Tag: theory of equations

Questions Related to theory of equations

If $ax^{3}+bx^{2}+cx+d=0$ is a reciprocal equation of the first type, then 

reciprocal equations theory of equations maths

  1. $a=d,b=c$

  2. $a=c,b=d$

  3. $a=-d,b=-c$

  4. $a=-c,b=-d$

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Correct Option: A
Explanation:

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$

If $f(x)=1+\displaystyle \int^{x} _{0}t^{2}f(t)dt$, then the number of solution of $f(x)=x^{2}+1$ is

reciprocal equations theory of equations maths

  1. $1$

  2. $2$

  3. $3$

  4. $4$

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Correct Option: A

The equation of the line, reciprocal of whose intercepts on the axes are $a$ and $b$ given by

reciprocal equations theory of equations maths

  1. $\dfrac x 2$ + $\dfrac yb$ = $1$

  2. $ax + by = 1$

  3. $ax + by = ab$

  4. $ax = by = 1$

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Correct Option: B
Explanation:

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

$\Rightarrow A= \left(\cfrac {1}{a},0\right)$
$\Rightarrow B= \left(0,\cfrac {1}{b}\right)$
$\therefore$ Equation of line= $\left(y-\cfrac {1}{b}\right)=\left(\cfrac {\cfrac {1}{b}-0}{0-\cfrac {1}{a}}\right)$
$\Rightarrow \left(\cfrac {1}{a}\right)\left(y-\cfrac {1}{b}\right)=\cfrac {1}{b}x$
$\Rightarrow \cfrac {-y}{a}+\cfrac {1}{ab}= \cfrac {x}{b}$
$\Rightarrow \cfrac {x}{b}+\cfrac {y}{a}=\cfrac {1}{ab}$
$\Rightarrow ax+by=1$

The equation $\sin^{-1}x-3\sin^{-1}a=0$ has real solutions for x if?

reciprocal equations theory of equations maths

  1. $a \in R$

  2. $a \in [-1, 1]$

  3. $a \in \left[0, \dfrac{1}{2}\right]$

  4. $a \in \left[-\dfrac{1}{2}, \dfrac{1}{2}\right]$

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Correct Option: A

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. $x^{2}$

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

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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

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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. $\mathrm{x}+1$

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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.