Tag: theory of equations

Quizzes (6) Questions (104)

Questions Related to theory of equations

Multiple choice reciprocal equations theory of equations maths

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

  1. $a=d,b=c$
  2. $a=c,b=d$
  3. $a=-d,b=-c$
  4. $a=-c,b=-d$
Reveal answer Fill a bubble to check yourself
A Correct answer
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$

Multiple choice reciprocal equations theory of equations maths

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

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given f(x) = 1 + integral(0 to x) t^2 * f(t) dt. Differentiating both sides gives f'(x) = x^2 * f(x). This is a separable differential equation: df/f = x^2 dx. Integrating gives ln|f(x)| = x^3/3 + C. Since f(0) = 1, we find C = 0, so f(x) = exp(x^3/3). We want to solve exp(x^3/3) = x^2 + 1. At x=0, both sides are 1. For x > 0, exp(x^3/3) grows much faster than x^2 + 1. For x < 0, exp(x^3/3) < 1 while x^2 + 1 > 1. Thus, only x=0 is a solution.

Multiple choice reciprocal equations theory of equations maths

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

  1. $\dfrac x 2$ + $\dfrac yb$ = $1$
  2. $ax + by = 1$
  3. $ax + by = ab$
  4. $ax = by = 1$
Reveal answer Fill a bubble to check yourself
B Correct answer
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= $\left(y-\cfrac {1}{b}\right)=\left(\cfrac {\cfrac {1}{b}-0}{0-\cfrac {1}{a}}\right)$
$\Rightarrow \left(\cfrac {1}{a}\right)\left(y-\cfrac {1}{b}\right)=\cfrac {1}{b}x$
$\Rightarrow \cfrac {-y}{a}+\cfrac {1}{ab}= \cfrac {x}{b}$
$\Rightarrow \cfrac {x}{b}+\cfrac {y}{a}=\cfrac {1}{ab}$
$\Rightarrow ax+by=1$

Multiple choice reciprocal equations theory of equations maths

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

  1. $a \in R$
  2. $a \in [-1, 1]$
  3. $a \in \left[0, \dfrac{1}{2}\right]$
  4. $a \in \left[-\dfrac{1}{2}, \dfrac{1}{2}\right]$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice reciprocal equations theory of equations maths

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

  1. $x=1$
  2. $x=-1$
  3. $x=\pm1$
  4. $x=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
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. 
Multiple choice reciprocal equations theory of equations maths

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

  1. $x+1$
  2. $x-1$
  3. $x^{2}-1$
  4. $x^{2}$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

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

Multiple choice reciprocal equations theory of equations maths

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

  1. $5x^2+3x-2=0$
  2. $2x^2+3x-5=0$
  3. $3x^2-3x+2=0$
  4. $2x^2+5x -3 = 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $x^2-(\alpha+\beta)x+\alpha\beta=0$

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

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

Multiple choice reciprocal equations theory of equations maths

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

  1. $x= 1$
  2. $x=-1$
  3. $x=\pm1$
  4. $x=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
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. 
Multiple choice reciprocal equations theory of equations maths

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

  1. reciprocal equation of first type

  2. reciprocal equation of second type

  3. reciprocal equation

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

Multiple choice reciprocal equations theory of equations maths

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:

  1. $\mathrm{x}-2$
  2. $\mathrm{x}-1$
  3. $\mathrm{x}$
  4. $\mathrm{x}+1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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