Tag: maths

Questions Related to maths

The root(s) of the reciprocal equation of second type and of even degree is/are

reciprocal equations theory of equations maths

  1. $x=1$

  2. $x=-1$

  3. $x=\pm1$

  4. $x=0$

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Correct Option: C
Explanation:
$x = -1$ is a root of the reciprocal equation of first type and of odd degree. 
$x = 1$ is a root of the reciprocal equation of second type and of odd degree.
$x=±1$ are two roots of reciprocal equation of second type and of even degree. 

lf $\mathrm{f}({x})=0$ is a reciprocal equation of second type and even degree, then a factor of $\mathrm{f}({x})$  is:

reciprocal equations theory of equations maths

  1. $x+1$

  2. $x-1$

  3. $x^{2}-1$

  4. $x^{2}$

Show answer
Correct Option: A,B,C
Explanation:

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

The equation whose roots are the reciprocal of the roots of $2x^2 - 3x -5=0$, is:

reciprocal equations theory of equations maths

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

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

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

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

Show answer
Correct Option: A
Explanation:

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


$\Rightarrow (x+1)(2x-5)=0$

$\therefore \alpha=-1$ and $\beta=\dfrac{5}{2}$

Reciprocal of these roots, $\alpha=-1$ and $\beta=\dfrac{2}{5}$

General form of quadratic equation $x^2-(\alpha+\beta)x+\alpha\beta=0$

$\Rightarrow x^2-(-1+\dfrac{2}{5})x-\dfrac{2}{5}=0$

$\Rightarrow 5x^2+3x-2=0$

The root of the reciprocal equation of first type and of odd degree is:

reciprocal equations theory of equations maths

  1. $x= 1$

  2. $x=-1$

  3. $x=\pm1$

  4. $x=0$

Show answer
Correct Option: B
Explanation:
$x = -1$ is a root of the reciprocal equation of first type and of odd degree. 
$x = 1$ is a root of the reciprocal equation of second type and of odd degree.
$x=±1$ are two roots of reciprocal equation of second type and of even degree. 

If the reciprocal of every root of an equation is also a root of it, then the equation is said to be a

reciprocal equations theory of equations maths

  1. reciprocal equation of first type

  2. reciprocal equation of second type

  3. reciprocal equation

  4. None of these

Show answer
Correct Option: C
Explanation:

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.

lf $\mathrm{f}(\mathrm{x})=0$ is a reciprocal equation of first type and odd degree, then a factor of $\mathrm{f}(\mathrm{x})$ is:

reciprocal equations theory of equations maths

  1. $\mathrm{x}-2$

  2. $\mathrm{x}-1$

  3. $\mathrm{x}$

  4. $\mathrm{x}+1$

Show answer
Correct Option: D
Explanation:

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

The root of the reciprocal equation of second type and of odd degree is:

reciprocal equations theory of equations maths

  1. $x=-1$

  2. $x=+1$

  3. $x=\pm1$

  4. $x=0$

Show answer
Correct Option: B
Explanation:
$x = -1$ is a root of the reciprocal equation of first type and of odd degree. 
$x = 1$ is a root of the reciprocal equation of second type and of odd degree.
$x=±1$ are two roots of reciprocal equation of second type and of even degree. 

lf $\mathrm{f}({x})=0$ is a reciprocal equation of second type and fifth degree, then a root of $\mathrm{f}({x})=0$  is:

reciprocal equations theory of equations maths

  1. $0$

  2. $1$

  3. $-1$

  4. $2$

Show answer
Correct Option: B
Explanation:

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

The roots equation $x^4-3x^3+4x^2-3x+1=0$ is 

reciprocal equations theory of equations maths

  1. $0$

  2. $1$

  3. $2$

  4. $3$

Show answer
Correct Option: B
Explanation:

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

If the coefficients from one end of an equation are equal in magnitude and sign to the coefficients from the other end, then the equation is said to be 

reciprocal equations theory of equations maths

  1. reciprocal equation of second type

  2. reciprocal equation of first type

  3. reciprocal equation

  4. None of these

Show answer
Correct Option: B
Explanation: