Tag: reciprocal equations

Questions Related to reciprocal equations

A ............ equation is one which remains the same when $x$ is replaced by $\dfrac{1}{x}$.

reciprocal equations theory of equations maths

  1. Reciprocal equation

  2. Radical equation

  3. Exponential equation

  4. Linear equation

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Correct Option: A
Explanation:

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

(Equivalently) an equation which is unchanged if the variable $x$ is replaced by its reciprocal $\dfrac{1}{x}$ is known as reciprocal equation.
Hence, option A is correct.

The roots of equation $2x^4-9x^3+14x^2-9x+2=0$ are

reciprocal equations theory of equations maths

  1. $(1,2,3,4)$

  2. $\left(1,1,\dfrac{1}{2},2\right)$

  3. $\left(1,\dfrac{1}{3},3,1\right)$

  4. $\left(0,1,1,\dfrac{1}{2}\right)$

Show answer
Correct Option: B
Explanation:

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

i.e multiplication of roots $=\dfrac{e}{a}$ in the equation $ax^4 +bx^3 + cx^2 +dx  +e  =0$  
$\Rightarrow $ $\dfrac{e}{a}  = \dfrac{2}{2}  = 1 $
Now, sum of the roots is $\dfrac{-b}{a}  = -(\dfrac{-9}{2}) =  \dfrac{9}{2}$

$\Rightarrow $ it is only possible in option B.
Hence, option B is correct.

The domain of reciprocal equation is :

reciprocal equations theory of equations maths

  1. $R$

  2. $R-{0}$

  3. $Q$

  4. $R^+$

Show answer
Correct Option: B
Explanation:

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

But, if $a = 0$ then $\dfrac{1}{a}$ does not exist  

Therefore, $0$ can not be in the domain of a reciprocal equation.
So, domain of reciprocal equation will be set of all real numbers excluding $0$.
Hence, option B is correct.

The range of reciprocal equation is:

reciprocal equations theory of equations maths

  1. $R$

  2. $R-{0}$

  3. $R^+$

  4. $Q$

Show answer
Correct Option: B
Explanation:

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. 

The domain and range of a reciprocal function are all the real number except for zero. 
This is because reciprocal of $0$ i.e. is undefined.
The range of reciprocal equation is $R-0$

The solution set of the equation 
$x^{2/3} + x^{1/3} = 2 $ is

reciprocal equations theory of equations maths

  1. ${-8, 1}$

  2. ${8, 1}$

  3. ${1, -1}$

  4. ${2, -2}\$

Show answer
Correct Option: A
Explanation:

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

Identify which of the following are reciprocal equations of 1st type.

reciprocal equations theory of equations maths

  1. $2x^4+5x^3+2x^2+5x-2=0$

  2. $2x^4-5x^3+2x^2-5x+2=0$

  3. $2x^4-5x^3+2x^2+5x-2=0$

  4. None of the above

Show answer
Correct Option: B
Explanation:

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.

Now $1st$ type = where cofficients  $a=e$ in 4th order equation  $ax^4 +bx^3 +cx^{2}  + dx+e = 0 $

In option [A]  $a =2$  and $e = -2$  not $1st$ type

In option [B]  $a =2$  and $e = 2$  this is a $1st$ type reciprocal equation 

In option [C]  $a =2$  and $e = -2$  not a $1st$ type reciprocal equation.
Hence, B is correct.

Identify if the following equation is a reciprocal equation by rearranging.

reciprocal equations theory of equations maths

  1. $2(x^4+1)+89x^2= 56x(x^2+1)$

  2. $2(x^4+1)+89x^2= 56x(x+1)$

  3. $2(x^4+1)+89x^2= 56x^2(x+1)$

  4. None of these

Show answer
Correct Option: D
Explanation:

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

(A)
After rearranging the equation 
$\Rightarrow $   $2x^4 -56x^3 +89x^2 -56x +2 = 0 $
$\Rightarrow $  $\dfrac{e}{a} = \dfrac{2}{2}  = 1 $
So, multiplication of roots is $1$  
Thus, it is an reciprocal equation 

(B)
$2x^4 -33x^2 -56x +2 = 0 $

$\Rightarrow $  $\dfrac{e}{a} = \dfrac{2}{2}  = 1 $

So, multiplication of roots is $1$  
Thus, it is an reciprocal equation 


(C)

$\Rightarrow $   $2x^4 -56x^3 +33x^2 +2 = 0 $

$\Rightarrow $  $\dfrac{e}{a} = \dfrac{2}{2}  = 1 $

So, multiplication of roots is $1$  
Thus, it is an reciprocal equation.

Hence, the answer is option D.

$2x^4-3x^3+7x^2+3x-2=0$ is not a reciprocal equation, because

reciprocal equations theory of equations maths

  1. The coefficients from beginning to end and vice versa are not the same.

  2. All the coefficients of terms are not same

  3. The coefficients from beginning to end and vice versa are same.

  4. None of these

Show answer
Correct Option: A
Explanation:

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 Equation $5x^4-3x^3+7x^2-4x+2=0$ is of the type

reciprocal equations theory of equations maths

  1. Quadratic

  2. Linear

  3. Reciprocal

  4. None

Show answer
Correct Option: D
Explanation:

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

Thus, it is neither linear nor quadratic.
Now to be reciprocal equation the multiplication of roots should be $1$  and in the given equation 
Multiplication of roots is $\dfrac{e}{a}=\dfrac{2}{5}$
So this not a reciprocal equation 
Hence, option D is correct.

The inverse of the function $f(x) = \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}$ is

reciprocal equations theory of equations maths

  1. $\dfrac{1}{2}\ell n\dfrac{{1 + x}}{{1 - x}}$

  2. $\dfrac{1}{2}\ell n\dfrac{{2 + x}}{{2 - x}}$

  3. $\dfrac{1}{2}\ell n\dfrac{{1 - x}}{{1 + x}}$

  4. $2\ell n(1 + x)$

Show answer
Correct Option: A
Explanation:

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

$y=\cfrac { \cfrac { { e }^{ 2x }-1 }{ { e }^{ x } }  }{ \cfrac { { e }^{ 2x }+1 }{ { e }^{ x } }  }$
 $y=\cfrac { { e }^{ 2x }-1 }{ { e }^{ 2x }+1 }$
 for universe replace;$x$ with $y$
$\cfrac { x }{ 1 } =\cfrac { { e }^{ 2y }-1 }{ { e }^{ 2y }+1 }$
 using componendo divodendo
 $\cfrac { x+1 }{ x-1 } =\cfrac { { e }^{ 2y }-1+{ e }^{ 2y }+1 }{ { e }^{ 2y }-1-{ e }^{ 2y }-1\quad  }$
 $\cfrac { x+1 }{ x-1 } =-\cfrac { 2{ e }^{ 2y } }{ 2 }$
 $\cfrac { 1+x }{ 1-x } ={ e }^{ 2y }$
$\left(\cfrac { 1+x }{ 1-x } \right)=2y$
$ y=\cfrac { 1 }{ 2 } ln\left(\cfrac { 1+x }{ 1-x } \right)$