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#### Which of the following is equal to x?(A)$x^{\frac{12}{7}}+x^{\frac{5}{7}}$(B)$\sqrt[12]{\left ( x^{4} \right )^{\frac{1}{3}}}$(C)$\left ( \sqrt{x^{3}} \right )^{\frac{2}{3}}$(D)$x^{\frac{12}{7}}\times x^{\frac{7}{12}}$

Solution.
(A) We have,

$x^{\frac{12}{7}}+x^{\frac{5}{7}}= x^{\frac{1}{7}\left ( 12 \right )}+x^{\frac{1}{7}\left ( 5 \right )}$
$= x^{\frac{1}{7}}\left ( x^{12} +x^{5}\right )\neq x$
(B) We have,

$\sqrt[12]{\left ( x^{4} \right )^{\frac{1}{3}}}=\left ( \left ( x^{4} \right ) ^{\frac{1}{3}}\right )^{\frac{1}{12}}$ $\left ( \sqrt[n]{a} = \left ( a \right )^{\frac{1}{n}}\right )$

$= x^{4\times ^{\frac{1}{3}\times \frac{1}{12}}}= x^{\frac{1}{9}}$             $\because \left ( a^{m} \right )^{n}= a^{m\times n}$

$\neq x$

(C) We have,

$\left ( \sqrt{x^{3}} \right )^{\frac{2}{3}}= \left ( \left ( x^{3} \right ) ^{\frac{2}{3}}\right )^{\frac{1}{2}}$       $\left ( \sqrt[n]{a} = \left ( a \right )^{\frac{1}{n}}\right )$

$= x^{3\times \frac{2}{3}\times \frac{1}{2}}$         $\because \left ( a^{m} \right )^{n}= a^{m\times n}$

= x

(D) We have,

$x^{\frac{12}{7}}\times x^{\frac{7}{12}}= x^{\frac{12}{7}+\frac{7}{12}}= x^{\frac{144+49}{84}}\neq x$
$\because a^{m}\times a^{n}= a^{m+n}$

Hence option C is correct.

#### Value of (256)0.16 × (256)0.09 is :(A) 4 (B) 16 (C) 64 (D) 256.25

Solution.
We have,

(256)0.16 × (256)0.09           [$\because$ am × an = am+n]

= (256)0.16 + 0.09
= (256)0.25

= $\left ( 256 \right )^{\frac{25}{100}}$

= $\left ( 256 \right )^{\frac{1}{4}}$

Now 256 = 28 = (22)4 = 44

= (44)1/4              [$\because$ (am)n = amn]

= 4

Hence option A is correct.

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#### The value of $\sqrt[4]{\left ( 81 \right )^{-2}}$is :(A)$\frac{1}{9}$ (B)$\frac{1}{3}$ (C)9 (D)$\frac{1}{81}$

Solution.

We have, $\sqrt[4]{\left ( 81 \right )^{-2}}$
We know that $\sqrt[n]{a}= \left ( a \right )^{\frac{1}{n}}$
So,
$\sqrt[4]{\left ( 81 \right )^{-2}}= \left ( \left ( 81 \right )^{-2} \right )^{\frac{1}{4}}$

$=\left ( 81 \right )^{-2\times \frac{1}{4}}$                    $\because \left ( a^{m} \right )^{n}= a^{m\times n}$

$= 81^{-\frac{1}{2}}= \left ( \frac{1}{81} \right )^{\frac{1}{2}}$            $\because \left ( a^{-m}= \left ( \frac{1}{a} \right )^{m} \right )$

$= \sqrt{\frac{1}{81}}$
$=\frac{1}{9}$

Hence option A is correct

#### The product $\sqrt[3]{2}.\sqrt[4]{2}.\sqrt[12]{32}$ equals:(A)$\sqrt{2}$ (B)2 (C)$12\sqrt{2}$ (D)$12\sqrt{32}$

Solution.
We have, $\sqrt[3]{2}.\sqrt[4]{2}.\sqrt[12]{32}$

We know that $\sqrt[n]{a}= \left ( a \right )^{\frac{1}{n}}$

$\sqrt[3]{2}.\sqrt[4]{2}.\sqrt[12]{32}$=$\left ( 2 \right )^{\frac{1}{3}}\left ( 2 \right )^{\frac{1}{4}}\left ( 2\cdot 2\cdot 2\cdot 2\cdot 2 \right )^{\frac{1}{12}}$

$= \left ( 2 \right )^{\frac{1}{3}}\left ( 2 \right )^{\frac{1}{4}}\left ( 2^{5} \right )^{\frac{1}{12}}$

$= \left ( 2 \right )^{\frac{1}{3}}\left ( 2 \right )^{\frac{1}{4}}\left ( 2 \right )^{\frac{5}{12}}$     $\because \left ( a^{m} \right )^{n}= a^{m\times n}$
$= \left ( 2 \right )^{\frac{1}{3}+ \frac{1}{4}+\frac{5}{12}}$
$\because a^{m}\times a^{n}= a^{m+n}$

Now,  $\frac{1}{3}+\frac{1}{4}+\frac{5}{12}= \frac{4+3+5}{12}= \frac{12}{12}= 1$

So, $\left ( 2 \right )^{\frac{1}{3}+\frac{1}{4}+\frac{5}{12}}=2^{1}= 2$

Hence option B is correct.

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#### $\sqrt[4]{\sqrt[3]{2^{2}}}$ equals :- (A) $2^{-\frac{1}{6}}$ (B)2-6 (C) $2^{\frac{1}{6}}$ (D) $2^{6}$

Solution.

We have, $\sqrt[4]{\sqrt[3]{2^{2}}}$   =$\left ( \left ( 2^{2} \right )^{}\frac{1}{3} \right )^{\frac{1}{4}}$

$\because \sqrt[n]{a}= \left ( a \right )^{\frac{1}{n}}$
$\because \left ( a^{m} \right )^{n}= a^{m\times n}$
So we get,
$\left ( 2 \right )^{2\times \frac{1}{3}\times \frac{1}{4}}$
$= \left ( 2 \right )^{\frac{1}{3}\times \frac{1}{2}}$
$= \left ( 2 \right )^{\frac{1}{6}}$

Hence option C is correct.

#### If $\sqrt{2}= 1\cdot 4142$ = 1.4142 then $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$  is equal to :(A)2.4142 (B)5.8282 (C)0.4142 (D)0.1718

Solution

We have, $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$
We have to rationalize it
$\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}\times \frac{\sqrt{2}-1}{\sqrt{2}-1}}$           [Multiplying numerator and denominator by $\sqrt{2}-1$]

= $\frac{\sqrt{\left ( \sqrt{2}-1 \right )\times\left ( \sqrt{2} -1\right ) }}{\sqrt{\left ( \sqrt{2} \right )^{2}-\left ( 1 \right )^{2}}}$      [$\because$ (a – b) (a + b) = a2 – b2]

= $\frac{\sqrt{\left ( \sqrt{2}-1 \right )^{2}}}{1}$

$\sqrt{\left ( \sqrt{2}-1 \right )^{2}}$
=$\sqrt{2}-1$
=$1\cdot 4142-1$
=0.4142

Hence option C is correct.

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#### The value of $\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}$ is equal to :(A) $\sqrt{2}$ (B) 2 (C)4 (D) 8

Solution.
$\sqrt{32}+\sqrt{48}= \sqrt{16\times 2}+\sqrt{16\times3}= 4\left ( \sqrt{2}+\sqrt{3} \right )$
$\sqrt{8}+\sqrt{12}= \sqrt{4\times 2}+\sqrt{4\times3}= 2\left ( \sqrt{2}+\sqrt{3} \right )$
$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}= \frac{4\left ( \sqrt{2}+\sqrt{3} \right )}{2\left ( \sqrt{2} \right )+\sqrt{3}}= 2$

Hence, (B) is the correct option.

#### After rationalizing the denominator of $\frac{7}{3\sqrt{3}-2\sqrt{2}}$ we get the denominator as:(A)13 (B)19 (C)5 (D)35

Solution.

We have,$\frac{7}{3\sqrt{3}-2\sqrt{2}}$
We have to rationalize it

$\frac{7}{3\sqrt{3}-2\sqrt{2}}\times \frac{3\sqrt{3}+2\sqrt{2}}{3\sqrt{3}+2\sqrt{2}}$
[Multiplying numerator and denominator by $3\sqrt{3}+2\sqrt{2}$]

= $\frac{7\left ( 3\sqrt{3}+2\sqrt{2} \right )}{\left ( 3\sqrt{3} \right )^{2}-\left ( 2\sqrt{2} \right )^{2}}$                    [$\because$ (a – b) (a + b) = a2 – b2]

$\frac{7\left ( 3\sqrt{3} \right )+2\sqrt{2}}{27-8}$

$\frac{7\left ( 3\sqrt{3} \right )+2\sqrt{2}}{19}$
Therefore we get the denominator as 19.
Hence (B) is the correct option.

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#### $\frac{1}{\sqrt{9}-\sqrt{8}}$ is equal to :(A) $\frac{1}{2}\left ( 3-2\sqrt{2} \right )$ (B) $\frac{1}{3+2\sqrt{2}}$(C)$3-2\sqrt{2}$ (D) $3+2\sqrt{2}$

Solution.
We have, $\frac{1}{\sqrt{9}-\sqrt{8}}$
We have to rationalize it
$\frac{1}{\sqrt{9}-\sqrt{8}}\times \frac{\sqrt{9}+\sqrt{8}}{\sqrt{9}+\sqrt{8}}$            [Multiplying and dividing by $\sqrt{9}+\sqrt{8}$]

= $\frac{\sqrt{9}+\sqrt{8}}{\left ( \sqrt{9} \right )^{2}-\left ( \sqrt{8} \right )^{2}}$      [$\because$ (a – b) (a + b) = a2 – b2]

=$\frac{3+2\sqrt{2}}{9-8}$
$= 3+2\sqrt{2}$

Hence option D is correct.

#### The number obtained on rationalizing the denominator of $\frac{1}{\sqrt{7}-2}$ is : (A) $\frac{\sqrt{7}+2}{3}$ (B)$\frac{\sqrt{7}-2}{3}$ (C)$\frac{\sqrt{7}+2}{5}$ (D)$\frac{\sqrt{7}+2}{45}$

Solution
.

We have, $\frac{1}{\sqrt{7}-2}$
We have to rationalize it

$\frac{1}{\sqrt{7}-2}\times \frac{\sqrt{7}+2}{\sqrt{7}+2}$       [Multiplying numerator and denominator by $\sqrt{7}+2$]

= $\frac{\sqrt{7}+2}{\left ( \sqrt{7} \right )^{2}-\left ( 2 \right )^{2}}$         [$\because$ (a – b) (a + b) = a2 – b2]

$= \frac{\sqrt{7}+2}{7-4}$
$= \frac{\sqrt{7}+2}{3}$

Hence option A is correct.

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