Tag: maths

Quizzes (518) Questions (9324)

Questions Related to maths

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

If the sum of two roots of the equation $x^{4}+px^{3}+qx^{2}+rx+8=0$ is equal to the sum of the other two, then $p^{3}+8r=$

  1. $p^2 - 4pq$
  2. $2pq$
  3. $p^2 - pq$
  4. $4pq$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the roots of the equation be $a,b,c,d$. 
As per the question,
$a+b = c+d$ 
From the theory of polynomials, 
$a+b+c+d = -p$
$ \Rightarrow a+b=c+d= \displaystyle \frac{-p}{2} $

Also,
$ab+ac+ad+bd+bc+cd  = q $
$ \Rightarrow (a+b)(c+d) +ab+cd  =q $
$ \Rightarrow ab+cd = q - \displaystyle \frac{p^2}{4} $

Also, 
$abc+abd+bcd+adc = -r $
$ \Rightarrow ab(c+d) +cd(a+b) = -r $
$ \Rightarrow \displaystyle \frac {-p}{2} (ab+cd) = -r $
$ \Rightarrow \displaystyle \frac {-p}{2} ( q - \displaystyle \frac{p^2}{4} ) = -r $
$ \Rightarrow -4pq + p^3 = -8r $
$ \Rightarrow p^3 + 8r = 4pq $

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 $

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

Which of the following is an irrational number? 

  1. $0.14$
  2. $0.14 \overline{16}$
  3. $1.1 {416}$
  4. $0.4014001400014....$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The decimal expansion of a rational number is either terminating or non-terminating repeating.

(A) $0.14$ is terminating, so it is a rational number

(B) $0.14\bar{16}=0.141616....$ 

is also rational ( non-terminating repeating ), where digits $16$ are repeating.

(C) $0.1416$ is terminating, so it is a rational number

(D) $0.4014001400014.....$

is an irrational number because it is neither terminating nor repeating.

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

State whether the following statement is true or false.

The following number is irrational
$7\sqrt {5}$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
An Irrational Number is a real number that cannot be written as a simple fraction.

Let us assume $7\sqrt{5}$ is rational.
Hence, $7\sqrt{5}$ can be written in form $\dfrac{a}{b}$

Where, $a$ and $b$ $(n\ne 0)$ are co-prime.
Hence, $7\sqrt{5}=\dfrac{a}{b}$
$\Rightarrow$  $\sqrt{5}=\dfrac{1}{7}\times\dfrac{a}{b}$

Here, $\dfrac{a}{7b}$ is a rational number, but $\sqrt{5}$ is irrational.

Since, Rational $\ne$ Irrational
This is a contriadition
$\therefore$  Our assumption is incorrect.
$\therefore$  $7\sqrt{5}$ is irrational number. 
Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

State whether the following statement is true or false.

The following number is irrational
$6+\sqrt {2}$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
An Irrational Number is a real number that cannot be written as a simple fraction.

Let us assume $6+\sqrt{2}$ is rational.
Hence, $6+\sqrt{2}$ can be written in form $\dfrac{a}{b}$

Where, $a$ and $b$ $(n\ne 0)$ are co-prime.
Hence, $6+\sqrt{2}=\dfrac{a}{b}$
$\Rightarrow$  $\sqrt{2}=\dfrac{a}{b}-6$

$\Rightarrow$  $\sqrt{2}=\dfrac{a-6b}{b}$

Here, $\dfrac{a-6b}{b}$ is a rational number, but $\sqrt{2}$ is irrational.

Since, Rational $\ne$ Irrational
This is a contriadition
$\therefore$  Our assumption is incorrect.
$\therefore$  $6+\sqrt{2}$ is irrational number. 
Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

Which of the following is always true 

  1. $irrational + irrational =irrational $
  2. $\dfrac{rational }{rational }=rational $
  3. $\dfrac{integer }{integer}=integer$

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

Counter-example for A: $(\sqrt{2}) + (4-\sqrt{2}) = 4$

Counter-example for C: $\dfrac{1}{2}=0.5$

Proof for B:

Let $q _1, q _2$ be two rational numbers such that $q _2\neq0$.


As they are rational, they can be written as $a/b, c/d$ respectively for some integers $a, b, c, d$. $(b,c,d\neq0)$

$\dfrac{q _1}{q _2}=\dfrac{a/b}{c/d}=\dfrac{ad}{bc}$

Since, $a,b,c,d$ were integers, even $ad$ and $bc(\neq0)$ are integers and therefore, the above expression is rational.