Tag: complex numbers and linear inequations

Questions Related to complex numbers and linear inequations

Is the following quadratic polynomial reducible or irreducible?
$f(x) = x^2 - \sqrt2$

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

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Correct Option: B
Explanation:

Let $\sqrt 2=a^2\Rightarrow a^2=2^{1/2}\Rightarrow a=2^{1/4}$

So $f(x)=x^2-a^2=(x+a)(x-a)=(x+2^{1/4})(x-2^{1/4})$, using $a^2-b^2=(a+b)(a-b)$
So given quadratic is reducible to two real roots 

If $x=2+\sqrt{3}$, $xy=1$, then $\cfrac { x }{ \sqrt { 2 } +\sqrt { x }  } +\cfrac { y }{ \sqrt { 2 } +\sqrt { y }  } =$.......

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

  1. $\sqrt{2}$

  2. $\sqrt{3}$

  3. $1$

  4. $2$

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Correct Option: A
Explanation:
Given $x=2+\sqrt{3}$ and $xy=1$
$\Rightarrow\,y=\dfrac{1}{x}=\dfrac{1}{2+\sqrt{3}}=\dfrac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$
Let $\sqrt{x}=\sqrt{a}+\sqrt{b}$
then $x=a+b+2\sqrt{ab}$ by squaring both sides
We have $x=2+\sqrt{3}=a+b+2\sqrt{ab}$
$\Rightarrow\,a+b=2,\,\sqrt{ab}=\dfrac{\sqrt{3}}{2}$ or $ab=\dfrac{3}{4}$
$\Rightarrow\,{\left(a-b\right)}^{2}={\left(a+b\right)}^{2}-4ab=4-4\times \dfrac{3}{4}=4-3=1$
$\Rightarrow\,a-b=1$
$\Rightarrow\,a+b=2,\,a-b=1$
$\Rightarrow\,2a=3$
$\Rightarrow\,a=\dfrac{3}{2}$
Put $a=\dfrac{3}{2}$ in $a+b=2$
$b=2-a=2-\dfrac{3}{2}=\dfrac{4-3}{2}=\dfrac{1}{2}$
$\therefore\,a=\dfrac{3}{2},b=\dfrac{1}{2}$
So,$\sqrt{x}=\dfrac{\sqrt{3}+1}{\sqrt{2}}$
$\sqrt{y}=\dfrac{1}{\sqrt{x}}=\dfrac{\sqrt{2}}{\sqrt{3}+1}\\$
$\sqrt{y}=\dfrac{\sqrt{2}}{\sqrt{3}+1}\times\dfrac{\sqrt{3}-1}{\sqrt{3}-1}\\$
$=\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{3-1}=\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{2}\times\dfrac{\sqrt{2}}{\sqrt{2}}\\$
$=\dfrac{2\left(\sqrt{3}-1\right)}{2\sqrt{2}}=\dfrac{\sqrt{3}-1}{\sqrt{2}}\\$
Now substitute in the given expression:
$\dfrac{x}{\sqrt{2}+\sqrt{x}}=\dfrac{\left(2+\sqrt{3}\right)\times\sqrt{2}}{\sqrt{2}\times \sqrt{2}+\sqrt{3}+1}\\$
$=\dfrac{\left(2+\sqrt{3}\right)\times\sqrt{2}}{2+\sqrt{3}+1}=\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{3+\sqrt{3}}\\$
$=\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{3+\sqrt{3}}\times\dfrac{3-\sqrt{3}}{3-\sqrt{3}}=\dfrac{\sqrt{2}\left(3+\sqrt{3}\right)}{6}\\$
$\dfrac{y}{\sqrt{2}-\sqrt{y}}=\dfrac{\left(2+\sqrt{3}\right)\times\sqrt{2}}{\sqrt{2}\times \sqrt{2}-\sqrt{3}+1}\\$
$=\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{3}+1}=\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{3-\sqrt{3}}\\$
$=\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{3-\sqrt{3}}\times\dfrac{3+\sqrt{3}}{3+\sqrt{3}}=\dfrac{\sqrt{2}\left(3-\sqrt{3}\right)}{6}\\$
Now,$\dfrac{x}{\sqrt{2}+\sqrt{x}}+\dfrac{y}{\sqrt{2}-\sqrt{y}}=\dfrac{\sqrt{2}\left(3+\sqrt{3}\right)}{6}+\dfrac{\sqrt{2}\left(3-\sqrt{3}\right)}{6}$
$=\dfrac{\sqrt{2}\left(3+\sqrt{3}+3-\sqrt{3}\right)}{6}$
$=\dfrac{6\sqrt{2}}{6}=\sqrt{2}$

The method of finding solution by trying out various values for the variable is called

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

  1. Error method

  2. Trial and error method

  3. Testing method

  4. Checking method

Show answer
Correct Option: B
Explanation:

$\Rightarrow$  The method of finding solution by trying out various values for the variable is called $Trial\,and\,error\,method.$

$\Rightarrow$  In this method, we often make a guess of the root of the equation.
$\Rightarrow$  We fine the values of L.H.S and R.H.S of the given equations for different values of the variable.
$\Rightarrow$  The values of the variable for which L.H.S = R.H.S is the root of the equation.

If $f\left( {x,y} \right) = \sqrt {{x^2} + {y^2}}  + \sqrt {{{\left( {x - 1} \right)}^2} + {y^2}}  + \sqrt {{x^2} + {{\left( {y - 1} \right)}^2}}  + \sqrt {{{\left( {x - 3} \right)}^2} + {{\left( {y - 4} \right)}^2}} $ where $x,y \in R$, then the minimum value of $f\left( {x,y} \right)$ is

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

  1. $2 + \sqrt 5 $

  2. $5 + \sqrt 2 $

  3. $5 - \sqrt 2 $

  4. $\sqrt 5 - 2$

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Correct Option: A

$|x - 1| + |x + 3| + |x - 5| = k$
How many values does $k$ have.

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

  1. only one solution

  2. two solution

  3. no solution

  4. infinite solutions

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Correct Option: A

Consider the equation $(1 + a + b)^{2} = 3(1 + a^{2} + b^{2})$, where a, b are real numbers.
Then

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

  1. there is no solution pair (a, b)

  2. there are infinitely many solution pairs (a, b)

  3. there are exactly two solution pairs (a, b)

  4. there is exactly one solution pair (a, b)

Show answer
Correct Option: D
Explanation:

$ (1+a+b)^{2} = 3(1+a^{2}+b^{2}) $

$ \Rightarrow (1+a^{2}+b^{2} + 2a + 2b + 2ab) = 3 + 3a^{2} +3b^{2} $
$ \Rightarrow a^{2}+b^{2} -ab-a-b+1 = 0 $
$ \Rightarrow (a-b)^{2} + ((a-1)(b-1)) = ((a-1)-(b-1))^{2} + ((a-1)(b-1)) = 0 $
Let $ a-1 = x $ and $ b-1 = y $

$ \Rightarrow (x-y)^{2} + (xy) = 0 $
$ \Rightarrow x^{2} - xy + y^{2} = 0 $                                 ...($1$)
Assume $ y \neq 0 $ and divide by $ y^{2} $

$ \Rightarrow \left ( \dfrac{x}{y} \right )^{2} - \dfrac{x}{y} + 1 = 0 $
Substitute $ t = \dfrac{x}{y} $
$ \Rightarrow t^{2} - t + 1 = 0 $
Discriminant $ D = 1-4 = -3 < 0 $. No solution here.

Assume $ y = 0 $ in equation ($1$). It satisfies the equation for $ x = 0 $. Hence the solution is $x=0$ and $y=0$

$ \Rightarrow a=1$ and $b=1$

The equations of the plane through the points $(1,-1,2),(-3,2,-2)$ and perpendicular to the plane $x+2y+3z+7=0$ is  

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

  1. $x+16y+11z-7=0$

  2. $17x+8y-11z+13=0$

  3. $x+y+z-2=0$

  4. $x-5y-3z=0$

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Correct Option: B