Tag: algebraic functions, equations and inequalities

Quizzes (2) Questions (62)

Questions Related to algebraic functions, equations and inequalities

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

Suppose $f(x) =3x^3-13x^2+14x-2$, it is assumed that $f(x)=0$ will have 3 root say $\alpha, \beta$ and $\gamma$, where $\alpha < \beta < \gamma$

$[\alpha], [\beta], [\gamma]$ (where, [-] denotes the greatest function) will be in

  1. AP

  2. GP

  3. HP

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

$f\left( x \right) =3{ x }^{ 3}-13{ x }^{ 2 }+14x-2$


$f\left( 0 \right) =-2=-ve$

$f\left( 1 \right) =3-13+14-2=2=+ve$

$f\left( 2 \right) =24-52+28-2=-2=-ve$

$f\left( 3 \right) =81-117+42-2=4=+ve$

$\therefore $ One root lies between 0 & 1

One root lies between 1 & 2

One root lies between 2 & 3

$\therefore \alpha \in \left( 0,1 \right) \Rightarrow \left[ \alpha  \right] =0$

$\beta \in  \left( 1,2 \right) \Rightarrow \left[ \beta  \right] =1$

$\gamma \in \left( 2,3 \right) \Rightarrow \left[ \gamma  \right] =2$

$\left[ \alpha  \right] ,\left[ \beta  \right] ,\left[ \gamma  \right] $ are in AP

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

lf one root of the equation $ax^{2}+bx+c=0$ is the square of the other, then

  1. $b^{2}+ac^{2}+a^{2}c=3abc$
  2. $b^{3}+ac^{2}+a^{2}c=3abc$
  3. $b^{2}+ac^{2}+a^{2}c+3abc=0$
  4. $b^{3}+ac^{2}+a^{2}c+3abc=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given equation $a{ x }^{ 2 }+bx+c$
Given that, one root of the equation is square of another.
So, lets assume $\alpha$ , ${ \alpha  }^{ 2 }$ are roots of the given equation
We know that,
Sum of roots $=$ $\alpha +{ \alpha  }^{ 2 }=\dfrac { -b }{ a }$ 
Product of roots $=$ $ \alpha \times { \alpha  }^{ 2 }=\dfrac { c }{ a }$
$\alpha (1+\alpha )=\dfrac { -b }{ a } \longrightarrow 1  $
${ \alpha  }^{ 3 }=\dfrac { c }{ a } \longrightarrow 2 $
Cubing equation (1) on both sides and substitute the value from equation (2).
${ \alpha  }^{ 3 }{ (1+\alpha ) }^{ 3 }=\dfrac { -{ b }^{ 3 } }{ { a }^{ 3 } } \ { \alpha  }^{ 3 }({ \alpha  }^{ 3 }+1+3{ \alpha  }(1+\alpha ))=\dfrac { -{ b }^{ 3 } }{ { a }^{ 3 } } \ \dfrac { c }{ a } \left (\dfrac { c }{ a } +1+3\left (\dfrac { -b }{ a } \right)\right)=\dfrac { -{ b }^{ 3 } }{ { a }^{ 3 } } \ \dfrac { ({ c }^{ 2 }+ac-3bc) }{ { a }^{ 2 } } =\dfrac { -{ b }^{ 3 } }{ { a }^{ 3 } } \ a({ c }^{ 2 }+ac-3bc)=-{ b }^{ 3 }\ { b }^{ 3 }+a{ c }^{ 2 }+{ a }^{ 2 }c=3abc $