Tag: fundamental theorem of algebra

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

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

  1. $\sqrt{2}$
  2. $\sqrt{3}$
  3. $1$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
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,$\dfrac{x}{\sqrt{2}+\sqrt{x}}+\dfrac{y}{\sqrt{2}-\sqrt{y}}=\dfrac{\sqrt{2}\left(3+\sqrt{3}\right)}{6}+\dfrac{\sqrt{2}\left(3-\sqrt{3}\right)}{6}$
$=\dfrac{\sqrt{2}\left(3+\sqrt{3}+3-\sqrt{3}\right)}{6}$
$=\dfrac{6\sqrt{2}}{6}=\sqrt{2}$
Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

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

  1. Error method

  2. Trial and error method

  3. Testing method

  4. Checking method
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  1. $2 + \sqrt 5 $
  2. $5 + \sqrt 2 $
  3. $5 - \sqrt 2 $
  4. $\sqrt 5 - 2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The function represents the sum of distances from (x, y) to four points: (0, 0), (1, 0), (0, 1), and (3, 4). This is a Fermat point problem. The minimum distance sum for these points is found by connecting the diagonals, resulting in 2 + sqrt(5).

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

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

  1. only one solution

  2. two solution

  3. no solution

  4. infinite solutions
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The function f(x) = |x - 1| + |x + 3| + |x - 5| is a continuous, strictly increasing function for x > 5 and strictly decreasing for x < -3. It has a minimum value at x = 1. Depending on the value of k, there is either one solution, no solution, or a range of solutions.

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

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

  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)
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $b=1$

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

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  

  1. $x+16y+11z-7=0$
  2. $17x+8y-11z+13=0$
  3. $x+y+z-2=0$
  4. $x-5y-3z=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The normal vector to the plane is the cross product of the vector between the two points and the normal vector of the given plane. Points: A(1, -1, 2), B(-3, 2, -2). Vector AB = (-4, 3, -4). Normal to given plane = (1, 2, 3). Cross product = (17, 8, -11). Equation: 17(x - 1) + 8(y + 1) - 11(z - 2) = 0, which simplifies to 17x + 8y - 11z + 13 = 0.