x^2-3x+4-\dfrac3x+\dfrac{1}{x^2}=0$
$x^2+\dfrac{1}{x^2}-3x-\dfrac3x+4=0$
$\left(x+\dfrac1x\right)^2-2-3\left(x+\dfrac1x\right)+4=0$
$\left(x+\dfrac1x\right)^2-3\left(x+\dfrac1x\right)+2=0$
Let $x+\dfrac1x=y$
$y^2-3y+2=0$
$(y-1)(y-2)=0$
$y=1$ or $y=2$
For $y=1$:
 $x+\dfrac1x=1$
$x^2-x+1=0$
$D=(-1)^2-4(1)(1)=-3<0$
Hence no real value of x exists for this case.
For $y=2$:
$x+\dfrac1x=2$
$x^2-2x+1=0$
$(x-1)^2=0$
$x=1$
Hence solution of the given equation is x=1.

Given, $(1-a^2)(x+a)-2a(1-x^2)=0$

We see that it is a reciprocal equation, so we divide it by $x^2$: 

As 'The coefficients from beginning to end and vice versa are the same'
therefore given equation is a reciprocal equation.

$\cfrac{8\sqrt{x-5}}{3x-7}=\cfrac{\sqrt{3x-7}}{x-5}$

Given, $2^{x^2}:2^{2x}=8:1$

Given, $x^{-2}-2x^{-1}=8$

Given equation is $\sqrt { 4{ x }^{ 2 }-7x-15 } =\sqrt { { x }^{ 2 }-9 } +\sqrt { { x }^{ 2 }-3x } $