Tag: maths

Questions Related to maths

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 $p^{3}+8r=$

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

  1. $p^2 - 4pq$

  2. $2pq$

  3. $p^2 - pq$

  4. $4pq$

Show answer
Correct Option: D
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 $

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

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

  1. AP

  2. GP

  3. HP

  4. None of these

Show answer
Correct Option: A
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

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

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

  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. $b^{3}+ac^{2}+a^{2}c+3abc=0$

Show answer
Correct Option: B
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 $

State whether true or false: 

 $3+\sqrt{6}$ is an irrational number.

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

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

AS we know 3 is an irrational number and $\sqrt{6}$ is also a rational number 

Addition of two rational numbers are always a rational number 
So,  3 + $\sqrt{6}$ is also a rational number

Which of the following is an irrational number? 

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

  1. $0.14$

  2. $0.14 \overline{16}$

  3. $1.1 {416}$

  4. $0.4014001400014....$

Show answer
Correct Option: D
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.

Each of the following numbers is irrational
i) $(5 + 3\sqrt{2})$
ii) $3 \sqrt{7}$
iii) $\dfrac{3}{\sqrt{5}}$
iv) $(2 - 3\sqrt{5})$
v) $(\sqrt{3} + \sqrt{5})$

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

  1. True

  2. False

Show answer
Correct Option: A

State whether the following statement is true or false.

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

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

  1. True

  2. False

Show answer
Correct Option: A
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. 

 $2-\sqrt {3}$ is an irrational number.

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

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

$\sqrt{3}$ is an irrational number.

Since, the difference of a rational and irrational number is irrational, hence, $2-\sqrt{3}$ is an irrational number.

State whether the following statement is true or false.

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

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

  1. True

  2. False

Show answer
Correct Option: A
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. 

Which of the following is always true 

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

  1. $irrational + irrational =irrational $

  2. $\dfrac{rational }{rational }=rational $

  3. $\dfrac{integer }{integer}=integer$

  4. None of these

Show answer
Correct Option: B
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.