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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}}

Answer. [C]
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.

 

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Value of (256)0.16 × (256)0.09 is :

(A) 4
(B) 16
(C) 64
(D) 256.25

Answer.   [A]
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}

Answer.     [A]
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

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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}

Answer.        [B]
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}

Answer.     [C]
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.

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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

Answer.        [C]
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

Answer.     [B]
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.

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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

Answer.    [B]
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}

Answer.[D]
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.

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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}

Answer. [A]
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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