Tag: theory of equations

Quizzes (6) Questions (104)

Questions Related to theory of equations

Multiple choice reciprocal equations theory of equations maths

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

  1. Reciprocal equation

  2. Radical equation

  3. Exponential equation

  4. Linear equation
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice reciprocal equations theory of equations maths

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

  1. $(1,2,3,4)$
  2. $\left(1,1,\dfrac{1}{2},2\right)$
  3. $\left(1,\dfrac{1}{3},3,1\right)$
  4. $\left(0,1,1,\dfrac{1}{2}\right)$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice reciprocal equations theory of equations maths

The domain of reciprocal equation is :

  1. $R$
  2. $R-\{0\}$
  3. $Q$
  4. $R^+$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice reciprocal equations theory of equations maths

The range of reciprocal equation is:

  1. $R$
  2. $R-{0}$
  3. $R^+$
  4. $Q$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $R-0$

Multiple choice reciprocal equations theory of equations maths

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

  1. $\{-8, 1\}$
  2. $\{8, 1\}$
  3. $\{1, -1\}$
  4. $\{2, -2}\$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

Multiple choice reciprocal equations theory of equations maths

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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice reciprocal equations theory of equations maths

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

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

Multiple choice reciprocal equations theory of equations maths

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

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice reciprocal equations theory of equations maths

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

  1. Quadratic

  2. Linear

  3. Reciprocal

  4. None
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

Multiple choice reciprocal equations theory of equations maths

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

  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. $2\ell n(1 + x)$
Reveal answer Fill a bubble to check yourself
A Correct answer
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)$