Tag: proofs of irrationality

Questions Related to proofs of irrationality

$\sqrt{5}\left{(\sqrt{5}+1)^{50}-(\sqrt{5}-1)^{50}\right}$ is?

  1. An irrational number

  2. $0$

  3. A natural number

  4. A prime number


Correct Option: A

Find x if $\dfrac{\sqrt{3x+1}+\sqrt{3x-6}}{\sqrt{3x+1}-\sqrt{3x-6}}=7$.

  1. $2$

  2. $5$

  3. $3$

  4. $7$


Correct Option: B
Explanation:
$\dfrac { \sqrt { 3x+1 } +\sqrt { 3x-6 }  }{ \sqrt { 3x+1 } -\sqrt { 3x-6 }  } =7$

Rotational give :-

$\dfrac { \left( 3x+1 \right) +\left( 3x-6 \right) +2\sqrt { \left( 3x+1 \right) \left( 3x-6 \right)  }  }{ \left( 3x+1 \right) -\left( 3x-6 \right)  } =7$

$\Rightarrow 6x-5+2\sqrt { \left( 3x+1 \right) \left( 3x-6 \right)  } =49$

$\Rightarrow 2\sqrt { \left( 3x+1 \right) \left( 3x-6 \right)  } =-6x+54$

$\Rightarrow \sqrt { \left( 3x+1 \right) \left( 3x-6 \right)  } =-3x+27$
which gives, $x=5$

Evaluate $\sqrt[3]{\left(\dfrac{1}{64}\right)^{-2}}$.

  1. $4$

  2. $16$

  3. $32$

  4. $64$


Correct Option: B
Explanation:
$\sqrt[3]{(\frac{1}{64})^{-2}}=\sqrt[3]{(64)^{2}}=(\sqrt[3]{64})^{2}=(4)^{2}=16$     $[a^{b}=(\frac{1}{a})^{-b}]$

Find the square root :

$14+6\sqrt 5$

  1. $\pm (3+\sqrt 7)$

  2. $\pm (3+\sqrt 5)$

  3. $\pm (7+\sqrt 5)$

  4. $\pm (2+\sqrt 5)$


Correct Option: B
Explanation:

$\sqrt{14+6\sqrt 5}$

$=\sqrt{9+5+6\sqrt 5}$
$=\sqrt{(3+\sqrt 5)^2}$
$=\pm (3+\sqrt 5)$

$\dfrac {5+\sqrt {2}}{3}$ is an irrational number.

  1. True

  2. False


Correct Option: A

The simplified form of the expression $\sqrt { \sqrt [ 3 ]{ 729{ x }^{ 12 } }  } -\dfrac { { x }^{ -2 }-{ x }^{ -3 } }{ { x }^{ -4 }-{ x }^{ -5 } } $ is

  1. ${ 3x }^{ 2 }$

  2. ${ 3x }^{ 3 }$

  3. ${ 2x }^{ 2 }$

  4. ${ 4x }^{ 2 }$


Correct Option: C
Explanation:

$\sqrt{(729x^{12})^{\frac{1}{3}}}-\cfrac{x^{-2}-x^{-3}}{x^{-4}-x^{-5}}$


$=\sqrt{(3^6x^{12})^{\frac{1}{3}}}-\cfrac{x^{-2}-x^{-3}}{x^{-4}-x^{-5}}$


$=\sqrt{(3^2x^4)}-\cfrac{x^{-2}(1-x^{-1})}{x^{-4}(1-x^{-1})}$

$=3x^2-x^2$

$\=2x^2$