Tag: maths

Questions Related to maths

 $\sqrt{3}-\sqrt{5}$ is an rational number.

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

  1. True

  2. False

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Correct Option: B

State true or false. 
$\sqrt { 3 } + \sqrt { 4 }$ is an rational number.

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

  1. True

  2. False

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Correct Option: B
Explanation:
A rational number is a number that can be written as a ratio. That means it can be written as a fraction, in which both the numerator (the number on top) and the denominator (the number on the bottom) are whole numbers.
$\Rightarrow$  All numbers that are not rational are considered irrational. An irrational number can be written as a decimal, but not as a fraction.

$\sqrt{3}=1.732$ is an irrational.
$\sqrt{4}=2$ is rational.
Now,
$\Rightarrow$  $ \sqrt{3}+\sqrt{4}=1.732+2$
                        $=3.732$
$3.732$ cannot be converted into fraction.
$\therefore$  $\sqrt{3}+\sqrt{4}$ is an irrational number.
$\therefore$  Given statement is false.

$\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+...}}}}$ up to $\infty$ is?

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

  1. $2$

  2. $3$

  3. $30$

  4. $5$

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Correct Option: B
Explanation:

We have,

$y=\sqrt { 6+\sqrt { 6+....\infty  }  } =\sqrt { 6+y } $

On squaring both sides, we get
$\Rightarrow { y }^{ 2 }-y-6=0$

$\Rightarrow y=3,-2$


But $y$ can not be $-ve$.

Hence, $y=3$ is the answer.

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

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

  1. An irrational number

  2. $0$

  3. A natural number

  4. A prime number

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Correct Option: A

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

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

  1. $2$

  2. $5$

  3. $3$

  4. $7$

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

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

  1. $4$

  2. $16$

  3. $32$

  4. $64$

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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 :

$14+6\sqrt 5$

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

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

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

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

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

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Correct Option: B
Explanation:

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

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

 $\dfrac {\surd 2}{3}$ is irrational number.

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

  1. True

  2. False

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Correct Option: A

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

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

  1. True

  2. False

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Correct Option: A

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

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

  1. True

  2. False

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Correct Option: A