Tag: maths

Questions Related to maths

Is the following quadratic polynomial reducible or irreducible?
$f(x) = -2x^2-2x-1$

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

  1. Reducible with one real root

  2. Reducible with two real roots

  3. Irreducible

  4. None of these

Show answer
Correct Option: C
Explanation:

To check whether the given quadratic polynomial is reducible or irreducible, we need to calculate the discriminant

Calculate the discriminant for the equation, $-2x^2-2x-1=0$

$D=b^{2}-4ac=(-2)^2-4(-2)(-1)=-4<0$
Quadratic equation is irreducible if $D<0$
$\therefore$ The quadratic polynomial is irreducible.

Correct option is C

If $\dfrac {1}{x}-\dfrac {1}{y}=\dfrac {1}{z}$, then z is equal to

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

  1. y-x

  2. x-y

  3. $\dfrac {y-x}{xy}$

  4. $\dfrac {xy}{y-x}$

Show answer
Correct Option: D
Explanation:

$\dfrac {1}{x}-\dfrac {1}{y}=\dfrac {1}{z}$ or $\dfrac {y-x}{xy}=\dfrac {1}{z}$ or $\dfrac {z}{1}=\dfrac {xy}{y-x}$

Which of the following quadratics is irreducible?

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

  1. $2x^2 - 5x + 3$

  2. $2x^2 - 5x - 3$

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

  4. $5x^2 - 2x - 3$

Show answer
Correct Option: C
Explanation:

We can check by comparing the D of quadratic equations with their relation with 0


$2x^2-5x+3=0$
$D=(-5)^2-4(2)(3)=25-24=1>0$
Reducible

$2x^2-5x-3=0$
$D=(-5)^2-4(2)(-3)=25+24=49>0$
Reducible

$5x^2-2x+3=0$
$D=(-2)^2-4(5)(3)=4-60=-56<0$
Irreducible


$5x^2-2x-3=0$
$D=(-2)^2-4(5)(-3)=4+60=64>0$
Rreducible


Therefore correct option is C


x is ........... variable

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

  1. Dependent

  2. Independent

  3. None

  4. both

Show answer
Correct Option: B
Explanation:

Value of x does not depend on any other variable. so it is independent variable.

When multiplicity of a polynomial exist?

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

  1. when a factor appears in conjugate

  2. when a factor appears more than once

  3. when a factor appears more than twice

  4. None of the above

Show answer
Correct Option: B
Explanation:

"multiplicity of a polynomial exist when a factor of a polynomial repeats more than once"

For eg. In the polynomial $(x -2)^3(x-3)^2(x-1 )$ 
The root $2$ have multiplicity $3$, the root  $3$ have $2$, and the root $1$ have $1$
Hence, B is correct.

Let $R=gS-4$. When $S=8,R= 16$. When $S= 10$, R is

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

  1. 11

  2. 14

  3. 21

  4. 20

Show answer
Correct Option: C
Explanation:

$R=gS-4, S=8, R=16$
or $16=g\times 8-4$ or $16+4=g\times 8$
or $g=\frac {20}{8}=\frac {5}{2}$
Therefore, when $S=10$
$R=\frac {5}{2}\times 10-4=21$

A three-digit number beginning from the left is abc. The number is

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

  1. cba

  2. a+10b+100c

  3. 100a+10b+c

  4. none of these

Show answer
Correct Option: C
Explanation:

Here, c is the digit in the units' place; b standing in the tens' place represents b tens; similarly, a represents a hundreds.
The number is, therefore, equal to a hundreds + b tens + c units $= 100a + 10b + c$.

Condition for an irreducible quadratic equation is-

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

  1. discriminant is positive

  2. discriminant is negative

  3. discriminant is zero

  4. None of the above

Show answer
Correct Option: B
Explanation:

Irreducible quadratic equation can not be reduced more i.e the quadratic equation which do not have real roots   

This means the roots are imaginary
So if roots are imaginary, then discriminant $D  <  0$
Hence, option B is correct.

The factorized form of $x^5 +x^4-x-1$ is

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

  1. $(x+1)^2(x+1)^2(x-1)$

  2. $(x^2+1)(x^2+1)^2(x-1)$

  3. $(x^2+1)(x+1)^2(x-1)$

  4. $(x^2+1)^2(x-1)$

Show answer
Correct Option: C
Explanation:

$x^5 +x^4-x-1$
$=x^4(x + 1)-1(x + 1)$
$=(x +1)(x^4-1)$
$= (x + 1) (x^2+ 1) (x^2-1)$
$ = (x + 1) (x^2+ 1) (x + 1) (x -1)$
$ = (x^2+ 1) (x + 1)^2(x -1)$

Solve the simultaneous equations using the convergent iterations:
$5x$ + $y$ + $2z$ = $19$
$2x$ + $3y$ +$8z$ = $39$
$x$ + $4y$ -$2z$ = $-2$

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

  1. $x=2$, $y=2$ and $\text z =1$

  2. $x=2$, $y=1$ and $\text z =4$

  3. $x=3$, $y=2$ and $\text z =2$

  4. None of these

Show answer
Correct Option: B
Explanation:

$5x+y+2z=19$

$x+4y-2z=-2$

Adding above two equations we get,
$5x+y+2z+x+4y-2z=19-2$
$\implies 6x+5y=17$

Taking another equation 
$2x+3y+8z=39$

Multiplying the given last equation with $4$
$4x+16y-8z=-8$

Adding above two equations 
$2x+3y+8z+4x+16y-8z=31$
$\implies 6x+19y=31$

Now,
$6x+19y-(6x+5y)=31-17$
$\implies 19y-5y=14$
$\implies y=1$

By substituting the value of $y$ into the above equations 
we get,
$x=2$ and $z=4$
$\therefore\ x=2,y=1$ and $z=4$.